Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables

go-math-grade-6-chapter-9-independent-and-dependent-variables-answer-key

Download Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables pdf for free of cost. It is very important for the students to learn the fundamentals at the secondary level. In order to help you guys, we are providing the answers to all the questions in HMH Go Math Grade 6 Chapter 9 Solution Key Independent and Dependent Variables.

Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables

The Independent and Dependent Variables chapter consists of the topics like equations and tables, analyze relationships, graphs etc. It is essential for students to know the relationship between the graphs and tables in this chapter. You can know different methods of solving the problems by using Go Math Grade 6 Solution Key Chapter 9 Independent and Dependent Variables. All you have to do is to tap the below-given links.

Lesson 1: Independent and Dependent Variables

Lesson 2: Equations and Tables

Lesson 3: Problem Solving • Analyze Relationships

Mid-Chapter Checkpoint

Lesson 4: Graph Relationships

Lesson 4: Graph Relationships

Lesson 5: Equations and Graphs

Chapter 9 Review/Test

Share and Show – Page No. 493

Identify the independent and dependent variables. Then write an equation to represent the relationship between them.

Question 1.
An online store lets customers have their name printed on any item they buy. The total cost c in dollars is the price of the item p in dollars plus $3.99 for the name.
Type below:
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Answer: c = p + $3.99

Explanation:
The independent variable is c, the price of the item because it is not going to depend on anything else.
The dependent variable is p because the total cost depends on how many items there are, whether your name is marked on it, etc.
The equation would be:
c = p + $3.99

Question 2.
A raft travels downriver at a rate of 6 miles per hour. The total distance d in miles that the raft travels is equal to the rate times the number of hours h.
Type below:
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Answer: d = 6 × h

Explanation:
Speed of the raft= 6 miles per hour
Total distance (d) of the raft = rate × number of hours h
The dependent variable is the number of hours h
The independent variable is distance d.
The equation would be:
d = 6 × h

Question 3.
Apples are on sale for $1.99 a pound. Sheila buys p pounds of apples for a total cost of c dollars.
Type below:
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Answer: c = p × $1.99

Explanation:
Apples are on sale for $1.99 a pound
p = pounds of apples
c = total cost of dollars
The equation would be:
c = p × $1.99
c is the independent variable.
p is the dependent variable.

On Your Own

Identify the independent and dependent variables. Then write an equation to represent the relationship between them.

Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables Page 493 Q4

Question 5.
Billy has $25. His father is going to give him more money. The total amount t Billy will have is equal to the amount m his father gives him plus the $25 Billy already has.
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Answer: t = m + $25

Explanation:
Billy has $25. His father is going to give him more money.
The total amount t Billy will have is equal to the amount m his father gives him plus the $25 Billy already has.
The equation would be:
t = m + $25
t is the independent variable
m is the dependent variable.

Question 6.
Connect Symbols and Words Describe a situation that can be represented by the equation c = 12b.
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Answer:
Melinda is making necklaces. She uses 12 beads for each necklace. The total number of beads b depends on the cost of the necklace c.
The equation is c = 12n

Question 7.
Belinda pays $4.25 for each glass she buys. The total cost c is equal to the price per glass times the number of glasses n plus $9.95 for shipping and handling. Write an equation and use it to find how much it will cost Belinda to buy 12 glasses.
Type below:
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Answer:
Belinda pays $4.25 for each glass she buys. The total cost c is equal to the price per glass times the number of glasses n plus $9.95 for shipping and handling.

Belinda pays $4.25 for each glass she buys.
The equation is: c = 4.25n + 9.95
Now we have to find the cost of 12 glasses.
c = 4.25(12) + 9.95
c = 60.95
It takes $60.95 to buy 12 glasses.

Unlock the Problem – Page No. 494

Question 8.
Benji decides to save $15 per week to buy a computer program. Write an equation that models the total amount t in dollars Benji will have saved in w weeks.
a. What does the variable t represent?
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Answer: t represents the total amount in dollars Banji saved.

Question 8.
b. Which is the dependent variable? Which is the independent variable? How do you know?
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Answer:
w is the dependent variable.
t is the independent variable.
w is dependent because it represents the number of weeks. So, we have to multiply 15 by w.
t is an independent variable because t determines the value of a dependent variable.

Question 8.
c. How can you find the total amount saved in w weeks?
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Answer: We can find the total amount saved in w weeks by multiplying 15 with w.

Question 8.
d. Write an equation for the total amount that Benji will have saved.
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Answer: t = 15w

Question 9.
Coach Diaz is buying hats for the baseball team. The total cost c is equal to the number of hats n that he buys times the sum of the price per hat h and a $2 charge per hat to have the team name printed on it. Write an equation that can be used to find the cost of the hats.
Type below:
________________

Answer: c = n + 2h

Explanation:
Coach Diaz is buying hats for the baseball team.
The total cost c is equal to the number of hats n that he buys times the sum of the price per hat h and a $2 charge per hat to have the team name printed on it.
c represents the total cost.
n is the number of hats
h is the price per hat.
The equation is c is equal to the number of hats plus price per hat and $2.
c = n + 2h

Question 10.
A steel cable that is \(\frac{1}{2}\) inch in diameter weighs 0.42 pound per foot. The total weight in pounds w is equal to 0.42 times of the number of feet f of steel cable. Choose the letter or equation that makes each sentence true.
The independent variable is ______________ .
The dependent variable is ______________ .
The equation that represents the relationship between the variables is ______________ .

Answer:
A steel cable that is \(\frac{1}{2}\) inch in diameter weighs 0.42 pound per foot.
The total weight in pounds w is equal to 0.42 times of the number of feet f of steel cable.
The equation would be:
w = 0.2f
f is the dependent variable
w is the dependent variable.

Independent and Dependent Variables – Page No. 495

Identify the independent and dependent variables. Then write an equation to represent the relationship between them.

Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables Page 495 Q1

Question 2.
An online clothing store charges $6 for shipping, no matter the price of the items. The total cost c in dollars is the price of the items ordered p plus $6 for shipping.
Type below:
________________

Answer: c = p + 6

Explanation:
Given,
An online clothing store charges $6 for shipping, no matter the price of the items.
The total cost c in dollars is the price of the items ordered p plus $6 for shipping.
The equation would be:
c = p + $6
where c = cost in dollars
p is the price of items
The independent variable is c.
The dependent variable is p

Question 3.
Melinda is making necklaces. She uses 12 beads for each necklace. The total number of beads b depends on the number of necklaces n.
Type below:
________________

Answer: b = 12n

Explanation:
Melinda is making necklaces. She uses 12 beads for each necklace.
The total number of beads b depends on the number of necklaces n.
b = total number of beads
n = number of necklaces
The equation would be:
b = 12n
b is the independent variable
n is the dependent variable.

Problem Solving

Question 4.
Maria earns $45 for every lawn that she mows. Her earnings e in dollars depend on the number of lawns n that she mows. Write an equation that represents this situation.
Type below:
________________

Answer: e = 45n

Explanation:
Maria earns $45 for every lawn that she mows.
Her earnings e in dollars depend on the number of lawns n that she mows.
e = earnings in dollars
n = number of lawns
The equation would be:
e = 45n
e is the independent variable.
n is the dependent variable.

Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables Page 495 Q5

Question 6.
Write a situation in which one unknown is dependent on another unknown. Write an equation for your situation and identify the dependent and independent variables.
Type below:
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Answer:
Byron is playing a game. He earns 10 points for each question he answers correctly. His total score s equals the number of correct answers a time a.
Answer:
Dependent variable: s
Independent variable: a
Equation: s = 10a

Lesson Check – Page No. 496

Question 1.
There are 12 boys in a math class. The total number of students s depends on the number of girls in the class g. Write an equation that represents this situation.
Type below:
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Answer: s = 12 + g

Explanation:
There are 12 boys in a math class.
The total number of students s depends on the number of girls in class g.
The equation would be:
s = 12 + g
s is the independent variable.
g is the dependent variable.

Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables Page 496 Q2

Spiral Review

Question 3.
The formula F = \(\frac{9}{5}\)C + 32 gives the Fahrenheit temperature for a Celsius temperature of C degrees. Gwen measured a Celsius temperature of 35 degrees. What is this temperature in degrees Fahrenheit?
______ °F

Answer: 95 degrees

Explanation:
The formula F = \(\frac{9}{5}\)C + 32 gives the Fahrenheit temperature for a Celsius temperature of C degrees.
C = 35
F = 9C ÷ 5 + 32
F = 9(35) ÷ 5 + 32
F = 315 ÷ 5 + 32
F = 63 + 32
F = 95 degrees

Question 4.
Write an equation to represent this sentence. The difference of a number n and 1.8 is 2.
Type below:
________________

Answer: n – 1.8 = 2

Explanation:
The difference of a number n and 1.8 is 2.
The phrase difference is nothing but subtraction.
The equation would be:
n – 1.8 = 2

Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables Page 496 Q5

Question 6.
Graph x ≤ 4.5 on a number line.
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Answer:
GO Math Grade 6 Chapter 9 answer key img-1

Share and Show – Page No. 499

Use the equation to complete the table.

Question 1.
y = x + 3
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 1
Type below:
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Answer:
Substitute the value of x in the above equation.
The equation is x + 3.
Go-Math-Grade-6-Answer-Key-Chapter-9-Independent-and-Dependent-Variables-img-1

Question 2.
y = 2x + 1
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 2
Type below:
________________

Answer:
Substitute the value of x in the above equation.
The equation is y = 2x + 1
Go-Math-Grade-6-Answer-Key-Chapter-9-Independent-and-Dependent-Variables-img-2

On Your Own

Write an equation for the relationship shown in the table. Then find the unknown value in the table.

Question 3.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 3
Type below:
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Answer:
The equation is y = 2x
The output is multiple of 2 and x
For x = 10
The output is y = 2x
y = 2 × 10 = 20

Question 4.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 4
Type below:
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Answer:
y = x ÷ 2
The output is the quotient of x and 2.
The output for x = 40 is
y = 40 ÷ 2
y = 20

Question 5.
The table shows the current cost of buying apps for a cell phone. Next month, the price of each app will double. Write an equation you can use to find the total cost y of buying x apps next month.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 5
Type below:
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Answer: y = 3x

Explanation:
The equation is multiple of 3 and x.
The equation is y = 3x

Question 6.
A beach resort charges $1.50 per hour plus $4.50 to rent a bicycle. The equation c = 1.50x + 4.50 gives the total cost c of renting a bicycle for x hours. Use numbers and words to explain how to find the cost c of renting a bicycle for 6 hours.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 6
Type below:
________________

Answer:
A beach resort charges $1.50 per hour plus $4.50 to rent a bicycle.
The equation c = 1.50x + 4.50 gives the total cost c of renting a bicycle for x hours.
For x = 1
c = 1.50(1) + 4.50
c = 1.50 + 4.50
c = $6.00
For x = 2
c = 1.50(2) + 4.50
c = 3.00 + 4.50
c = $7.50
For x = 3
c = 1.50(3) + 4.50
c = 4.50 + 4.50
c = $9.00
For x = 4
c = 1.50(4) + 4.50
c = 6.00 + 4.50
c = $10.50

</aCause and Effect – Page No. 500

The reading skill cause and effect can help you understand how a change in one variable may cause a change in another variable.

In karate, a person’s skill level is often shown by the color of his or her belt. At Sara’s karate school, students must pass a test to move from one belt level to the next. Each test costs $23. Sara hopes to move up 3 belt levels this year. How will this affect her karate expenses?
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 7

Question 7.
Write an equation to show the relationship between cause and effect. Then use the equation to solve the problem.
Type below:
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Answer: y = 23x

Explanation:
Let x represent the number of belt levels Sara moves up and let y represent the increase in dollars in her karate expenses.
Write the equation:
y = 23x
Sara plans to move up 3 levels, so replace x with 3
y = 23 × 3
y = 69
So, if Sara moves up 3 belt levels this year, her karate expenses will increase by $69.

Write an equation to show the relationship between cause and effect. Then use the equation to solve the problem.

Question 8.
Classes at Tony’s karate school cost $29.50 per month. This year he plans to take 2 more months of classes than he did last year. How will this affect Tony’s karate expenses?
Type below:
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Answer:
The equation is y = 29.50x
where x is the number of additional classes
y is the increase in dollars in expenses.
Tony plans to take 2 more months of classes so his expenses will increase by y = 29.5 × 2 = $59

Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables Page 500 Q9

Equations and Tables – Page No. 501

Use the equation to complete the table.

Question 1.
y = 6x
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 8
Type below:
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Answer:
The equation is y = 6x
Substitute the value of x in the above equation.
Go-Math-Grade-6-Answer-Key-Chapter-9-Independent-and-Dependent-Variables-img-8

Question 2.
y = x − 7
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 9
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Answer:
The equation is y = x – 7
Substitute the value of x in the equation.
Use the equation to get the output y.
Go-Math-Grade-6-Answer-Key-Chapter-9-Independent-and-Dependent-Variables-img-9

Question 3.
y = 3x + 4
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 10
Type below:
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Answer:
The equation is y = 3x + 4
Substitute the value of x in the above equation.
Use the equation to get the output y.
Go-Math-Grade-6-Answer-Key-Chapter-9-Independent-and-Dependent-Variables-img-10 (1)

Write an equation for the relationship shown in the table. Then find the unknown value in the table.

Question 4.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 11
Type below:
________________

Answer: y = 8x

Explanation:
The equation is the multiple of 8.
The equation is y = 8x
Substitute x = 3 in the equation.
y = 8(3) = 24
Thus the unknown value is 24.

Question 5.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 12
Type below:
________________

Answer: y = x ÷ 2

Explanation:
The equation is divisible by 2.
The equation is y = x ÷ 2
Substitute x = 22 in the equation.
y = x ÷ 2
y = 22 ÷ 2
y = 11
Therefore the unknown value is 11.

Problem Solving

Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables Page 501 Q6

Question 7.
Write an equation for the relationship shown in the table. Then use the equation to find the estimated number of shrimp in a 5-pound bag.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 13
Type below:
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Answer: y = 24x

Explanation:
The equation is the multiple of 24.
The equation is y = 24x

Question 8.
Write a word problem that can be represented by a table and equation. Solve your problem and include the table and equation.
Type below:
________________

Answer:
Susie ran a race. She ran 5 miles an hour and the race took her x hours to complete.
y = 5x
Use the equation to get the output y.
Go-Math-Grade-6-Answer-Key-Chapter-9-Independent-and-Dependent-Variables-img-10

Lesson Check – Page No. 502

Question 1.
Write an equation that represents the relationship shown in the table.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 14
Type below:
________________

Answer: y = x – 4

Explanation:
The relationship between x and y is y = x – 4.
We get the output when we subtract 4 from x.

Question 2.
There is a one-time fee of $27 to join a gym. The monthly cost of using the gym is $18. Write an equation for the relationship that gives the total cost y in dollars of joining the gym and using it for x months.
Type below:
________________

Answer: y = 18x + 27

Explanation:
Given,
There is a one-time fee of $27 to join a gym. The monthly cost of using the gym is $18.
Here y represents the total coast in dollars of joining the gym.
x represents months.
So, the equation would be: y = 18x + 27

Spiral Review

Question 3.
Mindy wants to buy several books that each cost $10. She has a coupon for $6 off her total cost. Write an expression to represent her total cost in dollars for b books.
Type below:
________________

Answer: 10b – 6

Explanation:
Given,
Mindy wants to buy several books that each cost $10.
She has a coupon for $6 off her total cost.
b represents the total cost in dollars for b books.
So, the equation to represent the total cost is 10b – 6.

Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables Page 502 Q4
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables Page 502 Q4.1

Question 5.
Which of the following are solutions to the inequality n > 7?
n = 7 n = 6.9 n = 7.2 n = 6\(\frac{1}{2}\)
Type below:
________________

Answer: n = -7

Explanation:
Substitute the value of n in the inequality.
n > 7
n = -7
-7 > -7
Thus -7 is the solution.
n = 6.9
-6.9 > -7
-6.9 is not the solution.
n = 7.2
-7.2 > -7
-7.2 is less than – 7
Thus -7.2 is not the solution.
n = 6\(\frac{1}{2}\)
6\(\frac{1}{2}\) > -7
6\(\frac{1}{2}\) is not the solution.

Question 6.
Marcus sold brownies at a bake sale. He sold d dollars worth of brownies. He spent $5.50 on materials, so his total profit p in dollars can be found by subtracting $5.50 from his earnings. Write an equation that represents this situation.
Type below:
________________

Answer: p = d – 5.50

Explanation:
Marcus sold brownies at a bake sale. He sold d dollars worth of brownies.
He spent $5.50 on materials, so his total profit p in dollars can be found by subtracting $5.50 from his earnings.
p represents the total profit in dollars.
d is the dollars worth of brownies.
The equation is p = d – 5.50

Share and Show – Page No. 505

Question 1.
A soccer coach is ordering shirts for the players. The table shows the total cost based on the number of shirts ordered. How much will it cost the coach to order 18 shirts?
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 15
$ _______

Answer: 270

Explanation:
First, find a pattern and write an equation.
The cost is $15 multiplied by the number of shirts.
c = $15 × n
Next, use the equation to find the cost of 18 shirts.
c = $15 × n
c = $15 × 18
c = $270
So, the cost of 18 shirts is $270.

Question 2.
What if the coach spent $375 to purchase a number of shirts? Could you use the same equation to find how many shirts the coach bought? Explain.
Type below:
________________

Answer:
Yes, I could use the same equation.
I could substitute 375 for the variable c and solve for n.

Question 3.
The table shows the number of miles the Carter family drove over time. If the pattern continues, will the Carter family have driven more than 400 miles in 8 hours? Explain.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 16
Type below:
________________

Answer: 376 miles

Explanation:
First, find a pattern and write an equation.
The distance is 47 miles multiplied by the number of hours.
y = 47 × x
Next, use the equations to find the distance for 8 hours.
y = 47x
y = 47 × 8
y = 376
So, the family will have driven 376 miles in 8 hours, which is less than 400 miles.

Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables Page 505 Q4

On Your Own – Page No. 506

Question 5.
A group of dancers practiced for 4 hours in March, 8 hours in April, 12 hours in May, and 16 hours in June. If the pattern continues, how many hours will they practice in November?
_______ hours

Answer: 36 hours

Explanation:
Given that, a group of dancers practiced for 4 hours in March, 8 hours in April, 12 hours in May, and 16 hours in June.
The equation would be h = 4m
m = 9
h = 4 × 9 = 36
Thus the group practiced 36 hours in the month of November.

Question 6.
The table shows the number of hours Jacob worked and the amount he earned each day.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 17
At the end of the week, he used his earnings to buy a new pair of skis. He had $218 left over. How much did the skis cost?
$ _______

Answer: 142

Explanation:
First, add the total amount he earned.
60 + 84 + 72 + 96 + 48 = 360
Jacob earned $360 for the week.
If he has $218 leftover, this means that the cost of the skis is 360 – 218 = 14
Therefore the cost of the skis is $142.

Question 7.
Pose a Problem Look back at Problem 6. Use the data in the table to write a new problem in which you could use the strategy to find a pattern. Then solve the problem.
Type below:
________________

Answer:
How much money would Jacob earn if he worked for 10 hours?
From the table, we can see that the pattern is that Jacob earns $12 per hour.
The equation is s = 12h
Where s is the total pay and h is the number of hours worked.
s = 12h
s = 12 × 10
s = 120
Thus Jacob earned $120 for 10 hours.

Question 8.
Draw Conclusions Marlon rode his bicycle 9 miles the first week, 18 miles the second week, and 27 miles the third week. If the pattern continues, will Marlon ride exactly 100 miles in a week at some point? Explain how you determined your answer.
Type below:
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Answer: No, Marlon will not ride exactly 100 miles in a week at some point.
Each number in the pattern is a multiple of 9 and 100 is not a multiple of 9.

Question 9.
A diving instructor ordered snorkels. The table shows the cost based on the number of snorkels ordered.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 18
If the diving instructor spent $1,024, how many snorkels did he order? Use numbers and words to explain your answer.
_______ snorkels

Answer: 32

Explanation:
Use the table to find the equation.
c represents the cost based on the number of snorkels.
s represents the number of snorkels
The equation would be:
c = 32s
The diving instructor spent $1,024
c = 1024
1024 = 32s
s = 1024/32
s = 32
Thus the diving instructor gets 32 snorkels for $1024.

Problem Solving Analyze Relationships – Page No. 507

The table shows the number of cups of yogurt needed to make different amounts of a fruit smoothie. Use the table for 1–3.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 19

Question 1.
Write an equation to represent the relationship.
Type below:
________________

Answer: c = 3b

Explanation:
c represents number of cups of yogurt
b represents the batches
From the table, we can observe that b is multiplied with 3 to get cups of yogurt.
So, the equation to find the number of cups of yogurt is c = 3b

Question 2.
How much yogurt is needed for 9 batches of smoothie?
_______ cups

Answer: 27

Explanation:
Given that there are 9 batches of smoothie.
By using the above equation we can find the number of cups.
c = 3b
c = 3 × 9 = 27 cups
Thus 27 cups of yogurt is need to make 9 batches of smoothie.

Question 3.
Jerry used 33 cups of yogurt to make smoothies. How many batches did he make?
_______ batches

Answer: 11 batches

Explanation:
Jerry used 33 cups of yogurt to make smoothies.
Use the equation to find the batches.
c = 3b
33 = 3b
b = 33/3
b = 11
Therefore jerry made 11 batches of smoothie.

The table shows the relationship between Winn’s age and his sister’s age. Use the table for 4–5.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 20

Question 4.
Write an equation to represent the relationship.
Type below:
________________

Answer: s = w + 4

Explanation:
By using the table we can find the relationship between wine’s age and wine’s sister’s age.
Winn’s sister’s age will be the sum of Winn’s age and 4.
So, the equation is s = w + 4

Question 5.
When Winn is 14 years old, how old will his sister be?
_______ years old

Answer: 18

Explanation:
Use the equation s = w + 4
W = 14 years
s = 14 + 4
s = 18 years
Thus winn’s sister’s age is 18 years.

Question 6.
Write a problem for the table. Use a pattern and an equation to solve your problem.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 21
Type below:
________________

Answer: m = 16h

Explanation:
Jerry runs 16 miles per hour. How many miles he can run in 5 hours?
The equation is m = 16h
m = 16 × 5 = 80 miles
Therefore jerry runs 80 miles in 5 hours.

Lesson Check – Page No. 508

Question 1.
The table shows the total cost c in dollars of n gift baskets. What will be the cost of 9 gift baskets?
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 22

Answer: $108

Explanation:
By seeing the above we can say that the equation is
c = 12n
n = 9
Use the equation to find the cost of 9 gift baskets.
c = 12 × 9
c = $108
Thus the cost of 9 gift baskets is $108.

Question 2.
The table shows the number of minutes m that Tara has practiced after d days. If Tara has practiced for 70 minutes, how many days has she practiced?
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 23
_______ days

Answer: 2 days

Explanation:
The table shows the number of minutes m that Tara has practiced after d days.
The equation would be
m = 35d
If Tara has practiced for 70 minutes
m = 70
Use the equation to find the number of days she practiced.
70 = 35d
d = 70/35
d = 2 days
Thus Tara has practiced 2 days.

Spiral Review

Question 3.
Soccer shirts cost $15 each, and soccer shorts cost $18 each. The expression 15n + 18n represents the total cost in dollars of n uniforms. Simplify the expression by combining like terms.
Type below:
________________

Answer: 33n

Explanation:
Soccer shirts cost $15 each, and soccer shorts cost $18 each.
The expression 15n + 18n represents the total cost in dollars of n uniforms.
Now combine the like terms.
15n + 18n = 33n

Question 4.
What is an equation that represents the relationship in the table?
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 24
Type below:
________________

Answer: y = x ÷ 2

Explanation:
By seeing the above table we can find the relationship between x and y.
y is the quotient of x and 2.
We get the value of y when you divide x by 2.
The equation is y = x ÷ 2

Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables Page 508 Q5

Question 6.
Marisol plans to make 9 mini-sandwiches for every 2 people attending her party. Write a ratio that is equivalent to Marisol’s ratio.
Type below:
________________

Answer: 9:2

Explanation:
Given that, Marisol plans to make 9 mini-sandwiches for every 2 people attending her party.
The ratio will be 9:2
Now we need to write the equivalent ratio for the 9 sandwiches for every 2 people i.e, 9:2
We know that the equivalent ratio can be written as
9/2 × 3/3 = 27/6
9/2 × 5/5 = 45/6
Thus the equivalent fractions are 27/6 and 45/6.

Mid-Chapter Checkpoint – Vocabulary – Page No. 509

Choose the best term from the box to complete the sentence.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 25

Question 1.
A(n) _____ has a value that determines the value of another quantity.
Type below:
________________

Answer: Independent variable
An Independent variable has a value that determines the value of another quantity.

Question 2.
A variable whose value is determined by the value of another quantity is called a(n) _____.
Type below:
________________

Answer: Dependent variable
A variable whose value is determined by the value of another quantity is called a Dependent variable.

Concepts and Skills

Identify the independent and dependent variables.

Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables Page 509 Q3

Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables Page 509 Q4

Write an equation for the relationship shown in the table. Then find the unknown value in the table.

Question 5.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 26
Type below:
________________

Answer: 49

Explanation:
The equation is y = 7x
x = 7
y = 7 × 7 = 49
Thus the unknown value y is 49.

Question 6.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 27
Type below:
________________

Answer: 12

Explanation:
The equation for the above table is
y = x ÷ 5
Use the equation to find the value of y where x = 60
y = 60 ÷ 5
y = 12
Thus the unknown value is 12.

Write an equation that describes the pattern shown in the table.

Question 7.
The table shows how the number of pepperoni slices used depends on the number of pizzas made.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 28
Type below:
_______________

Answer: y = 17x

Explanation:
The table shows how the number of pepperoni slices used depends on the number of pizzas made.
y is 17 times of x.
The equation for the above table is y = 17x

Question 8.
Brayden is training for a marathon. The table shows how the number of miles he runs depends on which week of training he is in.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 29
Type below:
________________

Answer: m = w + 5

Explanation:
Brayden is training for a marathon. The table shows how the number of miles he runs depends on which week of training he is in.
m is equal to the sum of w and 5.
Thus the equation is m = w + 5.

Page No. 510

Question 9.
The band has a total of 152 members. Some of the members are in the marching band, and the rest are in the concert band. Write an equation that models how many marching band members m there are if there are c concert band members.
Type below:
________________

Answer: m = 152 – c

Explanation:
Given,
The band has a total of 152 members. Some of the members are in the marching band, and the rest are in the concert band.
m is equal to the difference of 152 and c.
The equation is m = 152 – c

Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables Page 510 Q10

Question 11.
Amy volunteers at an animal shelter. She worked 10 hours in March, 12 hours in April, 14 hours in May, and 16 hours in June. If the pattern continues, how many hours will she work in December?
_______ hours

Answer: 28 hours

Explanation:
Amy volunteers at an animal shelter.
She worked 10 hours in March, 12 hours in April, 14 hours in May, and 16 hours in June.
As she started working from the march. December will be the 10th month.
Keep on adding 2 hours for each month you get 28 hours for December.
Thus she worked 28 hours in December.

Question 12.
Aaron wants to buy a new snowboard. The table shows the amount that he has saved. If the pattern in the table continues, how much will he have saved after 1 year?
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 30
$ _______

Answer: $540

Explanation:
Aaron wants to buy a new snowboard. The table shows the amount that he has saved.
The equation will be s = 45m
s is the money saved
m is the number of months
1 year = 12 months
s = 45 × 12
s = 540
Thus he saved $540 after 1 year.

Share and Show – Page No. 513

Graph the relationship represented by the table.

Question 1.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 31
Type below:
________________

Answer: y = 50x
Go Math Grade 6 Chapter 9 answer key img-19

Question 2.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 32
Type below:
________________

Answer: y = 5x
Go Math Grade 6 Chapter 9 answer key img-20

Graph the relationship represented by the table to find the unknown value of y.

Question 3.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 33
Type below:
________________

Answer: 3
Go Math Grade 6 Chapter 9 answer key img-15

Question 4.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 34
Type below:
________________

Answer: 6
Go Math Grade 6 Chapter 9 answer key img-16

On Your Own

Practice: Copy and Solve Graph the relationship represented by the table to find the unknown value of y.

Question 5.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 35
Type below:
________________

Answer: 5
Go Math Grade 6 Chapter 9 answer key img-17

Question 6.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 36
Type below:
________________

Answer: 7
Go Math Grade 6 Chapter 9 answer key img-18

Problem Solving + Applications – Page No. 514

The table at the right shows the typical price of a popular brand of corn cereal over time. Use the table for 7–8.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 37

Question 7.
Use Graphs Complete the table below to show the cost of buying 1 to 5 boxes of corn cereal in 1988. Then graph the relationship on the coordinate plane at right.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 38
Type below:
________________

Answer:
Go-Math-Grade-6-Answer-Key-Chapter-9-Independent-and-Dependent-Variables-img-38
Go Math Grade 6 Chapter 9 answer key img-26

Question 8.
Suppose you graphed the cost of buying 1 to 5 boxes of corn cereal using the 1968 price and the 2008 price. Explain how those graphs would compare to the graph you made using the 1988 price.
Type below:
________________

Answer:
The points on both graphs would lie on a line, but the line for the 1968 costs would rise less steeply than the line for 1988 costs and the line for the 2008 costs would rise more steeply than the line for 1988 costs.

Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables Page 514 Q9

Question 10.
Graph the relationship represented by the table to find the unknown value of y.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 39
Type below:
________________

Answer: 3
Go Math Grade 6 Chapter 9 answer key img-25

Graph Relationships – Page No. 515

Graph the relationship represented by the table.

Question 1.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 40
Type below:
________________

Answer: y = 25x

Go Math Grade 6 Chapter 9 answer key img-10

Question 2.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 41
Type below:
________________

Answer:
Go Math Grade 6 Chapter 9 answer key img-11

Graph the relationship represented by the table to find the unknown value of y.

Question 3.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 42
Type below:
________________

Answer: 6

Go Math Grade 6 Chapter 9 answer key img-12

Question 4.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 43
Type below:
________________

Answer: 2

Go Math Grade 6 Chapter 9 answer key img-13

Problem Solving

Question 5.
Graph the relationship represented by the table.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 44
Type below:
________________

Answer: y = 15x

Go Math Grade 6 Chapter 9 answer key img-14

Question 6.
Use the graph to find the cost of purchasing 5 DVDs.
$ ______

Answer:
The above graph shows that the cost of 5 DVDs is $75.

Question 7.
Both tables and graphs can be used to represent relationships between two variables. Explain how tables and graphs are similar and how they are different.
Type below:
________________

Answer:
Tables and graphs can be useful tools for helping people make decisions. However, they only provide part of a story. Inferences often have to be made from the data shown. As well as being able to identify clearly what the graph or table is telling us, it is important to identify what parts of the story are missing.

Lesson Check – Page No. 516

Question 1.
Mei wants to graph the relationship represented by the table. Write an ordered pair that is a point on the graph of the relationship.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 45
Type below:
________________

Answer: y = 8x
Go Math Grade 6 Chapter 9 answer key img-27

Question 2.
An online bookstore charges $2 to ship any book. Cole graphs the relationship that gives the total cost y in dollars to buy and ship a book that costs x dollars. Name an ordered pair that is a point on the graph of the relationship.
Type below:
________________

Answer:
An online bookstore charges $2 to ship any book.
Cole graphs the relationship that gives the total cost y in dollars to buy and ship a book that costs x dollars.
y = x + 2
x = 4
y = 4 + 2
y = 6
The ordered pair is (4,6)

Spiral Review

Question 3.
Write an expression that is equivalent to 6(g + 4).
Type below:
________________

Answer:
6(g + 4)
6 × g + 6 × 4
6g + 24

Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables Page 516 Q4

Question 5.
Graph n > 2 on a number line.
Type below:
________________

Answer:
HMH Go Math Grade 6 Key Chapter 9 img-28

Question 6.
Sam is ordering lunch for the people in his office. The table shows the cost of lunch based on the number of people. How much will lunch cost for 35 people?
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 46
$ _____

Answer: 280

Explanation:
Sam is ordering lunch for the people in his office.
The table shows the cost of lunch based on the number of people.
The equation is c = 8n
c = 8 × 35
c = 280
Thus the lunch cost for 35 people is $280.

Share and Show – Page No. 519

Graph the linear equation.

Question 1.
y = x + 2
Type below:
________________

Answer:
Go math grade 6 chapter 9 answer key img-28

Question 2.
y = 3x
Type below:
________________

Answer:
Go math grade 6 chapter 9 answer key img-29

Write the linear equation for the relationship shown by the graph.

Question 3.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 47
Type below:
________________

Answer: y = x – 1

Question 4.
Type below:
________________

On Your Own

Graph the linear equation.

Question 5.
y = x + 1
Type below:
________________

Answer:
Go math grade 6 chapter 9 answer key img-30

Question 6.
y = 2x − 1
Type below:
________________

Answer:
Go math grade 6 chapter 9 answer key img-31

Question 7.
Identify Relationships The graph shows the number of loaves of bread y that Kareem bakes in x hours. Write the linear equation for the relationship shown by the graph.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 48
Type below:
________________

Answer:
The ordered pairs are (1,1), (2,2), (4,4), (5,5)
Look for a pattern among the pairs: each y value is the same as the corresponding x-value.
The equation is y = x
y = x

Problem Solving + Applications – Page No. 520

The graph shows the growth of a bamboo plant. Use the graph for 8–9.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 49

Question 8.
Write a linear equation for the relationship shown by the graph. Use your equation to predict the height of the bamboo plant after 7 days.
Type below:
________________

Answer:
Write the ordered pairs from the graph: (1,50), (2,100), (3,150), (4,200), (5,250).
Look for a pattern among the pairs: each y value is 50 times the corresponding x value.
The equation is y = 50x
For x = 7, the solution is y = 50 × 7 = 350.
So, the height of the bamboo plant after 7 days will be 350 centimeters.

Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables Page 520 Q9

Question 10.
Maria graphed the linear equation y = x + 3. Then she used her ruler to draw a vertical line through the point (4, 0). At what point do the two lines intersect?
Type below:
________________

Answer:
y = x + 3
y = 4 + 0 = 4
y = 4 + 3 = 7
The coordinate is (4, 7)
Go math grade 6 chapter 9 answer key img-32

Question 11.
Antonio claims the linear equation for the relationship shown by the graph is y = \(\frac{1}{2}\)x + 2. Use numbers and words to support Antonio’s claim.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 50
Type below:
________________

Answer:
The ordered pairs (2,3) and (6,5) on the line make the equation.
y = \(\frac{1}{2}\)x + 2
y = 1/2 × 2 + 2
y = 1 + 2 = 3
y = 1/2 × 6 + 2
y = 3 + 2 = 5

Equations and Graphs – Page No. 521

Graph the linear equation.

Question 1.
y = x − 3
Type below:
________________

Answer:
Go Math Grade 6 Chapter 9 answer key img-22

Question 2.
y = x ÷ 3
Type below:
________________

Answer:
Go Math Grade 6 Chapter 9 answer key img-23

Write a linear equation for the relationship shown by the graph.

Question 3.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 51
Type below:
________________

Answer:
By seeing the above graph we can say that the equation is
y = x + 1

Question 4.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 52
Type below:
________________

Answer:
The ordered pairs are (1,4), (1.5,6), (2,8)
By seeing the above pairs we can say that the equation is y = 4x

Problem Solving

Question 5.
Dee is driving at an average speed of 50 miles per hour. Write a linear equation for the relationship that gives the distance y in miles that Dee drives in x hours.
Type below:
________________

Answer: y = 50x

Explanation:
Dee is driving at an average speed of 50 miles per hour.
y represents the distance in miles
x is the number of hours.
y is equal to the product of 50 and x.
y = 50x

Question 6.
Graph the relationship from Exercise 5.
Type below:
________________

Answer:
Go Math Grade 6 Chapter 9 answer key img-24

Question 7.
Explain how to write a linear equation for a line on a graph.
Type below:
________________

Answer:
To write an equation in slope-intercept form, given a graph of that equation, pick two points on the line and use them to find the slope.

Lesson Check – Page No. 522

Question 1.
A balloon rises at a rate of 10 feet per second. What is the linear equation for the relationship that gives the height y in feet of the balloon after x seconds?
Type below:
________________

Answer: The linear equation for the relationship is y = 10x

Question 2.
Write the linear equation that is shown by the graph.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 53
Type below:
________________

Answer:
Write the ordered pairs from the graph: (3,3), (5,5), (8,8)
Look for a pattern among the pairs: each y value is the same as the corresponding x-value.
The equation is y = x

Spiral Review

Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables Page 522 Q3

Question 4.
Which of the following are solutions of j ≥ 0.6?
j = 1      j = 0.6       j = \(\frac{3}{5}\)       j = 0.12        j = 0.08
Type below:
________________

Answer: j = \(\frac{3}{5}\)

Explanation:
Substitute the values of j in the inequality.
j = 1
1 ≥ 0.6
1 is greater than 0.6 but not equal.
Thus 1 is not the solution of j ≥ 0.6.
j = 0.6
-0.6 ≥ 0.6
-0.6 is less than 0.6
Thus -0.6 is not the solution of j ≥ 0.6.
j = \(\frac{3}{5}\)
\(\frac{3}{5}\) ≥ 0.6
\(\frac{3}{5}\) = 0.6
0.6 ≥ 0.6
Thus \(\frac{3}{5}\) is the solution.
j = 0.12
0.12 ≥ 0.6
0.12 is less than 0.6.
Thus 0.12 is not the solution of j ≥ 0.6.
j = 0.08
0.08 ≥ 0.6
0.08 is less than 0.6.
Thus 0.08 is not the solution of j ≥ 0.6.

Question 5.
Red grapes cost $2.49 per pound. Write an equation that shows the relationship between the cost c in dollars and the number of pounds of grapes p.
Type below:
________________

Answer: c = 2.49p

Explanation:
Given,
Red grapes cost $2.49 per pound.
c is the cost in dollars.
p is the number of pounds of grapes.
The equation c is equal to the product of the number of pounds of grapes and $2.49
c = 2.49p

Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables Page 522 Q6

Chapter 9 Review/Test – Page No. 523

Question 1.
A box of peanut butter crackers contains 12 individual snacks. The total number of individual snacks s is equal to 12 times the number of boxes of crackers b.
The independent variable is _____.
The dependent variable is _____.
The equation that represents the relationship between the variables is _____.

Answer:
The independent variable is b.
The dependent variable is s.
The equation that represents the relationship between the variables is s = 12b.

Question 2.
A stationery store charges $8 to print logos on paper purchases. The total cost c is the price of the paper p plus $8 for printing the logo.
For numbers 2a–2d, select True or False for each statement.
2a. The total cost c depends on the price of the paper.
2b. c is the dependent variable.
2c. p is the independent variable.
2d. The equation that represents the relationship between the variables is c = 8p.
2a. ____________
2b. ____________
2c. ____________
2d. ____________

Answer:
2a. True
2b. True
2c. True
2d. False

Explanation:
2a. c represents the relationship between the two quantities.
So, the statement “The total cost c depends on the price of the paper” is true.
2b. c is the total cost so the statement “c is the dependent variable” is true.
2c. p represents the price to print logos
So, the statement “p is the independent variable” is true.
2d. The total cost c is the price of the paper p plus $8 for printing the logo.
The equation would be:
c = 8 + p
Thus the statement “The equation that represents the relationship between the variables is c = 8p” is false.

Question 3.
An electrician charges $75 an hour for labor and an initial fee of $65. The total cost c equals 75 times the number of hours x plus 65. Write an equation for the relationship and use the equation to complete the table.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 54
Type below:
________________

Answer: c = 75x + 65
Substitute the value of x in the equation.
Go-Math-Grade-6-Answer-Key-Chapter-9-Independent-and-Dependent-Variables-img-54

Page No. 524

Question 4.
The community center offers classes in arts and crafts. There is a registration fee of $125 and each class costs $79. The total cost c in dollars equals 79 times the number of classes n plus 125.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 55
For numbers 4a–4d, select True or False for each statement.
4a. The registration fee is $120.
4b. n is the independent variable.
4c. c is the dependent variable.
4d. The cost for 7 classes is $678.
4a. ____________
4b. ____________
4c. ____________
4d. ____________

Answer:
4a. False
4b. True
4c. True
4d. True

Explanation:
4a. The registration fee is $120.
The registration fee is $125, not $120.
So, the statement is false.
4b. n is the independent variable.
n represents the number of classes.
The statement is true.
4c. c is the dependent variable.
c depends on the registration fee.
Thus the statement is true.
4d. The cost for 7 classes is $678
79 × 7 + 125 = $678
Thus the statement is true.

Question 5.
Ms. Walsh is buying calculators for her class. The table shows the total cost based on the number of calculators purchased.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 56
If Ms. Walsh spent a total of $525, how many calculators did she buy? Use numbers and words to explain your answer.
Type below:
________________

Answer:
She bought 35 calculators. I found a pattern and wrote the equation c = 15n.
Since I know that Mrs.Walsh spent a total of $525, I can substitute 525 for c and solve for n
525 = 15n
n = 35

Chapter 9 Review/Test – Page No. 525

Question 6.
The table shows the number of cups of lemonade that can be made from cups of lemon juice.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 57
Mary Beth says the number of cups of lemon juice j depends on the number of cups of lemonade l. She says the equation j = 7l represents the relationship between the cups of lemon juice j and the cups of lemonade l. Is Mary Beth correct? Use words and numbers to explain why or why not.
Type below:
________________

Answer:
Mary Beth is not correct. The number of cups of lemonade l depends on the number of cups of lemon juice j.
So l is the dependent variable and j is the independent variable.
The equation showing the relationship is l = 7j

Question 7.
For numbers 7a–7d, choose Yes or No to indicate whether the points, when graphed, would lie on the same line.
7a. (1, 6), (2, 4), (3, 2), (4, 0)
7b. (1, 1), (2, 4), (3, 9), (4, 16)
7c. (1, 3), (2, 5), (3, 7), (4, 9)
7d. (1, 8), (2, 10), (3, 12), (4, 14)
7a. ____________
7b. ____________
7c. ____________
7d. ____________

Answer:
7a. Yes
Go Math Grade 6 Chapter 9 answer key img-2
7b. No
Go Math Grade 6 Chapter 9 answer key img-3
7c. Yes
Go Math Grade 6 Chapter 9 answer key img-4
7d. Yes
Go Math Grade 6 Chapter 9 answer key img-5

Question 8.
Graph the relationship represented by the table to find the unknown value.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 58
Type below:
________________

Answer: 10

Go Math Grade 6 Chapter 9 answer key img-6

Chapter 9 Review/Test – Page No. 526

Question 9.
Graph the relationship represented by the table.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 59
Type below:
________________

Answer:
Go Math Grade 6 Chapter 9 answer key img-7

Question 10.
Miranda’s wages are $15 per hour. Write a linear equation that gives the wages w in dollars that Miranda earns in h hours.
Type below:
________________

Answer: w = 15h
Go Math Grade 6 Chapter 9 answer key img-8

Question 11.
The table shows the number of miles m that Lucinda could walk in h hours.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 60
Graph the relationship between hours h and miles m. Then write the equation that shows the relationship.
Type below:
________________

Answer: m = 4h
Go Math Grade 6 Chapter 9 answer key img-9

Chapter 9 Review/Test – Page No. 527

Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables Page 527 Q12

Question 13.
Lacy is staying at a hotel that costs $85 per night. The total cost c in dollars of Lacy’s stay is 85 times the number of nights n she stays.
For numbers, 13a–13d, select True or False for each statement.
13a. The number of nights n is dependent on the cost c.
13b. n is the independent variable.
13c. c is the dependent variable.
13d. The equation that represents the total cost is c = 85n.
13a. ____________
13b. ____________
13c. ____________
13d. ____________

Answer:
13a. False
13b. True
13c. True
13d. True

Explanation:
13a. The number of nights n is dependent on the cost c.
n is independent on the cost c.
So, the statement is false.
13b. n is the independent variable.
The statement is true.
13c. c is the dependent variable.
c is dependent because it depends on the cost c.
So, the statement is true.
13d. The equation that represents the total cost is c = 85n.
The equation is true.

Question 14.
A taxi cab company charges an initial fee of $5 and then $4 per mile for a ride. Use the equation c = 4x + 5 to complete the table.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 61
Type below:
________________

Answer:
Substitute the value of x in the equation.
We get,
Go-Math-Grade-6-Answer-Key-Chapter-9-Independent-and-Dependent-Variables-img-61

Chapter 9 Review/Test – Page No. 528

Question 15.
A grocery display of cans is arranged in the form of a pyramid with 1 can in the top row, 3 in the second row from the top, 5 in the third row, and 7 in the fourth row. The total number of cans c equals 2 times the row r minus 1. Use the equation c = 2r − 1 to complete the table.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 62
Type below:
________________

Answer:
A grocery display of cans is arranged in the form of a pyramid with 1 can in the top row, 3 in the second row from the top, 5 in the third row, and 7 in the fourth row.
c = 2r − 1
Substitute r in the equation.
Go-Math-Grade-6-Answer-Key-Chapter-9-Independent-and-Dependent-Variables-img-62

Question 16.
The graph shows the number of words Mason read in a given amount of minutes. If Mason continues to read at the same rate, how many words will he have read in 5 minutes?
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 63
______ words

Answer: 1000 words
By seeing the above graph we can say that Mason can read 1000 words in 5 minutes.

Question 17.
Casey claims the linear equation for the relationship shown by the graph is c = 25j. Use numbers and words to support Casey’s claim.
Go Math Grade 6 Answer Key Chapter 9 Independent and Dependent Variables img 64
Type below:
________________

Answer: The ordered pairs (1,25), (3,75), (5,125) and (7,175) each make the equation c = 25j

Conclusion:

I wish the details prevailed in the Go Math Grade 6 Answer Key Chapter 9 is helpful for you. Share this pdf link with your friends and help them to overcome the difficulties. If you have any doubts regarding the solutions you can leave a comment in the comment section.

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Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures

Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures

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Big Ideas Math Book 6th Grade Answer Key Chapter 9 Statistical Measures

Know the concept of the statistical measures with the help of our Big Ideas Math Grade 6 Solution Key Chapter 9. This BIM 6th Grade Chapter 9 Statistical Measures download pdf will help the students to overcome the difficulties in maths and also to improve their performance in the exams. You can score good marks in exams by referring to our Big Ideas Math Book 6th Grade Answer Key Chapter 9 Statistical Measures.

Performance Task

Lesson: 1 Introduction to Statistics

Lesson: 1 Introduction to Statistics

Lesson: 2 Mean

Lesson: 3 Measures of Center

Lesson: 4 Measures of Variation

Lesson: 5 Mean Absolute Deviation

Lesson: 5 Mean Absolute Deviation

Chapter 9: Statistical Measures

Statistical Measures STEAM Video/Performance Task

STEAM Video
Daylight in the Big City
Averages can be used to compare different sets of data. How can you use averages to compare the amounts of day light in different cities? Can you think of any other real-life situations where averages are useful?
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 1
Watch the STEAM Video “Daylight in the Big City.” Then answer the following questions.
1. Why do different cities have different amounts of daylight throughout the year?

Answer:
Our amount of daylight hours depends on our latitude and how Earth orbits the sun. This causes a seasonal variation in the intensity of sunlight reaching the surface and the number of hours of daylight. The variation in intensity results because the angle at which the sun’s rays hit the Earth changes with the time of year.

2. Robert’s table includes the difference of the greatest amount of daylight and the least amount of daylight in Lagos, Nigeria, and in Moscow, Russia.
Lagos: 44 minutes
Moscow:633 minutes
Use these values to make a prediction about the difference between the greatest amount of daylight and the least amount of daylight in a city in Alaska.

Answer:
The least daylight in Alaska is 1092 minutes in Juneau
The greatest daylight in Alaska is 1320 minutes in Fairbanks

Performance Task
Which Measure of Center Is Best: Mean, Median, or Mode?
After completing this chapter, you will be able to use the concepts you learned to answer the questions in the STEAM Video Performance Task. You will be given the greatest and least amounts of daylight in the 15 cities in the United States with the greatest populations.
s
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 3
You will determine which measure of center best represents the data. Why might someone be interested in the amounts of daylight throughout the year in a city?

Statistical Measures Getting Ready for Chapter 9

Chapter Exploration
Work with a partner. Write the number of letters in each of your first names on the board.
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 4
1. Write all of the numbers on a piece of paper. The collection of numbers is called data.
2. Talk with your partner about how you can organize the data. What conclusions can you make about the numbers of letters in the first names of the students in your class?
3. Draw a grid like the one shown below. Then use the grid to draw a graph of the data.
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 5

Answer:
3,6,9,5,6,7,6,5,5,8,6,8,5,6,4,4,7,6,3,5,6,5,5

4. THE CENTER OF THE DATA Use the graph of the data in Exercise 3 to answer the following.
a. Is there one number that occurs more than any of the other numbers? If so, write a sentence that interprets this number in the context of your class.
b. Complete the sentence, “In my class, the average number of letters in a student’s first name is __________.” Justify your reasoning.
c. Organize your data using a different type of graph. Describe the advantages or disadvantages of this graph.

Answer:
a. Yes, 6, 5, 8 are more than other numbers given in the data.
b. “In my class, the average number of letters in a student’s first name is 5 and 6.

Vocabulary
The following vocabulary terms are defined in this chapter. Think about what each term might mean and record your thoughts.
statistical question
measure of center
measure of variation
mean
median
range

Lesson 9.1 Introduction to Statistics

EXPLORATION 1

Using Data to Answer a Question
Work with a partner.
a. Use your pulse to find your heart rate in beats per minute.
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.1 1
b. Collect the recorded heart rates of the students in your class, including yourself. How spread out are the data? Use a diagram to justify your answer.
c. REASONING How would you answer the following question by using only one value? Explain your reasoning.
“What is the heart rate of a sixth-grade student?”
Answer: Your pulse is measured by counting the number of times your heart beats in one minute. For example, if your heart contracts 72 times in one minute, your pulse would be 72 beats per minute (BPM).

EXPLORATION 2

Identifying Types of Questions
Work with a partner.
a. Answer each question on your own. Then compare your answers with your partner’s answers. For which questions should your answers be the same? For which questions might your answers be different?
1. How many states are in the United States?
Answer: There are 50 states in the United States.

2. How much does a movie ticket cost? Math Practice
Answer: $9.16
3. What color fur do bears have? Build Arguments How can comparing your answers help you support your conjecture?
Answer: The color white becomes visible to our eyes when an object reflects back all.

4. How tall is your math teacher?
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.1 2
b. CONJECTURE
Some of the questions in part(a) are considered statistical questions. Which ones are they? Explain.
Answer: 5.10 inches

Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.1 3

Statistics is the science of collecting, organizing, analyzing, and interpreting data. A statistical question is one for which you do not expect to get a single answer. Instead, you expect a variety of answers, and you are interested in the distribution and tendency of those answers.

Try It
Determine whether the question is a statistical question. Explain.
Question 1.
What types of cell phones do students have in your class?
Answer:
Smartphones, Cell phones give students access to tools and apps that can help them complete and stay on top of their class work. These tools can also teach students to develop better study habits, like time management and organization skills.

Question 2.
How many desks are in your classroom?
Answer: 25

Question 3.
How much do virtual-reality headsets cost?
Answer: $499

Question 4.
How many minutes are in your lunch period?
Answer: 45 minutes

A dot plot uses a number line to show the number of times each value in a data set occurs. Dot plots show the spread and the distribution of a data set.

Question 5.
Repeat parts (a)–(c)using the dot plot below that shows the times of students in a 100-meter race.
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.1 7
Answer:

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.
Question 6.
VOCABULARY
What is a statistical question? Give an example and a non-example.
Answer:
Eg for statistical question: a. How much do bags of pretzels cost at the grocery store?
Because you can anticipate that the prices will vary, it is a statistical question. table at the right may represent the prices of several bags of pretzels at a grocery store.
Eg for non-statistical question: b. How many days does your school have off for spring break this year?
Answer: Because there is only one answer, it is not a statistical question.

Question 7.
OPEN-ENDED
Write and answer a statistical question using the dot plot. Then find and interpret the number of data values.
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.1 8
Answer: There are 16 data values on the dot plot.

Question 8.
You record the amount of snowfall each day for several days. Then you create the dot plot.
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.1 11
a. Find and interpret the number of data values on the dot plot.
Answer: There are 13 data values on the dot plot.

b. How can you collect these data? What are the units?
Answer: We can collect the data by using the dots given in the above figure.
c. Write a statistical question that you can answer using the dot plot. Then answer the question.
Answer: dot plots are best used to show a distribution of data.

Question 9.
You conduct a survey to answer, “How many hours does a typical sixth-grade student spend exercising during a week?” Use the data in the table to answer the question.
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.1 12
Answer:
Given the data
5, 1, 5, 3, 5, 4, 5, 2, 5, 4, 3, 4, 6, 5, 6
The typical sixth-grade student spend exercising during a week is 6 hours.

Introduction to Statistics Homework & Practice 9.1

Review & Refresh

Solve the inequality. Graph the solution.
Question 1.
x – 16 > 8
Answer: x>3.

big ideas math answers grade 6 chapter 9 statistical measures img_1

Question 2.
p + 6 ≤ 8
Answer:   p ≤ 2

big ideas math answers grade 6 chapter 9 statistical measures img_2

Question 3.
54 > 6k
Answer: 9>k

big ideas math answers grade 6 chapter 9 statistical measures img_3

Question 4.
\(\frac{m}{12}\) ≥ 3
Answer: m ≤ 36

Tell whether the ordered pair is a solution of the equation.
Question 5.
y = 4x; (2, 8)
Answer: The given ordered pair is a solution of the equation.
Given : y = 4x;(2,8)
y=8;x=2
8=4 × 2
8=8 (satisfied)

Question 6.
y = 3x + 5; (3, 15)
Answer: Given order pair is not an absolute solution of ordered pair
Given: y = 3x + 5; (3, 15)
y=15;x=3
15=3(3)+5
15=9+5
15=14 (not satisfied)

Question 7.
y = 6x – 15; (4, 9)
Answer:
The given ordered pair is a solution of the equation.
Given: y = 6x – 15; (4, 9)
9=6(4)-15
9=24-15
9=9

Question 8.
A point is reflected in the x-axis. The reflected point is (4, −3). What is the original point?
A. (-3, 4)
B. (-4, 3)
C. (-4, -3)
D. (4, 3)
Answer: B,(-4,3)

Order the numbers from least to greatest.
Question 9.
24%, \(\frac{1}{4}\) , 0.2, \(\frac{7}{20}\) , 0.32
Answer:0.24,0.25,0.2.0.35,0.32
0.2,0.24,0.32,0.35

Question 10.
\(\frac{7}{8}\), 85%, 0.88, \(\frac{3}{4}\) , 78%
Answer:0.875,0.78,0.88,0.75,0.78
0.75,0.78,0.85,0.875,0.88

Concepts, Skills, &Problem Solving

IDENTIFYING TYPES OF QUESTIONS Answer the question. Tell whether your answer should be the same as your classmates’. (See Exploration 2, p. 413.)
Question 11.
How many inches are in 1 foot?
Answer: 12 inches

Question 12.
How many pets do you have?
Answer: none

Question 13.
On what day of the month were you born?
Answer: 27th April

Question 14.
How many senators are in Congress?
Answer: The Senate is composed of 100 Senators, 2 for each state. Until the ratification of the 17th Amendment in 1913, Senators were chosen by state legislatures, not by popular vote. Since then, they have been elected to six-year terms by the people of each state.

IDENTIFYING STATISTICAL QUESTIONS
Determine whether the question is a statistical question. Explain.
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.1 13
Question 15.
What are the eye colors of sixth-grade students?
Answer: brown

Question 16.
At what temperature (in degrees Fahrenheit) does water freeze?
Answer: 32 degrees Fahrenheit

Question 17.
How many pages are in the favorite books of students your age?
Answer: 200 pages

Question 18.
How many hours do sixth-grade students use the Internet each week?
Answer: 1.5 hour each

Question 19.
MODELING REAL LIFE
The vertical dot plot shows the heights of the players on a recent NBA championship team.
a. Find and interpret the number of data values on the dot plot.
b. How can you collect these data? What are the units?
c. Write a statistical question that you can answer using the dot plot. Then answer the question.
Answer:

Question 20.
MODELING REAL LIFE
The dot plot shows the lengths of earthworms.
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.1 14
a. Find and interpret the number of data values on the dot plot.
Answer: There are 21 data values on the plot.
b. How can you collect these data? What are the units?
Answer: Based on dot plots and units are measured in mm.
c. Write a statistical question that you can answer using the dot plot. Then answer the question.
Answer: Find the mode of the length of earthworms using the dot plot.
23 is repeated times.
So, the mode is 23.

DESCRIBING DATA
Display the data in a dot plot. Identify any clusters, peaks, or gaps in the data.
Question 21.
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.1 15
Answer:
bim grade 6 chapter 9 statictical measures answers key img_5

Data are clustered around 22 and around 25
Peak at 25
The gap between 16 and 21

Question 22.
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.1 16
Answer:
bim grade 6 chapter 9 statictical measures answers key img_6

No clusters
Peak at 83
No gaps

INTERPRETING DATA
The dot plot shows the speeds of cars in a traffic study. Estimate the speed limit. Explain your reasoning.
Question 23.
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.1 17
Answer: Most of the data clustered around 44 and 45 , hence the estimated speed is between 44-45 miles per hour

Question 24.
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.1 18
Answer: Most of the data clustered around 65 , there is a peak at 65 and gaps between”60-62″ and 63-65.

Question 25.
DIG DEEPER!
You conduct a survey to answer, “How many hours does a sixth-grade student spend on homework during a school night?” The table shows the results.
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.1 19
a. Is this a statistical question? Explain.
Answer: yes, it is a statistical question because students work in the different time zone based on individual student capacity.
b. Identify any clusters, peaks, or gaps in the data.
Answer: cluster is around 2. There is a peak at 2 and there is no gap.
c. Use the distribution of the data to answer the question.
Answer: A total of 29 data values are distributed.

RESEARCH
Use the Internet to research and identify the method of measurement and the units used when collecting data about the topic.
Question 26.
wind speed
Answer: The instruments used to measure wind are known as anemometers and can record wind speed, direction, and the strength of gusts. The normal unit of wind speed is the knot (nautical mile per hour = 0.51 m sec-1 = 1.15 mph).

Question 27.
amount of rainfall
Answer:
The standard instrument for the measurement of rainfall is the 203mm (8 inches) rain gauge. This is essentially a circular funnel with a diameter of 203mm which collects the rain into a graduated and calibrated cylinder. The measuring cylinder can record up to 25mm of precipitation

Question 28.
earthquake intensity
Answer: The Richter scale measures the largest wiggle (amplitude) on the recording, but other magnitude scales measure different parts of the earthquake. The USGS currently reports earthquake magnitudes using the Moment Magnitude scale, though many other magnitudes are calculated for research and comparison purposes.

Question 29.
REASONING
Write a question about letters in the English alphabet that is not a statistical question. Then write a question about letters that is a statistical question. Explain your reasoning.
Answer: Statistical Question: How many letters in the English alphabet are used to spell a student’s name in class?
Reasoning: The original question has one answer. This Question will have many answers.

Question 30.
REASONING
A bar graph shows the favorite colors of 30 people. Does it make sense to describe clusters in the data? peaks? gaps? Explain.
Answer: No, It doesn’t make sense to describe the distribution. Colors are not measures numerically.

Lesson 9.2 Mean

EXPLORATION 1

Finding a Balance Point
Work with a partner. The diagrams show the numbers of tokens brought to a batting cage. Where on the number line is the data set balanced ? Is this a good representation of the average? Explain.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures 9.2 1

EXPLORATION 2

Finding a Fair Share
Work with a partner. One token lets you hit 12 baseballs in a batting cage. The table shows the numbers of tokens six friends bring to the batting cage.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures 9.2 2
a. Regroup the tokens so that everyone has the same amount. How many times can each friend use the batting cage? Explain how this represents a “fair share. “Use Clear Definitions What does it mean for data to have an average? How does this help you answer the question?
b. how can you find the answer in part(a) algebraically?
c. Write a statistical question that can be answered using the value in part(a).
Answer:

Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures 9.2 3

Try It

Find the mean of the data.
Question 1.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures 9.2 6
Answer:
The sum of the data/no of values
The sum of the data=45+54+13+44+89+60+9+18;
no of values=8
The sum of the data=332:no of values=8; 332/8=41.5 is the mean of the data

Question 2.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures 9.2 7
Answer:
555 is mean for the above-given data.

Question 3.
WHA IT?
The monthly rainfall in May was 0.5 inch in City A and 2 inches in City B. Does this affect your answer in Example 2? Explain.
Answer:

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.
Question 4.
NUMBER SENSE
Is the mean always equal to a value in the data set? Explain.
Answer: It is the value that is most common. You will notice, however, that the mean is not often one of the actual values that you have observed in your data set. In addition, the mean is the only measure of central tendency where the sum of the deviations of each value from the mean is always zero.

Question 5.
WRITING
Explain why the mean describes a typical value in a data set.
Answer:
A measure of central tendency is a single value that attempts to describe a set of data by identifying the central position within that set of data. The mean (often called the average) is most likely the measure of central tendency that you are most familiar with, but there are others, such as the median and the mode.

Question 6.
NUMBER SENSE
What can you determine when the mean of one data set is greater than the mean of another data set? Explain your reasoning.
Answer:

Question 7.
COMPARING MEANS
Compare the means of the data sets.
Data set A: 43, 32, 16, 41, 24, 19, 30, 27
Data set B: 44, 18, 29, 24, 36, 22, 26, 21
Answer:
An outlier is a data value that is much greater or much less than the other values. When included in a data set, it can affect the mean.

Question 8.
DIG DEEPER!
The monthly numbers of customers at a store in the first half of a year are 282, 270, 320, 351, 319, and 252. The monthly numbers of customers in the second half of the year are 211, 185, 192, 216, 168, and 144. Compare the mean monthly customers in the first half of the year with the mean monthly customers in the second half of the year.
Answer:

Question 9.
The table shows tournament finishes for a golfer. What place does the golfer typically finish in tournaments? Explain how you found your answer.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures 9.2 12
Answer: Mean=sum of data/number of data values
Mean=118/16
Mean=7.375
a. The golfer’s mean finish was about 7th
b. The finishes 37 and 26 are much greater than other finishes. They are outliers

Mean Homework & Practice 9.2

Review & Refresh

Determine whether the question is a statistical question. Explain.
Question 1.
How tall are sixth-grade students?
Answer: The average height for a sixth grader (age 12) is about five feet. Girls tend to be about an inch taller on average. But there is a wide range. Any height from about 52 inches (4′4″) to 65 inches (5′5″) is in the normal range according to the CDC.

Question 2.
How many minutes are there in 1 Year?
Answer:
An average Gregorian year is 365.2425 days (52.1775 weeks, 8765.82 hours, 525949.2 minutes, or 31556952 seconds). For this calendar, a common year is 365 days (8760 hours, 525600 minutes, or 31536000 seconds), and a leap year is 366 days (8784 hours, 527040 minutes, or 31622400 seconds).

Question 3.
How many counties are in Tennessee?
Answer: Tennessee’s 95 counties are divided into four TDOT regions. Regional offices are located in Jackson (Region 4), Nashville (Region 3), Chattanooga (Region 2), and Knoxville (Region 1).

Question 4.
What is a student’s favorite sport?
Answer: cricket

Write the percent as a fraction or mixed number in simplest form.
Question 5.
84%
Answer:0.84

Question 6.
71%
Answer:0.71

Question 7.
353%
Answer:3.53

Question 8.
0.2%
Answer:0.002

Divide. Check your answer.
Question 9.
11.7 ÷ 9
Answer:1.3

Question 10.
\(\sqrt [ 5 ]{ 72.8 } \)
Answer: 2.35

Question 11.
\(\sqrt [ 6.8 ]{ 28.56 } \)
Answer: 1.63

Question 12.
93 ÷ 3.75
Answer:24.8

Concepts, Skills, & Problem Solving

FINDING A FAIR SHARE Regroup the amounts so that each person has the same amount. What is the amount? (See Exploration 2, p. 419.)
Question 13.
Dollars brought by friends to a fair: 11, 12, 12, 12, 12, 12, 13
Answer:
Given : 11,12,12,12,12,12,13.
Mean=Sum of data/number of data values
Mean=84/7
Mean=12
Answer = 12 dollars for each friend

Question 14.
Tickets earned by friends playing an arcade game: 0, 0, 0, 1, 1, 2, 3
Answer:
Given : 0,0,0,1,1,2,3.
Mean=Sum of data/number of data values
Mean= 7/7
Mean=1
Answer = 1 Tickets each friend

FINDING THE MEAN
Find the mean of the data.
Question 15.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures 9.2 13
Answer: 2 is the mean of the data.

Question 16.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures 9.2 14
Answer: 3 is the mean of the above-given data.

Question 17.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures 9.2 15
Answer: 103 is the mean of the above-given data

Question 18.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures 9.2 16
Answer: 14.8 is the mean of the above-given data.

Question 19.
MODELING REAL LIFE
You and your friends are watching a television show. One of your friends asks, “How long are the commercial breaks during this show?”Break Times (minutes)
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures 9.2 17
a. Is this a statistical question? Explain.
Answer: Yes it is a statistical question.

b.Use the mean of the values in the table to answer the question.
Answer:
Given the data,
4.2, 3.5, 4.55, 2.75, 2.25
x̄ = (4.2 + 3.5 + 4.55 + 2.75 + 2.25)/5
x̄ = 17.25/5
= 3.45

Question 20.
MODELING REAL LIFE
The table shows the monthly rainfall amounts at a measuring station.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures 9.2 18
a. What is the mean monthly rainfall?
Answer:
x̄ = (22.5 + 1.51 + 1.86 + 2.06 + 3.48 + 4.47 + 3.37 + 5.40 + 5.45 + 4.34 + 2.64 + 2.14)/12
= 33.54/12
= 2.795

b. Compare the mean monthly rainfall for the first half of the year with the mean monthly rainfall for the second half of the year.
Answer:
Mean:
x̄ = (22.5 + 1.51 + 1.86 + 2.06 + 3.48 + 4.47)/6
= 15.6/6
= 2.6
For second 6 months:
x̄ = (3.37 + 5.40 + 5.45 + 4.34 + 2.64 + 2.14)/6
= 23.34/6
= 3.89
The mean value of the second 6 months is greater than the first 6 months.

Question 21.
OPEN-ENDED
Create two different data sets that have six values and a mean of 21.
Answer:
Mean of 21:
Set 1:
12, 31, 21, 24, 13, 25 for these numbers we can calculate the mean we get 21
Set 2:
12, 31, 20, 30, 10, 18 for these numbers we can calculate the mean we get 21

Question 22.
MODELING REAL LIFE
The bar graph shows your cell phone data usage for five months. Describe how the outlier affects the mean. Then use the data to answer the statistical question, “How much cell phone data do you use in a month?”
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures 9.2 19
Answer: 288 is a lot less than the other data values so it is an outlier
Mean with outlier=10/5
Mean with outlier = 2
Mean without outlier = 6.18/5
Mean without outlier = 1.236
The outlier causes the mean to be about 0.76 data usage.

Question 23.
MODELING REAL LIFE
The table shows the heights of the volleyball players on two teams. Compare the mean heights of the two teams. Do outliers affect either mean? Explain.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures 9.2 20
Answer:
Dolphins=59+65+53+56+58+61+64+68+51+56+54+57=702
Total no of observations=12;Mean=702\12=58.5
Tigers=63+68+66+58+54+55+61+62+53+70+64+64=683
Total no of observations=12; Mean=683/12=56.9

Question 24.
REASONING
Use a dot plot to explain why the mean of the data set below is the point where the data set is balanced.
11, 13, 17, 15, 12, 18, 12
Answer:
mean = (11 + 13 + 17 + 15 + 18 + 12)/6
= 86/6
= 14.3

Question 25.
DIG DEEPER!
In your class, 7 students do not receive a weekly allowance, 5 students receive $3, 7 students receive $5, 3 students receive $6, and 2 students receive $8.
a. What is the mean weekly allowance? Explain how you found your answer.
b. A new student who joins your class receives a weekly allowance of $3.50. Without calculating, explain how this affects the mean.
Answer:
Given number of students receive no amount = 7
Number of students receive $3 = 5
Then, total amount 5 students receive = 5 × 3 = $15
Then, total amount 7 students receive = 5 × 7 = $35
Number of students receive $6 = 3
Then total amount 3 students receive = 6 × 3 = $18
Number of students receive $8 = 2
Then, total amount 2 students receive = 2 × 8 = $16
Now, the total amount all students receive =
15 + 35 + 18 + 6 = 84
The total students = 7 + 5 + 7 + 3 + 2 = 24
Mean = total amount/total amount = 84/24 = $3.5
Hence, the mean weekly allowance is $3.5

Question 26.
PRECISION
A collection of 8 geodes has a mean weight of 14 ounces. A different collection of 12 geodes has a mean weight of 14 ounces. What is the mean weight of the 20 geodes? Explain how you found your answer.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures 9.2 21
Answer:
Given,
A collection of 8 geodes has a mean weight of 14 ounces.
A different collection of 12 geodes has a mean weight of 14 ounces.
Total weight of the first 8 backpacks
8×14
112 pounds
Total weight of the second 12 backpacks
12×9
108
Total weight of the whole 20 backpacks
112+108
220
So the mean weight of the 20 backpacks
220 / 20
11

Lesson 9.3 Measures of Center

EXPLORATION 1

Finding the Median
Work with a partner.
a. Write the total numbers of letters in the first and last names of 15 celebrities, historical figures, or people you know. One person is already listed for you.
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures 9.3 1

Dr. B. R. Ambedkar-8
Otto von Bismarck-15
A. P. J. Abdul Kalam-10
Vallabhbhai Patel-16
Alexander Hamilton-17
Jawaharlal Nehru -15
Mother Teresa -12
Thomas Jefferson-15
J. R. D. Tata -4
Indira Gandhi -12
Sachin Tendulkar-15
Napoleon Bonaparte-17
John Adams-9
Karl Marx-8
Andrew Jackson-13
b. Order the values in your data set from least to greatest. Then write the data on a strip of grid paper with 15 boxes.
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures 9.3 2
c. The middle value of the data set is called the median. The value (or values) that occur most often is called the mode. Find the median and the mode of your data set. Explain how you found your answers.
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures 9.3 3
d. Why are the median and the mode considered averages of a data set?
Answer:

Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures 9.3 4

A measure of center is a measure that describes the typical value of a data set. The mean is one type of measure of center. Here are two others.

Try It

Question 1.
Find the median and mode of the data.1, 2, 20, 4, 17, 8, 12, 9, 5, 20, 13
Answer: Given the data,
1, 2, 20, 4, 17, 8, 12, 9, 5, 20, 13
First, write the numbers in the ascending or descending order.
1, 2, 4, 5, 8, 9, 12, 13, 17, 20, 20
The Median is 9.
The mode is 20 because it is repeated more than once.

Question 2.
100, 75, 90, 80, 110, 102
Answer:
Given the data,
100, 75, 90, 80, 110, 102
First, write the numbers in the ascending or descending order.
75, 80, 90, 100, 102, 110
= (90+100)/2
= 85
Mode:
No mode in the data.

Question 3.
One member of the class was absent and ends up voting for horror. Does this change the mode? Explain.
Answer: No

Question 4.
The times (in minutes) it takes six students to travel to school are 8, 10, 10, 15, 20, and 45. Find the mean, median, and mode of the data with and without the outlier. Which measure does the outlier affect the most?
Answer:
Median:
Write the numbers in ascending or descending order
8, 10, 10, 15, 20, and 45
= (10 + 15)/2 = 25/2 = 12.5
Mode:
10 is the mode. Because it is the most repeated number.
Mean:
Adding up the values and then dividing by the number of values.
= (8 + 10 + 10 + 15 + 20 + 45)/6
= 108/6
= 18

Question 5.
WHAT IF?
The store decreases the price of each video game by$3. How does this decrease affect the mean, median, and mode?
Answer:

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.
Question 6.
FINDING MEASURES OF CENTER
Consider the data set below.
15, 18, 13, 11, 12, 21, 9, 11
a. Find the mean, median, and mode of the data.

Answer:
Given the data,
15, 18, 13, 11, 12, 21, 9, 11
x̄ = (15 + 18 + 13 + 11 + 12 + 21 + 9 + 11)/8
x̄ = 110/8
x̄ = 13.75
Median:
Write the numbers in ascending order and descending order.
9, 11, 11, 12, 13, 15, 18, 21
= (12 + 13)/2
= 12.5
Mode:
11 is the mode because this is repeated more than one time.

b. Each value in the data set is decreased by 7. How does this change affect the mean, median, and mode?
Answer:
Each value is decreased by 7 in the given data
8, 11, 6, 4, 5, 14, 2, 4
x̄ = (8 + 11 + 6 + 4 + 5 + 14 + 2 + 4)/8
x̄ = 54/8
x̄ = 6.75

Question 7.
WRITING
Explain why a typical value in a data set can be described by the median or the mode.
Answer:
For data from skewed distributions, the median is better than the mean because it isn’t influenced by extremely large values. The mode is the only measure you can use for nominal or categorical data that can’t be ordered

Question 8.
How does removing the outlier affect your answer in Example 5?
Answer:

Question 9.
It takes 10 contestants on a television show 43, 41, 62, 40, 44, 43, 44, 46, 45, and 41 seconds to cross a canyon on a zipline. Find the mean, median, and mode of the data with and without the outlier. Which measure does the outlier affect the most?
Answer:

Question 10.
The table shows the weights of several great white sharks. Use the data to answer the statistical question, “What is the weight of a great white shark?”
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures 9.3 12
Answer:

Measures of Center Homework & Practice 9.3

Review & Refresh

Find the mean of the data.
Question 1.
1, 5, 8, 4, 5, 7, 6, 6, 2, 3
Answer: 4.7

Explanation:
Given the data,
1, 5, 8, 4, 5, 7, 6, 6, 2, 3
x̄ = ∑x/n
x̄ = (1 + 5 + 8 + 4 + 5 + 7 + 6 + 6 + 2 + 3)/16
x̄ = 49/16
x̄ = 3.06

Question 2.
9, 12, 11, 11, 10, 7, 4, 8
Answer: 9

Explanation:
Given the data,
9, 12, 11, 11, 10, 7, 4, 8
x̄ = ∑x/n
x̄ = (9 + 12 + 11 + 11 + 10 + 7 + 4 + 8)/8
x̄ = 72/8
x̄ = 9

Question 3.
26, 42, 31, 50, 29, 37, 44, 31
Answer: 36.25

Explanation:
Given the data,
26, 42, 31, 50, 29, 37, 44, 31
x̄ = ∑x/n
x̄ = (26+42+31+50+29+37+44+31)/8
x̄ = 290/8
x̄ = 36.25

Question 4.
53, 45, 43, 55, 28, 21, 61, 29, 24, 40, 27, 42
Answer: 39

Explanation:
Given the data,
53, 45, 43, 55, 28, 21, 61, 29, 24, 40, 27, 42
x̄ = ∑x/n
x̄ = (53+45+43+55+28+21+61+29+24+40+27+42)/12
x̄ = 468/12
x̄ = 39

Question 5.
A shelf in your room can hold at most 30 pounds.  ere are 12 pounds of books already on the shelf. Which inequality represents the number of pounds you can add to the shelf?
A. x < 18
B. x ≥ 18
C. x ≤ 42
D. x ≤ 18
Answer: x ≤ 18

Explanation:
12+x ≤ 30
12+x -12 ≤ 30-12
x ≤ 18

Find the missing values in the ratio table. Then write the equivalent ratios.
Question 6.
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures 9.3 13
Answer:
Big-Ideas-Math-Answers-Grade-6-Chapter-9-Statistical-Measures-9.3-13

Question 7.
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures 9.3 14
Answer:
Big-Ideas-Math-Answers-Grade-6-Chapter-9-Statistical-Measures-9.3-14

Find the surface area of the prism.

Question 8.
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures 9.3 15
Answer:
Given,
l = 6m
w = 5m
h = 5m
We know that,
Surface Area of the Prism = 2lw + 2lh + 2hw
= 2(6 × 5) + 2(6 × 8) + 2(8 × 5)
= 60 + 96 + 80
= 236 sq. meters

Question 9.
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures 9.3 16
Answer:
Given,
l = 4.5 ft
w = 2ft
h = 3.5ft
We know that,
Surface Area of the Prism = 2lw + 2lh + 2hw
= 2(4.5 × 2) + 2(4.5 × 3.5) + 2(2 × 3.5)
= 18 + 31.5 + 14
= 63.5 sq. ft

Question 10.
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures 9.3 17
Answer:
Given,
l = 6 yd
w = 4 yd
h = 2 yd
We know that,
Surface Area of the Prism = bh + 2lh + lb
= 2 × 4 + 2(6 × 5) + 6 × 2
= 8 + 60 + 12
= 80 sq. yards

Concepts, Skills, & Problem Solving

FINDING THE MEDIAN Use grid paper to find the median of the data. (See Exploration 1, p. 425.)
Question 11.
9, 7, 2, 4, 3, 5, 9, 6, 8, 0, 3, 8
Answer:
First, arrange the numbers in ascending or descending order.
= 0, 2, 3, 3, 4, 5, 6, 7, 8, 8, 9, 9
= (5 + 6)/2
= 11/2
= 5.5

Question 12.
16, 24, 13, 36, 22, 26, 22, 28, 25
Answer:
First, arrange the numbers in ascending or descending order.
13, 16, 22, 22, 24, 25, 26, 28, 36
24 is the median.
The median is the middle score in a set of given data.

FINDING THE MEDIAN AND MODE
Find the median and mode of the data.
Question 13.
3, 5, 7, 9, 11, 3, 8
Answer: The Median is 7; The Mode is 3.
Given: 3, 5, 7, 9, 11, 3, 8
Sorted list: 3,3,5,7,8,9,11
Median is the middle number in a sorted list of numbers = 7
The mode is the value that appears most frequently in a data set = 3

Question 14.
14, 19, 16, 13, 16, 14
Answer: The Median is 15; The Modes are 14 and 16.
Given: 13,14,14,16,16,19
Sorted list: 14, 19, 16, 13, 16, 14
Median is the middle number in a sorted list of numbers = 15
The mode is the value that appears most frequently in a data set = 14,16

Question 15.
16. 93, 81, 94, 71, 89, 92, 94, 99
Answer: The Median is 90.5; The Mode is 94.
Given: 16, 93, 81, 94, 71, 89, 92, 94, 99
Sorted list: 16,71,81,89,92,93,94,94,99
Median is the middle number in a sorted list of numbers = 92
The mode is the value that appears most frequently in a data set = 94

Question 16.
44, 13, 36, 52, 19, 27, 33
Answer: The Median is 33; There are no modes.
Given: 44, 13, 36, 52, 19, 27, 33
Sorted list: 13,19,27,33,36,44,52
Median is the middle number in a sorted list of numbers = 33
The mode is the value that appears most frequently in a data set = no mode

Question 17.
12, 33, 18, 28, 29, 12, 17, 4, 2
Answer: The Median is 17; The Modes are 12.
Given: 12, 33, 18, 28, 29, 12, 17, 4, 2
Sorted list: 2,4,12,12,17,18,28,29,33
Median is the middle number in a sorted list of numbers = 17
The mode is the value that appears most frequently in a data set = 12

Question 18.
55, 44, 40, 55, 48, 44, 58, 67
Answer:
The Median is 51.5
The Modes are 44 and 55.
Given: 55, 44, 40, 55, 48, 44, 58, 67
Sorted list: 40,44,44,48,55,55,58,67
Median is the middle number in a sorted list of numbers = 51.5
The mode is the value that appears most frequently in a data set = 44,55

Question 19.
YOU BE THE TEACHER
Your friend finds the median of the data. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures 9.3 18
Answer: No, first the given data is arranged in ascending order then after median is to be found. The median is 55

FINDING THE MODE
Find the mode of the data.
Question 20.
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures 9.3 19
Answer: The modes are Black and Blue.

Question 21.
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures 9.3 20
Answer: The modes are singing, dancing, comedy.

Question 22.
REASONING
In Exercises 20 and 21, can you find the mean and median of the data? Explain.
Answer: You can’t find the mean and median in exercises 20 and 21.
The data set is not made up of numbers

FINDING MEASURES OF CENTER
Find the mean, median, and mode of the data.
Question 23.
4.7, 8.51, 6.5, 7.42, 9.64, 7.2, 9.3
Answer: Given: 4.7, 8.51, 6.5, 7.42, 9.64, 7.2, 9.3
Sorted list: 4.7, 6.5, 7.2, 7.42, 8.51, 9.64
Mean: x̄ = ∑x/n
x̄ = (4.7+6.5+7.2+7.42+8.51+9.64)/6
x̄ = 43.97/6
x̄ =7.32
Median: 7.42.
Mode: no mode.

Question 24.
8\(\frac{1}{2}\), 6\(\frac{5}{8}\), 3\(\frac{1}{8}\), 5\(\frac{3}{4}\), 6\(\frac{5}{8}\), 5\(\frac{1}{4}\), 10\(\frac{5}{8}\), 4\(\frac{1}{2}\)
Answer: Given: 8.5, 6.62, 3.12, 5.75, 6.62, 5.25, 10.62, 4.5
Sorted list: 3.12, 4.5, 5.25, 5.75, 6.62, 6.62, 8.5, 10.62
Mean: x̄ = ∑x/n
x̄ = (3.12, 4.5, 5.25, 5.75, 6.62, 6.62, 8.5, 10.62)/8
x̄ =
x̄ =
Median: 6.18
Mode: 6.62

Question 25.
MODELING REAL LIFE
The weights (in ounces) of several moon rocks are shown in the table. Find the mean, median, and mode of the weights.
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures 9.3 21
Answer:
Mean
x̄ = (2.2 + 2.2 + 3.2 + 2.4 + 2.8 + 3.4 + 2.6 + 3.0 + 2.5)/9
Median:
Write the moon rock weights in ascending or descending order.
2.6 is the median
Mode:
2.2 is repeated move times
So, 2.2 is the mode.

REMOVING AN OUTLIER Find the mean, median, and mode of the data with and without the outlier. Which measure does the outlier affect the most?
Question 26.
45, 52, 17, 63, 57, 42, 54, 58
Answer:
Outliners means removing of the small data value
17 is the outliner
x̄ = ∑x/n
= (45 + 52 + 17 + 63 + 57 + 42 + 54 + 58)/8
= 388/8 = 48.5
Mean without outliner:
= (45 + 52 + 63 + 57 + 42 + 54 + 58)/7
= 371/7 = 53
Median with outliner:
17, 42, 45, 52, 54, 57, 58, 63
= (52 + 54)/2
= 106/2
= 53
Median without outliner:
42, 45, 52, 54, 57, 58, 63
54 is the median
Mode:
There is no change of value in the without outliner and with the outliner.
So, there is no mode in the data values.

Question 27.
85, 77, 211, 88, 91, 84, 85
Answer:
77 is the outliner
Mean with outliner:
x̄ = (85 + 77 + 211 + 88 + 91 + 84 + 85)/7
=721/7
= 103
Mean without outliner:
x̄ = (85 + 211 + 88 + 91 + 84 + 85)/6
= 644/6
= 107
Median with outliner:
Write the data values in ascending or descending order.
77, 84, 85, 88, 91, 211
85 is the median.
Median without outliner:
84, 85, 85, 88, 91, 211
= (85 + 88)/2
= 173/2
= 86.5
Mode:
There is no change of value in the without outliner and with the outliner.
85 is the mode.

Question 28.
23, 73, 45, 27, 23, 25, 43, 45
Answer:
73 is the outliner
Mean with outliner:
Mean = (23 + 45 + 27 + 23 + 25 + 43 + 45)
= 231/7
= 33
Mean with outliner:
Mean = (23 + 45 + 27 + 23 + 25 + 43 + 45+ 73)
= 304/8
= 38

Question 29.
101, 110, 99, 100, 64, 112, 110, 111, 102
Answer:
64 is the outliner
Mean with outliner:
x̄ = (101 + 110 + 99 + 100 + 64 + 112 + 110 + 111 + 102)/9
= 901/9 = 101
Mean with outliner:
x̄ = (101 + 110 + 99 + 100 + 112 + 110 + 111 + 102)/8
= 755/8
= 94.37
Median:
Write the data values in ascending or descending order
64, 99, 100, 101, 102, 110, 111, 112
Median without outliner:
= (101 + 102)/2
= 203/2
= 101.5
Mode:
Mode with and without outliner = 110

Question 30.
REASONING
The table shows the monthly salaries for employees at a company.
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures 9.3 22
a. Find the mean, median, and mode of the data.
b. Each employee receives a 5% raise. Find the mean, median, and mode of the data with the raise. How does this increase affect the mean, median, and mode of the data?
c. How are the mean, median, and mode of the monthly salaries related to the mean, median, and mode of the annual salaries?
Answer:

CHOOSING A MEASURE OF CENTER
Find the mean, median, and mode of the data. Choose the measure that best represents the data. Explain your reasoning.
Question 31.
48, 12, 11, 45, 48, 48, 43, 32
Answer:
Write the data in ascending order or descending order.
11, 12, 32, 43, 45, 48, 48, 48
= (32 + 43)/2
= 75/2
= 37.5
48 is the mode of the data

Question 32.
12, 13, 40, 95, 88, 7, 95
Answer:
Mean:
x̄ = ∑x/n
= (12 + 13 + 40 + 95 + 88 + 7 + 95)/7
= 350/7 = 50
Median:
7, 12, 13, 40, 88, 95, 95
40 is the median
mode:
95 is the mode.

Question 33.
2, 8, 10, 12, 56, 9, 5, 2, 4
Answer:
Mean:
x̄ = ∑x/n
= (2 + 8 + 10 + 12 + 56 + 9 + 5 + 2 + 4)/9
= 108/9
= 12
Median:
2, 2, 4, 5, 8, 9, 10, 12, 56
8 is the median
Mode:
2 is the mode.

Question 34.
126, 62, 144, 81, 144, 103
Answer:
Mean:
x̄ = ∑x/n
= (126 + 62 + 144 + 81 + 144 + 103)6
= 660/60
= 11
Median:
62, 81, 103, 126, 144, 144
= (103 + 126)/2
= 114.5

Question 35.
MODELING REAL LIFE
The weather forecast for a week is shown. Which measure of center best represents the high temperatures? the low temperatures? Explain your reasoning.
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures 9.3 23
Answer:

Question 36.
RESEARCH
Find the costs of 10 different boxes of cereal. Choose one cereal whose cost will be an outlier.
a. Which measure of center does the outlier affect the most? Justify your answer.
b. Use the data to answer the statistical question, “How much does a box of cereal cost?”
Answer:

Question 37.
PROBLEM SOLVING
The bar graph shows the numbers of hours you volunteered at an animal shelter. What is the minimum number of hours you need to volunteer in the seventh week to justify that you volunteered an average of 10 hours per week for the 7 weeks? Explain your answer using measures of center.
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures 9.3 24
Answer:

Question 38.
REASONING
Why is the mode the least frequently used measure of center to describe a data set? Explain.
Answer:
The mode can be helpful in some analyses, but generally it does not contain enough accurate information to be useful in determining the shape of a distribution. When it is not a “Normal Distribution” the Mode can be misleading, although it is helpful in conjunction with the Mean for defining the amount of skewness in a distribution.

Question 39.
DIG DEEPER!
The data are the prices of several fitness wristbands at a store.
$130 $170 $230 $130
$250 $275 $130 $185
a. Does the price shown in the advertisement represent the prices well? Explain.
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures 9.3 25
b. Why might the store use this advertisement?
c. In this situation, why might a person want to know the mean? the median? the mode? Explain.
Answer:

Question 40.
CRITICAL THINKING
The expressions 3x, 9x, 4x, 23x, 6x, and 3x form a data set. Assume x> 0.
a. Find the mean, median, and mode of the data.
b. Is there an outlier? If so, what is it?
Answer:
Mean: This is an average of all the numbers. Add up the numbers and then divide by how many numbers there are.
(3 + 9 + 4 + 23 + 6 + 3)/6 = 48/6 = 8
Median: The number in the middle, when the numbers are in order. If there are 2 middle numbers, average them together.
3, 3, 4, 6, 9, 23 : 4 and 6 are the middle numbers. 4+6/2 = 10/2 = 5
Mode: What value occurs most frequently? 3 is the only duplicate
Outlier: What value is abnormal to our set of data? All of our numbers are small (single digits), except for 23. That makes it an outlier.

Lesson 9.4 Measures of Variation

EXPLORATION 1

Interpreting Statements
Work with a partner. There are 24 students in your class. Your teacher makes the following statements.
Big Ideas Math Solutions Grade 6 Chapter 9 Statistical Measures 9.4 1
• “The exam scores range from 75% to 96%.”
a. What does each statement mean? Explain.
b. Use your teacher’s statements to make a dot plot that can represent the distribution of the exam scores of the class.
c. Compare your dot plot with other groups’. How are they alike? different?

EXPLORATION 2

Grouping Data
Work with a partner. The numbers of U.S.states visited by students in a sixth-grade class are shown.
Big Ideas Math Solutions Grade 6 Chapter 9 Statistical Measures 9.4 2
a. Represent the data using a dot plot. Between what values do the data range?
b. Use the dot plot to make observations about the data.
c. How can you describe the middle half of the data?

A measure of variation is a measure that describes the distribution of a data set. A simple measure of variation to find is the range. The range of a data set is the difference of the greatest value and the least value.

Big Ideas Math Solutions Grade 6 Chapter 9 Statistical Measures 9.4 3

Try It
Question 1.
The ages of people in line for a roller coaster are 15, 17, 21, 32, 41, 30, 25, 52, 16, 39, 11, and 24. Find and interpret the range of the ages.
Answer:
Given,
The ages of people in line for a roller coaster are 15, 17, 21, 32, 41, 30, 25, 52, 16, 39, 11, and 24.
Range = (upper value – lower value)/2
= (52 – 11)/2
= 41/2
= 20.5

Question 2.
The data are the number of pages in each of an author’s novels. Find and interpret the interquartile range of the data.
356, 364, 390, 468, 400, 382, 376, 396, 350
Answer:
Given,
The data are the number of pages in each of an author’s novels.
356, 364, 390, 468, 400, 382, 376, 396, 350
Lower quartile = 360
Upper quartile = 398
Interquartile range = 38

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.
Question 3.
WRITING
Explain why the variability of a data set can be described by the range or the interquartile range.
Answer:
The interquartile range is the third quartile (Q3) minus the first quartile (Q1). But the IQR is less affected by outliers: the 2 values come from the middle half of the data set, so they are unlikely to be extreme scores. The IQR gives a consistent measure of variability for skewed as well as normal distributions.

Question 4.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Solutions Grade 6 Chapter 9 Statistical Measures 9.4 8
Answer:

Question 5.
The table shows the distances traveled by a paper airplane. Find and interpret the range and interquartile range of the distances.
Big Ideas Math Solutions Grade 6 Chapter 9 Statistical Measures 9.4 11
Answer: Given: 13.5, 12.5, 21, 16.75, 10.25, 19, 32, 26.5, 29,16.25, 28.5, 18.5.

Question 6.
The table shows the years of teaching experience of math teachers at a school. How do the outlier or outliers affect the variability of the data?
Big Ideas Math Solutions Grade 6 Chapter 9 Statistical Measures 9.4 12
Answer:
Given the data
5, 10, 7, 8, 10, 11, 22, 8, 6, 35
22 is added to the data set
22 is the outliner
so there is no effect to measure of center and the measure of variability.

Measures of Variation Homework & Practice 9.4

Review & Refresh

Find the mean, median, and mode of the data.
Question 1.
4, 8, 11, 6, 4, 5, 9, 10, 10, 4
Answer:
Mean = x̄ = (4 + 8 + 11 + 6 + 4 + 5 + 9 + 10 + 10 + 4)/10
= 71/10
= 7.1
Median:
Write the data in ascending or descending order.
4, 4, 4, 5, 6, 8, 9, 10, 10, 11
= (5 + 8)/2
= 13/2
=6.5
Mode:
More number if data repeated is called mode.
4 is the mode.

Question 2.
74, 78, 86, 67, 80
Answer:
Mean = x̄ = (74 + 78 + 86 + 67 + 80)/5
= 385/5
= 77
Median:
Write the data in ascending or descending order.
67, 74, 78, 80, 86
78 is the median
Mode:
There is no mode in the data.

Question 3.
15, 18, 17, 17, 15, 16, 14
Answer:
Mean = x̄ = (15 + 18 + 17 + 17 + 15 + 16 + 14)/7
= 112/7 = 16
Median:
Write the data in ascending or descending order.
14, 15, 15, 16, 17, 17, 18
16 is the median
Mode:
17, 15 are the median.

Question 4.
31, 14, 18, 26, 17, 32
Answer:
Mean:
x̄ = (31 + 14 + 18 + 26 + 17 + 32)/6
Median:
Write the data in ascending or descending order.
14, 17, 18, 26, 31, 32
= (18 + 26)/2
= 44/2
= 22
Mode:
There is no mode in the data.

Copy and complete the statement using < or >.
Question 5.
Big Ideas Math Solutions Grade 6 Chapter 9 Statistical Measures 9.4 13
Answer:
A negative number is less than the positive number
6 > -7

Question 6.
Big Ideas Math Solutions Grade 6 Chapter 9 Statistical Measures 9.4 14
Answer:
A negative number is less than the positive number
-3 < 0

Question 7.
Big Ideas Math Solutions Grade 6 Chapter 9 Statistical Measures 9.4 15
Answer:
A negative number is less than the positive number
14 > -14

Question 8.
Big Ideas Math Solutions Grade 6 Chapter 9 Statistical Measures 9.4 16
Answer:
A negative number is less than the positive number
8 > -10

Find the surface area of the pyramid.
Question 9.
Big Ideas Math Solutions Grade 6 Chapter 9 Statistical Measures 9.4 17
Answer:
Given,
Length = 12 mm
Height = 14 mm
A = a² + 2a √a²/4 + h²
Area = 509.56 sq. mm

Question 10.
Big Ideas Math Solutions Grade 6 Chapter 9 Statistical Measures 9.4 18
Answer:
Given,
Length = 5 in
Height = 8.5 in
A = a² + 2a √a²/4 + h²
Area = 113.6 sq. inches

Question 11.
Big Ideas Math Solutions Grade 6 Chapter 9 Statistical Measures 9.4 19
Answer:
Given,
Length = 6 ft
Height = 9 ft
A = a² + 2a √a²/4 + h²
Area = 149.84 sq.ft

Concepts, Skills, &Problem Solving

INTERPRETING STATEMENTS There are 20 students in your class. Your teacher makes the two statements shown. Use your teacher’s statements to make a dot plot that can represent the distribution of the scores of the class. (See Exploration 1, p. 433.)
Question 12.
“The quiz scores range from 65% to 95%.”
“The scores were evenly spread out.”
Answer:

Question 13.
“The project scores range from 78% to 93%.”
“Most of the students received low scores.”
Answer:

FINDING THE RANGE Find the range of the data.
Question 14.
4, 8, 2, 9, 5, 3
Answer: 7

Explanation:
Range is the difference of higher value and lower value
lowest value = 2
highest value = 9
R = 9 – 2
R = 7

Question 15.
28, 42, 36, 23, 14, 47, 40
Answer: 33

Explanation:
The range is the difference between higher value and lower value
Lowest value: 14
Highest value: 47
Range = 47 – 14
R = 33

Question 16.
26, 21, 27, 33, 24, 29
Answer: 12

Explanation:
The range is the difference between higher value and lower value
Lowest value: 21
Highest value: 33
Range = 33 – 21
R = 12

Question 17.
52, 40, 49, 48, 62, 54, 44, 58, 39
Answer: 23

Explanation:
The range is the difference between higher value and lower value
Lowest value: 39
Highest value: 62
Range = 62 – 39
R = 23

Question 18.
133, 117, 152, 127, 168, 146, 174
Answer: 57

Explanation:
The range is the difference between higher value and lower value
Lowest value: 117
Highest value: 174
Range = 174 – 117
R = 57

Question 19.
4.8, 5.5, 4.2, 8.9, 3.4, 7.5, 1.6, 3.8
Answer: 7.3

Explanation:
The range is the difference of higher value and lower value
Lowest value: 1.6
Highest value: 8.9
Range = 8.9 – 1.6
R = 7.3

Question 20.
YOU BE THE TEACHER
Your friend finds the range of the data. Is your friend correct? Explain your reasoning.
Big Ideas Math Solutions Grade 6 Chapter 9 Statistical Measures 9.4 20
Answer:
The range is the difference between higher value and lower value
Lowest value: 28
Highest value: 59
Range =  59 – 28
Range = 31

FINDING THE INTERQUARTILE RANGE Find the interquartile range of the data.
Question 21.
4, 6, 4, 2, 9, 1, 12, 7
Answer: 6

Explanation:
This simple formula is used for calculating the interquartile range:
IQR = Xu – Xl
Lower quartile (xL): 2.5
Upper quartile (xU): 8.5
IQR = 8.5 – 2.5
IQR = 6

Question 22.
18, 22, 15, 16, 15, 13, 19, 18
Answer: 3.75

Explanation:
This simple formula is used for calculating the interquartile range:
IQR = Xu – Xl
Lower quartile (xL): 15
Upper quartile (xU): 18.75
IQR = 18.75 – 15
= 3.75

Question 23.
40, 33, 37, 54, 41, 34, 27, 39, 35
Answer: 7

Explanation:
This simple formula is used for calculating the interquartile range:
IQR = Xu – Xl
Lower quartile (xL): 33.5
Upper quartile (xU): 40.5
IQR = 40.5 – 33.5
= 7

Question 24.
84, 75, 90, 87, 99, 91, 85, 88, 76, 92, 94
Answer: 8

Explanation:
This simple formula is used for calculating the interquartile range:
IQR = Xu – Xl
Lower quartile (xL): 84
Upper quartile (xU): 92
IQR = 92 – 84
= 8

Question 25.
132, 127, 106, 140, 158, 135, 129, 138
Answer: 12

Explanation:
This simple formula is used for calculating the interquartile range:
IQR = Xu – Xl
Lower quartile (xL): 127.5
Upper quartile (xU): 139.5
IQR = 139.5 – 127.5
= 12

Question 26.
38, 55, 61, 56, 46, 67, 59, 75, 65, 58
Answer: 12.75

Explanation:
This simple formula is used for calculating the interquartile range:
IQR = Xu – Xl
Lower quartile (xL): 52.75
Upper quartile (xU): 65.5
IQR = 65.5  – 52.75
= 12.75

Question 27.
MODELING REAL LIFE
The table shows the number of tornadoes in Alabama each year for several years. Find and interpret the range and interquartile range of the data. Then determine whether there are any outliers.
Big Ideas Math Solutions Grade 6 Chapter 9 Statistical Measures 9.4 21
Answer:
The data is 65, 32, 54, 23, 55, 145,37, 80, 94, 42, 69, 77
Range:
Lowest value: 23
Highest value: 145
R = Highest value – Lowest value
R = 145 – 23
R = 122
IQR:
This simple formula is used for calculating the interquartile range:
IQR = Xu – Xl
Lower quartile (xL): 38.25
Upper quartile (xU): 79.25
IQR = 79.25 – 38.25
= 41

Question 28.
WRITING
Consider a data set that has no mode. Which measure of variation is greater, the range or the interquartile range? Explain your reasoning.
Answer:
It would be based on the set of numbers you have, but in most cases, it is the interquartile range, because the mode is usually closer to the median. This leaves the interquartile range as a larger number.

Question 29.
CRITICAL THINKING
Is it possible for the range of a data set to be equal to the interquartile range? Explain your reasoning.
Answer:
The interquartile range (IQR) is a measure of variability, based on dividing a data set into quartiles. Quartiles divide a rank-ordered data set into four equal parts.

Question 30.
REASONING
How does an outlier affect the range of a data set? Explain.
Answer:
Outlier An extreme value in a set of data that is much higher or lower than the other numbers. Outliers affect the mean value of the data but have little effect on the median or mode of a given set of data.

Question 31.
MODELING REAL LIFE
The table shows the numbers of points scored by players on a sixth-grade basketball team in a season.
Big Ideas Math Solutions Grade 6 Chapter 9 Statistical Measures 9.4 22
a. Find the range and interquartile range of the data.
b. Identify the outlier(s) in the data set. Find the range and interquartile range of the data set without the outlier(s). Which measure does the outlier or outliers affect more?
Answer:

Question 32.
DIG DEEPER!
Two data sets have the same range. Can you assume that the interquartile ranges of the two data sets are about the same? Give an example to justify your answer.
Answer:
Yes,
A data set with the least value of 2 and the greatest value of 20 will have the same range as a data set with the least value of 82 and the greatest value of 100 will have the same range of 18.

Question 33.
MODELING REAL LIFE
The tables show the ages of the finalists for two reality singing competitions.
Big Ideas Math Solutions Grade 6 Chapter 9 Statistical Measures 9.4 23
a. Find the mean, median, range, and interquartile range of the ages for each show. Compare the results.

Answer:
18, 15, 22, 18, 24, 17, 21, 16, 28, 21
Mean:
x̄ = ∑x/n = (18 + 15 + 22 + 18 + 24 + 17 + 21 + 16 + 28 + 21)/10
=200/10 = 20
Median:
15, 16,  17,  18,  18, 21, 22, 24, 28
= (18 + 21)/2
= 39/2
= 19.5
Range:
(28 – 15)/2
= 13/2
= 6.5
interquartile range:
Number of observations: 10
Xl = 16.75
Xu = 22.5
Xu – Xl = 5.75
Ages of show B:
Mean:
x̄ = ∑x/n = (21 + 20 + 23 + 13 + 15 + 18 + 17 + 22 + 36 + 25)/10
= 210/10 = 21
Median:
13, 15, 17, 18, 20, 21, 22, 23, 25, 36
= (20 + 21)/2 = 41/2 = 20.5
Range:
(36 – 13)/2
= 23/2
= 11.5
Interquartile Range:
Samples = 10
Xl = 16.5
Xu = 23.5

b. A 21-year-old is voted off Show A, and the 36-year-old is voted off Show B. How do these changes affect the measures in part(a)? Explain.
Answer:
Mean:
x̄ = ∑x/n = (18 + 17 + 15 + 22 + 16 + 18 + 28 + 24)/8
= 158/8
= 79
Median: 15, 16, 17, 18, 18, 22, 24, 28
(18 + 18)/2
= 36/2
= 18
Range:
(28 – 15)/2
= 13/2
= 6.5
Interquartile Range:
Samples = 8
Xl = 16.25
Xu = 23.5
Interquartile Range = 23.5 – 16.25
= 7.25
21, 20, 23, 13, 15, 18, 17, 22, 25
Mean = (21 + 20 + 23 + 13 + 15 + 18 + 17 + 22 + 25)/9
= 174/2
= 87
Median:
13, 15, 17, 18, 21, 20, 22, 23, 25
21 is the median
Range:
(25 – 13)/2
= 12/2
= 6
Interquartile Range:
data = 9
Xl = 16
Xu = 22.5
(Xu – Xl) = 22.5 – 16
= 6.5
In Part A there is no effect on the range and it affects the mean, median, interquartile.

Question 34.
OPEN-ENDED
Create a set of data with 7 values that has a mean of 30, a median of 26, a range of 50, and an interquartile range of 36.
Answer:
The first thing we need to do is to put the data in increasing order. This is needed to calculate the median:
30,31,32,33,34,35,35,36,37,39

Lesson 9.5 Mean Absolute Deviation

Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.5 1

EXPLORATION 1

Finding Distances from the Mean
Work with a partner. The table shows the exam scores of 14 students in your class.
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.5 2
a. Which exam score deviates the most from the mean? Which exam score deviates the least from the mean? Explain how you found your answers.
b. How far is each data value from the mean?
c. Divide the sum of the values in part(b) by the number of values. In your own words, what does this represent?
d. REASONING Ina data set, what does it mean when the value you found in part(c) is close to 0? Explain.

Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.5 3

Another measure of variation is the mean absolute deviation. The mean absolute deviation is an average of how much data values differ from the mean.

Try It
Question 1.
Find and interpret the mean absolute deviation of the data.
5, 8, 8, 10, 13, 14, 16, 22
Answer: Number of observations : 8
Mean: 12

Question 2.
WHAT IF?
The pitcher allows 4 runs in the next game. How would you expect the mean absolute deviation to change? Explain.
Answer:

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.
Question 3.
WRITING
Explain why the variability of a data set can be described by the mean absolute deviation.
Answer:

Question 4.
FINDING THE MEAN ABSOLUTE DEVIATION
Find and interpret the mean absolute deviation of the data. 8, 12, 4, 3, 14, 1, 9, 13
Answer: number of observations:8
Mean: 8
mean absolute deviation: 4

Question 5.
WHICH ONE do DOESN’T BELONG?
Which one does not belong with the other three? Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.5 6
Answer: MEAN
A mean is different from all the above-given factors
A mean is the simple mathematical average of a set of two or more numbers.
The mean for a given set of numbers can be computed in more than one way, including the arithmetic mean method, which uses the sum of the numbers in the series, and the geometric mean method, which is the average of a set of products.

Question 6.
The tables show the numbers of questions answered correctly by members of two teams on a game show. Compare the mean, median, and mean absolute deviation of the numbers of correct answers for each team. What can you conclude?
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.5 9
Answer:
Tiger sharks
3, 6, 5, 4, 4, 2
Mean: (3 + 6 + 5 + 4 + 4 + 2)/6
= 24/6
= 4
Median:
2, 3, 4, 4, 5, 6
= (4 + 4)/2
= 4
MAD:
Number of observations: 6
Mean = 4
MAD = 1
Bear Cats:
Mean:
6, 1, 4, 1, 8, 4
(6 + 1 + 4 + 1 + 8 + 4)/6
= 24/6
= 4
Median:
1, 1, 4, 4, 6, 8
= (4 + 4)/2
= 4
MAD:
Number of observations: 6
Mean = 4
MAD = 2
The mean, Median, Mean Absolute Deviation of both tiger sharks and Bear Cats are the same.

Question 7.
DIG DEEPER!
The data set shows the numbers of books that students in your book club read last summer.
8, 6, 11, 12, 14, 12, 11, 6, 15, 9, 7, 10, 9, 13, 5, 8
A new student who read 18 books last summer joins the club. Is18 an outlier? How does including this value in the data set affect the measures of center and variation? Explain.
Answer: 8 is added to the dataset.
Yes, 18 is an outliner
No, it does not affect the measures of the center and variation by removing the outliner.
If the outliner is not removed then it affects the measures of center and variation.

Mean Absolute Deviation Homework & Practice 9.5

Review & Refresh

Find the range and interquartile range of the data.
Question 1.
23, 45, 39, 34, 28, 41, 26, 33
Answer:
Number of observations:8
Lower quartile (xL): 26.5
Upper quartile (xU): 40.5
interquartile range = 14
Range:
Number of observations:8
Lowest value: 23
Highest value: 45
Range = 45 – 23
= 22

Question 2.
63, 53, 48, 61, 69, 63, 57, 72, 46
Answer:
Number of observations:9
Lower quartile (xL): 50.5
Upper quartile (xU): 66
interquartile range = 15.5
Range:
Number of observations:9
Lowest value: 46
Highest value: 72
Range = 26

Graph the integer and its opposite.
Question 3.
15
Answer:
Big Ideas Math Grade 6 Chapter 9 Statistics Answer Key img_5

Question 4.
17
Answer:
Big Ideas Math Grade 6 Chapter 9 Statistics Answer Key img_6

Question 16.
– 22
Answer:
Big Ideas Math Grade 6 Chapter 9 Statistics Answer Key img_7

Question 7.
Find the numbers of faces, edges, and vertices of the solid.
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.5 10
Answer:
The name of the solid is a pentagon.
Number of vertices = 5
Number of faces = 5
Numver of edges = 5

Write the word sentence as an equation.
Question 8.
17 plus a number q is 40.
Answer:
We have to write the equation for the word sentence.
The phrase ‘plus’ indicates ‘+’
17 + q = 40

Question 9.
The product of a number s and 14 is 49.
Answer:
We have to write the equation for the word sentence.
The phrase product indicates ‘×’
s × 14 = 49

Question 10.
The difference of a number b and 9 is 32.
Answer:
We have to write the equation for the word sentence.
The phrase difference indicates ‘-‘
b – 9 = 32

Question 11.
The quotient of 36 and a number g is 9.
Answer:
We have to write the equation for the word sentence.
The phrase quotient indicates ‘÷’
36 ÷ g = 9

Concepts, Skills, &Problem Solving

FINDING DISTANCES FROM THE MEAN Find the average distance of each data value in the set from the mean. (See Exploration 1, p. 439.)
Question 12.
Model years of used cars on a lot: 2014, 2006, 2009, 2011, 2005
Answer:

Question 13.
Prices of kites at a shop: $7, $20, $9, $35, $12, $15, $7, $10, $20, $25
Answer:

FINDING THE MEAN ABSOLUTE DEVIATION Find and interpret the mean absolute deviation of the data.
Question 14.
69, 51, 71, 77, 71, 80, 75, 63, 73
Answer:
Given the data
69, 51, 71, 77, 71, 80, 75, 63, 73
Number of samples = 9
Mean Absolute Deviation = 70

Question 15.
94, 86, 95, 99, 88, 90
Answer:
Given the data
94, 86, 95, 99, 88, 90
Number of samples = 6
Mean Absolute Deviation = 92

Question 16.
46, 54, 43, 57, 50, 62, 78, 42
Answer:
Given the data
46, 54, 43, 57, 50, 62, 78, 42
Number of samples = 8
Mean Absolute Deviation = 54

Question 17.
25, 28, 20, 22, 32, 28, 35, 34, 30, 36
Answer:
Given the data
25, 28, 20, 22, 32, 28, 35, 34, 30, 36
Number of samples = 10
Mean Absolute Deviation = 29

Question 18.
101, 115, 124, 125, 173, 165, 170
Answer:
Given the data
101, 115, 124, 125, 173, 165, 170
Number of samples = 7
Mean Absolute Deviation = 139

Question 19.
1.1, 7.5, 4.9, 0.4, 2.2, 3.3, 5.1
Answer:
Given the data
1.1, 7.5, 4.9, 0.4, 2.2, 3.3, 5.1
Number of samples = 7
Mean Absolute Deviation = 3.5

Question 20.
\(\frac{1}{4}, \frac{5}{8}, \frac{3}{8}, \frac{3}{4}, \frac{1}{2}\)
Answer:
Number of observations:5
Mean (x̄): 0.5
Mean Absolute Deviation (MAD): 0.15

Question 21.
4.6, 8.5, 7.2, 6.6, 5.1, 6.2, 8.1, 10.3
Answer:
Number of observations:8
Mean (x̄): 7.075
Mean Absolute Deviation (MAD): 1.45

Question 22.
YOU BE THE TEACHER
Your friend finds and interprets the mean absolute deviation of the data set 35, 40, 38, 32, 42, and 41. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.5 11
Answer:
x̄ = ∑x/n = (35 + 40 + 38)/3
= 113/3
= 37.6
Yes, the data values are different from the mean by an average of 3.

Question 23.
MODELING REAL LIFE
The data set shows the admission prices at several glass-blowing workshops.
$20, $20, $16, $12, $15, $25, $11
Find and interpret the range, interquartile range, and mean absolute deviation of the data.
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.5 12
Answer:
Range = (25 – 11)
= 14/2
= 7
Interquartile range:
Samples = 7
Xl = 12
Xu = 20
Xu – Xl = 20 – 12
= 8
Absolute Deviation of the data:
Data = 7
Mean = 17
Mean Absolute Deviation = 4

Question 24.
MODELING REAL LIFE
The table shows the prices of the five most-expensive and least-expensive dishes on a menu. Find the MAD of each data set. Then compare their variations.
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.5 13
Answer:
Five expensive dishes
$28, $30, $28, $39, $25
MAD:
Dishes = 5
Mean $30
MAD = $3.6
First leasr expensive dishes:
$7, $7, $10, $8, $12
MAD:
Dishes = 5
Mean $8.8
MAD = $1.76
Mean Absolute Deviation of five most expensive dishes is greater than Mean Absolute Deviation of five least expensive dishes.

Question 25.
REASONING
The data sets show the years of the coins in two collections.
Your collection: 1950, 1952, 1908, 1902, 1955, 1954, 1901, 1910
Your friend’s collection: 1929, 1935, 1928, 1930, 1925, 1932, 1933, 1920
Compare the measures of center and the measures of variation for each data set. What can you conclude?
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.5 14
Answer:
The measure of center is a value of the center or middle of a data set.
There are 4 measures of center they are
Mean
Median
Mode
Midrange
four measures of variations
Range
Interquartile range
Variance
Standard deviation
your collection:
Mean: (1950 + 1952 + 1908 + 1902 + 1955 + 1954 + 1901 + 1910)/8
= 1,929
Median: 1901, 1902, 1908, 1910, 1950, 1952, 1954, 1955
= (1910 + 1952)/2
= 1930
Mode: There is no mode
Midrange:
(1955 + 1901)/2
= 3856/2
= 1928
Range:
(1955 – 1901)/2
= 54/2
= 27
Interquartile range:
Number of observations = 8
Xl = 1903.5
Xu = 1953.5
Interquartile range = 50
Variance = 655.14
Standard deviation = 25.59

Question 26.
MODELING REAL LIFE
You survey students in your class about the numbers of movies they watched last month. A new student joins the class who watched 22 movies last month. Is22 an outlier? How does including this value affect the measures of center and the measures of variation? Explain.
Big Ideas Math Answer Key Grade 6 Chapter 9 Statistical Measures 9.5 15
Answer:

REASONING
Which data set would have the greater mean absolute deviation? Explain your reasoning.
Question 27.
guesses for number of gumballs in a jar
guesses for number of baseballs in a jar
Answer:
Gumballs in the jar have a greater mean absolute deviation because baseballs are larger than baseballs.

Question 28.
monthly rainfall amounts in a city
monthly amounts of water used in a home
Answer:

Question 29.
REASONING
Range, interquartile range, and mean absolute deviation are all measures of variation. Which measure of variation is most reliable? Explain your reasoning.
Answer:

Question 30.
DIG DEEPER!
Add and subtract the MAD from the mean in the original data set in Exercise 26.
a. What percent of the values are within one MAD of the mean? two MADs of the mean? Which values are more than twice the MAD from the mean?
b. What do you notice as you get more and more MADs away from the mean? Explain.
Answer:

Statistical Measures Connecting Concepts

Using the Problem-Solving Plan

Question 1.
Six friends play a carnival game in which a person throws darts at balloons. Each person throws the same number of darts and then records the portion of the balloons that pop. Find and interpret the mean, median, and MAD of the data.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures cc 1
Understand the problem.
You know that each person throws the same number of darts. You are given the portion of balloons popped by each person as a fraction, a decimal, or a percent.

Make a plan.
First, write each fraction and each decimal as a percent. Next, order the percents from least to greatest. Then find and interpret the mean, median, and MAD of the data.

Solve and check.
Use the plan to solve the problem. Then check your solution.
Answer:

Question 2.
The cost c (in dollars) to rent skis at a resort for n days is represented by the equation c = 22n. The durations of several ski rentals are shown in the table. Find the range and interquartile range of the costs of the ski rentals. Then determine whether any of the costs are outliers.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures cc 2
Answer:
Given the equation c = 22n
c = 22(1) = 22
c = 22(5) = 1100
c = 22(1) = 22
c = 22(3) = 66
c = 22(5) = 110
c = 22(4) = 88
c = 22(3) = 66
c = 22(12) = 264
c = 22(1) = 22
c = 22(12) = 264
c = 22(5) = 110
c = 22(7) = 154
c = 22(4) = 88
c = 22(1) = 22
22, 110, 22, 66, 110, 88, 66, 264, 22, 264, 110, 154, 88, 22
Range = (264 – 22)/2 = 242/2
= 141
Interquartile range:
Number of observations: 14
lower quartile = 22
upper quartile = 121
Interquartile range = upper quartile – lower quartile
= 121 – 22
= 99

Performance Task
Which Measure of Center Is Best: Mean, Median, or Mode?
At the beginning of this chapter, you watched a STEAM Video called “Daylight in the Big City.“ You are now ready to complete the performance task related to this video, available at BigIdeasMath.com. Be sure to use the problem-solving plan as you work through the performance task.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures cc 3

Statistical Measures Chapter Review

Review Vocabulary

Write the definition and give an example of each vocabulary term.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures cr 1

Graphic Organizers

You can use a Definition and Example Chart to organize information about a concept. Here is an example of a Definition and Example Chart for the vocabulary term statistical question.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures cr 2

Choose and complete a graphic organizer to help you study the concept.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures cr 3
1. mean
2. outlier
3. median
4. mode
5. range
6. quartiles
7. interquartile range

Chapter Self-Assessment

As you complete the exercises, use the scale below to rate your understanding of the success criteria in your journal.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures crr 1

9.1 Introduction to Statistics (pp. 413–418)
Learning Target: Identify statistical questions and use data to answer statistical questions.

Determine whether the question is a statistical question. Explain.
Question 1.
How many positive integers are less than 20?
Answer: There are only 19 numbers in that group

Question 2.
In what month were the students in a sixth-grade class born?
Answer: February

Question 3.
The dot plot shows the number of televisions owned by each family on a city block.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures crr 3
a. Find and interpret the number of data values on the dot plot.
b. Write a statistical question that you can answer using the dot plot. Then answer the question.
Answer:

Display the data in a dot plot. Identify any clusters, peaks, or gaps in the data
Question 4.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures crr 4
Answer:

Question 5.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures crr 5
Answer:

Question 6.
You conduct a survey to answer, “What is the heart rate of a typical sixth-grade student?” e table shows the results. Use the distribution of the data to answer the question.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures crr 6
Answer:

9.2 Mean (pp. 419–424)
Learning Target: Find and interpret the mean of a data set.

Question 7.
Find the mean of the data.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures crr 7
Answer:
x̄ = ∑x/n =(1112+1409+675+536+1398+162)/6
x̄ = ∑x/n=6751/6
x̄ = ∑x/n=1125.16

Question 8.
The double bar graph shows the monthly profit for two toy companies over a four-month period. Compare the mean monthly profits.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures crr 8
Answer:
Company A:
3.6, 3, 3.4, 4
Mean: (3.6 + 3 + 3.4 + 4)/4 = 14/4 = 3.5
Company B:
3, 4.3, 2.2, 4.1
Mean: (3 + 4.3 + 2.2 + 4.1)/4
= 13.6/4
= 3.4

Question 9.
The table shows the test scores for a class of sixth-grade students. Describe how the outlier affects the mean. Then use the data to answer the statistical question, “What is the typical test score for a student in the class?”
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures crr 9
Answer:

9.3 Measures of Center (pp. 425–432)
Learning Target: Find and interpret the median and mode of a data set.

Find the median and mode of the data.
Question 10.
8, 8, 6, 8, 4, 5, 6
Answer:
Median:
write the given data in ascending order or descending order.
4, 5, 6, 8, 8, 8
= (6 + 8)/2
= 14/2
= 7
Mode:
8 is the mode.

Question 11.
24, 74, 61, 29, 38, 27, 68, 54
Answer:
Median:
write the given data in ascending order or descending order.
24, 74, 61, 29, 38, 27, 68, 54
= 24, 27, 29, 38, 54, 61, 68, 74
= (38 + 54)/2
= 92/2
= 48
Mode:
There is no mode in the data.

Question 12.
Find the mean, median, and mode of the data set 67, 52, 50, 99, 66, 50, and 57 with and without the outlier. Which measure does the outlier affect the most?
Answer:
Given the data,
67, 52, 50, 99, 66, 50, and 57
Mean with outliner:
(67 + 52 + 50 + 99 + 66 + 50 + 57)/7
= 441/7
= 63
Mean without outliner:
66 is the median
Mode with outliner: 50
Mode without outliner:
No mode
Outliners affect the mean value of the data but have little effect on the median or mode of a given set of data.

Question 13.
The table shows the lengths of several movies. Which measure of center best represents the data? Explain your reasoning.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures crr 13
Answer:

Question 14.
Give an example of a data set that does not have a median. Explain why the data set does not have a median.
Answer:

9.4 Measures of Variation (pp. 433–438)
Learning Target: Find and interpret the range and interquartile range of a data set.

Find the range of the data.
Question 15.
45, 76, 98, 21, 52, 39
Answer:
Lowest value = 21
Highest value = 98
Range = (98 – 21)/2
= 77/2
= 38.5

Question 16.
95, 63, 52, 8, 93, 16, 42, 37, 62
Answer:
Lowest value = 8
Highest value = 95
Range = (95 – 8)/2
= 87/2
= 43.5

Find the interquartile range of the data.
Question 17.
28, 46, 25, 76, 18, 25, 47, 83, 44
Answer:
Given the data
28, 46, 25, 76, 18, 25, 47, 83, 44
Number of observations: 9
lower quartile: 25
upper quartile: 61.5
Interquartile range (Xu – Xl) = 36.5

Question 18.
14, 25, 97, 55, 66, 28, 92, 38, 94
Answer:
Given the data
14, 25, 97, 55, 66, 28, 92, 38, 94
Number of observations: 9
lower quartile: 26.5
upper quartile: 93
Interquartile range (Xu – Xl) = 66.5

Question 19.
The table shows the weights of several adult emperor penguins. Find and interpret the range and interquartile range of the data. Then determine whether there are any outliers.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures crr 19
Answer:
25, 27, 36, 23.5, 33.5, 31.25, 30.75, 32, 24, 29.25
Yes there are outliner
Range: (36  – 25)/2
= 11/2
= 5.5
Interquartile range:
Number of observations = 10
Mean = 29.225
MAD = 3.98

Question 20.
Two data sets have the same interquartile range. Can you assume that the ranges of the two data sets are about the same? Give an example to justify your answer.
Answer:
23
Yes, a data set with the least value of 2 and the greatest value of 20 will have the same range as a data set with the least value of 82 and the greatest value of 100 will have the same range of 18.

9.5 Mean Absolute Deviation (pp. 439–444)
Learning Target: Find and interpret the mean absolute deviation of a data set.

Find and interpret the mean absolute deviation of the data.
Question 21.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures crr 21
Answer:
Given data,
6, 8.5, 6, 9, 10, 7, 8, 9.5
No. of observations: 8
Mean = 8
Mean Absolute Deviation: 1.25

Question 22.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures crr 22
Answer:
Given data,
130, 150, 190, 100, 175, 120, 165, 140, 180, 190
No. of observations: 10
Mean = 154
Mean Absolute Deviation: 26

Question 23.
The table shows the prices of the five most-expensive and least-expensive manicures given by a salon technician on a particular day. Find the MAD of each data set. Then compare their variations.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures crr 23
Answer:
five most-expensive:
$58, $52, $70, $49, $56
No. of observations: 5
Mean = 57
Mean Absolute Deviation: 5.6
5 least-expensive manicures:
$10, $10, $15, $10, $15
No. of observations: 5
Mean = 12
Mean Absolute Deviation: 2.4
The Mean Absolute Deviation of the five most-expensive is greater than the Mean Absolute Deviation of the 5 least-expensive manicures.

Question 24.
You record the lengths of songs you stream. The next song is 276 seconds long. Is 276 an outlier? How does including this value affect the measures of center and the measures of variation? Explain.
Big Ideas Math Answers 6th Grade Chapter 9 Statistical Measures crr 24
Answer:
Given the data,
233, 219, 163, 213, 224, 208, 225, 220, 222, 240, 228, 219, 260, 249, 209, 236,  206
The next song is 276 seconds long.
276 is the outliner.
We can remove 276 from the given data set.
So, there is no effect on the center and the measure of variations.

Statistical Measures Practice Test

Find the mean, median, mode, range, and interquartile range of the data.
Question 1.
5, 6, 4, 24, 10, 6, 9, 8
Answer:
Mean = (5 + 6 + 4 + 24 + 10 + 6 + 9 + 8)/8
= 72/8
= 9
Median:
4, 5, 6, 6, 8, 9, 10, 24
= (6 + 8)/2 = 14/2
= 7
Mode:
6 is the mode
range = (24 – 4)/2
= 20/2
= 10
Range:
Lowest value: 4
Highest value: 24
Range: 20
Interquartile range:
Lower quartile (xL): 5.25
Upper quartile (xU): 9.75
Interquartile range (xU-xL): 4.5

Question 2.
46, 27, 94, 56, 53, 65, 43
Answer:
Given the data,
46, 27, 94, 56, 53, 65, 43
Mean = (46 + 27 + 94 + 56 + 53 + 65 + 43)/7
= 16.75
Median = 15.5
Mode: There is no mode
Range:
Number of observations = 7
Lowest value: 27
Highest value: 94
Range: 67
Interquartile range:
Lower quartile (xL): 43
Upper quartile (xU): 65
Interquartile range (xU-xL): 22

Question 3.
32, 58, 19, 36, 44, 57, 11, 26, 74
Answer:
Given the data,
32, 58, 19, 36, 44, 57, 11, 26, 74
Mean = (32 + 58 + 19 + 36 + 44 + 57 + 11 + 26 + 74)/9
= 357/9
= 39.66
Median:
Arrange the data in ascending or descending order.
11, 19, 26, 32, 36, 44, 57, 58, 74
Median = 36
Mode: There is no mode in the data
Range:
Lowest value: 11
Highest value: 74
Range: 63
Interquartile range:
Lower quartile (xL): 22.5
Upper quartile (xU): 57.5
Interquartile range (xU-xL): 35

Question 4.
36, 24, 49, 32, 37, 28, 38, 40, 39
Answer:
Given the data
36, 24, 49, 32, 37, 28, 38, 40, 39
Arrange the data in ascending or descending order.
24, 28, 32, 36, 37, 38, 39, 40, 49
Mean = (24 + 28 + 32 + 36 + 37 + 28 + 38 + 40 + 49)/9
= 34.66
Median: 37
Mode: There is no mode
Range:
Lowest value: 24
Highest value: 49
Range: 25
Interquartile range:
Lower quartile (xL): 30
Upper quartile (xU): 39.5
Interquartile range (xU-xL): 9.5

Find and interpret the mean absolute deviation of the data.
Question 5.
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures pt 5
Answer:
Given the data,
312, 286, 196, 201, 158, 225, 206, 192
Mean (x̄): 0.5
Mean Absolute Deviation (MAD): 0.15

Question 6.
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures pt 6
Answer:
Given the data,
15, 8, 19, 20, 18, 20, 22, 14, 10, 15
Mean (x̄): 16.1
Mean Absolute Deviation (MAD): 3.7

Question 7.
You conduct a survey to answer, “How many Times (minutes)minutes does it take a typical sixth-grade student to run a mile?” The table shows the results. Use the distribution of the data to answer the question.
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures pt 7
Answer:

Question 8.
The table shows the weights of Alaskan malamute 8181808281dogs at a veterinarian’s office. Which measure of center best represents the weight of an Alaskan malamute? Explain your reasoning.
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures pt 8
Answer:

Question 9.
The table shows the numbers of guests Numbers of Guests at a hotel on different days.
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures pt 9
a. Find the range and interquartile range of the data.
b. Use the interquartile range to identify the outlier(s) in the data set. Find the range and interquartile range of the data set without the outlier(s). Which measure did the outlier or outliers affect more?
Answer:

Question 10.
The data sets show the numbers of hours worked each week by two people for several weeks.
Person A: 9, 18, 12, 6, 9, 21, 3, 12
Person B: 12, 18, 15, 16, 14, 12, 15, 18
Compare the measures of center and the measures of variation for each data set. What can you conclude?
Answer:

Question 11.
The table shows the lengths of several bearded dragons captured for a study. Find the mean, median, and mode of the data in centimeters and in inches. How does converting to inches affect the mean, median, and mode?
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures pt 11
Answer:

Statistical Measures Cumulative Practice

Question 1.
Which statement can be represented by a negative integer?
A. The temperature rises 15 degrees.
B. A hot-air balloon ascends 450 yards.
C. You earn $50 completing chores.
D. A submarine submerges 260 feet.
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures cp 1
Answer: D. A submarine submerges 260 feet.

Question 2.
What is the height h (in inches) of the prism?
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures cp 2
Answer:
h = v/lw
h = 5850/30(12 1/4)
h = 5850/(30 × 12.25)
h = 5850/367.50
h = 15.91 inches

Question 3.
Which is the solution of the inequality \(\frac{2}{3}\)x < 6?
F. x < 4
G. x < 5\(\frac{1}{3}\)
H. x < 6\(\frac{2}{3}\)
I. x < 9
Answer: I. x < 9

Question 4.
The number of hours that each of six students spent reading last week is shown in the bar graph.
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures cp 4
For the data in the bar graph, which measure is the?
A. mean
B. median
C. mode
D. range
Answer: C. mode

Explanation:
In the above bar graph, 10 is repeated two ways.
Thus the correct answer is option C.

Question 5.
Which list of numbers is in order from least to greatest?
F. – 5.41, – 3.6, – 3.2, – 3.06, – 1
G. – 1, – 3.06, – 3.2, – 3.6, – 5.41
H. – 5.41, – 3.06, – 3.2, – 3.6, – 1
I. – 1, – 3.6, – 3.2, – 3.06, – 5.41
Answer: F. – 5.41, – 3.6, – 3.2, – 3.06, – 1

Explanation:
We have to write the numbers from least to greatest
The negative sign with the highest number will be the least.
– 5.41, – 3.6, – 3.2, – 3.06, – 1
Thus the correct answer is option F.

Question 6.
What is the mean absolute deviation of the data shown in the dot plot, rounded to the nearest tenth?
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures cp 6
A. 1.4
B. 3
C. 3.2
D. 57.
Answer:
Data from the dot plot
5, 5, 4, 4, 6, 1
Number of observations: 6
Mean = 4.166
Mean absolute deviation = 1.66
Thus the correct answer is option A.

Question 7.
A family wants to buy tickets to a theme park. There are separate ticket prices for adults and children.
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures cp 7
Which expression represents the total cost (in dollars) for adult tickets c and child tickets?
F. 600 (a + c)
G. 50(a × c)
H. 30a + 20c
I. 30a × 20c
Answer: H. 30a + 20c

Question 8.
The dot plot shows the leap distances (in feet) of a tree frog. How many leaps were recorded?
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures cp 8
Answer: 7 leaps were recorded

Question 9.
What is the value of the expression when a = 6 and b = 14?
0.8a + 0.02b
A. 0.4828
B. 0.8814
C. 5.08
D. 16.4
Answer:
Given the expression,
0.8a + 0.02b
a = 6
b = 14
0.8(6) + 0.02(14)
4.8 + 0.28
= 5.08
Thus the correct answer is option C.

Question 10.
Which property was not used to simplify the expression?
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures cp 10
F. Distributive Property
G. Associative Property of Addition
H. Multiplication Property of One
I. Commutative Property of Multiplication
Answer: I. Commutative Property of Multiplication

Question 11.
What are the coordinates of Point P?
Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures cp 11
A. (- 3, – 2)
B. (3, – 2)
C. (- 2, – 3)
D. (-2, 3)
Answer: B. (3, – 2)

Explanation:
By seeing the above graph we can write the ordered pair P.
the x-axis is on 3 and the y-axis is on -2
Thus the correct answer is option B.

Question 12.
Create a data set with 5 numbers that has the following measures.
Think
Solve
Explain
• a mean of 7
• a median of 9
Explain how you created your data set.
Answer:
The data set is 3, 2, 9, 1, 20

Final Words:

I hope the article regarding the Big Ideas Math Answers Grade 6 Chapter 9 Statistical Measures is helpful for the students who are lagging in this concept. Feel free to post the comments if you have any doubts regarding the methods or answers. We will try to clarify your doubts as early as possible.

Go Math Grade 4 Answer Key Homework Practice FL Chapter 4 Divide by 1-Digit Numbers

go-math-grade-4-chapter-4-divide-by-1-digit-numbers-pages-67-93-answer-key

Detailed and Step-by-step explanation of Chapter 5 concepts is provided in this Go Math Grade 4 Answer Key Homework Practice FL Chapter 4 Divide by 1-Digit Numbers. Assure that you should practice with the help of Go math HMH grade 4 chapter 5 solution key and improve mathematical and logical skills. Learning & practicing the fundamentals of math chapter 5 concepts is very important to score more marks in the exams. So, download online Go Math Grade 4 Solution Key Homework Practice FL Chapter 4 Divide by 1-Digit Numbers pdf and overcome all the difficulties in math.

Go Math Grade 4 Answer Key Homework Practice FL Chapter 4 Divide by 1-Digit Numbers

Consistent practice helps students to gain more knowledge and overcome their weak points. So, grab these Concept-wise Chapter 4 Go math Grade 4 Answer Key pdf and practice regularly for securing good scores in the exams. Some of the topics covered in Go Math Solution Key Grade 4 Homework Practice FL Chapter 4 Divide by 1-Digit Numbers are Estimate Quotients Using Multiples, Remainders, Divide Tens, Hundreds, and Thousands, etc. Solve the questions provided at the end of the page and test your subject knowledge.

Lesson: 1 – Estimate Quotients Using Multiples

Lesson: 2 – Remainders

Lesson: 3 – Interpret the Remainder

Interpret the Remainder

Lesson: 4 – Divide Tens, Hundreds, and Thousands

Lesson: 5 – Estimate Quotients Using Compatible Numbers

Lesson: 6 – Division and the Distributive Property

Lesson: 7 – Divide Using Repeated Subtraction

Lesson: 8 – Divide Using Partial Quotients

Divide Using Partial Quotients

Lesson: 9 – Model Division with Regrouping

Lesson: 10 – Place the First Digit

Lesson: 11 – Divide by 1-Digit Numbers

Lesson: 12 – Problem Solving Multistep Division Problems

Lesson: 13

Common Core – Divide by 1-Digit Numbers – Page No. 69

Estimate Quotients Using Multiples

Find two numbers the quotient is between. Then estimate the quotient.

Question 1.
175 ÷ 6
Think: 6 × 20 = 120 and 6 × 30 = 180. So, 175 ÷ 6 is between 20 and 30. Since 175 is closer to 180 than to 120, the quotient is about 30.
between 20 and 30
about 30

Question 2.
53 ÷ 3
between ____ and ____
about ____

Answer: About 18

Explanation:
17 × 3= 51 and 18 × 3 = 54. 53 is between 51 and 54. 53 ÷ 3 is closest to 17 and 18. So, 53 ÷ 3 is between 17 and 18. So, 53 ÷ 3 will be about 18.

Question 3.
75 ÷ 4
between ____ and ____
about ____

Answer: About 19

Explanation:
18 × 4= 72 and 19 × 4= 76. 75 is between 72 and 76. 75 ÷ 4 is closest to 18 and 19. So, 75÷ 4 is between 18 and 19. So, 75 ÷ 4 will be about 19.

Question 4.
215 ÷ 9
between ____ and ____
about ____

Answer: About 24

Explanation:
23 × 9= 207 and 24 × 9 = 216. 24 is between 207 and 216. 215 ÷ 9 is closest to 23 and 24. So, 215 ÷ 9 is between 23 and 24. So, 215 ÷ 9 will be about 24.

Question 5.
284 ÷ 5
between ____ and ____
about ____

Answer: About 57

Explanation:
56 × 5 = 280 and 57 × 5 = 285. 284 is between 280 and 285. 284 ÷ 5 is closest to 56 and 57. So, 284 ÷ 5 is between 56 and 57. So, 175 ÷ 6 will be about 57.

Question 6.
191 ÷ 3
between ____ and ____
about ____

Answer: About 64

Explanation:
63 × 3 = 189 and 64 × 3 = 192. 191 is between 189 and 192. 191 ÷ 3 is closest to 63 and 64. So, 191 ÷ 3 is between 63 and 64. So, 175 ÷ 6 will be about 64.

Question 7.
100 ÷ 7
between ____ and ____
about ____

Answer: About 14

Explanation:
14 × 7 = 98 and 15 × 7 = 105. 100 is between 98 and 105. 100 ÷ 7 is closest to 14 and 15.
So, 100 ÷ 7 is between 14 and 15. So, 100 ÷ 7 will be about 14.

Question 8.
438 ÷ 7
between ____ and ____
about ____

Answer: About 63

Explanation:
63 × 7 = 441 and 62 × 7 = 434. 438 is between 434 and 441. 438 ÷ 7 is closest to 62 and 63. So, 438 ÷ 7 is between 62 and 63. So, 438 ÷ 7 will be about 63.

Question 9.
103 ÷ 8
between ____ and ____
about ____

Answer: About 13

Explanation:
13 × 8 = 104 and 12 ×8 = 96. 103 is between 96 and 104. 103 ÷ 8 is closest to 12 and 13. So, 103 ÷ 8 is between 12 and 13. So, 103 ÷ 8 will be about 13.

Question 10.
255 ÷ 9
between ____ and ____
about ____

Answer: About 28

Explanation:
28 × 9 = 252 and 29 × 9 = 261. 255 is between 252 and 261. 255 ÷ 9 is closest to 28 and 29.
So, 255 ÷ 9 is between 28 and 29. So, 255 ÷ 9 will be about 28.

Problem Solving

Question 11
Joy collected 287 aluminum cans in 6 hours. About how many cans did she collect per hour?
about ____ cans

Answer: About 48 cans

Explanation:
47 × 6 = 282 and 48 × 6 = 288. 287 is between 282 and 288. 287 ÷ 6 is closest to 47 and 48. So, 287 ÷ 6 is between 47 and 48. So, 287 ÷6 will be about 48.

Question 12.
Paul sold 162 cups of lemonade in 5 hours. About how many cups of lemonade did he sell each hour?
about ____ cups

Answer: About 32 cups of lemonade he sold in each hour

Explanation:
32 × 5 = 160 and 33 × 5 = 165. 162 is between 160 and 165. 162 ÷ 5 is closest to 32 and 33. So, 162 ÷ 5 is between 32 and 33. So, 162 ÷ 5 will be about 32.

Common Core – Divide by 1-Digit Numbers – Page No. 70

Divide by 1-Digit Numbers

Lesson Check

Question 1.
Abby did 121 sit-ups in 8 minutes. Which is the best estimate of the number of sit-ups she did in 1 minute?
Options:
a. about 12
b. about 15
c. about 16
d. about 20

Answer: About 15

Explanation:
15 × 8 = 120 and 16 × 8 = 128. 121 is between 120 and 128. 121 ÷ 8 is closest to 120 and 128. So, 121 ÷ 8 is between 15 and 16.
So, 121 ÷ 8 will be about 15.
Thus the correct answer is option b.

Question 2.
The Garibaldi family drove 400 miles in 7 hours. Which is the best estimate of the number of miles they drove in 1 hour?
Options:
a. about 40 miles
b. about 50 miles
c. about 60 miles
d. about 70 miles

Answer: about 60 miles

Explanation:
Given,
The Garibaldi family drove 400 miles in 7 hours.
57 × 7 = 399 and 58 × 7 = 406.
400 is between 399 and 406. 400 ÷ 7 is closest to 57 and 58.
So, 400 ÷ 7 is between 57 and 58.
So, 400 ÷ 7 will be about 57.
Thus the correct answer is option c.

Spiral Review

Question 3.
Twelve boys collected 16 aluminum cans each. Fifteen girls collected 14 aluminum cans each. How many more cans did the girls collect than the boys?
Options:
a. 8
b. 12
c. 14
d. 18

Answer: 18

Explanation:
Given that,
Twelve boys collected 16 aluminum cans each.
Fifteen girls collected 14 aluminum cans each.
Number of aluminium cans boys had= 12 × 16=192
Number of aluminium cans girls had = 15 × 14=210
Girls collected more cans compared to boys,
Number of more cans collected by girls= 210 – 192=18
Thus the correct answer is option d.

Question 4.
George bought 30 packs of football cards. There were 14 cards in each pack. How many cards did George buy?
Options:
a. 170
b. 320
c. 420
d. 520

Answer: 420

Explanation:
Given,
George bought 30 packs of football cards.
There were 14 cards in each pack.
Number of packs of football cards= 30
Number of cards in each pack= 14
Total number of cards George bought=30×14=420
Thus the correct answer is option c.

Question 5.
Sarah made a necklace using 5 times as many blue beads as white beads. She used a total of 30 beads. How many blue beads did Sarah use?
Options:
a. 5
b. 6
c. 24
d. 25

Answer: 25

Explanation:
Given,
Sarah made a necklace using 5 times as many blue beads as white beads.
She used a total of 30 beads.
Let the number of white beads be x while the number of blue beads are 5x.
Total number of beads in the necklace=30 beads
According to the problem,
5x+x=30
6x=30
x=30/6=5
Therefore the number of blue beads in the necklace are 5x= 5×5=25
Thus the correct answer is option d.

Question 6.
This year, Ms. Webster flew 145,000 miles on business. Last year, she flew 83,125 miles on business. How many more miles did Ms. Webster fly on business this year?
Options:
a. 61,125 miles
b. 61,875 miles
c. 61,985 miles
d. 62,125 miles

Answer: 61,875 miles

Explanation:
Given,
This year, Ms. Webster flew 145,000 miles on business.
Last year, she flew 83,125 miles on business.
Number of miles Ms Webster flew in this year = 145,000 miles
Number of miles Ms Webster flew in the last year = 83,125 miles
Number of more miles travelled by Ms Webster =145,000 – 83,125 = 61,875
Thus the correct answer is option b.

Common Core – Divide by 1-Digit Numbers – Page No. 71

Remainders

Use counters to find the quotient and remainder.

Question 1.
13 ÷ 4
3 r1

Explanation:
Quotient:
A. Use 13 counters to represent the 13 dominoes. Then draw 4 circles to represent the divisor.
B. Share the counters equally among the 4 groups by placing them in the circles.
C. Number of counters formed in each group = quotient of 13 ÷ 4
D. Number of circles are equally filled with 4 counters, therefore, the quotient is 3.
Remainder:
The number of counters left over is the remainder. The number of counters leftover= 1
For 13 ÷ 4, the quotient is 3 and the remainder is 1, or 3 r1.

Question 2.
24 ÷ 7
_____ R _____

Answer: 3 r3

Explanation:
Quotient:
A. Use 24 counters to represent the 24 dominoes. Then draw 7 circles to represent the divisor.
B. Share the counters equally among the 7 groups by placing them in the circles.
C. Number of counters formed in each group = quotient of 24 ÷ 7
D. Number of circles are equally filled with 3 counters, therefore, the quotient is 3
Remainder:
The number of counters left over is the remainder. The number of counters leftover= 3
For 24 ÷ 7, the quotient is 3 and the remainder is 3, or 3 r3.

Question 3.
39 ÷ 5
_____ R _____

Answer: 7 r4

Explanation:
Quotient:
A. Use 39 counters to represent the 39dominoes. Then draw 5 circles to represent the divisor.
B. Share the counters equally among the 5 groups by placing them in the circles.
C. Number of counters formed in each group = quotient 39 ÷ 5
D. Number of circles are equally filled with 7 counters, therefore, the quotient is 7
Remainder:
The number of counters left over is the remainder. The number of counters leftover= 4
For 39 ÷ 5, the quotient is 7 and the remainder is 4, or 7 r4.

Question 4.
36 ÷ 8
_____ R _____

Answer: 4 r4

Explanation:
Quotient:
A. Use 36 counters to represent the 36 dominoes. Then draw 8 circles to represent the divisor.
B. Share the counters equally among the 8 groups by placing them in the circles.
C. Number of counters formed in each group = quotient of 36 ÷ 8
D. Number of circles are equally filled with 4 counters, therefore, the quotient is 4
Remainder:
The number of counters left over is the remainder. The number of counters leftover= 4
For 36 ÷ 8, the quotient is 4 and the remainder is 4, or 4 r4.

Question 5.
6)\(\overline { 27 } \)
_____ R _____

Answer: 4 r3

Explanation:
Quotient:
A. Use 27 counters to represent the 27 dominoes. Then draw 6 circles to represent the divisor.
B. Share the counters equally among the 6 groups by placing them in the circles.
C. Number of counters formed in each group = quotient of 27 ÷6
D. Number of circles are equally filled with 4 counters, therefore, the quotient is 4
Remainder:
The number of counters left over is the remainder. The number of counters leftover= 3
For 27 ÷ 6, the quotient is 4 and the remainder is 3, or 4 r3.

Question 6.
25 ÷ 9
_____ R _____

Answer: 2 r7

Explanation:
Quotient:
A. Use 25 counters to represent the 25 dominoes. Then draw 9 circles to represent the divisor.
B. Share the counters equally among the 9 groups by placing them in the circles.
C. Number of counters formed in each group = quotient of 25 ÷ 9
D. Number of circles are equally filled with 2 counters, therefore, the quotient is 2
Remainder:
The number of counters left over is the remainder. The number of counters leftover= 7
For 25 ÷ 7, the quotient is 2 and the remainder is 7, or 2 r7.

Question 7.
3)\(\overline { 17 } \)
_____ R _____

Answer: 5 r2

Explanation:
Quotient:
A. Use 17 counters to represent the 17 dominoes. Then draw 3 circles to represent the divisor.
B. Share the counters equally among the 3 groups by placing them in the circles.
C. Number of counters formed in each group = quotient of 17 ÷ 3
D. Number of circles are equally filled with 5 counters, therefore, the quotient is 5
Remainder:
The number of counters left over is the remainder. The number of counters leftover= 2
For 17 ÷ 3, the quotient is 5 and the remainder is 2, or 5 r2.

Question 8.
26 ÷ 4
_____ R _____

Answer: 6 r2

Explanation:
Quotient:
A. Use 26 counters to represent the 26 dominoes. Then draw 4 circles to represent the divisor.
B. Share the counters equally among the 4 groups by placing them in the circles.
C. Number of counters formed in each group = quotient of 26 ÷ 4
D. Number of circles are equally filled with 6 counters, therefore, the quotient is 6
Remainder:
The number of counters left over is the remainder. The number of counters leftover= 2
For 26 ÷ 4, the quotient is 6 and the remainder is 2, or 6 r2.

Divide. Draw a quick picture to help.

Question 9.
14 ÷ 3
_____ R _____

Answer: Quotient: 4 Remainder: 2

Explanation:
Quotient:
A. Use 14 counters to represent the 14 dominoes. Then draw 3 circles to represent the divisor.
B. Share the counters equally among the 3 groups by placing them in the circles.
C. Number of circles filled= quotient of 14 ÷ 3 = 4
Remainder:
The number of counters left over is the remainder. The number of counters leftover= 2

Question 10.
5)\(\overline { 29 } \)
_____ R _____

Answer: Quotient: 5 Remainder: 4

Explanation:
Quotient:
A. Use 29 counters to represent the 29 dominoes. Then draw 5 circles to represent the divisor.
B. Share the counters equally among the 5 groups by placing them in the circles.
C. Number of circles filled= quotient of 29 ÷ 5 = 5
Remainder:
The number of counters left over is the remainder. The number of counters leftover= 4

Problem Solving

Question 11.
What is the quotient and remainder in the division problem modeled below?
Go Math Grade 4 Answer Key Homework Practice FL Chapter 4 Divide by 1-Digit Numbers Common Core - Divide by 1-Digit Numbers img 1
_____ R _____

Answer: quotient:6 remainder2

Explanation:
Quotient:
A. Use 20 counters to represent the 20 dominoes. Then draw 3 circles to represent the divisor.
B. Share the counters equally among the 3 groups by placing them in the circles.
C. Number of counters formed in each group = quotient of 20 ÷ 3
D. Number of circles are equally filled with 6 counters, therefore, the quotient is 6
Remainder:
The number of counters left over is the remainder. The number of counters leftover= 2
For 20 ÷ 3, the quotient is 6 and the remainder is 2, or 6 r2.

Question 12.
Mark drew the following model and said it represented the problem 21 ÷ 4. Is Mark’s model correct? If so, what is the quotient and remainder? If not, what is the correct quotient and remainder?
Go Math Grade 4 Answer Key Homework Practice FL Chapter 4 Divide by 1-Digit Numbers Common Core - Divide by 1-Digit Numbers img 2
_____ : _____ r _____

Answer: 4 r5

Explanation:
Quotient:
A. Use 21 counters to represent the 21 dominoes. Then draw 4 circles to represent the divisor.
B. Share the counters equally among the 4 groups by placing them in the circles.
C. Number of counters formed in each group = quotient of 21 ÷ 4
D. Number of circles are equally filled with 4 counters, therefore, the quotient is 4
Remainder:
The number of counters left over is the remainder. The number of counters leftover= 5
For 21 ÷ 4, the quotient is 4 and the remainder is 5, or 4 r5.

Common Core – Divide by 1-Digit Numbers – Page No. 72

Lesson Check

Question 1.
What is the quotient and remainder for 32 ÷ 6?
Options:
a. 4 r3
b. 5 r1
c. 5 r2
d. 6 r1

Answer: 5 r2

Explanation:
Quotient:
A. Use 32 counters to represent the 32 dominoes. Then draw 6 circles to represent the divisor.
B. Share the counters equally among the 5 groups by placing them in the circles.
C. Number of counters formed in each group = quotient of 32 ÷ 6
D. Number of circles are equally filled with 5 counters, therefore, the quotient is 5
Remainder:
The number of counters left over is the remainder. The number of counters leftover= 2
For 32 ÷ 6, the quotient is 5 and the remainder is 2, or 5 r2.
Thus the correct answer is option c.

Question 2.
What is the remainder in the division problem modeled below?
Go Math Grade 4 Answer Key Homework Practice FL Chapter 4 Divide by 1-Digit Numbers Common Core - Divide by 1-Digit Numbers img 3
Options:
a. 8
b. 4
c. 3
d. 1

Answer: 3

Explanation:
When a number cannot be divided evenly, the amount left over is called the remainder.
The number of counters that are left = remainder = 3
Thus the correct answer is option c.

Spiral Review

Question 3.
Each kit to build a castle contains 235 parts. How many parts are in 4 of the kits?
Options:
a. 1,020
b. 940
c. 920
d. 840

Answer: 940

Explanation:
Number of parts used to build a castle in each kit=235 parts
Number of kits= 4
Total number of parts in 4 of the kits= 235 x 4=940 parts

Thus the correct answer is option b.

Question 4.
In 2010, the population of Alaska was about 710,200. What is this number written in word form?
Options:
a. seven hundred ten thousand, two
b. seven hundred twelve thousand
c. seventy-one thousand, two
d. seven hundred ten thousand, two hundred

Answer: seven hundred ten thousand, two hundred

Explanation:
The ones and tens place of the number are zeroes, so the next place which is hundreds is considered and the value is 7 so, it can be written as seven hundred and in the thousands period, it can be written as seven hundred ten thousand.
Thus the correct answer is option d.

Question 5.
At the theater, one section of seats has 8 rows with 12 seats in each row. In the center of the first 3 rows are 4 broken seats that cannot be used. How many seats can be used in the section?
Options:
a. 84
b. 88
c. 92
d. 96

Answer: 92

Explanation:
Given,
Number of rows at the theatre = 8
Number of seats in each row= 12
Number of seats broken and that cannot be used to sit= 4
Total number of seats that can be used= 12 × 8 – 4 = 96 – 4 = 92
Thus the correct answer is option c.

Question 6.
What partial products are shown by the model below?
Go Math Grade 4 Answer Key Homework Practice FL Chapter 4 Divide by 1-Digit Numbers Common Core - Divide by 1-Digit Numbers img 4
Options:
a. 300, 24
b. 300, 600, 40, 60
c. 300, 60, 40, 24
d. 300, 180, 40, 24

Answer: 300, 180, 40, 24

Explanation:
The whole rectangle is divided into four small rectangles the areas of these rectangles are:

Area of yellow rectangle= 30 x 10=300
Area of green rectangle= 4 x 10 = 40
Area of pink rectangle= 6 x 30= 180
Area of blue rectangle= 4 x 6= 24
Thus the correct answer is option d.

Common Core – Divide by 1-Digit Numbers – Page No. 73

Interpret the Remainder

Interpret the remainder to solve.

Question 1.
Hakeem has 100 tomato plants. He wants to plant them in rows of 8. How many full rows will he have?
Think: 100 ÷ 8 is 12 with a remainder of 4. The question asks “how many full rows,” so use only the quotient.
12 full rows

Explanation:
Quotient:
A. Use 100 counters to represent the 100 dominoes. Then draw 8 circles to represent the divisor.
B. Share the counters equally among the 8 groups by placing them in the circles.
C. Number of counters formed in each group = quotient of 100 ÷ 8
D. Number of circles are equally filled with 12 counters, therefore, the quotient is 12
Therefore, the tomatoes placed in full rows are 12

Question 2.
A teacher has 27 students in her class. She asks the students to form as many groups of 4 as possible. How many students will not be in a group?
______ students

Answer: 3 students will not be in the group

Explanation:
Total number of students in the class= 27
Number of students who make a group=4
Number of groups that can be made =Quotient of 27÷ 4=6
Number of students who do not come under a group= Remainder of 27÷ 4=3

Question 3.
A sporting goods company can ship 6 footballs in each carton. How many cartons are needed to ship 75 footballs?
______ cartons

Answer: 12 full cartons and 0.5 or 1/2 carton to ship all the 75 footballs

Explanation:
Total number of footballs that should be shipped= 75
Number of footballs placed in each carton = 6
Number of cartons required=Quotient of 75÷ 6=12

Since each carton carries 6 balls, half carton contains 3 balls because 6÷3=2, therefore, each half of the carton contains 3 balls.

Question 4.
A carpenter has a board that is 10 feet long. He wants to make 6 table legs that are all the same length. What is the longest each leg can be?
______ foot

Answer: The length of the longest leg = 4 foot-long

Explanation:
According to the question,
Length of the board the carpenter has= 10 foot long
Number of table legs that are to be made = 6
Length of the 6 table legs are equal
then,
Length of each table leg = Quotient of 10 ÷ 6 =1 foot
Length of the longest table leg = Remainder of 10 ÷ 6 = 4 foot.

Question 5.
Allie wants to arrange her flower garden in 8 equal rows. She buys 60 plants. What is the greatest number of plants she can put in each row?
______ plants

Answer: 7

Explanation:
Total number of plants Allie bought = 60
Number of rows = 8
Number of plants in each row= Quotient of 60 ÷ 8 = 7
Thus the greatest number of plants she can put in a row is 7.

Problem Solving

Question 6.
Joanna has 70 beads. She uses 8 beads for each bracelet. She makes as many bracelets as possible. How many beads will Joanna have left over?
______ beads

Answer: 6 beads

Explanation:
Total number of beads Joanna has= 70 beads
Number beads used for each bracelet= 8 beads
Number of bracelets made with these beads= Quotient of 70÷8= 7 bracelets
then,
The number of beads leftover= Remainder of 70÷8= 6 beads

Question 7.
A teacher wants to give 3 markers to each of her 25 students. Markers come in packages of 8. How many packages of markers will the teacher need?
______ packages

Answer: 10 packages

Explanation:
Total number of students= 25
Number of markers each student got= 3
Total number of markers the teacher needs to distribute= 25 x 3= 75
Number of markers in each package= 8
Number of packages the teacher required= Quotient of 75÷8=9
While the remainder = 3
Therefore the total number of packages = 10

Common Core – Divide by 1-Digit Numbers – Page No. 74

Lesson Check

Question 1.
Marcus sorts his 85 baseball cards into stacks of 9 cards each. How many stacks of 9 cards can Marcus make?
Options:
a. 4
b. 8
c. 9
d. 10

Answer: 10

Explanation:
Total number of baseball cards=85
Number of cards in each stack=9
Number of stacks sorted= Quotient of 85÷9=9
While the remainder=4
So the total number of stacks required= 10
Thus the correct answer is option d.

Question 2.
A minivan can hold up to 7 people. How many minivans are needed to take 45 people to a basketball game?
Options:
a. 3
b. 5
c. 6
d. 7

Answer: 7

Explanation:
A minivan can hold up to 7 people.
Total number of people who want to hire the minivan= 45 people
Number of minivans required= Quotient of 45÷7= 6 vans
While the remainder is 3.
Total number of minivans required to take the people to the baseball game= 7 minivans
Thus the correct answer is option d.

Spiral Review

Question 3.
Mrs. Wilkerson cut some oranges into 20 equal pieces to be shared by 6 friends. How many pieces did each person get and how many pieces were left over?
Options:
a. 2 pieces with 4 pieces left over
b. 3 pieces with 2 pieces left over
c. 3 pieces with 4 pieces left over
d. 4 pieces with 2 pieces left over

Answer: 3 pieces with 2 pieces left over

Explanation:
Total number of orange pieces= 20
Number of friends= 6
Number of pieces each friend got= Quotient of 20÷6= 3 pieces
Number of pieces leftover= Remainder of 20÷6= 2 pieces
Thus the correct answer is option b.

Question 4.
A school bought 32 new desks. Each desk cost $24. Which is the best estimate of how much the school spent on the new desks?
Options:
a. $500
b. $750
c. $1,000
d. $1,200

Answer: $750

Explanation:
Total number of desks= 32
Cost of each desk= $24
Total cost spent on the desks= 32 x 24=$768

So the estimated value can be $768.
Thus the correct answer is option b.

Question 5.
Kris has a box of 8 crayons. Sylvia’s box has 6 times as many crayons as Kris’s box. How many crayons are in Sylvia’s box?
Options:
a. 48
b. 42
c. 36
d. 4

Answer: 48

Explanation:
Number of crayons in Kris box=8
Number of crayons in Sylvia’s box= 6 times as many crayons as Kris’s box= 6 x 8=48
Thus the correct answer is option a.

Question 6.
Yesterday, 1,743 people visited the fair. Today, there are 576 more people at the fair than yesterday. How many people are at the fair today?
Options:
a. 1,167
b. 2,219
c. 2,319
d. 2,367

Answer: 2,319

Explanation:
Number of people in the fair yesterday= 1,743
Number of more people at the fair than yesterday= 576
Total number of people in the fair today=2,319

Thus the correct answer is option c.

Common Core – Divide by 1-Digit Numbers – Page No. 75

Divide Tens, Hundreds, and Thousands

Use basic facts and place value to find the quotient.

Question 1.
3,600 ÷ 4 = 900
Think: 3,600 is 36 hundreds.
Use the basic fact 36 ÷ 4 = 9.
So, 36 hundreds ÷ 4 = 9 hundreds, or 900.

Question 2.
240 ÷ 6 = ______

Answer: 40

Explanation:
STEP 1 Identify the basic fact. 24 ÷ 6
STEP 2 Use place value. 240 = 24 tens
STEP 3 Divide. 24 tens ÷ 6 = 4 tens
240 ÷ 6 = 40

Question 3.
5,400 ÷ 9 = ______

Answer: 600

Explanation:
STEP 1 Identify the basic fact. 54 ÷ 9
STEP 2 Use place value. 5,400 = 54 hundreds
STEP 3 Divide. 54 hundreds ÷ 9 = 6 hundreds
5,400 ÷ 9 = 600

Question 4.
300 ÷ 5 = ______

Answer: 60

Explanation:
STEP 1 Identify the basic fact. 30 ÷ 5
STEP 2 Use place value. 300 = 30 tens
STEP 3 Divide. 30 tens ÷ 5 = 60 tens
300 ÷ 5 = 60

Question 5.
4,800 ÷ 6 = ______

Answer: 800

Explanation:
STEP 1 Identify the basic fact. 48 ÷ 6
STEP 2 Use place value. 4,800 = 48 hundreds
STEP 3 Divide. 48 hundreds ÷ 6 = 80 hundreds
4,800 ÷ 6 = 800

Question 6.
420 ÷ 7 = ______

Answer: 60

Explanation:
STEP 1 Identify the basic fact. 42 ÷ 7
STEP 2 Use place value. 420 = 42 tens
STEP 3 Divide. 42 tens ÷ 7 = 60 tens
420 ÷ 7 = 60

Question 7.
150 ÷ 3 = ______

Answer: 50

Explanation:
STEP 1 Identify the basic fact. 15 ÷ 3
STEP 2 Use place value. 150 = 15 tens
STEP 3 Divide. 15 tens ÷ 3 = 5 tens
150 ÷ 3 = 50

Question 8.
6,300 ÷ 7 = ______

Answer: 900

Explanation:
STEP 1 Identify the basic fact. 63 ÷ 7
STEP 2 Use place value. 6,300 = 63 hundreds
STEP 3 Divide. 63 hundreds ÷ 7 = 9 hundreds
6,300 ÷ 7 = 900

Question 9.
1,200 ÷ 4 = ______

Answer: 300

Explanation:
STEP 1 Identify the basic fact. 12 ÷ 4
STEP 2 Use place value. 1,200 = 12 hundreds
STEP 3 Divide. 12 hundreds ÷ 4 = 3 hundreds
1,200 ÷ 4 = 300

Question 10.
360 ÷ 6 = ______

Answer: 60

Explanation:
STEP 1 Identify the basic fact. 36 ÷ 6
STEP 2 Use place value. 360 = 36 tens
STEP 3 Divide. 36 tens ÷ 6 = 6 tens
360 ÷ 6 = 60

Find the quotient.

Question 11.
28 ÷ 4 = ______
280 ÷ 4 = ______
2,800 ÷ 4 = ______

Answer: 7, 70, 700

Explanation:
Quotient:
A. Use 28 counters to represent the 28 dominoes. Then draw 4 circles to represent the divisor.
B. Share the counters equally among the 4 groups by placing them in the circles.
C. Number of counters formed in each group = quotient of 28 ÷ 4
D. Number of circles are equally filled with 7 counters, therefore, the quotient is 7

STEP 1 Identify the basic fact. 28 ÷ 4
STEP 2 Use place value. 280 = 28 tens
STEP 3 Divide. 28 tens ÷ 4 = 7 tens
280 ÷ 4 = 70

STEP 1 Identify the basic fact. 28 ÷ 4
STEP 2 Use place value. 2,800 = 28 hundreds
STEP 3 Divide. 28 hundreds ÷ 4 = 7 hundreds
2,800 ÷ 4 = 700

Question 12.
18 ÷ 3 = ______
180 ÷ 3 = ______
1,800 ÷ 3 = ______

Answer: 6, 60, 600

Explanation:
Quotient:
A. Use 18 counters to represent the 18 dominoes. Then draw 3 circles to represent the divisor.
B. Share the counters equally among the 3 groups by placing them in the circles.
C. Number of counters formed in each group = quotient of 18 ÷ 3
D. Number of circles are equally filled with 6 counters, therefore, the quotient is 6

STEP 1 Identify the basic fact. 18 ÷ 3
STEP 2 Use place value. 180 = 18 tens
STEP 3 Divide. 18 tens ÷ 3 = 6 tens
180 ÷ 6 = 60

STEP 1 Identify the basic fact. 18 ÷ 3
STEP 2 Use place value. 1,800 = 18 hundreds
STEP 3 Divide. 18 hundreds ÷ 3 = 6 hundreds
1,800 ÷ 3 = 600

Question 13.
45 ÷ 9 = ______
450 ÷ 9 = ______
4,500 ÷ 9 = ______

Answer: 5, 50, 500

Explanation:
Quotient:
A. Use 45 counters to represent the 45 dominoes. Then draw 9 circles to represent the divisor.
B. Share the counters equally among the 9 groups by placing them in the circles.
C. Number of counters formed in each group = quotient of 45 ÷ 9
D. Number of circles are equally filled with 5 counters, therefore, the quotient is 5

STEP 1 Identify the basic fact. 45 ÷ 9
STEP 2 Use place value. 450 = 45 tens
STEP 3 Divide. 45 tens ÷ 9 = 5 tens
450 ÷ 9 = 50

STEP 1 Identify the basic fact. 45 ÷ 9
STEP 2 Use place value. 4,500 = 45 hundreds
STEP 3 Divide. 45 hundred ÷ 9 = 5 hundred
4,500 ÷ 9 = 500

Problem Solving

Question 14.
At an assembly, 180 students sit in 9 equal rows. How many students sit in each row?
______ students

Answer: 20

Explanation:
Total number of students= 180
Number of rows= 9
Number of students in each row= 180 ÷ 9 = 20

Question 15.
Hilary can read 560 words in 7 minutes. How many words can Hilary read in 1 minute?
______ words

Answer: 80

Explanation:
Total number of words Hilary can read in 7 minutes = 560
Number of words Hilary can read in 1 minute= 560 ÷ 7= 80
Therefore Hilary can read 80 words in 1 minute.

Question 16.
A company produces 7,200 gallons of bottled water each day. The company puts 8 one-gallon bottles in each carton. How many cartons are needed to hold all the one-gallon bottles produced in one day?
______ cartons

Answer: 900

Explanation:
Total number of gallons bottled in each day= 7,200
Number of gallons bottled in each carton= 8
Number of cartons used= 7,200 ÷ 8= 900

Question 17.
An airplane flew 2,400 miles in 4 hours. If the plane flew the same number of miles each hour, how many miles did it fly in 1 hour?
______ miles

Answer: 600

Explanation:
Total number of miles flew in 4 hours= 2,400
Number of miles flew in 1 hour= 2,400 ÷ 4 = 600

Common Core – Divide by 1-Digit Numbers – Page No. 76

Lesson Check

Question 1.
A baseball player hits a ball 360 feet to the outfield. It takes the ball 4 seconds to travel this distance. How many feet does the ball travel in 1 second?
Options:
a. 9 feet
b. 40 feet
c. 90 feet
d. 900 feet

Answer: 90 feet

Explanation:
The height to which the player hits a ball=360 feet
Height to which the ball travels in 1 second= 360÷4= 90 feet
The correct answer is option c.

Question 2.
Sebastian rides his bike 2,000 meters in 5 minutes. How many meters does he bike in 1 minute?
Options:
a. 4 meters
b. 40 meters
c. 50 meters
d. 400 meters

Answer: 400 meters

Explanation:
Total number of meters travelled in 5 minutes= 2,000
Number of meters travelled in 1 minute= 2,000÷5= 400
The correct answer is option d.

Spiral Review

Question 3.
A full container of juice holds 64 ounces. How many 7-ounce servings of juice are in a full container?
Options:
a. 1
b. 8
c. 9
d. 10

Answer: 9

Explanation:
A full container of juice holds= 63 ounces
Quantity of servings of juice in one glass=7 ounce
The number of servings of the juice are= 63÷7=9
The correct answer is option c.

Question 4.
Paolo pays $244 for 5 identical calculators. Which is the best estimate of how much Paolo pays for one calculator?
Options:
a. $40
b. $50
c. $60
d. $245

Answer: $50

Explanation:
Amount Paolo pays for the identical calculators = $244
Number of identical calculators=5
The best-estimated value of each identical calculator=$244 ÷ 5is approximately $50.
The correct answer is option b.

Question 5.
A football team paid $28 per jersey. They bought 16 jerseys. How much money did the team spend on jerseys?
Options:
a. $44
b. $196
c. $408
d. $448

Answer: $448

Explanation:
Cost of each jersey=$28
Number of jerseys= 16
Total cost of the jerseys= $28 x 16= $448
The correct answer is option d.

Question 6.
Suzanne bought 50 apples at the apple orchard. She bought 4 times as many red apples as green apples. How many more red apples than green apples did Suzanne buy?
Options:
a. 10
b. 25
c. 30
d. 40

Answer: 40

Explanation:
Let the number of green apples be x and the number of red apples be 4x
4x + x = 50
x = 50 ÷ 5= 10
Number of red balls = 4x = 4 x 10 = 40
The correct answer is option d.

Common Core – Divide by 1-Digit Numbers – Page No. 77

Estimate Quotients Using Compatible Numbers

Use compatible numbers to estimate the quotient.

Question 1.
389 ÷ 4
400 ÷ 4 = 100

Question 2.
358 ÷ 3
_____ ÷ 3 = _____

Answer: 120

Explanation:
What number close to358 is easy to divide by 3?
360 is close to 358. What basic fact can you use?
360 ÷ 3
Choose 360 because it is close to 358 and can easily be divided by 3.
36 ÷3 = 12
360 ÷ 3 =120
358 ÷ 3 is about 120

Question 3.
784 ÷ 8
_____ ÷ 8 = _____

Answer: 100

Explanation:
What number close to 784 is easy to divide by 8?
800 is close to 784. What basic fact can you use?
800 ÷ 8
Choose 800 because it is close to 784 and can easily be divided by 8.
80 ÷ 8 = 10
800 ÷ 8 = 100
784 ÷ 8 is about 100.

Question 4.
179 ÷ 9
_____ ÷ 9 = _____

Answer: 20

Explanation:
What number close to 179 is easy to divide by 9?
180 is close to 179. What basic fact can you use?
180 ÷ 9
Choose 180 because it is close to 179 and can easily be divided by 9.
18 ÷ 9 = 2
180 ÷ 9 = 20
179 ÷ 9 is about 20

Question 5.
315 ÷ 8
_____ ÷ 8 = _____

Answer: 40

Explanation:
What number close to 315 is easy to divide by 8?
320 is close to 315. What basic fact can you use?
320 ÷ 8
Choose 320 because it is close to 315 and can easily be divided by 8.
32 ÷ 8 = 4
320 ÷ 8 =40
315 ÷ 8 is about 40.

Question 6.
2,116 ÷ 7
_____ ÷ 7 = _____

Answer: 300

Explanation:
What number close to 2,116 is easy to divide by 7?
2,100 is close to 2,116. What basic fact can you use?
2,100 ÷ 7
Choose 2,100 because it is close to 2,116 and can easily be divided by 7.
21 ÷ 7= 3
2,100 ÷ 7 = 300
2,116 ÷ 7 is about 300

Question 7.
4,156 ÷ 7
_____ ÷ 7 = _____

Answer: 600

Explanation:
What number close to 4,156 is easy to divide by 7?
4,200 is close to 4,156. What basic fact can you use?
4,200 ÷7
Choose 4,200 because it is close to 4,156 and can easily be divided by 7.
42 ÷ 7 = 6
4,200 ÷ 7 = 600
4,156 ÷ 7 is about 600.

Question 8.
474 ÷ 9
_____ ÷ 9 = _____

Answer: 50

Explanation:
What number close to 474 is easy to divide by 9?
450 is close to 474. What basic fact can you use?
450 ÷ 9
Choose 450 because it is close to 474 and can easily be divided by 9.
45 ÷ 9 = 5
450 ÷ 9 = 50
474 ÷ 9 is about 50.

Use compatible numbers to find two estimates that the quotient is between.

Question 9.
1,624 ÷ 3
_____ ÷ 3 = _____
_____ ÷ 3 = _____

Answer: The quotient is between 500 and 600

Explanation:
What number close to 1,624 is easy to divide by 3?
1,500 is close to 1,624. What basic fact can you use?
1,500 ÷ 3
Choose 1,500 because it is close to 1,624 and can easily be divided by 3.
15 ÷ 3 = 5
1,500 ÷ 3 = 500
1,624 ÷ 3 is about 500

What number close to 1,624 is easy to divide by 3?
1,800 is close to 1,624. What basic fact can you use?
1,800 ÷ 3
Choose 1,800 because it is close to 1,624 and can easily be divided by 3.
18 ÷ 3 = 6
1,800 ÷ 3 = 600
1,624 ÷ 3 is about 600

Question 10.
2,593 ÷ 6
_____ ÷ 6 = _____
_____ ÷ 6 = _____

Answer: The quotient is between 400 and 500

Explanation:
What number close to 2,593 is easy to divide by 6?
2,400 is close to 2,593. What basic fact can you use?
2,400 ÷ 6
Choose 2,400 because it is close to 2,593 and can easily be divided by 6.
24 ÷ 6 = 4
2,400 ÷ 6 = 400
2,593 ÷ 6 is about 400

What number close to 2,593 is easy to divide by 6?
3,000 is close to 2,593. What basic fact can you use?
3000 ÷ 6
Choose 3,000 because it is close to 2,593 and can easily be divided by 6.
30 ÷ 6 = 5
3,000 ÷ 6 = 500
2,593 ÷ 6 is about 500

Question 11.
1,045 ÷ 2
_____ ÷ 2 = _____
_____ ÷ 2 = _____

Answer: The quotient is between 520 and 525

Explanation:
What number close to 1,045 is easy to divide by 2?
1,040 is close to 1,045. What basic fact can you use?
1,040 ÷ 2
Choose 1,040 because it is close to 1,045 and can easily be divided by 2.
1,04 ÷ 2 = 52
1,040 ÷ 2 = 520
1,045 ÷ 2 is about 520

What number close to 1,045 is easy to divide by 2?
1,050 is close to 1,045. What basic fact can you use?
1,050 ÷ 2
Choose 1,050 because it is close to 1,045 and can easily be divided by 2.
1,050 ÷ 2 = 525
1,045 ÷ 2 is about 525

Question 12.
1,754 ÷ 9
_____ ÷ 9 = _____
_____ ÷ 9 = _____

Answer: The quotient is between 195 and 200

Explanation:
What number close to 1,754 is easy to divide by 9?
1,755 is close to 1,754. What basic fact can you use?
1,755 ÷ 9
Choose 1,755 because it is close to 1,754 and can easily be divided by 9.
1,755 ÷ 9 = 195
1,754 ÷ 9 is about 195

What number close to 1,754 is easy to divide by 9?
1,800 is close to 1,754. What basic fact can you use?
1,800 ÷ 9
Choose 1,800 because it is close to 1,754 and can easily be divided by 9.
18 ÷ 9 = 2
1,800 ÷ 9 = 200
1,754 ÷ 9 is about 200

Question 13.
2,363 ÷ 8
_____ ÷ 8 = _____
_____ ÷ 8 = _____

Answer: The quotient is between 295 and 300

Explanation:
What number close to 2,363 is easy to divide by 8?
2,360 is close to 2,363. What basic fact can you use?
2,360 ÷ 8
Choose 2,360 because it is close to 2,363 and can easily be divided by 8.
2,360 ÷ 8 = 295
2,363 ÷ 8 is about 295

What number close to 2,363 is easy to divide by 8?
2,400 is close to 2,363. What basic fact can you use?
2,400 ÷ 8
Choose 2,400 because it is close to 2,363 and can easily be divided by 8.
24 ÷ 8 = 3
2,400 ÷ 8= 300
2,363 ÷ 8 is about 300.

Question 14.
1,649 ÷ 5
_____ ÷ 5 = _____
_____ ÷ 5 = _____

Answer: The quotient is between 329 and 330

Explanation:
What number close to 1,649 is easy to divide by 5?
1,645 is close to 1,649. What basic fact can you use?
1,645 ÷ 5
Choose 1,645 because it is close to 1,649 and can easily be divided by 5.
1,645 ÷ 5 = 329
1,649 ÷ 5 is about 329

What number close to 1,650 is easy to divide by 5?
1,650 is close to 1,649. What basic fact can you use?
1,650 ÷ 5
Choose 1,650 because it is close to 1,649 and can easily be divided by 5.
1,650 ÷ 5 = 330
1,649 ÷ 5 is about 330

Question 15.
5,535 ÷ 7
_____ ÷ 7 = _____
_____ ÷ 7 = _____

Answer: The quotient is between 790 and 791

Explanation:
What number close to 5,535 is easy to divide by 7?
5,530 is close to 5,535. What basic fact can you use?
5,530 ÷ 7
Choose 5,530 because it is close to 5,535 and can easily be divided by 7.
553 ÷ 7 = 79
5,530 ÷ 7 = 790
5,535 ÷ 7 is about 790

What number close to 5,535 is easy to divide by 7?
5,537 is close to 5,535. What basic fact can you use?
5,537 ÷ 7
Choose 5,537 because it is close to 5,535 and can easily be divided by 7.
553 ÷ 7 = 79
5,537 ÷ 7 = 791
5,535 ÷ 7 is about 791

Question 16.
3,640 ÷ 6
_____ ÷ 6 = _____
_____ ÷ 6 = _____

Answer: The quotient is between 606 and 607

Explanation:
What number close to 3,640 is easy to divide by 6?
3,636 is close to 3,640. What basic fact can you use?
3,636 ÷ 6
Choose 3,636 because it is close to 3,640 and can easily be divided by 6.
36 ÷ 6 = 6
3,636 ÷ 6 = 606
3,640 ÷ 6 is about 606

What number close to 3,640 is easy to divide by 6?
3,642 is close to 3,640. What basic fact can you use?
3,642 ÷ 6
Choose 3,642 because it is close to 3,640 and can easily be divided by 6.
3,642 ÷ 6 = 607
3,640 ÷ 6 is about 607

Problem Solving

Question 17.
A CD store sold 3,467 CDs in 7 days. About the same number of CDs were sold each day. About how many CDs did the store sell each day?
about _____ CDs

Answer: 495(approx)

Explanation:
Total number of CDs in the store= 3,467
Number of days= 7
Number of CDs sold on one day= 3,467 ÷ 7=495(approx)

Question 18.
Marcus has 731 books. He puts about the same number of books on each of 9 shelves in his a bookcase. About how many books are on each shelf?
about _____ books

Answer: 81 books(approx)

Explanation:
Total number of books Marcus has= 731
Number of shelves= 9
Number of books on each shelf= 731÷9= 81 (approx)

Common Core – Divide by 1-Digit Numbers – Page No. 78

Lesson Check

Question 1.
Jamal is planting seeds for a garden nursery. He plants 9 seeds in each container. If Jamal has 296 seeds to plant, about how many containers will he use?
Options:
a. about 20
b. about 30
c. about 200
d. about 300

Answer: about 30

Explanation:
Total number of seeds Jamal has= 296
Number of seeds placed in each container= 9
Number of containers Jamal used= 296÷9= 32.8=33 (approx)
Therefore, the number of containers used is about 30
The correct answer is option b.

Question 2.
Winona purchased a set of vintage beads. There are 2,140 beads in the set. If she uses the beads to make bracelets that have 7 beads each, about how many bracelets can she make?
Options:
a. about 30
b. about 140
c. about 300
d. about 14,000

Answer: about 300

Explanation:
Total number of beads Winona has= 2,140
Number of beads in each bracelet= 7
Number of bracelets made= 2,140÷7=305.7=306(approx)
Therefore, the number of bracelets made are about 30
The correct answer is option c.

Spiral Review

Question 3.
A train traveled 360 miles in 6 hours. How many miles per hour did the train travel?
Options:
a. 60 miles per hour
b. 66 miles per hour
c. 70 miles per hour
d. 600 miles per hour

Answer: 60 miles per hour

Explanation:
Total number of miles traveled by train= 360
Time taken by the train to cover 360 miles= 6 hours
Number of miles traveled in each hour= 360÷6=60 miles
The correct answer is option a.

Question 4.
An orchard has 12 rows of pear trees. Each row has 15 pear trees. How many pear trees are there in the orchard?
Options:
a. 170
b. 180
c. 185
d. 190

Answer: 180

Explanation:
Number of rows of pear trees in an orchard= 12
Number of pear trees in each row=15
Total number of pear trees in the orchard= 12 x 15=180
The correct answer is option b.

Question 5.
Megan rounded 366,458 to 370,000. To which place did Megan round the number?
Options:
a. hundred thousands
b. ten thousands
c. thousands
d. hundreds

Answer: ten thousands

Explanation:
The given number is 366,458, the ten thousand place digit has 6 which while rounding off should be changed to the next consecutive number and the digits in the other places should be written as zeroes.
The correct answer is option b.

Question 6.
Mr. Jessup, an airline pilot, flies 1,350 miles a day. How many miles will he fly in 8 days?
Options:
a. 1,358 miles
b. 8,400 miles
c. 10,800 miles
d. 13,508 miles

Answer: 10,800 miles

Explanation:
Number of miles flew by Mr.Jessup in one day= 1,350 miles
Number of days=8
Total number of miles flew by Mr.Jessup in 8 days= 1,350 x 8= 10,800 miles.
The correct answer is option c.

Common Core – Divide by 1-Digit Numbers – Page No. 79

Division and the Distributive Property

Find the quotient.

Question 1.
54 ÷ 3 = ( 30 ÷ 3) + ( 24 ÷ 3)
= 10 + 8
= 18
Go Math Grade 4 Answer Key Homework Practice FL Chapter 4 Divide by 1-Digit Numbers Common Core - Divide by 1-Digit Numbers img 5

Question 2.
81 ÷ 3 = ______

Answer: 27

Explanation:
81 ÷ 3
STEP1 Find the nearest estimates of the number 81
STEP2 We can break the number 81 into 21 + 60
STEP3 We must divide the two parts of the number (dividend) with the divisor.
STEP4 (60 ÷ 3) + (21 ÷ 3)
STEP5 Add quotients of the above 20 +7= 27

Question 3.
232 ÷ 4 = ______

Answer: 58

Explanation:
232 ÷ 4
STEP1 Find the nearest estimates of the number 232
STEP2 We can break the number 232 into 200 + 32
STEP3 We must divide the two parts of the number (dividend) with the divisor.
STEP4 (200 ÷ 4) + (32 ÷ 4)
STEP5 Add quotients of the above 50 +8= 58

Question 4.
305 ÷ 5 = ______

Answer: 61

Explanation:
305 ÷ 5
STEP1 Find the nearest estimates of the number 305
STEP2 We can break the number 305 into 300 + 5
STEP3 We must divide the two parts of the number (dividend) with the divisor.
STEP4 (300 ÷ 5) + (5 ÷ 5)
STEP5 Add quotients of the above 60 +1= 61

Question 5.
246 ÷ 6 = ______

Answer: 41

Explanation:
246 ÷ 6
STEP1 Find the nearest estimates of the number 246
STEP2 We can break the number 246 into 240 + 6
STEP3 We must divide the two parts of the number (dividend) with the divisor.
STEP4 (240 ÷ 6) + (6 ÷ 6)
STEP5 Add quotients of the above 40 +1= 41

Question 6.
69 ÷ 3 = ______

Answer: 23

Explanation:
69 ÷ 3
STEP1 Find the nearest estimates of the number 69
STEP2 We can break the number 69 into 60 + 9
STEP3 We must divide the two parts of the number (dividend) with the divisor.
STEP4 (60 ÷ 3) + (9 ÷ 3)
STEP5 Add quotients of the above 20 +3= 23

Question 7.
477 ÷ 9 = ______

Answer: 53

Explanation:
477 ÷ 9
STEP1 Find the nearest estimates of the number 477
STEP2 We can break the number 477 into 450 + 27
STEP3 We must divide the two parts of the number (dividend) with the divisor.
STEP4 (450 ÷ 9) + (27 ÷ 9)
STEP5 Add quotients of the above 50 +3= 53

Question 8.
224 ÷ 7 = ______

Answer: 32

Explanation:
224 ÷ 7
STEP1 Find the nearest estimates of the number 224
STEP2 We can break the number 224 into 210 + 14
STEP3 We must divide the two parts of the number (dividend) with the divisor.
STEP4 (210 ÷ 7) + (14 ÷ 7)
STEP5 Add quotients of the above 30 +2= 32

Question 9.
72 ÷ 4 = ______

Answer: 18

Explanation:
72 ÷ 4
STEP1 Find the nearest estimates of the number 72
STEP2 We can break the number 72 into 40 + 32
STEP3 We must divide the two parts of the number (dividend) with the divisor.
STEP4 (40 ÷ 4) + (32 ÷ 4)
STEP5 Add quotients of the above 10 +8= 18

Question 10.
315 ÷ 3 = ______

Answer: 105

Explanation:
315 ÷ 3
STEP1 Find the nearest estimates of the number 315
STEP2 We can break the number 315 into 300 + 15
STEP3 We must divide the two parts of the number (dividend) with the divisor.
STEP4 (300 ÷ 3) + (15 ÷3)
STEP5 Add quotients of the above 100 +5= 105

Problem Solving

Question 11.
Cecily picked 219 apples. She divided the apples equally into 3 baskets. How many apples are in each basket?
______ apples

Answer: 73 apples

Explanation:
The total number of apples Cecily picked= 219 apples
Number of parts into which she wanted to divide the apples= 3
Number of apples in each part = Quotient of 147 ÷ 7
STEP1 Find the nearest estimates of the number 219
STEP2 We can break the number 219 into 210 + 9
STEP3 We must divide the two parts of the number (dividend) with the divisor.
STEP4 (210 ÷ 3) + (9 ÷ 3)
STEP5 Add quotients of the above 70 +3= 73

Question 12.
Jordan has 260 basketball cards. He divides them into 4 equal groups. How many cards are in each group?
______ cards

Answer: 65 cards

Explanation:
The total number of basketball cards Jordan has= 260 basketball cards
Number of parts into which he wanted to divide the cards= 4
Number of apples in each part = Quotient of 260 ÷ 4
STEP1 Find the nearest estimates of the number 260
STEP2 We can break the number 260 into 240 + 20
STEP3 We must divide the two parts of the number (dividend) with the divisor.
STEP4 (240 ÷ 4) + (20 ÷ 4)
STEP5 Add quotients of the above 60 +5= 65

Question 13.
The Wilsons drove 324 miles in 6 hours. If they drove the same number of miles each hour, how many miles did they drive in 1 hour?
______ miles

Answer: 54 miles

Explanation:
The total number of miles drove by Wilson= 324 miles
Number of hours he drove = 6
Number of miles drove in each hour = Quotient of 324 ÷ 6
STEP1 Find the nearest estimates of the number 324
STEP2 We can break the number 324 into 300 + 24
STEP3 We must divide the two parts of the number (dividend) with the divisor.
STEP4 (300 ÷ 6) + (24 ÷ 6)
STEP5 Add quotients of the above 50 +4= 54

Question 14.
Phil has 189 stamps to put into his stamp album. He puts the same number of stamps on each of 9 pages. How many stamps does Phil put on each page?
______ stamps

Answer: 21 stamps

Explanation:
The total number of stamps Phil has= 189 stamps
Number of pages= 9
Number of stamps put on each page = Quotient of 189 ÷ 9
STEP1 Find the nearest estimates of the number 189
STEP2 We can break the number 189 into 180 + 9
STEP3 We must divide the two parts of the number (dividend) with the divisor.
STEP4 (180 ÷ 9) + (9 ÷ 9)
STEP5 Add quotients of the above 20 +1= 21

Common Core – Divide by 1-Digit Numbers – Page No. 80

Lesson Check

Question 1.
A landscaping company planted 176 trees in 8 equal rows in the new park. How many trees did the company plant in each row?
Options:
a. 18
b. 20
c. 22
d. 24

Answer: 22

Explanation:
The total number of trees in the landscaping= 176 trees
Number of rows= 8
Number of trees in each row = Quotient of 176 ÷ 8
STEP1 Find the nearest estimates of the number 176
STEP2 We can break the number 176 into 160 + 16
STEP3 We must divide the two parts of the number (dividend) with the divisor.
STEP4 (160 ÷ 8) + (16 ÷ 8)
STEP5 Add quotients of the above 20 +2= 22

Question 2.
Arnold can do 65 pushups in 5 minutes. How many pushups can he do in 1 minute?
Options:
a. 11
b. 13
c. 15
d. 17

Answer: 13

Explanation:
The total number of pushups done by Arnold = 65
Number of minutes spent on pushups= 5
Number of pushups done in each minute = Quotient of 65 ÷ 5
STEP1 Find the nearest estimates of the number 65
STEP2 We can break the number 65 into 60 + 5
STEP3 We must divide the two parts of the number (dividend) with the divisor.
STEP4 (60 ÷ 5) + (5 ÷ 5)
STEP5 Add quotients of the above 12 +1= 13

Spiral Review

Question 3.
Last Saturday, there were 1,486 people at the Cineplex. There were about the same number of people in each of the 6 theaters. Which is the best estimate of the number of people in each theater?
Options:
a. between 20 and 30
b. between 80 and 90
c. between 100 and 200
d. between 200 and 300

Answer: between 200 and 300

Explanation:
Total number of people at the Cineplex= 1,486 people
Number of theatres = 6
Number of people at each theatre= estimate of the number of people 1,486 ÷ 6

What number close to 1,486 is easy to divide by 6?
1,488 is close to 1,486. What basic fact can you use?
1,488 ÷ 6
Choose 1,488 because it is close to 1,486 and can easily be divided by 6.
1,488 ÷ 6 = 248
1,486 ÷ 6 is about 248

What number close to 1,486 is easy to divide by 6?
1,482 is close to 1,486 . What basic fact can you use?
1,482 ÷ 6
Choose 1,482 because it is close to 1,486 and can easily be divided by 6.
1,482 ÷ 6 = 247
1,486 ÷ 6 is about 247

Question 4.
Nancy walked 50 minutes each day for 4 days last week. Gillian walked 35 minutes each day for 6 days last week. Which statement is true?
Options:
a. Gillian walked 10 minutes more than Nancy.
b. Gillian walked 20 minutes more than Nancy.
c. Nancy walked 10 minutes more than Gillian.
d. Nancy walked 15 minutes more than Gillian.

Answer: Nancy walked 15 minutes more than Gillian.

Explanation:
Time walked by Nancy= 50 minutes
Time walked by Gillian= 35 minutes
Nancy walked more time compared to Gillian
50-35=15 minutes
Therefore, Nancy walked 15 minutes more than Gillian.

Question 5.
Three boys share 28 toy cars equally. Which best describes how the cars are shared?
Options:
a. Each gets 3 cars with 1 left over.
b. Each gets 8 cars with 2 left over.
c. Each gets 9 cars with 1 left over.
d. Each gets 10 cars with 2 left over.

Answer: Each gets 9 cars with 1 left over.

Explanation:
Total number of toys three boys have= 28
Number of toys each boy got= 28 ÷3=9.33
Therefore we can say that each gets 9 cars with 1 leftover.

Question 6.
An airplane flies at a speed of 474 miles per hour. How many miles does the plane fly in 5 hours?
Options:
a. 2,070 miles
b. 2,140 miles
c. 2,370 miles
d. 2,730 miles

Answer: 2,370 miles

Explanation:
Number of miles flew by airplane in one hour= 474
Number of hours the airplane flew= 5 hours
Total number of miles flew in 5 hours= 474 x 5= 2,370 miles

Common Core – Divide by 1-Digit Numbers – Page No. 81

Divide Using Repeated Subtraction

Use repeated subtraction to divide.

Question 1.
42 ÷ 3 = 14
3)\(\overline { 42 } \)
-30 ← 10 × 3 | 10
——-
12
-12 ← 4 × 3 | +4
——-             ———
0                     14

Question 2.
72 ÷ 4 = ______

Answer: 18

Explanation:
A. Begin with 72 counters. Subtract 4 counters.
B. Subtract 4 counters from 72 and repeat the processes until the remainder cannot be subtracted from the divisor.
C. Record the number of counters left and the number of times you subtracted.
D. The number of times you subtracted is the quotient is 18

Question 3.
93 ÷ 3 = ______

Answer: 31

Explanation:
A. Begin with 93 counters. Subtract 3 counters.
B. Subtract 3 counters from 93 and repeat the processes until the remainder cannot be subtracted from the divisor.
C. Record the number of counters left and the number of times you subtracted.
D. The number of times you subtracted is the quotient is 31

Question 4.
35 ÷ 4 = ______ r ______

Answer: 8r3

Explanation:
Quotient:
A. Use 35 counters to represent the 35 dominoes. Then draw 4 circles to represent the divisor.
B. Share the counters equally among the 4 groups by placing them in the circles.
C. Number of counters formed in each group = quotient of 35 ÷ 4
D. Number of circles are equally filled with 4 counters, therefore, the quotient is 8
Remainder:
The number of counters left over is the remainder. The number of counters leftover= 3
For 35 ÷ 4, the quotient is 8 and the remainder is 3, or 8 r3.

Question 5.
93 ÷ 10 = ______ r ______

Answer: 9r3

Explanation:
Quotient:
A. Use 93 counters to represent the 93 dominoes. Then draw 10 circles to represent the divisor.
B. Share the counters equally among the 10 groups by placing them in the circles.
C. Number of counters formed in each group = quotient of 93 ÷ 10
D. Number of circles are equally filled with 10 counters, therefore, the quotient is 9
Remainder:
The number of counters left over is the remainder. The number of counters leftover= 3
For 93 ÷ 10, the quotient is 9 and the remainder is 3, or 9 r3.

Question 6.
86 ÷ 9 = ______ r ______

Answer: 9r5

Explanation:
Quotient:
A. Use 86 counters to represent the 86 dominoes. Then draw 9 circles to represent the divisor.
B. Share the counters equally among the 9 groups by placing them in the circles.
C. Number of counters formed in each group = quotient of 86 ÷ 9
D. Number of circles are equally filled with 9 counters, therefore, the quotient is 9
Remainder:
The number of counters left over is the remainder. The number of counters leftover= 5
For 86 ÷ 9, the quotient is 9 and the remainder is 5, or 9 r5.

Draw a number line to divide.

Question 7.
70 ÷ 5 = ______

Answer: 14

Explanation:
A. Draw a number line with 5 as each interval.
B. Draw up to 70 and count the intervals, it gives the quotient.
C. The quotient is 14

Problem Solving

Question 8.
Gretchen has 48 small shells. She uses 2 shells to make one pair of earrings. How many pairs of earrings can she make?
______ pairs

Answer: 24 pairs

Explanation:
Total number of small shells= 48
Number of shells used to make one pair of earrings = 2
Number of pair of earrings made = 48 ÷ 2 =24

Question 9.
James wants to purchase a telescope for $54. If he saves $3 per week, in how many weeks will he have saved enough to purchase the telescope?
______ weeks

Answer: $18

Explanation:
Cost of the telescope=$54
Amount saved each week = $3
Number of weeks he has to save the money to purchase the telescope = $54 ÷ $3 = $18

Common Core – Divide by 1-Digit Numbers – Page No. 82

Lesson Check

Question 1.
Randall collects postcards that his friends send him when they travel. He can put 6 cards on one scrapbook page. How many pages does Randall need to fit 42 postcards?
Options:
a. 3
b. 4
c. 6
d. 7

Answer: 7

Explanation:
Total number of postcards Randall has = 42 postcards
Number of postcards on one scrapbook page = 6 cards
Number of pages needed to fit the postcards = 42 ÷ 6=7
The correct answer is option d.

Question 2.
Ari stocks shelves at a grocery store. He puts 35 cans of juice on each shelf. The shelf has 4 equal rows and another row with only 3 cans. How many cans are in each of the equal rows?
Options:
a. 6
b. 7
c. 8
d. 9

Answer: 8

Explanation:
Total number of cans of juice on each shelf = 35
Number of rows = 4
Number of cans on the other shelf = 3
Number of cans placed on the first shelf = 35 – 3 = 32
Number of juice cans in the first row = 32 ÷ 4 = 8 cans
The correct answer is option c.

Spiral Review

Question 3.
Fiona sorted her CDs into separate bins. She placed 4 CDs in each bin. If she has 160 CDs, how many bins did she fill?
Options:
a. 4
b. 16
c. 40
d. 156

Answer: 40

Explanation:
Total number of CD’s in Fiona has = 160 CD’s
Number of CD’s placed in each bin = 4
Number of bins required to place the CD’s = 160 ÷ 4 = 40
The correct answer is option c.

Question 4.
Eamon is arranging 39 books on 3 shelves. If he puts the same number of books on each shelf, how many books will there be on each shelf?
Options:
a. 11
b. 12
c. 13
d. 14

Answer: 13

Explanation:
Total number of books Eamon has = 39 books
Number of shelves = 3
Number of books in each shelf = 39 ÷ 3 = 13
The correct answer is option c.

Question 5.
A newborn boa constrictor measures 18 inches long. An adult boa constrictor measures 9 times the length of the newborn plus 2 inches. How long is the adult?
Options:
a. 142 inches
b. 162 inches
c. 164 inches
d. 172 inches

Answer: 164 inches

Explanation:
Length of newborn boa constrictor = 18 inches
Length of an adult boa constrictor = 9 x Length of newborn boa constrictor = 9 x 18 = 162
Total length of an adult boa constrictor = 162 + 2 = 164 inches
The correct answer is option c.

Question 6.
Madison has 6 rolls of coins. Each roll has 20 coins. How many coins does Madison have in all?
Options:
a. 110
b. 120
c. 125
d. 130

Answer: 120

Explanation:
Number of rolls of coins = 6
Number of coins in each roll = 20
Total number of coins Madison has = 20 x 6 = 120
The correct answer is option b.

Common Core – Divide by 1-Digit Numbers – Page No. 83

Divide Using Partial Quotients

Divide. Use partial quotients.

Question 1.
8)\(\overline { 184 } \)
-80 ← 10 × 8 10
——-
104
-80 ← 10 × 8 +10
-24
-24 ← 3 × 8  +3
——-              ———
0                      23

Question 2.
6)\(\overline { 258 } \)
______

Answer: 43

Explanation:
STEP 1
Start by subtracting a greater multiple, such as 40 times the divisor.
Continue subtracting until the remaining number is less than the multiple, 6.
STEP 2
Subtract smaller multiples, such as 3 times the divisor until the remaining number is less than the divisor. In other words, keep going until you no longer a remainder is left in the place of the remainder. Then add the partial quotients to find the quotient.
So, there are 40 x 6 = 240 : 258 – 240 = 18
3 x 6 = 18 : 18 – 18 = 0
Therefore the quotient is 43 ( 40 + 3)

Question 3.
5)\(\overline { 630 } \)
______

Answer: 126

Explanation:
STEP 1
Start by subtracting a greater multiple, such as 100 times the divisor.
Continue subtracting until the remaining number is less than the multiple, 5.
STEP 2
Subtract smaller multiples, such as 20 times the divisor until the remaining number is less than the divisor. In other words, keep going until you no longer a remainder is left in the place of the remainder. Then add the partial quotients to find the quotient.
So, there are 100 x 5 = 500 : 630 – 500 = 130
5 x 20 = 100 : 130 – 100 = 30 : 5 x 6 = 30 : 30 – 30 = 0
Therefore the quotient is 126 ( 100 + 20 + 6)

Divide. Use rectangular models to record the partial quotients.

Question 4.
246 ÷ 3 = ____

Answer: 82

Explanation:
STEP 1
Start by subtracting a greater multiple, such as 80 times the divisor.
Continue subtracting until the remaining number is less than the multiple, 3.
STEP 2
Subtract smaller multiples, such as 80 times the divisor until the remaining number is less than the divisor. In other words, keep going until you no longer a remainder is left in the place of the remainder. Then add the partial quotients to find the quotient.
So, there are 80 x 3 = 240 : 246 – 240 = 6
3 x 2 = 6 : 6 – 6 = 0
Therefore the quotient is 82 ( 80 + 2)
The rectangle models are given below :

Question 5.
126 ÷ 2 = ____

Answer: 63

Explanation:
STEP 1
Start by subtracting a greater multiple, such as 60 times the divisor.
Continue subtracting until the remaining number is less than the multiple,2.
STEP 2
Subtract smaller multiples, such as 60 times the divisor until the remaining number is less than the divisor. In other words, keep going until you no longer a remainder is left in the place of the remainder. Then add the partial quotients to find the quotient.
So, there are 60 x 2 = 120 : 126 – 120 = 6
2 x 3 = 6 : 6 – 6 = 0
Therefore the quotient is 63 ( 60 +3)
The rectangle models are given below :

Question 6.
605 ÷ 5 = ____

Answer: 121

Explanation:
STEP 1
Start by subtracting a greater multiple, such as 100 times the divisor.
Continue subtracting until the remaining number is less than the multiple, 5.
STEP 2
Subtract smaller multiples, such as 20 times the divisor until the remaining number is less than the divisor. In other words, keep going until you no longer a remainder is left in the place of the remainder. Then add the partial quotients to find the quotient.
So, there are 100 x 5 = 500 : 605 – 500 = 105
5 x 20 = 100 : 105 – 100 = 5 : 5 x 1 = 5 : 5 – 5 = 0
Therefore the quotient is 121 ( 100 + 20 + 1)
The rectangle models are given below :

Divide. Use either way to record the partial quotients.

Question 7.
492 ÷ 3 = ____

Answer: 164

Explanation:
STEP 1
Start by subtracting a greater multiple, such as 100 times the divisor.
Continue subtracting until the remaining number is less than the multiple, 3.
STEP 2
Subtract smaller multiples, such as 50 times the divisor until the remaining number is less than the divisor. In other words, keep going until you no longer a remainder is left in the place of the remainder. Then add the partial quotients to find the quotient.
So, there are 100 x 3 = 300 : 492 – 300 = 192
50 x 3 = 150 : 192 – 150 = 42 : 3 x 14 = 42 : 42 – 42 = 0
Therefore the quotient is 164 ( 100 + 50 + 14)

Question 8.
224 ÷ 7 = ____

Answer: 32

Explanation:
STEP 1
Start by subtracting a greater multiple, such as 30 times the divisor.
Continue subtracting until the remaining number is less than the multiple, 7.
STEP 2
Subtract smaller multiples, such as 30 times the divisor until the remaining number is less than the divisor. In other words, keep going until you no longer a remainder is left in the place of the remainder. Then add the partial quotients to find the quotient.
So, there are 30 x 7 = 210 : 224 – 210 = 14
7 x 2 = 14 : 14 – 14 = 0
Therefore the quotient is 32 ( 30 + 2)

Question 9.
692 ÷ 4 = ____

Answer: 173

Explanation:
STEP 1
Start by subtracting a greater multiple, such as 100 times the divisor.
Continue subtracting until the remaining number is less than the multiple, 4.
STEP 2
Subtract smaller multiples, such as 100 times the divisor until the remaining number is less than the divisor. In other words, keep going until you no longer a remainder is left in the place of the remainder. Then add the partial quotients to find the quotient.
So, there are 100 x 4 = 400 : 692 – 400 = 392
4 x 50 = 200 : 392 – 200 = 192 : 4 x 48 = 192 : 192 – 192 = 0
Therefore the quotient is 198 ( 100 + 50 + 48)

Problem Solving

Question 10.
Allison took 112 photos on vacation. She wants to put them in a photo album that holds 4 photos on each page. How many pages can she fill?
____ pages

Answer: 28 pages

Explanation:
STEP 1
Start by subtracting a greater multiple, such as 20 times the divisor.
Continue subtracting until the remaining number is less than the multiple, 4.
STEP 2
Subtract smaller multiples, such as 20 times the divisor until the remaining number is less than the divisor. In other words, keep going until you no longer a remainder is left in the place of the remainder. Then add the partial quotients to find the quotient.
So, there are 20 x 4 = 80 : 112 – 80 = 32
4 x 8 = 32 : 32 – 32 = 0
Therefore the quotient is 28 ( 20 + 8)

Question 11.
Hector saved $726 in 6 months. He saved the same amount each month. How much did Hector save each month?
$ ____

Answer: $121

Explanation:
STEP 1
Start by subtracting a greater multiple, such as 100 times the divisor.
Continue subtracting until the remaining number is less than the multiple, 6.
STEP 2
Subtract smaller multiples, such as 100 times the divisor until the remaining number is less than the divisor. In other words, keep going until you no longer a remainder is left in the place of the remainder. Then add the partial quotients to find the quotient.
So, there are 100 x 6 = 600 : 726 – 600 = 126
6 x 20 = 120 : 126 – 120 = 6 : 6 x 1 = 6 : 6 – 6 = 0
Therefore the quotient is 121 ( 100 + 20 +1)

Common Core – Divide by 1-Digit Numbers – Page No. 84

Lesson Check

Question 1.
Annaka used partial quotients to divide 145 ÷ 5. Which shows a possible sum of partial quotients?
Options:
a. 50 + 50 + 45
b. 100 + 40 + 5
c. 10 + 10 + 9
d. 10 + 4 + 5

Answer: 10 + 10 + 9

Explanation:
STEP 1
Start by subtracting a greater multiple, such as 100 times the divisor.
Continue subtracting until the remaining number is less than the multiple, 4.
STEP 2
Subtract smaller multiples, such as 10 times the divisor until the remaining number is less than the divisor. In other words, keep going until you no longer a remainder is left in the place of the remainder. Then add the partial quotients to find the quotient.
So, there are 10 x 5 = 50 : 145 – 50 = 95
5 x 10 = 50 : 95 – 50 = 45 : 5 x 9 = 45 : 45 – 45 = 0
Therefore the quotient is 29 ( 10 + 10 +9)

Question 2.
Mel used partial quotients to find the quotient 378 ÷ 3. Which might show the partial quotients that Mel found?
Options:
a. 100, 10, 10, 9
b. 100, 10, 10, 6
c. 100, 30, 30, 6
d. 300, 70, 8

Answer: 100, 10, 10, 6

Explanation:
STEP 1
Start by subtracting a greater multiple, such as 100 times the divisor.
Continue subtracting until the remaining number is less than the multiple, 3.
STEP 2
Subtract smaller multiples, such as 10 times the divisor until the remaining number is less than the divisor. In other words, keep going until you no longer a remainder is left in the place of the remainder. Then add the partial quotients to find the quotient.
So, there are 100 x 3 = 300 : 378 – 300 = 78
10 x 3 =30 : 78 – 30 = 48 : 3 x 16 = 48 : 48 – 48 = 0
Therefore the quotient is 126 ( 100 + 10 +10 + 6)

Spiral Review

Question 3.
What are the partial products of 42 × 5?
Options:
a. 9 and 7
b. 20 and 10
c. 200 and 7
d. 200 and 10

Answer: 200 and 10

Explanation:
STEP1
42 x 5
Start by multiplying the digit five with the units digit 2 = 5 x 2 =10
Multiply the digit 5 with 4 in the tens place = 4 x 5 = 20
Since 4 is in the tens place when we multiply 4 and 5 we must place it in the hundreds place by assuming the units digit to be zero.
Therefore, the partial product of 42 x 5 = 200

Question 4.
Mr. Watson buys 4 gallons of paint that cost $34 per gallon. How much does Mr. Watson spend on paint?
Options:
a. $38
b. $126
c. $136
d. $1,216

Answer: $136

Explanation:
Cost of each gallon of paint = $34
Number of gallons = 4
The total cost of the gallons = $ 34 x 4 = $136

Question 5.
Use the area model to find the product 28 × 32.
Go Math Grade 4 Answer Key Homework Practice FL Chapter 4 Divide by 1-Digit Numbers Common Core - Divide by 1-Digit Numbers img 6
Options:
a. 840
b. 856
c. 880
d. 896

Answer: 896

Explanation:
The whole rectangle is divided into four small rectangles the areas of these rectangles are:

Area of yellow rectangle= 30 x 20=600
Area of green rectangle= 2 x 20 = 40
Area of pink rectangle= 8 x 30= 240
Area of blue rectangle= 2 x 8= 16
Product of 32 and 28 = Area of yellow rectangle + Area of green rectangle + Area of pink rectangle + Area of the blue rectangle = 600+40+240+16 = 896

Question 6.
An adult male lion eats about 108 pounds of meat per week. About how much meat does an adult male lion eat in one day?
Options:
a. about 14 pounds
b. about 15 pounds
c. about 16 pounds
d. about 17 pounds

Answer: about 15 pounds

Explanation:
Mass of meat an adult lion eats in one week = 108
Number of days in a week = 7
Mass of meat ate by the lion in one day = 108 ÷ 7 = 15.4 pounds = about 15 pounds

Common Core – Divide by 1-Digit Numbers – Page No. 85

Model Division with Regrouping

Divide. Use base-ten blocks.

Question 1.
63 ÷ 4 = 15 r3
Go Math Grade 4 Answer Key Homework Practice FL Chapter 4 Divide by 1-Digit Numbers Common Core - Divide by 1-Digit Numbers img 7

Explanation:
A. draw 4 circles to represent the divisor. Then use base-ten blocks to model 63. Show 63 as 6 tens and 3 ones.
B. Share the tens equally among the 4 groups.
C. If there are any tens left, regroup them as ones. Share the ones equally among the 4 groups.
D. There are 1 ten(s) and 5 one(s) in each group. So, the quotient is 15.
E. After grouping, there are 3 blocks that weren’t grouped. So, the remainder is 3

Question 2.
83 ÷ 3
_____ R _____

Answer: 27 r 2

Explanation:
A. Draw 3 circles to represent the divisor. Then use base-ten blocks to model 83. Show 83 as 8 tens and 3 ones.
B. Share the tens equally among the 3 groups.
C. If there are any tens left, regroup them as ones. Share the ones equally among the 3 groups.
D. There are 2 ten(s) and 7 one(s) in each group. So, the quotient is 27.
E. After grouping, there are 2 blocks that weren’t grouped. So, the remainder is 2

Divide. Draw quick pictures. Record the steps.

Question 3.
85 ÷ 5
_____

Answer: 17

Explanation:
A. Draw 5 circles to represent the divisor. Then use base-ten blocks to model 85. Show 85 as 8 tens and 5 ones.
B. Share the tens equally among the 5 groups.
C. If there are any tens left, regroup them as ones. Share the ones equally among the 5 groups.
D. There are 1 ten(s) and 7 one(s) in each group. So, the quotient is 17.

Question 4.
97 ÷ 4
_____ R _____

Answer: 24 r 1

Explanation:
A. Draw 4 circles to represent the divisor. Then use base-ten blocks to model 97. Show 97 as 9 tens and 7 ones.
B. Share the tens equally among the 4 groups.
C. If there are any tens left, regroup them as ones. Share the ones equally among the 4 groups.
D. There are 2 ten(s) and 4 one(s) in each group. So, the quotient is 24.
E. After grouping, there is 1 block that wasn’t grouped. So, the remainder is 1.

Problem Solving

Question 5.
Tamara sold 92 cold drinks during her 2-hour shift at a festival food stand. If she sold the same number of drinks each hour, how many cold drinks did she sell each hour?
_____ cold drinks

Answer: 46 cold drinks

Explanation:
Total number of cold drinks Tamara sold = 92
The time in which she sold the drinks = 2 hours
Number of drinks she sold in each hour = 92 ÷ 2 = 46

Question 6.
In 3 days Donald earned $42 running errands. He earned the same amount each day. How much did Donald earn from running errands each day?
$ _____

Answer: $14

Explanation:
Total amount earned by Donald = $42
Number of days = 3
Amount earned on each day = $42 ÷ 3 = $14

Common Core – Divide by 1-Digit Numbers – Page No. 86

Lesson Check

Question 1.
Gail bought 80 buttons to put on the shirts she makes. She uses 5 buttons for each shirt. How many shirts can Gail make with the buttons she bought?
Options:
a. 14
b. 16
c. 17
d. 18

Answer: 16

Explanation:
Total number of buttons = 80
Number of buttons used for each shirt = 5
Number of shirts she can make = 80 ÷ 5 =16
The correct answer is option b.

Question 2.
Marty counted how many breaths he took in 3 minutes. In that time, he took 51 breaths. He took the same number of breaths each minute. How many breaths did Marty take in one minute?
Options:
a. 15
b. 16
c. 17
d. 19

Answer: 17

Explanation:
Total number of breaths Marty counted = 51
Time in which the breath was counted = 3 minutes
Number of breaths in one minute = 51 ÷ 3 = 17
The correct answer is option c.

Spiral Review

Question 3.
Kate is solving brain teasers. She solved 6 brain teasers in 72 minutes. How long did she spend on each brain teaser?
Options:
a. 12 minutes
b. 14 minutes
c. 18 minutes
d. 22 minutes

Answer: 12 minutes

Explanation:
Number of brain teasers solved = 6
Number of minutes spent on brain teasers = 72 minutes
Number of minutes spent on each problem = 72 ÷ 6 =12 minutes
The correct answer is option a.

Question 4.
Jenny works at a package delivery store. She puts mailing stickers on packages. Each package needs 5 stickers. How many stickers will Jenny use if she is mailing 105 packages?
Options:
a. 725
b. 625
c. 525
d. 21

Answer: 525

Explanation:
Number of packages = 105
Number of stickers on each package = 5
Total number of stickers on the packages = 105 x 5 = 525
The correct answer is option c.

Question 5.
The Puzzle Company packs standardized puzzles into boxes that hold 8 puzzles. How many boxes would it take to pack up 192 standard-sized puzzles?
Options:
a. 12
b. 16
c. 22
d. 24

Answer: 24

Explanation:
Total number of puzzles = 192
Number of puzzles in each box = 8
Number of boxes used = 192 ÷ 8 = 24 boxes
The correct answer is option d.

Question 6.
Mt. Whitney in California is 14,494 feet tall. Mt. McKinley in Alaska is 5,826 feet taller than Mt. Whitney. How tall is Mt. McKinley?
Options:
a. 21,310 feet
b. 20,320 feet
c. 20,230 feet
d. 19,310 feet

Answer: 20,320 feet

Explanation:
Height of Mt. Whitney in California = 14,494 feet
The height of Mt. McKinley in Alaska is 5,826 feet taller than Mt. Whitney.
Therefore the height of Mt. McKinley in Alaska = 14,494 feet + 5,826 feet = 20,320 feet
The correct answer is option b.

Common Core – Divide by 1-Digit Numbers – Page No. 87

Place the First Digit

Divide.

Question 1.
62
3)\(\overline { 186 } \)
-18
——–
06
-6
——–
0

Explanation:
STEP 1 Use place value to place the first digit. Look at the hundreds in 186. 180 hundred can be shared among 3 groups
without regrouping.
Now there is 18 tens and 6 ones to share among 3 groups.
The first digit of the quotient will be in the tens place.
STEP 2 Divide the tens.
Divide. 180 ÷ 3
Multiply. 3 × 60 = 180
Subtract. 186 − 180 = 6 ones
STEP 3 Divide the ones.
Now there are 6 ones to share among 3 groups.
Divide. 6 ones ÷ 3
Multiply. 2×3 ones
Subtract. 6 ones − 2 ones =0 one
So, the quotient is 62 (60 + 2) and the remainder is 0

Question 2.
4)\(\overline { 298 } \)
_____ R _____

Answer:
STEP 1 Use place value to place the first digit. Look at the hundreds in 298. 280 hundred can be shared among 4 groups
without regrouping.
Now there are 28 tens and 18 ones to share among 4 groups.
The first digit of the quotient will be in the hundreds place.
STEP 2 Divide the tens.
Divide. 280 ÷ 4
Multiply. 4 × 70 = 280
Subtract. 280 − 280 = 0 ones
STEP 3 Divide the ones.
Now there are 18 ones to share among 4 groups.
Divide. 18 ones ÷ 4
Multiply. 4×4 ones
Subtract. 18 ones − 16 ones = 2 ones
So, the quotient is 74 (70 + 4) and the remainder is 2.

Question 3.
3)\(\overline { 461 } \)
_____ R _____

Answer: 153

Explanation:
STEP 1 Use place value to place the first digit. Look at the hundreds in 461. 450 hundred can be shared among 3 groups
without regrouping.
Now there is 45 tens and 11 ones to share among 3 groups.
The first digit of the quotient will be in the hundreds place.
STEP 2 Divide the tens.
Divide. 450 ÷ 3
Multiply. 3 × 150 = 450
Subtract. 450 − 450 = 0 ones
STEP 3 Divide the ones.
Now there are 11 ones to share among 3 groups.
Divide. 11 ones ÷ 3
Multiply. 3×3 ones
Subtract. 11 ones − 9 ones = 2 ones
So, the quotient is 153 (150 + 3) and the remainder is 2

Question 4.
9)\(\overline { 315 } \)
_____ R _____

Answer: 35

Explanation:
STEP 1 Use place value to place the first digit. Look at the hundreds in 315. 310 hundred can be shared among 9 groups
without regrouping.
Now there is 31 tens and 5 ones to share among 9 groups.
The first digit of the quotient will be in the hundreds place.
STEP 2 Divide the tens.
Divide.310 ÷ 9
Multiply. 9 × 30 = 270
Subtract. 310 − 270 = 40 ones
STEP 3 Divide the ones.
Now there are 40 + 5 = 45 ones to share among 9 groups.
Divide. 45 ones ÷ 9
Multiply. 5×9 ones
Subtract. 45 ones − 45 ones = 0 ones
So, the quotient is 35 (30 + 5) and the remainder is 0

Question 5.
2)\(\overline { 766 } \)
_____ R _____

Answer: 383

Explanation:
STEP 1 Use place value to place the first digit. Look at the hundreds in 766. 760 hundred can be shared among 2 groups
without regrouping.
Now there is 76 tens and 6 ones to share among 2 groups.
The first digit of the quotient will be in the hundreds place.
STEP 2 Divide the tens.
Divide. 760 ÷ 2
Multiply. 2 × 380 = 760
Subtract. 760 − 760 = 0 ones
STEP 3 Divide the ones.
Now there are 6 ones to share among 2 groups.
Divide. 6 ones ÷ 2
Multiply. 2×3 ones
Subtract. 6 ones − 6 ones = 0 ones
So, the quotient is 383 (380 + 3) and the remainder is 0

Question 6.
4)\(\overline { 604 } \)
_____ R _____

Answer: 151

Explanation:
STEP 1 Use place value to place the first digit. Look at the hundreds in 604. 600 hundred can be shared among 4 groups
without regrouping.
Now there is 60 tens and 4 ones to share among 4 groups.
The first digit of the quotient will be in the hundreds place.
STEP 2 Divide the tens.
Divide. 600 ÷ 4
Multiply. 4 × 150 = 600
Subtract. 600 − 600 = 0 ones
STEP 3 Divide the ones.
Now there are 4 ones to share among 4 groups.
Divide. 4 ones ÷ 4
Multiply. 4×1 ones
Subtract. 4 ones − 4 ones = 0 ones
So, the quotient is 151 (150 + 1) and the remainder is 0

Question 7.
6)\(\overline { 796 } \)
_____ R _____

Answer: 132

Explanation:
STEP 1 Use place value to place the first digit. Look at the hundreds in 796. 790 hundred can be shared among 6 groups
without regrouping.
Now there is 79 tens and 6 ones to share among 6 groups.
The first digit of the quotient will be in the hundreds place.
STEP 2 Divide the tens.
Divide. 790 ÷ 6
Multiply. 6 × 131 = 786
Subtract. 790 − 786 = 4 ones
STEP 3 Divide the ones.
Now there are 4 + 6 = 10 ones to share among 6 groups.
Divide. 10 ones ÷ 6
Multiply. 6×1 ones
Subtract. 10 ones − 6 ones = 4 ones
So, the quotient is 132 (131 + 1) and the remainder is 4.

Question 8.
5)\(\overline { 449 } \)
_____ R _____

Answer: 89

Explanation:
STEP 1 Use place value to place the first digit. Look at the hundreds in 449. 440 hundred can be shared among 5 groups
without regrouping.
Now there is 44 tens and 9 ones to share among 5 groups.
The first digit of the quotient will be in the hundreds place.
STEP 2 Divide the tens.
Divide. 440 ÷ 5
Multiply. 5 × 88 = 440
Subtract. 440 − 440 = 0 ones
STEP 3 Divide the ones.
Now there are 9 ones to share among 5 groups.
Divide. 9 ones ÷ 5
Multiply. 5×1 ones
Subtract. 9 ones − 5 ones = 4 ones
So, the quotient is 89 (88 + 1) and the remainder is 4

Question 9.
6)\(\overline { 756 } \)
_____ R _____

Answer: 126

Explanation:
STEP 1 Use place value to place the first digit. Look at the hundreds in 756. 750 hundred can be shared among 6 groups
without regrouping.
Now there is 75 tens and 6 ones to share among 6 groups.
The first digit of the quotient will be in the hundreds place.
STEP 2 Divide the tens.
Divide. 750 ÷ 6
Multiply. 6 × 125 = 750
Subtract. 750 − 750 = 0 ones
STEP 3 Divide the ones.
Now there are 6 ones to share among 6 groups.
Divide. 6 ones ÷ 6
Multiply. 6×1 ones
Subtract. 6 ones − 6 ones = 0 ones
So, the quotient is 126 (125 + 1) and the remainder is 0

Question 10.
7)\(\overline { 521 } \)
_____ R _____

Answer: 74

Explanation:
STEP 1 Use place value to place the first digit. Look at the hundreds in 521. 520 hundred can be shared among 7 groups
without regrouping.
Now there is 52 tens and 1 one to share among 7 groups.
The first digit of the quotient will be in the hundreds place.
STEP 2 Divide the tens.
Divide. 520 ÷ 7
Multiply. 7 × 74 = 518
Subtract. 520 − 518 = 2 ones
STEP 3 Divide the ones.
Now there are 2 + 1 = 3 ones to share among 7 groups.
Divide. 3 ones ÷ 7 (not possible)
So, the quotient is 74 and the remainder is 3

Question 11.
5)\(\overline { 675 } \)
_____ R _____

Answer: 135

Explanation:
STEP 1 Use place value to place the first digit. Look at the hundreds in 675. 670 hundred can be shared among 5 groups
without regrouping.
Now there is 67 tens and 5 ones to share among 5 groups.
The first digit of the quotient will be in the hundreds place.
STEP 2 Divide the tens.
Divide. 670 ÷ 5
Multiply. 5 × 134 = 670
Subtract. 670 − 670 = 0 ones
STEP 3 Divide the ones.
Now there are 5 ones to share among 5 groups.
Divide. 5 ones ÷ 5
Multiply. 5×1 ones
Subtract. 5 ones − 5 ones = 0 ones
So, the quotient is 135 (134 + 1) and the remainder is 0.

Question 12.
8)\(\overline { 933 } \)
_____ R _____

Answer: 116

Explanation:
STEP 1 Use place value to place the first digit. Look at the hundreds in 933. 930 hundred can be shared among 8 groups
without regrouping.
Now there is 93 tens and 3 ones to share among 8 groups.
The first digit of the quotient will be in the hundreds place.
STEP 2 Divide the tens.
Divide. 930 ÷ 8
Multiply. 8 × 116 = 928
Subtract. 930 − 928 = 2 ones
STEP 3 Divide the ones.
Now there are 2 + 3 = 5 ones to share among 8 groups.
Divide. 5 ones ÷ 8 (not possible)
So, the quotient is 116 (100 + 3) and the remainder is 5.

Problem Solving

Question 13.
There are 132 projects in the science fair. If 8 projects can fit in a row, how many full rows of projects can be made? How many projects are in the row that is not full?
_____ full rows
_____ projects in the non-full row

Answer: 16 full rows and 4 projects in the non-full row

Explanation:
Total number of projects = 132
Number of projects placed in full row = 8
Number of rows having full projects =Quotient of 132 ÷ 8 = 16
Number of projects in the non-full row = Remainder of 132 ÷ 8 = 4

Question 14.
There are 798 calories in six 10-ounce bottles of apple juice. How many calories are there in one 10-ounce bottle of apple juice?
_____ R _____ calories in one 10-ounce bottles of juice

Answer: 133 calories

Explanation:
Number of calories in 6 bottles of apple juice = 798
Number of calories in each bottle = 798 ÷6 = 133 calories

Common Core – Divide by 1-Digit Numbers – Page No. 88

Lesson Check

Question 1.
To divide 572 ÷ 4, Stanley estimated to place the first digit of the quotient. In which place is the first digit of the quotient?
Options:
a. ones
b. tens
c. hundreds
d. thousands

Answer: hundreds

Explanation:
The quotient of 572÷ 4 is 143
STEP 1 Use place value to place the first digit. Look at the hundreds in 572. 560 hundred can be shared among 4 groups
without regrouping.
Now there is 1 ten to share among 4 groups.
The first digit of the quotient will be in the hundreds place.

Question 2.
Onetta biked 325 miles in 5 days. If she biked the same number of miles each day, how far did she bike each day?
Options:
a. 1,625 miles
b. 320 miles
c. 65 miles
d. 61 miles

Answer: 65 miles

Explanation:
Total number of miles biked = 325 miles
Number of days biked = 5
Number of miles biked on each day = Quotient of 325 ÷ 5 = 65

Spiral Review

Question 3.
Mort makes beaded necklaces that he sells for $32 each. About how much will Mort make if he sells 36 necklaces at the local art fair?
Options:
a. $120
b. $900
c. $1,200
d. $1,600

Answer: $1,200

Explanation:
Cost of each beaded necklace = $32
Number of necklaces = 36
The total cost of the necklaces = $32 x 36 = $1,200 (approx)

Question 4.
Which is the best estimate of 54 × 68?
Options:
a. 4,200
b. 3,500
c. 3,000
d. 350

Answer: 3,500

Explanation:

Taking the terms nearest to the 54 x 68 as 54 x 65 = 3510 = 3500 (approx)

Question 5.
Ms. Eisner pays $888 for 6 nights in a hotel. How much does Ms. Eisner pay per night?
Options:
a. $5,328
b. $882
c. $148
d. $114

Answer: $148

Explanation:
Total pays of Ms Eisner in a hotel = $888
Number of nights = 6
Amount Ms Eisner pay per night = $888 ÷ 6 = $148

Question 6.
Which division problem does the model show?
Go Math Grade 4 Answer Key Homework Practice FL Chapter 4 Divide by 1-Digit Numbers Common Core - Divide by 1-Digit Numbers img 8
Options:
a. 42 ÷ 3
b. 44 ÷3
c. 51 ÷ 3
d. 54 ÷ 3

Answer: 54 ÷ 3

Explanation:
Number of counters in each model = 18
Number of models = 3
Total number of counters = 18 x 3 = 54
Therefore the model displays = 54 ÷ 3

Common Core – Divide by 1-Digit Numbers – Page No. 89

Divide by 1-Digit Numbers

Divide and check.

Question 1.
318
\(\overline { 2)636 } \) 318
-6     × 2
———  ———
03 636
-2
———
16
-16
———
0

Question 2.
4)\(\overline { 631 } \)
_____ R _____

Answer:
STEP 1 Use place value to place the first digit. The first digit of the quotient will be in the hundreds place.
STEP 2 Divide the hundreds.
STEP 3 Divide the tens.
STEP 4 Divide the ones.

Question 3.
8)\(\overline { 906 } \)
_____ R _____

Answer:
STEP 1 Use place value to place the first digit. The first digit of the quotient will be in the hundreds place.
STEP 2 Divide the hundreds.
STEP 3 Divide the tens.
STEP 4 Divide the ones.

Question 4.
6)\(\overline { 6,739 } \)
_____ R _____

Answer:
STEP 1 Use place value to place the first digit. Look at the thousands in 6,739. 6 thousand can be shared among 6 groups without regrouping. The first digit of the quotient will be in the thousands place.
STEP 2 Divide the thousands.
STEP 3 Divide the hundreds.
STEP 4 Divide the tens.
STEP 5 Divide the ones.

Question 5.
4)\(\overline { 2,328 } \)
_____ R _____

Answer:
STEP 1 Use place value to place the first digit. Look at the thousands in 2,328. 2 thousand can be shared among 4 groups without regrouping. The first digit of the quotient will be in the thousands place.
STEP 2 Divide the thousands.
STEP 3 Divide the hundreds.
STEP 4 Divide the tens.
STEP 5 Divide the ones.

Question 6.
5)\(\overline { 7,549 } \)
_____ R _____

Answer:
STEP 1 Use place value to place the first digit. Look at the thousands in 7,549. 7 thousand can be shared among 5 groups without regrouping. The first digit of the quotient will be in the thousands place.
STEP 2 Divide the thousands.
STEP 3 Divide the hundreds.
STEP 4 Divide the tens.
STEP 5 Divide the ones.

Problem Solving

Use the table for 7 and 8.
Go Math Grade 4 Answer Key Homework Practice FL Chapter 4 Divide by 1-Digit Numbers Common Core - Divide by 1-Digit Numbers img 9

Question 7.
The Briggs rented a car for 5 weeks. What was the cost of their rental car per week?
$ _____

Answer: $197

Explanation:
Cost of the car of Briggs = $985
Number of weeks = 5
Cost of rent per week = $985 ÷ 5 =$ 197

Question 8.
The Lees rented a car for 4 weeks. The Santos rented a car for 2 weeks. Whose weekly rental cost was lower? Explain.
The rental cost of _________

Answer: Weekly rental cost was lower for Lees compared to Santos

Explanation:
Cost of the car of Lees = $632
Number of weeks = 4
Cost of rent per week = $632 ÷ 4 =$ 158

Cost of the car of Santos = $328
Number of weeks = 2
Cost of rent per week = $328 ÷ 2 =$ 164
Therefore weekly rental cost was lower for Lees compared to Santos.

Common Core – Divide by 1-Digit Numbers – Page No. 90

Lesson Check

Question 1.
Which expression can be used to check
the quotient 646 ÷ 3?
Options:
a. (251 × 3) + 1
b. (215 × 3) + 2
c. (215 × 3) + 1
d. 646 × 3

Answer: (215 × 3) + 1

Explanation:
Multiply 215 x 3 = 645
Then add 1 to 645
Then the dividend is 645 + 1 = 646
Thus the correct answer is option c.

Question 2.
There are 8 volunteers at the telethon. The goal for the evening is to raise $952. If each volunteer raises the same amount, what is the minimum amount each needs to raise to meet the goal?
Options:
a. $7,616
b. $944
c. $119
d. $106

Answer: $7,616

Explanation:
Number of volunteers = 8
Amount raised by each volunteer = $952
Total amount raised = $952 x 8 = $7,616

Thus the correct answer is option a.

Spiral Review

Question 3.
Which product is shown by the model?
Go Math Grade 4 Answer Key Homework Practice FL Chapter 4 Divide by 1-Digit Numbers Common Core - Divide by 1-Digit Numbers img 10
Options:
a. 5 × 15 = 75
b. 5 × 16 = 80
c. 5 × 17 = 75
d. 5 × 17 = 85

Answer: 5 × 17 = 85

Explanation:
By counting the number of counters we can give the expression.
Number of counters in one row = 17
Number of rows = 5
Therefore the expression = 5 × 17 = 85
Thus the correct answer is option d.

Question 4.
The computer lab at a high school ordered 26 packages of CDs. There were 50 CDs in each package. How many CDs did the computer lab order?
Options:
a. 1,330
b. 1,300
c. 1,030
d. 130

Answer: 1,300

Explanation:
Number of packages = 26
Number of CDs in each pack = 50
Total number of CDs the computer lab ordered = 26 x 50 = 1,300
Thus the correct answer is option b.

Question 5.
Which of the following division problems has a quotient with the first digit in the hundreds place?
Options:
a. 892 ÷ 9
b. 644 ÷ 8
c. 429 ÷ 5
d. 306 ÷ 2

Answer: 306 ÷ 2

Explanation:
Use place value to place the first digit. Look at the hundreds in 306. 300 hundred can be shared among 2 groups
without regrouping.
Now there is 30 tens and 6 ones to share among 2 groups.
The first digit of the quotient will be in the hundreds place.
Thus the correct answer is option d.

Question 6.
Sharon has 64 ounces of juice. She is going to use the juice to fill as many 6-ounce glasses as possible. She will drink the leftover juice. How much juice will Sharon drink?
Options:
a. 4 ounces
b. 6 ounces
c. 10 ounces
d. 12 ounces

Answer: 4 ounces

Explanation:
The total quantity of juice = 64 ounces
Quantity of juice she filled = 6 ounces
Quantity of juice she drank = Remainder of 64 ÷ 6 = 4

Thus the correct answer is option a.

Common Core – Divide by 1-Digit Numbers – Page No. 91

Problem Solving Multistep Division Problems

Solve. Draw a diagram to help you.

Question 1.
There are 3 trays of eggs. Each tray holds 30 eggs. How many people can be served if each person eats 2 eggs?
Go Math Grade 4 Answer Key Homework Practice FL Chapter 4 Divide by 1-Digit Numbers Common Core - Divide by 1-Digit Numbers img 11
Think: What do I need to find? How can I draw a diagram to help?
45 people can be served

Question 2.
There are 8 pencils in a package. How many packages will be needed for 28 children if each child gets 4 pencils?
______ packages

Answer: 14 packages

Explanation:
Number of pencils in each package = 8

Number of children = 28

Number of pencils each child needs = 4
Total number of pencils = 28 x 4 =112
Number of packages = 112 ÷ 8 = 14

Question 3.
There are 3 boxes of tangerines. Each box has 93 tangerines. The tangerines will be divided equally among 9 classrooms. How many tangerines will each classroom get?
______ tangerines

Answer: 31

Explanation:
Number of boxes = 3
Number of tangerines in each box = 93
Total number of tangerines = 93 x 3 = 279

Number of classrooms = 9
Number of tangerines in each classroom = 279 ÷ 9 = 31

Question 4.
Misty has 84 photos from her vacation and 48 photos from a class outing. She wants to put all the photos in an album with 4 photos on each page. How many pages does she need?
______ pages

Answer: 33 pages

Explanation:
Number of photos from her vacation = 84

Number of photos from her class outing = 48

Total number of photos = 84 + 48 = 132
Number of photos in each page = 4
Number of pages required = 132 ÷ 4 = 33

Common Core – Divide by 1-Digit Numbers – Page No. 93

Lessons 4.1, 4.5

Estimate the quotient.

Question 1.
67 ÷ 4
about ______

Answer: About 17

Explanation:
The number close to 67 is 70.
Divide 70 by 4 is 17.5
Thus the estimated quotient of 67 ÷ 4 is 17.

Question 2.
72 ÷ 5
about ______

Answer: About 14

Explanation:
The number close to 72 is 70.
Divide 70 by 5 is 14.
Thus the estimated quotient of 72 ÷ 5 is 14.

Question 3.

213 ÷ 3
about ______

Answer: About 70

Explanation:
The number close to 213 is 210.
Divide 210 by 3 is 70.
Thus the estimated quotient of 213 ÷ 3 is 70.

Question 4.
484 ÷ 6
about ______

Answer: About 80

Explanation:
The number close to 484 is 480.
Divide 480 by 6 is 80.
Thus the estimated quotient of 484 ÷ 6 is 80.

Question 5.
446 ÷ 7
about ______

Answer: About 60

Explanation:
The number close to 446 is 440.
Divide 440 by 7 is 60.
Thus the estimated quotient of 446 ÷ 7 is 60.

Question 6.
1,246 ÷ 4
about ______

Answer: About 300

Explanation:
The number close to 1246 is 1200.
Divide 1200 by 4 is 300.
Thus the estimated quotient of 1,246 ÷ 4 is 300.

Question 7.
708 ÷ 9
about ______

Answer: About 80

Explanation:
The number close to 708 is 700.
Divide 700 by 9 is 80 (approx).
Thus the estimated quotient of 708 ÷ 9 is 80.

Question 8.
2,657 ÷ 3
about ______

Answer: About 900

Explanation:
The number close to 2,657 is 2700.
Divide 2700 by 3 is 900.
Thus the estimated quotient of 2,657 ÷ 3 is 900.

Lesson 4.2

Use counters or quick pictures to find the quotient and remainder.

Question 9.
44 ÷ 5
______ R ______

Answer: 8R4

Explanation:
Quotient:
A. Use 44 counters to represent the 44 dominoes. Then draw 5 circles to represent the divisor.
B. Share the counters equally among the 5 groups by placing them in the circles.
C. Number of groups of counters formed = quotient of 44 ÷ 5
D. Number of circles equally filled is8, therefore, the quotient is 8.
Remainder:
The number of counters left over is the remainder. The number of counters leftover= 4
For 44 ÷ 5, the quotient is 8 and the remainder is 4, or 8R4.

Question 10.
8)\(\overline { 21 } \)
______ R ______

Answer: 2R5

Explanation:
Quotient:
A. Use 21 counters to represent the 21 dominoes. Then draw 8 circles to represent the divisor.
B. Share the counters equally among the 8 groups by placing them in the circles.
C. Number of groups of counters formed = quotient of 21 ÷ 8
D. Number of circles equally filled is 2, therefore, the quotient is 2.
Remainder:
The number of counters left over is the remainder. The number of counters leftover= 5
For 21 ÷ 8, the quotient is 2 and the remainder is 5, or 2R5.

Question 11.
4)\(\overline { 75 } \)
______ R ______

Answer: 18R3

Explanation:
Quotient:
A. Use 75 counters to represent the 75 dominoes. Then draw 4 circles to represent the divisor.
B. Share the counters equally among the 4 groups by placing them in the circles.
C. Number of groups of counters formed = quotient of 75 ÷ 4
D. Number of circles equally filled is 18, therefore, the quotient is 18.
Remainder:
The number of counters left over is the remainder. The number of counters leftover= 3
For 21 ÷ 8, the quotient is 18 and the remainder is 3, or 18R3.

Question 12.
76 ÷ 6
______ R ______

Answer: 12R4

Explanation:
Quotient:
A. Use 76 counters to represent the 76 dominoes. Then draw 6 circles to represent the divisor.
B. Share the counters equally among the 6 groups by placing them in the circles.
C. Number of groups of counters formed = quotient of 76 ÷ 6
D. Number of circles equally filled is 12, therefore, the quotient is 12.
Remainder:
The number of counters left over is the remainder. The number of counters leftover= 4
For 76 ÷ 6, the quotient is 12 and the remainder is 4, or 12R4.

Lesson 4.3

Interpret the remainder to solve.

Question 13.
Kelly divides 29 markers equally among 7 friends. If Kelly keeps the leftover markers, how many markers will she keep?
______ marker(s)

Answer: 1 marker

Explanation:
Given,
Kelly divides 29 markers equally among 7 friends.
1 because 4 markers for each friend (4 × 7) would be 28 and the last one would be leftover because it’s not enough for everyone.

Question 14.
Dave has a board that is 29 inches long. He cuts the board into 4 equal pieces. How long will each piece be?
______ inches

Answer: 7 inches

Explanation:
Dave has a board that is 29 inches long and want to cut it into 4 pieces.
You are asked the length of each piece.
To solve the question, you need to divide the total length of the board by the number of pieces Dave wants to make.
Then, the length of each piece would be: 29 inches/4= 7.25 inches

Question 15.
Eight students can ride in each van. How many vans are needed for 29 students?
______ vans

Answer: 4 vans

Explanation:
Given,
Eight students can ride in each van.
29/8 = 3.625 = 4(approx)
Therefore 4 vans are needed for 29 students.

Question 16.
Mac has 40 ounces of juice. He pours 6 ounces in each glass. How many glasses can he fill?
______ glasses

Answer: 6 glasses

Explanation:
Given,
Mac has 40 ounces of juice. He pours 6 ounces in each glass.
Divide 40 by 6
40/6 = 6.66 ≈ 6
Thus Mac can fill 6 glasses.

Lesson 4.4

Use basic facts and place value to find the quotient.

Question 17.
120 ÷ 4 = ______

Answer: 30

Explanation:
STEP 1 Identify the basic fact. 120 ÷ 4
STEP 2 Use place value. 120 = 12 tens
STEP 3 Divide. 12 tens ÷ 4 = 3 tens
120 ÷ 4 = 30

Question 18.
280 ÷ 7 = ______

Answer: 40

Explanation:
STEP 1 Identify the basic fact. 280 ÷ 7
STEP 2 Use place value. 280 = 28 tens
STEP 3 Divide. 28 tens ÷ 7 = 4 tens
280 ÷ 7 = 40

Question 19.
3,000 ÷ 5 = ______

Answer: 600

Explanation:
STEP 1 Identify the basic fact. 3000 ÷ 5
STEP 2 Use place value. 3000 = 300 tens
STEP 3 Divide. 300 tens ÷ 5 = 60 tens
3,000 ÷ 5 = 60 tens

Question 20.
4,800 ÷ 6 = ______

Answer: 800

Explanation:
STEP 1 Identify the basic fact. 4,800 ÷ 6
STEP 2 Use place value. 4800 = 480 tens
STEP 3 Divide. 480 tens ÷ 6 = 80 tens
4,800 ÷ 6 = 800

Question 21.
5,600 ÷ 8 = ______

Answer: 700

Explanation:
STEP 1 Identify the basic fact. 5,600 ÷ 8
STEP 2 Use place value. 5600 = 560 tens
STEP 3 Divide. 560 tens ÷ 8 = 70 tens
5,600 ÷ 8 = 700

Question 22.
6,300 ÷ 9 = ______

Answer: 700

Explanation:
STEP 1 Identify the basic fact. 6,300 ÷ 9
STEP 2 Use place value. 6300 = 630 tens
STEP 3 Divide. 630 tens ÷ 9 = 70 tens
6,300 ÷ 9 = 700

Common Core – Divide by 1-Digit Numbers – Page No. 94

Lessons 4.6–4.7

Choose a method and divide.

Question 1.
68 ÷ 4 = ______

Answer: 17

Explanation:
The number close to 68 is 70.
Divide 70 by 4 is 17 (approx).
Thus the estimated quotient of 68 ÷ 4 is 17.

Question 2.
48 ÷ 3 = ______

Answer: 16

Explanation:
The number close to 48 is 50.
Divide 50 by 3 is 16  (approx).
Thus the estimated quotient of 48 ÷ 3 is 16.

Question 3.
108 ÷ 9 = ______

Answer: 12

Explanation:
The number close to 108 is 100.
Divide 100 by 9 is 12 (approx).
Thus the estimated quotient of 108 ÷ 9 is 12.

Question 4.
74 ÷ 2 = ______

Answer: 37

Explanation:
The number close to 74 is 70.
Divide 70 by 2 is 37 (approx).
Thus the estimated quotient of 74 ÷ 2 is 37.

Question 5.
122 ÷ 5 = ______ R ______

Answer: 24R2

Explanation:
Quotient:
A. Use 122 counters to represent the 122 dominoes. Then draw 5 circles to represent the divisor.
B. Share the counters equally among the 5 groups by placing them in the circles.
C. Number of groups of counters formed = quotient of 122 ÷ 5
D. Number of circles equally filled are 24, therefore, the quotient is 24.
Remainder:
The number of counters left over is the remainder. The number of counters leftover= 2
For 122 ÷ 5, the quotient is 24 and the remainder is 2, or 24R2.

Question 6.
165 ÷ 6 = ______ R ______

Answer: 27R3

Explanation:
Quotient:
A. Use 165 counters to represent the 165 dominoes. Then draw 6 circles to represent the divisor.
B. Share the counters equally among the 6 groups by placing them in the circles.
C. Number of groups of counters formed = quotient of 165 ÷ 6.
D. Number of circles equally filled are 27, therefore, the quotient is 27.
Remainder:
The number of counters left over is the remainder. The number of counters leftover= 3
For 165 ÷ 6, the quotient is 27 and the remainder is 3, or 27R3.

Lessons 4.8–4.9

Divide.

Question 7.
4)\(\overline { 848 } \)
______

Answer: 212

Go Math Grade 4 Chapter 4 Answer Key

Question 8.
7)\(\overline { 287 } \)
______

Answer: 41

Go Math Grade 4 Answer Key Homework Practice FL Chapter 4 Divide by 1-Digit Numbers img-2

Question 9.
5)\(\overline { 405 } \)
______

Answer: 81
Go Math Grade 4 Answer Key Homework Practice FL Chapter 4 Divide by 1-Digit Numbers img-3

Question 10.
3)\(\overline { 696 } \)
______

Answer: 232
Go Math Grade 4 Answer Key Homework Practice FL Chapter 4 Divide by 1-Digit Numbers img-4

Question 11.
96 ÷ 6 = ______

Answer: 16
Go Math Grade 4 Answer Key Homework Practice FL Chapter 4 Divide by 1-Digit Numbers img-5

Question 12.
76 ÷ 5 = ______ R ______

Answer: 15R1
Go Math Grade 4 Answer Key Homework Practice FL Chapter 4 Divide by 1-Digit Numbers img-6

Question 13.
58 ÷ 4 = ______ R ______

Answer: 14R2
Go Math Grade 4 Answer Key Homework Practice FL Chapter 4 Divide by 1-Digit Numbers img-7

Question 14.
85 ÷ 2 = ______ R ______

Answer: 42R1
Go Math Grade 4 Answer Key Homework Practice FL Chapter 4 Divide by 1-Digit Numbers img-8

Lessons 4.10–4.11

Divide and check.

Question 15.
4)\(\overline { 896 } \)
______

Answer: 224

Explanation:
224
× 4
896

Question 16.
5)\(\overline { 833 } \)
______ R ______

Answer: 166r3

Explanation:
166
× 5
830
+ 3
833

Question 17.
6)\(\overline { 527 } \)
______ R ______

Answer: 87r5

Explanation:
87
×6
522
+ 5
527

Question 18.
3)\(\overline { 935 } \)
______ R ______

Answer: 311r2

Explanation:
311
× 3
933
+ 2
935

Question 19.
3)\(\overline { 1,976 } \)
______ R ______

Answer: 658R2

Explanation:
658
× 3
1974
+    2
1976

Question 20.
6)\(\overline { 1,042 } \)
______ R ______

Answer: 173r4

Explanation:
173
×   6
1038
+   4
1042

Lesson 4.12

Solve. Draw a diagram to help you.

Question 21.
Ellis has 2 dozen white baseballs and 4 dozen yellow baseballs. He needs to divide them into cartons that hold 6 each. How many cartons can he fill?
______ cartons

Answer: 6 cartons

Explanation:
Given,
Ellis has 2 dozen white baseballs and 4 dozen yellow baseballs.
He needs to divide them into cartons that hold 6 each.
6 2 Dozens and 4 Dozens are 12+24 = 36/6 = 6
Therefore he can fill 6 cartons.

Question 22.
A family of 2 adults and 3 children went out to dinner. The total bill was $42. Each child’s dinner cost $4. How much did each adult’s dinner cost?
$ ______

Answer: $15

Explanation:
Each child’s dinner – $4
3 child’s dinner – $4 x 3 = $12
$42 – 12 = $30
$30 divided by 2 = $15
Thus each adult’s dinner cost is $15.

Conclusion:

Find more questions for practice from here, Go Math Grade 4 Answer Key Chapter 4 Divide by 1-Digit Numbers and develop your mathematical skills. Drop your queries and feedback by posting the comment below and we’ll update if anything requires as well as we’ll answer your doubts Asap.

Go Math Grade 3 Answer Key Chapter 6 Understand Division Extra Practice

go-math-grade-3-chapter-6-understand-division-extra-practice-answer-key

If you looking to practice Go Math 3rd Grade Textbook Questions then take the help of the Go Math Grade 3 Answer Key Chapter 6 Understand Division Extra Practice. You need to have strong fundamentals in Maths in order to become a pro in the Subject. You can easily understand the basics of the division with the Go Math Grade 3 Answer Key Chapter 6 Understand Division Extra Practice. Solve as Many Questions as possible from the Extra Practice and Clear the Final Exams with better grades.

3rd Grade Go Math Answer Key Ch 6 Understand Division Extra Practice

3rd Grade Go Math Answer Key Ch 6 Understand Division Extra Practice

Repeated subtraction, Equal groups, Number line, related multiplication, and division facts are all the topics covered in the 3rd Grade Go Math Answer Key Ch 6. Before you begin your preparation firstly know the syllabus i.e. concepts in Chapter 6 Understand Division and prepare accordingly. Check out the Step by Step Solutions provided for 3rd Grade Go Math Answer Key Chapter 6 Understand Division Extra Practice and learn the concepts efficiently.

Common Core – Page No. 123000

Lessons 6.1–6.3 Make equal groups.

Complete the table.

Counters Number of Equal Groups Number in Each Group
1. 18 9 ________
2. 24 ________ 8
3. 12 6 ________
4. 35 7 ________
5. 32 ________ 4
6. 25 ________ 5

Answer:

Counters Number of Equal Groups Number in Each Group
1. 18 9 2
2. 24 3 8
3. 12 6 2
4. 35 7 5
5. 32 8 4
6. 25 5 5

Explanation:

1. Number of counters = 18
Number of equal groups = 9
Number in each group = x
x × 9 = 18
x= 18/9 = 2
Therefore number in each group = 2

2. Number of counters = 24
Number in each group = 8
Number of equal groups = x
x × 8 = 24
x = 24/8 = 3
Thus the number of equal groups = 3

3. Number of counters = 12
Number of equal groups = 6
Number in each group = x
x × 6 = 12
x = 12/6 = 2
So, the number in each group = 2

4. Number of counters = 35
Number of equal groups = 7
Number in each group = x
x × 7 = 35
x = 35/7 = 5
x = 5
Therefore number in each group = 5

5. Number of counters = 32
Number of equal groups = x
Number in each group = 8
x × 8 = 32
x = 32/8 = 4
Thus the number of equal groups = 4

6. Number of counters = 25
Number of equal groups = x
Number in each group = 5
x × 5 = 25
x = 25/5 = 5
So, the number of equal groups = 5

Lesson 6.4

Write a division equation for the picture.

Question 1.
Go Math Grade 3 Answer Key Chapter 6 Understand Division Extra Practice Common Core img 1
Type below:
__________

Answer: 27 ÷ 3 = 9 or 27 ÷ 9 = 3

Explanation:

Total number of counters = 27
Number of equal groups = 3
Number in each group = 9
The division equation is
Number of counters by number of groups = 27 ÷ 3 = 9
or
Number of counters by number in each group = 27 ÷ 9 = 3

Question 2.
Go Math Grade 3 Answer Key Chapter 6 Understand Division Extra Practice Common Core img 2
Type below:
__________

Answer: 24 ÷ 4 = 6 or 24 ÷ 6 = 4

Explanation:

Total number of counters = 24
Number of equal groups = 4
Number in each group = 6
The division equation is
Number of counters by number of groups = 24 ÷ 4 = 6
or
Number of counters by number in each group = 24 ÷ 6 = 4

Lesson 6.5

Write a division equation.

Question 3.
Go Math Grade 3 Answer Key Chapter 6 Understand Division Extra Practice Common Core img 3
Type below:
__________

Answer: 3 groups, 15 ÷ 5 = 3

Explanation:

Step 1:

Starts at 15

Step 2:

Count back by 5s as many times as you can.

Step 3:

Count the number of times you jumped back 5.
You jumped back by 15 three times
There are 3 jumps of 5 in 15.

Question 4.
Go Math Grade 3 Answer Key Chapter 6 Understand Division Extra Practice Common Core img 4
Type below:
__________

Answer: 24 ÷ 6 = 4

Explanation:

Step 1:

Begins at 24

Step 2:

Subtract with 6 until you get 0.

Step 3:

Count the number of times you subtract with 6.

You subtract 4 times
There are 4 groups of 6 with 24
So, 24 ÷ 6 = 4

Common Core – Page No. 124000

Lesson 6.6

Make an array. Then write a division equation.

Question 1.
12 tiles in 4 rows
______ ÷ ______ = ______

Answer: 12 ÷ 4 = 3

Explanation:

■ ■ ■
■ ■ ■
■ ■ ■
■ ■ ■
Total number of tiles = 12
Number of rows = 4
Number of tiles in each row = x
Divide the number of tiles by number of rows = 12 ÷ 4 = 3

Question 2.
18 tiles in 3 rows
______ ÷ ______ = ______

Answer: 18 ÷ 3 = 6

Explanation:

■ ■ ■ ■ ■ ■
■ ■ ■ ■ ■ ■
■ ■ ■ ■ ■ ■
Total number of tiles = 18
Number of rows = 3
Number of tiles in each row = y
Divide the number of tiles by no. of rows = 18 ÷ 3 = 6

Question 3.
35 tiles in 5 rows
______ ÷ ______ = ______

Answer: 35 ÷ 5 = 7

Explanation:

■ ■ ■ ■ ■ ■ ■
■ ■ ■ ■ ■ ■ ■
■ ■ ■ ■ ■ ■ ■
■ ■ ■ ■ ■ ■ ■
■ ■ ■ ■ ■ ■ ■

Total number of tiles = 35
Number of rows = 5
Number of tiles in each row = p
Divide the number of tiles by number of rows = 35 ÷ 5 = 7

Question 4.
28 tiles in 7 rows
______ ÷ ______ = ______

Answer: 28 ÷ 7 = 4

Explanation:

■ ■ ■ ■
■ ■ ■ ■
■ ■ ■ ■
■ ■ ■ ■
■ ■ ■ ■
■ ■ ■ ■
■ ■ ■ ■

Total number of tiles = 28
Number of rows = 7
Number of tiles in each row = x
Divide the number of tiles by number of rows = 28 ÷ 7 = 4

Lesson 6.7

Complete the equations.

Question 5.
8 × ______ = 40 40 ÷ 8 = ______

Answer: 5, 5

Explanation:

Let x be the unknown factor
8 × x = 40
x = 40/8
x = 5
Check whether the related multiplication and division facts are the same or not.
40 ÷ 8 = 5
The related facts of 40 and 8 are 5.

Question 6.
6 × ______ = 36 36 ÷ 6 = ______

Answer: 6, 6

Explanation:

Let y be the unknown factor
6 × y = 36
y = 36/6 = 6
Check if the related multiplication and division facts are the same or not.
36 ÷ 6 = 6
The related facts of 36 and 6 are 6.

Question 7.
3 × ______ = 21 21 ÷ 3 = ______

Answer: 7, 7

Explanation:

Let x be the unknown factor
3 × x = 21
x = 21
Check whether the related facts are the same or not.
21 ÷ 3 = 7
The quotient is 7.

Question 8.
2 × ______ = 18 18 ÷ 2 = ______

Answer: 9, 9

Explanation:

Let b be the unknown factor
2 × b = 18
b = 18/2 = 9
Check the related multiplication and division facts
18 ÷ 2 = 9
The related facts of 18 and 2 are 9.

Lesson 6.8 (pp. 239–243)

Write the related facts for the array.

Question 9.
■ ■ ■ ■ ■
■ ■ ■ ■ ■
■ ■ ■ ■ ■
______ × ______ = ______
______ × ______ = ______
______ ÷ ______ = ______
______ ÷ ______ = ______

Answer:

3 × 5 = 15
5 × 3 = 15
15 ÷ 3 = 5
15 ÷ 5 = 3

Explanation:

Total number of tiles = 15
Number of equal rows = 3
Number of rows in each group = 5
So, the related 5, 3 and 15 is 5× 3 = 15, 3×5 = 15, 15 ÷ 3 = 5 and 15÷ 5 = 3

Question 10.
■ ■ ■ ■ ■ ■
■ ■ ■ ■ ■ ■
■ ■ ■ ■ ■ ■
______ × ______ = ______
______ × ______ = ______
______ ÷ ______ = ______
______ ÷ ______ = ______

Answer:

3 × 6 = 18
6 × 3 = 18
18 ÷ 3 = 6
18 ÷ 6 = 3

Explanation:

Total number of tiles = 18
Number of equal rows = 3
Number of rows in each group = 6
So, the related 18, 3 and 6 is 3 × 6 = 18, 6 × 3 = 18, 18 ÷ 3 = 6 and 18 ÷ 6 = 3

Question 11.
■ ■ ■ ■ ■
■ ■ ■ ■ ■
______ × ______ = ______
______ × ______ = ______
______ ÷ ______ = ______
______ ÷ ______ = ______

Answer:

2 × 5 = 10
5 × 2 = 10
10 ÷ 2 = 5
10 ÷ 5 = 2

Explanation:

Total number of tiles = 10
Number of equal rows = 2
Number of rows in each group = 5
So, the related 2, 5 and 10 is 2 × 5 = 10, 5 × 2 = 10, 10 ÷ 2 = 5 and 10 ÷ 5 = 2

Lesson 6.9

Find the quotient.

Question 12.
7 ÷ 1 = ______

Answer: 7

Explanation:

Any number divided by 1 will be the same number. Thus the quotient is 7.

Question 13.
4 ÷ 4 = ______

Answer: 1

Explanation:

The number divided by the same number will be always 1. Thus the quotient is 1.

Question 14.
9 ÷ 1 = ______

Answer: 9

Explanation:

Any number divided by 1 will be always the same number. So, the quotient is 9.

Question 15.
0 ÷ 1 = ______

Answer: 0

Explanation:

0 divided by any number is always 0. So, the quotient is 0.

Question 16.
Anton has 8 flower pots. He plants 1 seed in each pot. How many seeds does Anton use?
______ seeds

Answer: 8 seeds

Explanation:

Anton has 8 flower pots.
He plants 1 seed in each pot.
Number of seeds Anton used = x
x × 1 = 8
x = 8/1
x = 8
Therefore there are 8 seeds in 8 flower pots.

All the Questions in Go Math Grade 3 Answer Key Chapter 6 Understand Division Extra Practice helps the students to be prepared for their exams. For any assistance needed you can always look upto  Go Math Grade 3 Answer Key Chapter 6 Understand Division.  You can get All Lessons Solutions in Chapter 6 Understand Division here.

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations

go-math-grade-6-chapter-8-solutions-of-equations-answer-key

Are you searching for the Go Math Grade 6 Solution Key for Chapter 8 Solutions of Equations? If my guess is correct then you are on the right page. We provide the solutions to all the questions in pdf format. So, Download Go Math 6th Grade Answer Key Chapter 6 Chapter 8 Solutions of Equations pdf for free. Our Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations is helpful for quick and easy learning.

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations

Enhance your performance in practice tests or assignments with the help of HMH Go Math 6th Grade Answer Key Chapter 8 Solutions of Equations. Get the solutions of Review Test and Mid Chapter Checkpoint in Go Math 6th Grade Chapter 8 Solutions of Equations. Scroll down this page to know the topics covered in this chapter. Make use of the links and Download Grade 6 Go Math Answer Key Chapter 8 Solutions of Equations.

Lesson 1: Solutions of Equations

Lesson 2: Write Equations

Lesson 3: Investigate • Model and Solve Addition Equations

Lesson 4: Solve Addition and Subtraction Equations

Lesson 5: Investigate • Model and Solve Multiplication Equations

Lesson 6: Solve Multiplication and Division Equations

Lesson 7: Problem Solving • Equations with Fractions

Mid-Chapter Checkpoint

Lesson 8: Solutions of Inequalities

Lesson 9: Write Inequalities

Lesson 10: Graph Inequalities

Lesson 10: Graph Inequalities

Chapter 8 Review/Test

Share and Show – Page No. 423

Determine whether the given value of the variable is a solution of the equation.

Question 1.
x + 12 = 29; x = 7
The variable is __________

Answer: not a solution

Explanation:
Substitute the value in the given equation
x + 12 = 29
If x = 7
7 + 12 = 29
19 ≠ 29
Thus the variable is not a solution.

Question 2.
n − 13 = 2; n = 15
The variable is __________

Answer: a solution

Explanation:
Substitute the value in the given equation
n = 15
n − 13 = 2
15 – 13 = 2
The variable is a solution.

Question 3.
\(\frac{1}{2}\)c = 14; c = 28
The variable is __________

Answer: a solution

Explanation:
Substitute the value in the given equation
c = 28
\(\frac{1}{2}\)c = 14
\(\frac{1}{2}\) × 28 = 14
14 = 14
Thus the variable is a solution.

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 423 Q4

Question 5.
d − 8.7 = 6; d = 14.7
The variable is __________

Answer: a solution

Explanation:
Substitute the value in the given equation
d = 14.7
d − 8.7 = 6
14.7 – 8.7 = 6
6 = 6
Thus the variable is a solution.

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 423 Q6

On Your Own

Determine whether the given value of the variable is a solution to the equation.

Question 7.
17.9 + v = 35.8; v = 17.9
The variable is __________

Answer: a solution

Explanation:
Substitute the value in the given equation
17.9 + v = 35.8
v = 17.9
17.9 + 17.9 = 35.8
35.8 = 35.8
Thus the variable is a solution.

Question 8.
c + 35 = 57; c = 32
The variable is __________

Answer: not a solution

Explanation:
Substitute the value in the given equation
c + 35 = 57
c = 32
32 + 35 = 57
67 ≠ 57
Thus the variable is not a solution.

Question 9.
18 = \(\frac{2}{3}\)h; h= 12
The variable is __________

Answer: not a solution

Explanation:
Substitute the value in the given equation
18 = \(\frac{2}{3}\)h
h = 12
\(\frac{2}{3}\) × 12 = 8
18 ≠ 8
Thus the variable is not a solution.

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 423 Q10
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 423 Q10.1

Question 11.
Antonio ran a total of 9 miles in two days. On the first day he ran 5 \(\frac{1}{4}\) miles. The equation 9 – d = 5 \(\frac{1}{4}\) can be used to find the distance d in miles Antonio ran the second day. Determine whether d = 4 \(\frac{3}{4}\), d = 4, or d = 3 \(\frac{3}{4}\) is a solution of the equation, and tell what the solution means.
The solution is ________ \(\frac{□}{□}\)

Answer: 3 \(\frac{3}{4}\)

Explanation:
9 – d = 5 \(\frac{1}{4}\)
Substitute d = 4 \(\frac{3}{4}\) in the above equation
9 – 4 \(\frac{3}{4}\) = 5 \(\frac{1}{4}\)
4 \(\frac{1}{4}\) ≠ 5 \(\frac{1}{4}\)
Not a solution
Substitute d = 4
9 – 4 = 5 \(\frac{1}{4}\)
5 ≠ 5 \(\frac{1}{4}\)
Not a solution
Substitute d = 3 \(\frac{3}{4}\)
9 – 3 \(\frac{3}{4}\) = 5 \(\frac{1}{4}\)
5 \(\frac{1}{4}\) = 5 \(\frac{1}{4}\)
9 – d = 5 \(\frac{1}{4}\); d = 3 \(\frac{3}{4}\) is a solution.

Problem Solving + Applications – Page No. 424

Use the table for 12–14.
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 1

Question 12.
Connect Symbols and Words The length of a day on Saturn is 14 hours less than a day on Mars. The equation 24.7 − s = 14 can be used to find the length in hours s of a day on Saturn. Determine whether s = 9.3 or s = 10.7 is a solution of the equation, and tell what the solution means.
Type below:
_____________

Answer: s = 10.7

Explanation:
The length of a day on Saturn is 14 hours less than a day on Mars.
The equation 24.7 − s = 14 can be used to find the length in hours s of a day on Saturn.
24.7 − s = 14
Substitute s = 9.3 in the equation
24.7 – 9.3 = 14
15.4 ≠ 14
Not a solution
Substitute s = 10.7 in the equation
24.7 – 10.7 = 14
14 = 14
Therefore s = 10.7 is a solution to the equation.

Question 13.
A storm on one of the planets listed in the table lasted for 60 hours, or 2.5 of the planet’s days. The equation 2.5h = 60 can be used to find the length in hours h of a day on the planet. Is the planet Earth, Mars, or Jupiter? Explain.
Type below:
_____________

Answer: Earth

Explanation:
A storm on one of the planets listed in the table lasted for 60 hours, or 2.5 of the planet’s days.
2.5h = 60
h = 60/2.5
h = 24 hours
By seeing the above table we can say that Earth is the answer.

Question 14.
A day on Pluto is 143.4 hours longer than a day on one of the planets listed in the table. The equation 153.3 − p = 143.4 can be used to find the length in hours p of a day on the planet. What is the length of a storm that lasts \(\frac{1}{3}\) of a day on this planet?
________ hours

Answer: 3.3 hours

Explanation:
A day on Pluto is 143.4 hours longer than a day on one of the planets listed in the table.
153.3 − p = 143.4
153.3 – 143.4 = p
p = 153.3 – 143.4
p = 9.9
Now p with \(\frac{1}{3}\) to find the length of a storm that lasts a day on this planet
9.9 × \(\frac{1}{3}\) = 3.3 hours

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 424 Q15

Question 16.
The marking period is 45 school days long. Today is the twenty-first day of the marking period. The equation x + 21 = 45 can be used to find the number of days x left in the marking period. Using substitution, Rachel determines there are _____ days left in the marking period.
Rachel determines there are _____________ days left.

Answer: 24

Explanation:
The marking period is 45 school days long. Today is the twenty-first day of the marking period.
The equation x + 21 = 45
x = 45 – 21 = 24 days
Using substitution, Rachel determines there are 24 days left in the marking period.
Thus Rachel determines there are 24 days left.

Solutions of Equations – Page No. 425

Determine whether the given value of the variable is a solution of the equation.

Question 1.
x − 7 = 15; x = 8
The variable is __________

Answer: not a solution

Explanation:
Substitute the value in the given equation.
x = 8
8 – 7 = 15
1 ≠ 15
Therefore the variable is not a solution.

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 425 Q2

Question 3.
\(\frac{1}{3}\)h = 6; h = 2
The variable is __________

Answer: not a solution

Explanation:
Substitute the value in the given equation.
\(\frac{1}{3}\)h = 6
h = 2
\(\frac{1}{3}\) × 2 = 6
\(\frac{2}{3}\) ≠ 6
Therefore the variable is not a solution.

Question 4.
16.1 + d = 22; d = 6.1
The variable is __________

Answer: not a solution

Explanation:
Substitute the value in the given equation.
16.1 + d = 22
d = 6.1
16.1 + 6.1 = 22
22.2 ≠ 22
Therefore the variable is not a solution.

Question 5.
9 = \(\frac{3}{4}\)e; e = 12
The variable is __________

Answer: a solution

Explanation:
Substitute the value in the given equation.
9 = \(\frac{3}{4}\)e
e = 12
9 = \(\frac{3}{4}\)(12)
9 = 3 × 3
9 = 9
Therefore the variable is a solution.

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 425 Q6

Problem Solving

Question 7.
Terrance needs to score 25 points to win a game. He has already scored 18 points. The equation 18 + p = 25 can be used to find the number of points p that Terrance still needs to score. Determine whether p = 7 or p = 13 is a solution of the equation, and tell what the solution means.
Type below:
_____________

Answer: p = 7

Explanation:
Terrance needs to score 25 points to win a game. He has already scored 18 points.
The equation is 18 + p = 25
Substitute p = 7 in the above equation.
18 + 7 = 25
25 = 25
The variable is a solution.
Substitute p = 13
18 + p = 25
18 + 13 = 25
31 ≠ 25
The variable is not a solution.
Therefore p = 7 is a solution for the equation.

Question 8.
Madeline has used 50 sheets of a roll of paper towels, which is \(\frac{5}{8}\) of the entire roll. The equation \(\frac{5}{8}\)s = 50 can be used to find the number of sheets s in a full roll. Determine whether s = 32 or s = 80 is a solution of the equation, and tell what the solution means.
Type below:
_____________

Answer:
Madeline has used 50 sheets of a roll of paper towels, which is \(\frac{5}{8}\) of the entire roll.
\(\frac{5}{8}\)s = 50
s = 50 × \(\frac{8}{5}\)
s = 80 because 80 × 5 = 400
400 ÷ 8 = 50

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 425 Q9

Lesson Check – Page No. 426

Question 1.
Sheena received a gift card for $50. She has already used it to buy a lamp for $39.99. The equation 39.99 + x = 50 can be used to find the amount x that is left on the gift card. What is the solution to the equation?
_____

Answer: 10.01

Explanation:
Given:
Sheena received a gift card for $50. She has already used it to buy a lamp for $39.99.
The equation 39.99 + x = 50
39.99 + x = 50
x = 50 – 39.99
x = 50.00 – 39.99
x = 10.01
Thus $10.01 is left on the gift card.

Question 2.
When Pete had a fever, his temperature was 101.4°F. After taking some medicine, his temperature was 99.2°F. The equation 101.4 – d = 99.2 can be used to find the number of degrees d that Pete’s temperature decreased. What is the solution of the equation?
_____

Answer: 2.2

Explanation:
Given,
When Pete had a fever, his temperature was 101.4°F.
After taking some medicine, his temperature was 99.2°F.
The equation 101.4 – d = 99.2
104.4 – 99.2 = d
d = 104.4 – 99.2
d = 2.2

Spiral Review

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 426 Q3

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 426 Q4

Question 5.
Andrew made p picture frames. He sold 2 of them at a craft fair. Write an expression that could be used to find the number of picture frames Andrew has left.
Type below:
_____________

Answer: p – 2

Explanation:
Andrew made p picture frames. He sold 2 of them at a craft fair.
The expression is the difference of 9 and 2
The equation is p – 2

Question 6.
Write an expression that is equivalent to 4 + 3(5 + x).
Type below:
_____________

Answer: 4 + 15 + 3x

Explanation:
4 + 3(5 + x) = 4 + 15 + 3x
3x + 19
Thus the expression 4 + 3(5 + x) is equivalent to 4 + 15 + 3x or 3x + 19

Share and Show – Page No. 429

Question 1.
Write an equation for the word sentence “25 is 13 more than a number.”
Type below:
_____________

Answer:
Let n represent the unknown number. The phrase ‘more than’ indicates an addition operation.
Thus the equation is 25 = 13 + n.

Write an equation for the word sentence.

Question 2.
The difference of a number and 2 is 3 \(\frac{1}{3}\).
Type below:
_____________

Answer:
Let n represents the unknown number.
The phrase “difference” indicates the subtraction operation.
The equation is n – 2 = 3 \(\frac{1}{3}\)

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 429 Q3

Write a word sentence for the equation.

Question 4.
x − 0.3 = 1.7
Type below:
_____________

Answer: The difference of x and 0.3 is 1.7

Question 5.
25 = \(\frac{1}{4}\)n
Type below:
_____________

Answer: 25 is n times \(\frac{1}{4}\)

Write an equation for the word sentence.

Question 6.
The quotient of a number and 20.7 is 9.
Type below:
_____________

Answer:
Let n represents the unknown number.
The phrase “quotient” indicates the division operation.
Thus the equation is n ÷ 20.7 = 9.

Question 7.
24 less than the number of snakes is 35.
Type below:
_____________

Answer:
Let n represents the unknown number.
The phrase “less than” indicates subtraction operation.
Thus the equation is n – 24 = 35

Question 8.
75 is 18 \(\frac{1}{2}\) more than a number.
Type below:
_____________

Answer:
Let n represents the unknown number.
The phrase “more than” indicates addition operation.
75 = 18 \(\frac{1}{2}\) + n

Question 9.
d degrees warmer than 50 degrees is 78 degrees.
Type below:
_____________

Answer:
Let n represents the unknown number.
The phrase “warmer than” indicates addition operation.
The equation is d + 50 = 78 degrees

Write a word sentence for the equation.

Question 10.
15g = 135
Type below:
_____________

Answer: g times 15 is 135

Question 11.
w ÷ 3.3 = 0.6
Type below:
_____________

Answer: The quotient of w and 3.3 is 0.6

Problem Solving + Applications – Page No. 430

To find out how far a car can travel on a certain amount of gas, multiply the car’s fuel efficiency in miles per gallon by the gas used in gallons. Use this information and the table for 12–13.
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 2

Question 12.
Write an equation that could be used to find how many miles a hybrid SUV can travel in the city on 20 gallons of gas.
Type below:
_____________

Answer:
From table 36 miles per gallon in the city.
A hybrid SUV uses 36 miles per gallon in the city.
So, no. of miles = y
x = no. of gallons
So, y = 36 × x
x = 20 gallons
Thus y = 36 × 20

Question 13.
A sedan traveled 504 miles on the highway on a full tank of gas. Write an equation that could be used to find the number of gallons the tank holds.
Type below:
_____________

Answer:
A sedan uses 28 miles per gallon on the highway.
The equation that could be used to find the number of gallons the tank holds is
504 = 28g

Question 14.
Connect Symbols to Words Sonya was born in 1998. Carmen was born 11 years after Sonya. If you wrote an equation to find the year in which Carmen was born, what operation would you use in your equation?
Type below:
_____________

Answer: In this equation, I would use addition or subtraction operation.

Question 15.
A magazine has 110 pages. There are 23 full-page ads and 14 half-page ads. The rest of the magazine consists of articles. Write an equation that can be used to find the number of pages of articles in the magazine.
Type below:
_____________

Answer:
The equation that can be used to find the number of pages of articles in the magazine is
23 + 14/2 + a = 110
where a represents the number of articles.

Question 16.
What’s the Error? Tony is traveling 560 miles to visit his cousins. He travels 313 miles the first day. He says that he can use the equation m − 313 = 560 to find the number of miles m he has left on his trip. Describe and correct Tony’s error.
Type below:
_____________

Answer:
Tony subtracted the number of miles traveled from the number of miles left.
Tony should have written m + 313 = 560

Question 17.
Jamie is making cookies for a bake sale. She triples the recipe in order to have enough cookies to sell. Jamie uses 12 cups of flour to make the triple batch. Write an equation that can be used to find out how much flour f is needed for one batch of cookies.
Type below:
_____________

Answer:
The equation that can be used to find out how much flour f is needed for one batch of cookies is 3f = 12

Write Equations – Page No. 431

Write an equation for the word sentence.

Question 1.
18 is 4.5 times a number.
Type below:
_____________

Answer:
Let n represents the unknown number.
The phrase “times” indicates the multiplication operation.
The equation is 18 = 4.5n

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 431 Q2

Question 3.
The difference of a number and \(\frac{2}{3}\) is \(\frac{3}{8}\).
Type below:
_____________

Answer:
Let n represents the unknown number.
The phrase “difference” indicates a subtraction operation.
The equation is n – \(\frac{2}{3}\) = \(\frac{3}{8}\)

Question 4.
A number divided by 0.5 is 29.
Type below:
_____________

Answer:
Let n represents the unknown number.
The phrase divided by indicates division operation.
The equation is n ÷ 0.5 = 29

Write a word sentence for the equation.

Question 5.
x − 14 = 52
Type below:
_____________

Answer:
14 less than x is 52
the difference of x and 14 is 52
14 fewer than a number is 52.

Question 6.
2.3m = 0.46
Type below:
_____________

Answer:
The product of 2.3 and m is 0.46
2.3 times m is .46
2.3 of m is 0.46

Question 7.
25 = k ÷ 5
Type below:
_____________

Answer: 25 is the quotient of k and 5.

Question 8.
\(4 \frac{1}{3}+q=5 \frac{1}{6}\)
Type below:
_____________

Answer:
The sum of \(4 \frac{1}{3}\) and q is [/latex]5 \frac{1}{6}[/latex]
q is more than \(4 \frac{1}{3}\) and [/latex]5 \frac{1}{6}[/latex]
\(4 \frac{1}{3}\) increased by a number is [/latex]5 \frac{1}{6}[/latex]

Question 9.
An ostrich egg weighs 2.9 pounds. The difference between the weight of this egg and the weight of an emu egg is 1.6 pounds. Write an equation that could be used to find the weight w, in pounds, of the emu egg.
Type below:
_____________

Answer: 2.9 – w = 1.6

Explanation:
An ostrich egg weighs 2.9 pounds.
The difference between the weight of this egg and the weight of an emu egg is 1.6 pounds.
The phrase “difference” indicates the subtraction operation.
The equation will be 2.9 – w = 1.6

Question 10.
In one week, the number of bowls a potter made was 6 times the number of plates. He made 90 bowls during the week. Write an equation that could be used to find the number of plates p that the potter made.
Type below:
_____________

Answer: 6p = 90

Explanation:
Given,
In one week, the number of bowls a potter made was 6 times the number of plates.
He made 90 bowls during the week.
The phrase “times” indicates the multiplication operation.
The equation to find the number of plates p that the potter made will be 6p = 90

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 431 Q11

Lesson Check – Page No. 432

Question 1.
Three friends are sharing the cost of a bucket of popcorn. The total cost of the popcorn is $5.70. Write an equation that could be used to find the amount ‘a’ in dollars that each friend should pay.
Type below:
_____________

Answer: 3a = 5.70

Explanation:
Three friends are sharing the cost of a bucket of popcorn.
The total cost of the popcorn is $5.70.
The expression will be “5.70 is the product of 3 and a.
The equation is 3a = 5.70

Question 2.
Salimah had 42 photos on her phone. After she deleted some of them, she had 23 photos left. What equation could be used to find the number of photos p that Salimah deleted?
Type below:
_____________

Answer: p + 23 = 42

Explanation:
Salimah had 42 photos on her phone. After she deleted some of them, she had 23 photos left.
The expression is the sum of p and 23 is 42.
Thus the equation is p + 23 = 42

Question 3.
A rope is 72 feet long. What is the length of the rope in yards?
______ yards

Answer: 24 yard

Explanation:
A rope is 72 feet long.
Convert from feet to yards.
1 yard = 3 feet
1 foot = 1/3 yards
72 feet = 72 × 1/3 = 24 yards
Thus the length of the rope is 24 yards.

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 432 Q4

Question 5.
The sides of a triangle have lengths s, s + 4, and 3s. Write an expression in the simplest form that represents the perimeter of the triangle.
Type below:
_____________

Answer: 5s + 4

Explanation:
The perimeter of the triangle is a + b + c
P = a + b + c
P = s + s + 4 + 3s
P = 5s + 4
Thus the perimeter of the triangle is 5s + 4

Question 6.
Gary knows that p = 2 \(\frac{1}{2}\) is a solution to one of the following equations. Which one has p = 2 \(\frac{1}{2}\) as its solution?
\(p+2 \frac{1}{2}=5\)        \(p-2 \frac{1}{2}=5\)
\(2+p=2 \frac{1}{2}\)       4 – p = 2 \(\frac{1}{2}\)
Type below:
_____________

Answer: p + 2 \(\frac{1}{2}\) = 5

Explanation:
\(p+2 \frac{1}{2}=5\)
p + 2 \(\frac{1}{2}\) = 5
p = 5 – 2 \(\frac{1}{2}\)
p = 2 \(\frac{1}{2}\)
\(p-2 \frac{1}{2}=5\)
p – 2 \(\frac{1}{2}\) = 5
p = 5 + 2 \(\frac{1}{2}\)
p = 7 \(\frac{1}{2}\)
\(2+p=2 \frac{1}{2}\)
2 + p = 2 \(\frac{1}{2}\)
p = 2 \(\frac{1}{2}\) – 2
p = \(\frac{1}{2}\)
4 – p = 2 \(\frac{1}{2}\)
p = 4 – 2 \(\frac{1}{2}\)
p = 1 \(\frac{1}{2}\)

Share and Show – Page No. 435

Model and solve the equation by using algebra tiles or iTools.

Question 1.
x + 5 = 7
x = ______

Answer: 2

Explanation:

  • Draw 2 rectangles on your MathBoard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model x + 5 in the left rectangle, and model 7 in the right rectangle.
  • To solve the equation, get the x tile by itself on one side. If you remove a tile from one side, you can keep the two sides equal by removing the same type of tile from the other side.
  • Remove five 1 tiles on the left side and five 1 tiles on the right side.
  • The remaining titles will be two 1 tiles on the right sides.

Question 2.
8 = x + 1
x = ______

Answer: 7

Explanation:

  • Draw 2 rectangles on your MathBoard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model x + 1 in the left rectangle, and model 8 in the right rectangle.
  • To solve the equation, get the x tile by itself on one side. If you remove a tile from one side, you can keep the two sides equal by removing the same type of tile from the other side.
  • Remove one 1 tiles on the left side and one 1 tiles on the right side.
  • The remaining titles will be seven 1 tiles on the right sides.

Question 3.
x + 2 = 5
x = ______

Answer: 3

Explanation:

  • Draw 2 rectangles on your MathBoard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model x + 2 in the left rectangle, and model 5 in the right rectangle.
  • To solve the equation, get the x tile by itself on one side. If you remove a tile from one side, you can keep the two sides equal by removing the same type of tile from the other side.
  • Remove two 1 tiles on the left side and five 1 tiles on the right side.
  • The remaining titles will be three 1 tiles on the right sides.

Question 4.
x + 6 = 8
x = ______

Answer: 2

Explanation:

  • Draw 2 rectangles on your MathBoard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model x + 6 in the left rectangle, and model 8 in the right rectangle.
  • To solve the equation, get the x tile by itself on one side. If you remove a tile from one side, you can keep the two sides equal by removing the same type of tile from the other side.
  • Remove six 1 tiles on the left side and six 1 tiles on the right side.
  • The remaining titles will be two 1 tiles on the right sides.

Question 5.
5 + x = 9
x = ______

Answer: 4

Explanation:

  • Draw 2 rectangles on your MathBoard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model x + 5 in the left rectangle, and model 9 in the right rectangle.
  • To solve the equation, get the x tile by itself on one side. If you remove a tile from one side, you can keep the two sides equal by removing the same type of tile from the other side.
  • Remove five 1 tiles on the left side and five 1 tiles on the right side.
  • The remaining titles will be four 1 tiles on the right sides.

Question 6.
5 = 4 + x
x = ______

Answer: 1

Explanation:

  • Draw 2 rectangles on your MathBoard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model x + 4 in the left rectangle, and model 5 in the right rectangle.
  • To solve the equation, get the x tile by itself on one side. If you remove a tile from one side, you can keep the two sides equal by removing the same type of tile from the other side.
  • Remove four 1 tiles on the left side and four 1 tiles on the right side.
  • The remaining titles will be one 1 tiles on the right sides.

Solve the equation by drawing a model.

Question 7.
x + 1 = 5
x = ______

Answer: 4

Explanation:

  • Draw 2 rectangles on your MathBoard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model x + 1 in the left rectangle, and model 5 in the right rectangle.
  • To solve the equation, get the x tile by itself on one side. If you remove a tile from one side, you can keep the two sides equal by removing the same type of tile from the other side.
  • Remove one 1 tiles on the left side and one 1 tiles on the right side.
  • The remaining titles will be four 1 tiles on the right sides.

Go Math Grade 6 Key Chapter 8 solution img-6

Question 8.
3 + x = 4
x = ______

Answer: 1

Explanation:

  • Draw 2 rectangles on your MathBoard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model x + 3 in the left rectangle, and model 4 in the right rectangle.
  • To solve the equation, get the x tile by itself on one side. If you remove a tile from one side, you can keep the two sides equal by removing the same type of tile from the other side.
  • Remove three 1 tiles on the left side and three 1 tiles on the right side.
  • The remaining titles will be one 1 tiles on the right sides.

Go Math Grade 6 Answer Key 8th chapter solution img-7

Question 9.
6 = x + 4
x = ______

Answer: 2

Explanation:

  • Draw 2 rectangles on your MathBoard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model x + 4 in the left rectangle, and model 6 in the right rectangle.
  • To solve the equation, get the x tile by itself on one side. If you remove a tile from one side, you can keep the two sides equal by removing the same type of tile from the other side.
  • Remove four 1 tiles on the left side and four 1 tiles on the right side.
  • The remaining titles will be two 1 tiles on the right sides.

HMH 6th Grade Go Math Answer Key solution img-8

Question 10.
8 = 2 + x
x = ______

Answer: 6

Explanation:

  • Draw 2 rectangles on your MathBoard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model x + 2 in the left rectangle, and model 8 in the right rectangle.
  • To solve the equation, get the x tile by itself on one side. If you remove a tile from one side, you can keep the two sides equal by removing the same type of tile from the other side.
  • Remove two 1 tiles on the left side and two 1 tiles on the right side.
  • The remaining titles will be six 1 tiles on the right sides.

6th Grade Go Math key solution img-9

Question 11.
Describe a Method Describe how you would draw a model to solve the equation x + 5 = 10.
Type below:
_____________

Answer: x = 5

Explanation:

  • Draw 2 rectangles on your MathBoard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model x + 5 in the left rectangle, and model 10 in the right rectangle.
  • To solve the equation, get the x tile by itself on one side. If you remove a tile from one side, you can keep the two sides equal by removing the same type of tile from the other side.
  • Remove five 1 tiles on the left side and five 1 tiles on the right side.
  • The remaining titles will be five 1 tiles on the right sides.

Go Math Answer Key Chapter 6th Grade solution img-10

Problem Solving + Applications – Page No. 436

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 3

Question 12.
Interpret a Result The table shows how long several animals have lived at a zoo. The giraffe has lived at the zoo 4 years longer than the mountain lion. The equation 5 = 4 + y can be used to find the number of years y the mountain lion has lived at the zoo. Solve the equation. Then tell what the solution means.
Type below:
_____________

Answer:
The table shows how long several animals have lived in a zoo.
The giraffe has lived at the zoo 4 years longer than the mountain lion.
5 = 4 + y
y = 5 – 4
y = 1
The solution is y = 1
The solution means that the mountain lion has lived at the zoo for 1 year.

Question 13.
Carlos walked 2 miles on Monday and 5 miles on Saturday. The number of miles he walked on those two days is 3 miles more than the number of miles he walked on Friday. Write and solve an addition equation to find the number of miles Carlos walked on Friday
Type below:
_____________

Answer:
Given that,
Carlos walked 2 miles on Monday and 5 miles on Saturday.
The number of miles he walked on those two days is 3 miles more than the number of miles he walked on Friday.
The equation is f + 3 = 2 + 5
f + 3 = 7
f = 7 – 3
f = 4
The solution is f = 4
The solution means that Carlos walked 4 miles on Friday.

Question 14.
Sense or Nonsense? Gabriela is solving the equation x + 1 = 6. She says that the solution must be less than 6. Is Gabriela’s statement sense or nonsense? Explain.
Type below:
_____________

Answer: Gabriela’s statement makes sense.
x + 1 = 6
x = 6 – 1
x = 5
Thus the solution is less than 6.

Question 15.
The Hawks beat the Tigers by 5 points in a football game. The Hawks scored a total of 12 points.
Use numbers and words to explain how this model can be used to solve the equation x + 5 = 12.
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 4
Type below:
_____________

Answer:
Remove 5 squares from each side. The rectangle is by itself on the left and 7 squares are on the right side.
So, the solution is x = 7

Model and Solve Addition Equations – Page No. 437

Model and solve the equation by using algebra tiles.

Question 1.
x + 6 = 9
x = ________

Answer: 3

Explanation:

  • Draw 2 rectangles on your MathBoard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model x + 6 in the left rectangle, and model 9 in the right rectangle.
  • To solve the equation, get the x tile by itself on one side. If you remove a tile from one side, you can keep the two sides equal by removing the same type of tile from the other side.
  • Remove six 1 tiles on the left side and six 1 tiles on the right side.
  • The remaining titles will be three 1 tiles on the right sides.

Thus x = 3

Question 2.
8 + x = 10
x = ________

Answer: 2

Explanation:

  • Draw 2 rectangles on your MathBoard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model x + 8 in the left rectangle, and model 10 in the right rectangle.
  • To solve the equation, get the x tile by itself on one side. If you remove a tile from one side, you can keep the two sides equal by removing the same type of tile from the other side.
  • Remove eight 1 tiles on the left side and eight 1 tiles on the right side.
  • The remaining titles will be two 1 tiles on the right sides.

8 + x = 10
x = 10 – 8 = 2
x = 2

Question 3.
9 = x + 1
x = ________

Answer: 8

Explanation:

  • Draw 2 rectangles on your MathBoard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model x + 1 in the left rectangle, and model 9 in the right rectangle.
  • To solve the equation, get the x tile by itself on one side. If you remove a tile from one side, you can keep the two sides equal by removing the same type of tile from the other side.
  • Remove 1 tile on the left side and 1 tile on the right side.
  • The remaining titles will be eight 1 tiles on the right sides.

Thus x = 8

Solve the equation by drawing a model.

Question 4.
x + 4 = 7
x = ________

Answer: 3

Go Math Answer Key Grade 6 Chapter 8 solution img-1

Question 5.
x + 6 = 10
x = ________

Answer: 4
Go Math Grade 6 Answer Key Chapter 8 solution img-2

Problem Solving

Question 6.
The temperature at 10:00 was 10°F. This is 3°F warmer than the temperature at 8:00. Model and solve the equation x + 3 = 10 to find the temperature x in degrees Fahrenheit at 8:00.
Type below:
_____________

Answer: x = 7

Explanation:
The temperature at 10:00 was 10°F. This is 3°F warmer than the temperature at 8:00.
The equation is x + 3 = 10
x = 10 – 3 = 7

Question 7.
Jaspar has 7 more checkers left than Karen does. Jaspar has 9 checkers left. Write and solve an addition equation to find out how many checkers Karen has left.
Type below:
_____________

Answer: c = 2

Explanation:
Jaspar has 7 more checkers left than Karen does. Jaspar has 9 checkers left.
The expression is c + 7 = 9
The equation to find out how many checkers Karen has left is c + 7 = 9.

Question 8.
Explain how to use a drawing to solve an addition equation such as x + 8 = 40.
Type below:
_____________

Answer: 32

Explanation:

  • Draw 2 rectangles on your MathBoard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model x + 8 in the left rectangle, and model 40 in the right rectangle.
  • To solve the equation, get the x tile by itself on one side. If you remove a tile from one side, you can keep the two sides equal by removing the same type of tile from the other side.
  • Remove eight 1 tile on the left side and eight 1 tile on the right side.
  • The remaining titles will be 32 1 tiles on the right side.

x + 8 = 40
x = 40 – 8
x = 32

Lesson Check – Page No. 438

Question 1.
What is the solution of the equation that is modeled by the algebra tiles?
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 5
x = ________

Answer: 1

The equation is x + 6 = 7
x = 7 – 6
x = 1

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 438 Q2

Spiral Review

Question 3.
A car’s gas tank has a capacity of 16 gallons. What is the capacity of the tank in pints?
________ pints

Answer: 128 pints

Explanation:
A car’s gas tank has a capacity of 16 gallons.
Convert from gallons to pints.
1 gallon = 8 pints
16 gallons = 16 × 8 = 128 pints
Thus the capacity of the tank is 128 pints.

Question 4.
Craig scored p points in a game. Marla scored twice as many points as Craig but 5 fewer than Nelson scored. How many points did Nelson score?
Type below:
_____________

Answer: 2p + 5

Explanation:
Craig scored p points in a game.
Marla scored twice as many points as Craig but 5 fewer than Nelson scored.
The equation will be 2p + 5.

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 438 Q5

Question 6.
The Empire State Building in New York City is 443.2 meters tall. This is 119.2 meters taller than the Eiffel Tower in Paris. Write an equation that can be used to find the height h in meters of the Eiffel Tower.
Type below:
_____________

Answer: 119.2 + h = 443.2

Explanation:
The Empire State Building in New York City is 443.2 meters tall.
This is 119.2 meters taller than the Eiffel Tower in Paris.
Here we have to use the addition operation.
The equation is 119.2 + h = 443.2

Share and Show – Page No. 441

Question 1.
Solve the equation n + 35 = 80.
n = ________

Answer: 45

Explanation:
The given equation is
n + 35 = 80
n = 80 – 35
n = 45

Solve the equation, and check the solution.

Question 2.
16 + x = 42
x = ________

Answer: 26

Explanation:
Given the equation 16 + x = 42
x + 16 = 42
x = 42 – 16
x = 26

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 441 Q3

Question 4.
m + \(\frac{3}{10}=\frac{7}{10}\)
m = \(\frac{□}{□}\)

Answer: \(\frac{4}{10}\)

Explanation:
The given equation is
m + \(\frac{3}{10}=\frac{7}{10}\)
m = \(\frac{7}{10}\) – \(\frac{3}{10}\)
The denominators are common so subtract the numerators
m = \(\frac{4}{10}\)

Question 5.
z – \(\frac{1}{3}=1 \frac{2}{3}\)
z = ________

Answer: 2

Explanation:
The given equation is
z – \(\frac{1}{3}=1 \frac{2}{3}\)
z = \(\frac{1}{3}\) + 1 \(\frac{2}{3}\)
z = 1 + \(\frac{1}{3}\) + \(\frac{2}{3}\)
z = 1 + \(\frac{3}{3}\)
z = 1 + 1 = 2
Thus the value of z is 2.

Question 6.
12 = x − 24
x = ________

Answer: 36

Explanation:
The given equation is
12 = x − 24
x – 24 = 12
x = 12 + 24
x = 36
Thus the value of x is 36.

Question 7.
25.3 = w − 14.9
w = ________

Answer: 40.2

Explanation:
The given equation is
25.3 = w − 14.9
w – 14.9 = 25.3
w = 25.3 + 14.9
w = 40.2
The value of w is 40.2

On Your Own

Practice: Copy and Solve Solve the equation, and check the solution.

Question 8.
y − \(\frac{3}{4}=\frac{1}{2}\)
y = _______ \(\frac{□}{□}\)

Answer: 1 \(\frac{1}{4}\)

Explanation:
The given equation is
y − \(\frac{3}{4}=\frac{1}{2}\)
y = \(\frac{1}{2}\) + \(\frac{3}{4}\)
y = 1 \(\frac{1}{4}\)
Therefore the value of y is 1 \(\frac{1}{4}\).

Question 9.
75 = n + 12
n = ________

Answer: 63

Explanation:
The given equation is
75 = n + 12
n + 12 = 75
n = 75 – 12
n = 63
The value of n is 63.

Question 10.
m + 16.8 = 40
m = ________

Answer: 23.2

Explanation:
The given equation is
m + 16.8 = 40
m = 40 – 16.8
m = 23.2
The value of m is 23.2

Question 11.
w − 36 = 56
w = ________

Answer: 92

Explanation:
The given equation is
w − 36 = 56
w = 56 + 36
w = 92
The value of w is 92.

Question 12.
8 \(\frac{2}{5}\) = d + 2\(\frac{2}{5}\)
d = ________

Answer: 6

Explanation:
The given equation is
8 \(\frac{2}{5}\) = d + 2\(\frac{2}{5}\)
d + 2\(\frac{2}{5}\) = 8 \(\frac{2}{5}\)
d = 8 \(\frac{2}{5}\) – 2\(\frac{2}{5}\)
d = 8 + \(\frac{2}{5}\) – 2 – \(\frac{2}{5}\)
d = 8 – 2 = 6
Thus the value of d is 6.

Question 13.
8.7 = r − 1.4
r = ________

Answer: 10.1

Explanation:
The given equation is
8.7 = r − 1.4
r − 1.4 = 8.7
r = 8.7 + 1.4
r = 10.1
The value of r is 10.1

Question 14.
The temperature dropped 8 degrees between 6:00 p.m. and midnight. The temperature at midnight was 26ºF. Write and solve an equation to find the temperature at 6:00 p.m.
________ ºF

Answer: 34ºF

Explanation:
The temperature dropped 8 degrees between 6:00 p.m. and midnight.
The temperature at midnight was 26ºF.
26ºF + 8ºF = 34ºF
The equation to find the temperature at 6:00 p.m is 34ºF

Question 15.
Reason Abstractly Write an addition equation that has the solution x = 9.
Type below:
_____________

Answer: x + 4 = 13

Explanation:
Let the equation be x + 4 = 13
x = 13 – 4
x = 9

Unlock the Problem – Page No. 442

Question 16.
In July, Kimberly made two deposits into her bank account. She made no withdrawals. At the end of July, her account balance was $120.62. Write and solve an equation to find Kimberly’s balance at the beginning of July.
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 6
a. What do you need to find?
Type below:
_____________

Answer: We need to find Kimberly’s balance at the beginning of July.

Question 16.
b. What information do you need from the bank statement?
Type below:
_____________

Answer: We need the information about the deposit on July 12 and July 25 from the bank statement.

Question 16.
c. Write an equation you can use to solve the problem. Explain what the variable represents.
Type below:
_____________

Answer:
x = bank account balance
y = deposit 1
z = deposit 2
x = y + z

Question 16.
d. Solve the equation. Show your work and describe each step.
Type below:
_____________

Answer: 120.62 = y + z
Where y is the deposit 1 and z represents the deposit 2.
y = $45.50, z = $43.24
45.50 + 43.24 = 88.74
x + 88.74 = 120.62

Question 16.
e. Write Kimberly’s balance at the beginning of July.
$ _______

Answer: 31.88

Explanation:
x + 88.74 = 120.62
x = 120.62 – 88.74
x = $31.88
Kimberly’s balance at the beginning of July is $31.88

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 442 Q17

Question 18.
Select the equations that have the solution n = 23. Mark all that apply.
Options:
a. 16 + n = 39
b. n – 4 = 19
c. 25 = n – 2
d. 12 = n – 11

Answer: A, B, D

Explanation:
a. 16 + n = 39
n = 23
16 + 23 = 39
39 = 39
The variable is a solution.
b. n – 4 = 19
n = 23
23 – 4 = 19
19 = 19
The variable is a solution.
c. 25 = n – 2
25 = 23 – 2
25 ≠ 21
The variable is not a solution.
d. 12 = n – 11
n = 23
12 = 23 – 11
12 = 12
The variable is a solution.
Thus the correct answers are options A, B, D.

Solve Addition and Subtraction Equations – Page No. 443

Solve Addition and Subtraction Equations - Page No. 443

Solve the equation, and check the solution.

Question 1.
y − 14 = 23
y = _______

Answer: 37

Explanation:
y − 14 = 23
y = 23 + 14
y = 37
Thus the solution is 37.

Question 2.
x + 3 = 15
x = _______

Answer: 12

Explanation:
The equation is x + 3 = 15
x = 15 – 3
x = 12
The solution is 12.

Question 3.
n + \(\frac{2}{5}=\frac{4}{5}\)
n = _______ \(\frac{□}{□}\)

Answer: \(\frac{2}{5}\)

Explanation:
The equation is n + \(\frac{2}{5}=\frac{4}{5}\)
n + \(\frac{2}{5}\) = \(\frac{4}{5}\)
n = \(\frac{4}{5}\) – \(\frac{2}{5}\)
n = (4 – 2)/5
n = \(\frac{2}{5}\)
Thus the solution is \(\frac{2}{5}\)

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 443 Q4

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 443 Q5

Question 6.
s + 55 = 55
s = _______

Answer: 0

Explanation:
The equation is s + 55 = 55
s = 55 – 55
s = 0
The solution is s = 0

Question 7.
23 = x − 12
x = _______

Answer: 35

Explanation:
The given equation is 23 = x – 12
x – 12 = 23
x = 23 + 12
x = 35
The solution is x = 35.

Question 8.
p − 14 = 14
p = _______

Answer: 28

Explanation:
The given equation is p − 14 = 14
p = 14 + 14
p = 28
The solution is p = 28.

Question 9.
m − \(2 \frac{3}{4}=6 \frac{1}{2}\)
m = _______ \(\frac{□}{□}\)

Answer: 9 \(\frac{1}{4}\)

Explanation:
The given equation is m − \(2 \frac{3}{4}=6 \frac{1}{2}\)
m – 2 \(\frac{3}{4}\) = 6 \(\frac{1}{2}\)
m = 6 \(\frac{1}{2}\) + 2 \(\frac{3}{4}\)
m = 6 + 2 + \(\frac{1}{2}\) + \(\frac{3}{4}\)
m = 8 + 1 \(\frac{1}{4}\)
m = 9 \(\frac{1}{4}\)

Problem Solving

Question 10.
A recipe calls for 5 \(\frac{1}{2}\) cups of flour. Lorenzo only has 3 \(\frac{3}{4}\) cups of flour. Write and solve an equation to find the additional amount of flour Lorenzo needs to make the recipe.
Type below:
_____________

Answer: 1 \(\frac{3}{4}\)

Explanation:
A recipe calls for 5 \(\frac{1}{2}\) cups of flour.
Lorenzo only has 3 \(\frac{3}{4}\) cups of flour.
x + 3 \(\frac{3}{4}\) = 5 \(\frac{1}{2}\)
x = 5 \(\frac{1}{2}\) – 3 \(\frac{3}{4}\)
x =  1 \(\frac{3}{4}\)

Question 11.
Jan used 22.5 gallons of water in the shower. This amount is 7.5 gallons less than the amount she used for washing clothes. Write and solve an equation to find the amount of water Jan used to wash clothes.
Type below:
_____________

Answer: 30

Explanation:
Jan used 22.5 gallons of water in the shower.
This amount is 7.5 gallons less than the amount she used for washing clothes.
Let the amount of water Jan used to wash clothes be x
x – 7.5 = 22.5
x = 22.5 + 7.5
x = 30
Therefore the amount of water Jan used to wash clothes is 30 gallons.

Question 12.
Explain how to check if your solution to an equation is correct.
Type below:
_____________

Answer:
i. Evaluate the left-hand side expression at the given value to get a number.
ii. Evaluate the right-hand side expression at the given value to get a number.
iii. See if the numbers match.

Lesson Check – Page No. 444

Question 1.
The price tag on a shirt says $21.50. The final cost of the shirt, including sales tax, is $23.22. The equation 21.50 + t = 23.22 can be used to find the amount of sales tax t in dollars. What is the sales tax?
$ _______

Answer: 1.72

Explanation:
The price tag on a shirt says $21.50.
The final cost of the shirt, including sales tax, is $23.22.
The equation is 21.50 + t = 23.22
t = 23.22 – 21.50
t = 1.72
Therefore the sales tax is $1.72 dollars.

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 444 Q2

Spiral Review

Question 3.
How would you convert a mass in centigrams to a mass in milligrams?
Type below:
_____________

Answer: The conversion factor is 10; so 1 centigram = 10 milligrams. In other words, the value in cg multiplies by 10 to get a value in mg.

Question 4.
In the expression 4 + 3x + 5y, what is the coefficient of x?
The coefficient is _______

Answer:
A numerical or constant quantity placed before and multiplying the variable in an algebraic expression.
Thus the coefficient of 3x is 3.

Question 5.
Write an expression that is equivalent to 10c.
Type below:
_____________

Answer:
-2(-5c) expand the brackets
-2 × -5c
= 10c

Question 6.
Miranda bought a $7-movie ticket and popcorn for a total of $10. The equation 7 + x = 10 can be used to find the cost x in dollars of the popcorn. How much did the popcorn cost?
$ _______

Answer: 3

Explanation:
Miranda bought a $7-movie ticket and popcorn for a total of $10.
The equation is 7 + x = 10
x = 10 – 7
x = 3
Therefore the cost of the popcorn is $3.

Share and Show – Page No. 447

Model and solve the equation by using algebra tiles.

Question 1.
4x = 16
x = _______

Answer: 4

Explanation:

  • Draw 2 rectangles on your Mathboard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model 4x in the left rectangle, and model 16 in the right rectangle.
  • There are four x tiles on the left side of your model. To solve the equation by using the model, you need to find the value of one x tile.
  • To do this, divide each side of your model into 4 equal groups.

Question 2.
3x = 12
x = _______

Answer: 4

Explanation:

  • Draw 2 rectangles on your Mathboard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model 3x in the left rectangle, and model 12 in the right rectangle.
  • There are three x tiles on the left side of your model. To solve the equation by using the model, you need to find the value of one x tile.
  • To do this, divide each side of your model into 3 equal groups.

Question 3.
4 = 4x
x = _______

Answer: 1

Explanation:

  • Draw 2 rectangles on your Mathboard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model 4x in the left rectangle, and model 4 in the right rectangle.
  • There are four x tiles on the left side of your model. To solve the equation by using the model, you need to find the value of one x tile.
  • To do this, divide each side of your model into 4 equal groups.

Question 4.
3x = 9
x = _______

Answer: 3

Explanation:

  • Draw 2 rectangles on your Mathboard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model 3x in the left rectangle, and model 9 in the right rectangle.
  • There are three x tiles on the left side of your model. To solve the equation by using the model, you need to find the value of one x tile.
  • To do this, divide each side of your model into 3 equal groups.

Question 5.
2x = 10
x = _______

Answer: 5

Explanation:

  • Draw 2 rectangles on your Mathboard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model 2x in the left rectangle, and model 10 in the right rectangle.
  • There are two x tiles on the left side of your model. To solve the equation by using the model, you need to find the value of one x tile.
  • To do this, divide each side of your model into two equal groups.

Question 6.
15 = 5x
x = _______

Answer: 3

Explanation:

  • Draw 2 rectangles on your Mathboard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model 5x in the left rectangle, and model 15 in the right rectangle.
  • There are five x tiles on the left side of your model. To solve the equation by using the model, you need to find the value of one x tile.
  • To do this, divide each side of your model into five equal groups.

Solve the equation by drawing a model.

Question 7.
4x = 8
x = _______

Answer: 2
Go Math Grade 6 Answer Key 8th chapter solution img-11

Question 8.
3x = 18
x = _______

Answer: 6
6th Grade Go Math Solution Key solution img-12

Problem Solving + Applications

Question 9.
Communicate Explain the steps you use to solve a multiplication equation with algebra tiles.
Type below:
_____________

Answer:
To solve an equation, model the terms of the equation on both sides of an equals sign.
Isolate the variable on one side by adding opposites and creating zero pairs.
To remove a factor from the variable, divide the sides into rows equal to the factor, and distribute the terms equally among all the rows.

Page No. 448

The bar graph shows the number of countries that competed in the first four modern Olympic Games. Use the bar graph for 10–11.
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 7

Question 10.
Naomi is doing a report about the 1900 and 1904 Olympic Games. Each page will contain info7rmation about 4 of the countries that competed each year. Write and solve an equation to find the number of pages Naomi will need.
_______ pages

Answer: 9 pages

Explanation:
By seeing the above table we can say that the equation is 4x = 36
The number of countries that competed in the 1900 summer Olympic games is 24.
The number of countries that competed in the 1904 summer Olympic games is 12.
The total number of countries competed in total is 36.
Each page of Naomi’s report contains information about 4 of the countries that competed each year.
4x = 36
x = 36/4
x = 9
Thus Naomi would require 9 pages to complete her report.

Question 11.
Pose a Problem Use the information in the bar graph to write and solve a problem involving a multiplication equation.
Type below:
_____________

Answer:
By seeing the above table we can say that the equation is 4x = 72
The number of countries that competed in the 1900 summer Olympic games is 24.
The number of countries that competed in the 1904 summer Olympic games is 12.
The number of countries that competed in the 1896 summer Olympic games is 14.
The number of countries that competed in the 1908 summer Olympic games is 22.
The total number of countries competed in total is 72.
4x = 72
x = 72/4
x = 18

Question 12.
The equation 7s = 21 can be used to find the number of snakes s in each cage at a zoo. Solve the equation. Then tell what the solution means.
s = _______

Answer: 3

Explanation:
The equation 7s = 21 can be used to find the number of snakes s in each cage at a zoo. Solve the equation.
7 × s = 21
s = 21/7 = 3
The solution s is 3.

Question 13.
A choir is made up of 6 vocal groups. Each group has an equal number of singers. There are 18 singers in the choir. Solve the equation 6p = 18 to find the number of singers in each group. Use a model.
_______ singers

Answer: 3 singers

Explanation:
A choir is made up of 6 vocal groups. Each group has an equal number of singers.
There are 18 singers in the choir.
The equation 6p = 18
p = 18/6 = 3
p = 3
The solution p is 3.

Model and Solve Multiplication Equations – Page No. 449

Model and solve the equation by using algebra tiles.

Question 1.
2x = 8
x = _______

Answer: 4

Explanation:

  • Draw 2 rectangles on your Mathboard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model 2x in the left rectangle, and model 8 in the right rectangle.
  • There are two x tiles on the left side of your model. To solve the equation by using the model, you need to find the value of one x tile.
  • To do this, divide each side of your model into two equal groups.

Question 2.
5x = 10
x = _______

Answer: 2

Explanation:

  • Draw 2 rectangles on your Mathboard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model 5x in the left rectangle, and model 10 in the right rectangle.
  • There are five x tiles on the left side of your model. To solve the equation by using the model, you need to find the value of one x tile.
  • To do this, divide each side of your model into five equal groups.

Question 3.
21 = 3x
x = _______

Answer: 7

Explanation:

  • Draw 2 rectangles on your Mathboard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model 3x in the left rectangle, and model 21 in the right rectangle.
  • There are three x tiles on the left side of your model. To solve the equation by using the model, you need to find the value of one x tile.
  • To do this, divide each side of your model into three equal groups.

Solve the equation by drawing a model.

Question 4.
6 = 3x

Answer: 2
HMH Go Math Grade 6 Key Chapter 8 solution img-13

Question 5.
4x = 12
x = _______

Answer: 3
Go Math 6th Grade Answer Key chapter 8 solution img-14

Problem Solving

Question 6.
A chef used 20 eggs to make 5 omelets. Model and solve the equation 5x = 20 to find the number of eggs x in each omelet.
_______ eggs

Answer: 4

Explanation:
A chef used 20 eggs to make 5 omelets.
The equation is 5x = 20
x = 50/5 = 4
Thus there are 4 eggs in each omelet.

Question 7.
Last month, Julio played 3 times as many video games as Scott did. Julio played 18 video games. Write and solve an equation to find the number of video games Scott played.
_______ video games

Answer: 6

Explanation:
Last month, Julio played 3 times as many video games as Scott did. Julio played 18 video games.
The equation will be 3x = 18
x = 18/3 = 6
x = 6
The number of video games Scott played is 6.

Question 8.
Write a multiplication equation, and explain how you can solve it by using a model.
Type below:
_____________

Answer:
15 = 5x
Explanation:

  • Draw 2 rectangles on your Mathboard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model 5x in the left rectangle, and model 15 in the right rectangle.
  • There are five x tiles on the left side of your model. To solve the equation by using the model, you need to find the value of one x tile.
  • To do this, divide each side of your model into five equal groups.

Lesson Check – Page No. 450

Question 1.
What is the solution of the equation that is modeled by the algebra tiles?
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 8
x = 1 _______

Answer: 1

Explanation:
The equation for the above figure is 3x = 3
Substitute x = 1
3(1) = 3
3/3 = 1
Thus the solution is 1.

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 450 Q2

Spiral Review

Question 3.
A rectangle is 12 feet wide and 96 inches long. What is the area of the rectangle?
_______ square feet

Answer: 1152

Explanation:
A rectangle is 12 feet wide and 96 inches long.
Area of rectangle is l × w
A = 12 × 96
A = 1152 square feet.
Thus the area of the rectangle is 1152 square feet.

Question 4.
Evaluate the algebraic expression 24 – x ÷ y for x = 8 and y = 2.
_______

Answer: 20

Explanation:
24 – x ÷ y for x = 8 and y = 2.
Substitute the value of x and y in the equation.
24 – (8 ÷ 2)
24 – 4 = 20

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 450 Q5

Question 6.
A pet store usually keeps 12 birds per cage, and there are 7 birds in the cage now. The equation 7 + x = 12 can be used to find the remaining number of birds x that can be placed in the cage. What is the solution to the equation?
x = _______

Answer: 5

Explanation:
A pet store usually keeps 12 birds per cage, and there are 7 birds in the cage now.
The equation is 7 + x = 12
x = 12 – 7
x = 5
Thus the solution of the equation is 5.

Share and Show – Page No. 453

Question 1.
Solve the equation 2.5m = 10.
m = _______

Answer: 4

Explanation:
2.5m = 10
m = 10/2.5
m = 4

Solve the equation, and check the solution.

Question 2.
3x = 210
x = _______

Answer: 70

Explanation:
3x = 210
x = 210/3
x = 70

Question 3.
2.8 = 4t
t = _______

Answer: 0.7

Explanation:
2.8 = 4t
4t = 2.8
t = 2.8/4
t = 0.7

Question 4.
\(\frac{1}{3}\)n = 15
n = _______

Answer: 45

Explanation:
\(\frac{1}{3}\)n = 15
n = 15 × 3
n = 45

Question 5.
\(\frac{1}{2}\)y = \(\frac{1}{10}\)
y = _______

Answer: \(\frac{1}{5}\)

Explanation:
\(\frac{1}{2}\)y = \(\frac{1}{10}\)
y = \(\frac{1}{10}\) × 2
y = \(\frac{1}{5}\)

Question 6.
25 = \(\frac{a}{5}\)
a = _______

Answer: 125

Explanation:
25 = \(\frac{a}{5}\)
a = 25 × 5
a = 125

Question 7.
1.3 = \(\frac{c}{4}\)
c = _______

Answer: 5.2

Explanation:
1.3 = \(\frac{c}{4}\)
c = 1.3 × 4
c = 5.2

On Your Own

Practice: Copy and Solve Solve the equation, and check the solution.

Question 8.
150 = 6m
m = _______

Answer: 25

Explanation:
6m = 150
m = 150/6
m = 25

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 453 Q9

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 453 Q10

Question 11.
There are 100 calories in 8 fluid ounces of orange juice and 140 calories in 8 fluid ounces of pineapple juice. Tia mixed 4 fluid ounces of each juice. Write and solve an equation to find the number of calories in each fluid ounce of Tia’s juice mixture.
_______ calories

Answer: 15 calories

Explanation:
Number of calories in 8 ounces of orange juice = 100
Number of calories in 1 ounce of juice = 100/8
Number of calories in 4 ounces of juice 100/8 × 4 = 50 calories
Number of calories in 8 ounces of pineapple juice = 140
Number of calories in 1 ounce of juice = 140/8
Number of calories in 4 ounces of pineapple juice = 140/8 × 4 =70 calories
Now the mixture has 50 + 70 calories = 120 calories in 8 ounces
So, 1 ounce of the mixture has 120/8 = 15 calories.

Question 12.
Write a division equation that has the solution x = 16.
Type below:
_____________

Answer:
2x = 32
x = 32/2
x = 16
Thus the equation is x = 16.

Problem Solving + Applications – Page No. 454

What’s the Error?

Question 13.
Melinda has a block of clay that weighs 14.4 ounces. She divides the clay into 6 equal pieces. To find the weight w in ounces of each piece, Melinda solved the equation 6w = 14.4.
Look at how Melinda solved the equation. Find her error.
6w = 14.4
\(\frac{6 w}{6}\) = 6 × 14.4
w = 86.4
Correct the error. Solve the equation, and explain your steps.
Describe the error that Melinda made
Type below:
_____________

Answer:
Melinda has a block of clay that weighs 14.4 ounces. She divides the clay into 6 equal pieces.
The equation is 6w = 14.4
Their error of Melinda is she used the multiplication equation to solve the equation.
She must have used the division equation to get the solution.
6w = 14.4
w = 14.4/6
w = 2.4

Question 14.
For numbers 14a−14d, choose Yes or No to indicate whether the equation has the solution x = 15.
14a. 15x = 30
14b. 4x = 60
14c. \(\frac{x}{5}\) = 3
14d. \(\frac{x}{3}\) = 5
14a. _____________
14b. _____________
14c. _____________
14d. _____________

Answer:
Given the value of x is 15
14a. 15x = 30
15 × 15 = 30
225 ≠ 30
The answer is No.
14b. 4x = 60
4 × 15 = 60
60 = 60
The answer is yes.
14c. \(\frac{x}{5}\) = 3
x/5 = 3
15/5 = 3
3 = 3
The answer is yes.
14d. \(\frac{x}{3}\) = 5
x/3 = 5
15/3 = 5
5 = 5
The answer is yes.

Solve Multiplication and Division Equations – Page No. 455

Solve the equation, and check the solution.

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 455 Q1

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 455 Q2

Question 3.
3.5x = 14.7
x = ________

Answer: 4.2

Explanation:
The given equation is
3.5x = 14.7
x = 14.7/3.5
x = 4.2
The solution x is 4.2

Question 4.
32 = 3.2c
c = ________

Answer: 10

Explanation:
The given equation is
32 = 3.2c
3.2 × c = 32
c = 32/3.2
c = 1/0.1 = 10
The solution c is 10.

Question 5.
\(\frac{2}{5}\)w = 40
w = ________

Answer: 100

Explanation:
The given equation is
\(\frac{2}{5}\)w = 40
\(\frac{2}{5}\) × w = 40
w = 40 × 5/2
w = 200/2
w = 100

Question 6.
\(\frac{a}{14}\) = 6.8
a = ________

Answer: 95.2

Explanation:
The given equation is
\(\frac{a}{14}\) = 6.8
a = 6.8 × 14
a = 95.2

Question 7.
1.6x = 1.6
x = ________

Answer: 1

Explanation:
The given equation is
1.6x = 1.6
x = 1.6/1.6
x = 1
The solution x is 1

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 455 Q8

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 455 Q9

Problem Solving

Question 10.
Anne runs 6 laps on a track. She runs a total of 1 mile, or 5,280 feet. Write and solve an equation to find the distance, in feet, that she runs in each lap.
________ feet

Answer: 880

Explanation:
Anne runs 6 laps on a track. She runs a total of 1 mile, or 5,280 feet.
Let the l represents the runs in each lap.
6 × l = 5280 feet
l = 5280/6
l = 880 feet
Therefore Anne runs 880 feets in each lap.

Question 11.
In a serving of 8 fluid ounces of pomegranate juice, there are 32.8 grams of carbohydrates. Write and solve an equation to find the amount of carbohydrates in each fluid ounce of the juice.
________ grams

Answer: 4.1

Explanation:
Given, In a serving of 8 fluid ounces of pomegranate juice, there are 32.8 grams of carbohydrates.
Let c represents the amount of carbohydrates in each fluid ounce of the juice
8 × c = 32.8 grams
c = 32.8/8
c = 4.1 grams

Question 12.
Write and solve a word problem that can be solved by solving a multiplication equation.
Type below:
_____________

Answer:
The quotient of 6 and p is 12
6 ÷ p = 12
p = 6/12
p = 1/2

Lesson Check – Page No. 456

Question 1.
Estella buys 1.8 pounds of walnuts for a total of $5.04. She solves the equation 1.8p = 5.04 to find the price p in dollars of one pound of walnuts. What does one pound of walnuts cost?
$ ________

Answer: 2.8

Explanation:
Given that, Estella buys 1.8 pounds of walnuts for a total of $5.04.
p represents the price in dollars of one pound of walnuts.
The equation to find one pound of walnuts cost is 1.8p = 5.04
1.8p = 5.04
p = 5.04/1.8
p = 2.8
Therefore the cost of one pound of walnuts is $2.8

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 456 Q2

Spiral Review

Question 3.
At top speed, a coyote can run at a speed of 44 miles per hour. If a coyote could maintain its top speed, how far could it run in 15 minutes?
________ miles

Answer: 11

Explanation:
At top speed, a coyote can run at a speed of 44 miles per hour.
Convert from minutes to hour.
60 minutes = 1 hour
15 minutes = 15 × 1/60 = 0.25 = 1/4
44 × 1/4 = 11 miles
A coyote can run at a speed of 11 miles for 15 minutes.

Question 4.
An online store sells DVDs for $10 each. The shipping charge for an entire order is $5.50. Frank orders d DVDs. Write an expression that represents the total cost of Frank’s DVDs.
Type below:
_____________

Answer: 10d + $5.50

Explanation:
An online store sells DVDs for $10 each.
The shipping charge for an entire order is $5.50. Frank orders d DVDs.
The expression will be the product of 10 and d more than 5.50
The expression is 10d + $5.50

Question 5.
A ring costs $27 more than a pair of earrings. The ring costs $90. Write an equation that can be used to find the cost c in dollars of the earrings.
Type below:
_____________

Answer: $90 – $27 = c

Explanation:
A ring costs $27 more than a pair of earrings.
The ring costs $90.
c represents the cost in dollars of the earrings.
Thus the equation is c + $27 = $90
c = $90 – $27.

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 456 Q6

Share and Show – Page No. 459

Question 1.
Connor ran 3 kilometers in a relay race. His distance represents \(\frac{3}{10}\) of the total distance of the race. The equation \(\frac{3}{10}\)d = 3 can be used to find the total distance d of the race in kilometers. What was the total distance of the race?
________ kilometers

Answer: 10

Explanation:
Connor ran 3 kilometers in a relay race.
His distance represents \(\frac{3}{10}\) of the total distance of the race.
\(\frac{3}{10}\)d = 3
3 × d = 3 × 10
3 × d = 30
d = 30/3 = 10 kilometers
Therefore the total distance of the race is 10 kilometers.

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 459 Q2

Question 3.
The lightest puppy in a litter weighs 9 ounces, which is \(\frac{3}{4}\) of the weight of the heaviest puppy. The equation \(\frac{3}{4}\)w = 9 can be used to find the weight w in ounces of the heaviest puppy. How much does the heaviest puppy weigh?
________ ounces

Answer: 12

Explanation:
The lightest puppy in a litter weighs 9 ounces, which is \(\frac{3}{4}\) of the weight of the heaviest puppy.
\(\frac{3}{4}\)w = 9
3 × w = 9 × 4
3 × w = 36
w = 36/3
w = 12
The heaviest puppy weighs 12 ounces.

Question 4.
Sophia took home \(\frac{2}{5}\) of the pizza that was left over from a party. The amount she took represents \(\frac{1}{2}\) of a whole pizza. The equation \(\frac{2}{5}\)p = \(\frac{1}{2}\) can be used to find the number of pizzas p left over from the party. How many pizzas were left over?
_______ \(\frac{□}{□}\) pizzas

Answer: 1 \(\frac{1}{4}\) pizzas

Explanation:
Sophia took home \(\frac{2}{5}\) of the pizza that was left over from a party.
The amount she took represents \(\frac{1}{2}\) of a whole pizza.
\(\frac{2}{5}\)p = \(\frac{1}{2}\)
p = \(\frac{1}{2}\) × \(\frac{5}{2}\)
p = \(\frac{5}{4}\)
p = 1 \(\frac{1}{4}\) pizzas
1 \(\frac{1}{4}\) pizzas were leftover.

Question 5.
A city received \(\frac{3}{4}\) inch of rain on July 31. This represents \(\frac{3}{10}\) of the total amount of rain the city received in July. The equation \(\frac{3}{10}\)r = \(\frac{3}{4}\) can be used to find the amount of rain r in inches the city received in July. How much rain did the city receive in July?
_______ \(\frac{□}{□}\) inches of rain

Answer: 2 \(\frac{1}{2}\) inches of rain

Explanation:
A city received \(\frac{3}{4}\) inch of rain on July 31.
This represents \(\frac{3}{10}\) of the total amount of rain the city received in July.
\(\frac{3}{10}\)r = \(\frac{3}{4}\)
r = \(\frac{3}{4}\) × \(\frac{10}{3}\)
r = \(\frac{30}{12}\)
r = \(\frac{5}{2}\)
r = 2 \(\frac{1}{2}\)
The city received 2 \(\frac{1}{2}\) inches of rain in July.

On Your Own – Page No. 460

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 460 Q6
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 460 Q6.1

Question 7.
A dog sled race is 25 miles long. The equation \(\frac{5}{8}\)k = 25 can be used to estimate the race’s length k in kilometers. Approximately how many hours will it take a dog sled team to finish the race if it travels at an average speed of 30 kilometers per hour?
_______ \(\frac{□}{□}\) hours

Answer: 1 \(\frac{1}{3}\) hours

Explanation:
A dog sled race is 25 miles long.
The equation \(\frac{5}{8}\)k = 25
k represents race length in kilometers.
\(\frac{5}{8}\)k = 25
5 × k = 25 × 8
5k = 200
k = 200/5 = 40
k = 40
Average speed is k/30
40/30 = 4/3
The average speed of 30 kilometers per hour is 1 \(\frac{1}{3}\) hours.

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 460 Q8

Question 9.
In a basket of fruit, \(\frac{5}{6}\) of the pieces of fruit are apples. There are 20 apples in the display. The equation \(\frac{5}{6}\)f = 20 can be used to find how many pieces of fruit f are in the basket. Use words and numbers to explain how to solve the equation to find how many pieces of fruit are in the basket.
_______ pieces of fruit

Answer: 24

Explanation:
In a basket of fruit, \(\frac{5}{6}\) of the pieces of fruit are apples.
There are 20 apples in the display.
\(\frac{5}{6}\)f = 20
5 × f = 20 × 6
5 × f = 120
f = 120/5
f = 24
There are 24 pieces of friut in the basket.

Problem Solving Equations with Fractions – Page No. 461

Read each problem and solve.

Question 1.
Stu is 4 feet tall. This height represents \(\frac{6}{7}\) of his brother’s height. The equation \(\frac{6}{7}\)h = 4 can be used to find the height h, in feet, of Stu’s brother. How tall is Stu’s brother?
______ \(\frac{□}{□}\) feet

Answer: 4 \(\frac{2}{3}\) feet

Explanation:
Stu is 4 feet tall. This height represents \(\frac{6}{7}\) of his brother’s height.
The equation \(\frac{6}{7}\)h = 4
6/7 × h = 4
6 × h = 4 × 7
6 × h =28
h = 28/6
h = 14/3
h = 4 \(\frac{2}{3}\) feet
Thus the height of Stu’s brother in feet is 4 \(\frac{2}{3}\) feet.

Question 2.
Bryce bought a bag of cashews. He served \(\frac{7}{8}\) pound of cashews at a party. This amount represents \(\frac{2}{3}\) of the entire bag. The equation \(\frac{2}{3}\)n = \(\frac{7}{8}\) can be used to find the number of pounds n in a full bag. How many pounds of cashews were in the bag that Bryce bought?
______ \(\frac{□}{□}\) pounds

Answer: 1 \(\frac{5}{16}\)

Explanation:
Bryce bought a bag of cashews.
He served \(\frac{7}{8}\) pound of cashews at a party.
This amount represents \(\frac{2}{3}\) of the entire bag.
\(\frac{2}{3}\)n = \(\frac{7}{8}\)
n = \(\frac{7}{8}\) × \(\frac{3}{2}\)
n = \(\frac{21}{16}\)
n = 1 \(\frac{5}{16}\)
Bryce bought 1 \(\frac{5}{16}\) pounds of cashews were in the bag.

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 461 Q3
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 461 Q3.1

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 461 Q4

Lesson Check – Page No. 462

Question 1.
Roger served \(\frac{5}{8}\) pound of crackers, which was \(\frac{2}{3}\) of the entire box. What was the weight of the crackers originally in the box?
\(\frac{□}{□}\) pounds

Answer: \(\frac{15}{16}\) pounds

Explanation:
Roger served \(\frac{5}{8}\) pound of crackers, which was \(\frac{2}{3}\)
\(\frac{2}{3}\) × p = \(\frac{5}{8}\)
p = \(\frac{5}{8}\) × \(\frac{3}{2}\)
p = \(\frac{15}{16}\) pounds
\(\frac{15}{16}\) was the weight of the crackers originally in the box.

Question 2.
Bowser ate 4 \(\frac{1}{2}\) pounds of dog food. That amount is \(\frac{3}{4}\) of the entire bag of dog food. How many pounds of dog food were originally in the bag?
______ pounds

Answer 6 pounds

Explanation:
Bowser ate 4 \(\frac{1}{2}\) pounds of dog food.
That amount is \(\frac{3}{4}\) of the entire bag of dog food.
4 \(\frac{1}{2}\) = \(\frac{9}{2}\)
\(\frac{3}{4}\) p = \(\frac{9}{2}\)
p = \(\frac{9}{2}\) × \(\frac{4}{3}\)
p = 6 pounds
6 pounds of dog food were originally in the bag.

Spiral Review

Question 3.
What is the quotient 4 \(\frac{2}{3}\) ÷ 4 \(\frac{1}{5}\)
_______ \(\frac{□}{□}\)

Answer: 1 \(\frac{1}{9}\)

Explanation:
4 \(\frac{2}{3}\) ÷ 4 \(\frac{1}{5}\)
\(\frac{14}{3}\) ÷ \(\frac{21}{5}\)
= \(\frac{70}{63}\)
The mixed fraction of \(\frac{70}{63}\) is 1 \(\frac{1}{9}\)
4 \(\frac{2}{3}\) ÷ 4 \(\frac{1}{5}\) = 1 \(\frac{1}{9}\)

Question 4.
Miranda had 4 pounds, 6 ounces of clay. She divided it into 10 equal parts. How heavy was each part?
_______ ounces

Answer: 7 ounces

Explanation:
Miranda had 4 pounds, 6 ounces of clay.
She divided it into 10 equal parts.
Convert from pounds to ounces
We know that
1 pound = 16 ounces
4 pounds = 4 × 16 ounces = 64 ounces
64 ounces + 6 ounces = 70 ounces
Now divide 70 ounces into 10 equal parts.
70 ÷ 10 = 7 ounces.
Thus each part was 7 ounces.

Question 5.
The amount Denise charges to repair computers is $50 an hour plus a $25 service fee. Write an expression to show how much she will charge for h hours of work.
Type below:
_____________

Answer: 50h + 25

Explanation:
The amount Denise charges to repair computers is $50 an hour plus a $25 service fee.
The expression will be the product of 50 and h more than 25.
The expression is 50h + 25.

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 462 Q6

Mid-Chapter Checkpoint – Vocabulary – Page No. 463

Choose the best term from the box to complete the sentence.
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 9

Question 1.
A(n) _____ is a statement that two mathematical expressions are equal.
Type below:
_____________

Answer: An equation is a mathematical statement that two expressions are equal.

Question 2.
Adding 5 and subtracting 5 are _____.
Type below:
_____________

Answer: Solution of an equation.

Concepts and Skills

Write an equation for the word sentence.

Question 3.
The sum of a number and 4.5 is 8.2.
Type below:
_____________

Answer:
The phrase “sum” indicates an addition operation.
So, the equation is n + 4.5 = 8.2

Question 4.
Three times the cost is $24.
Type below:
_____________

Answer:
The phrase “times” indicates multiplication.
Multiply 3 with c.
3c = 24

Determine whether the given value of the variable is a solution of the equation.

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 463 Q5

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 463 Q6

Solve the equation, and check the solution.

Question 7.
a + 2.4 = 7.8
a = _____

Answer: 5.4

Explanation:
Given the equation is a + 2.4 = 7.8
a + 2.4 = 7.8
a = 7.8 – 2.4
a = 5.4

Question 8.
\(b-\frac{1}{4}=3 \frac{1}{2}\)
b = _______ \(\frac{□}{□}\)

Answer: 3 \(\frac{3}{4}\)

Explanation:
Given the equation is \(b-\frac{1}{4}=3 \frac{1}{2}\)
b – \(\frac{1}{4}\) = 3 \(\frac{1}{2}\)
b = 3 \(\frac{1}{2}\) + \(\frac{1}{4}\)
b = 3 + \(\frac{1}{4}\) + \(\frac{1}{2}\)
b = 3 \(\frac{3}{4}\)

Question 9.
3x = 27
x = _______

Answer: 9

Explanation:
Given the equation is 3x = 27
x = 27/3
x = 9

Question 10.
\(\frac{1}{3} s=\frac{1}{5}\)
s = \(\frac{□}{□}\)

Answer: \(\frac{3}{5}\)

Explanation:
Given the equation is \(\frac{1}{3} s=\frac{1}{5}\)
\(\frac{1}{3}\)s = \(\frac{1}{5}\)
s = \(\frac{3}{5}\)

Question 11.
\(\frac{t}{4}\) = 16
t = _______

Answer: 64

Explanation:
Given the equation is \(\frac{t}{4}\) = 16
t = 16 × 4
t = 64

Question 12.
\(\frac{w}{7}\) = 0.3
w = _______

Answer: 2.1

Explanation:
\(\frac{w}{7}\) = 0.3
w/7 = 0.3
w = 0.3 × 7
w = 2.1

Page No. 464

Question 13.
A stadium has a total of 18,000 seats. Of these, 7,500 are field seats, and the rest are grandstand seats. Write an equation that could be used to find the number of grandstand seats s.
Type below:
_____________

Answer: s + 7500 = 18000

Explanation:
A stadium has a total of 18,000 seats.
Of these, 7,500 are field seats, and the rest are grandstand seats.
Let s be the number of grandstand seats.
s + 7,500 = 18,000

Question 14.
Aaron wants to buy a bicycle that costs $128. So far, he has saved $56. The equation a + 56 = 128 can be used to find the amount a in dollars that Aaron still needs to save. What is the solution of the equation?
The solution is _______

Answer: 72

Explanation:
Aaron wants to buy a bicycle that costs $128. So far, he has saved $56.
The equation a + 56 = 128
a = 128 – 56
a = 72
The solution of the equation a + 56 = 128 is 72.

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 464 Q15

Question 16.
Crystal is picking blueberries. So far, she has filled \(\frac{2}{3}\) of her basket, and the blueberries weigh \(\frac{3}{4}\) pound. The equation \(\frac{2}{3}\)w = \(\frac{3}{4}\) can be used to estimate the weight w in pounds of the blueberries when the basket is full. About how much will the blueberries in Crystal’s basket weigh when it is full?
______ \(\frac{□}{□}\) pounds

Answer: 1 \(\frac{1}{8}\) pounds

Explanation:
Crystal is picking blueberries. So far, she has filled \(\frac{2}{3}\) of her basket, and the blueberries weigh \(\frac{3}{4}\) pound.
The equation \(\frac{2}{3}\)w = \(\frac{3}{4}\)
w = \(\frac{3}{4}\) × \(\frac{3}{2}\)
w = \(\frac{9}{8}\)
The mixed fraction of \(\frac{9}{8}\) is 1 \(\frac{1}{8}\) pounds

Share and Show – Page No. 467

Determine whether the given value of the variable is a solution of the inequality.

Question 1.
a ≥ −6, a = −3
The variable is _____________

Answer: a solution

Explanation:
Substitute the solution a in the inequality.
a = -3
-3 ≥ -6
-3 is greater than -6
Thus the variable is a solution.

Question 2.
y < 7.8, y = 8 The variable is _____________

Answer: not a solution

Explanation:
Substitute the solution y in the inequality.
y = 8
8 is less than 7.8
8<7.8
The variable is not the solution.

Question 3.
c > \(\frac{1}{4}\), c = \(\frac{1}{5}\)
The variable is _____________

Answer: not a solution

Explanation:
Substitute the solution c in the inequality.
c = \(\frac{1}{5}\)
\(\frac{1}{5}\) > \(\frac{1}{4}\)
\(\frac{1}{5}\) is greater than \(\frac{1}{4}\)
\(\frac{1}{5}\) > \(\frac{1}{4}\)
Thus the variable is a solution.

Question 4.
x ≤ 3, x = 3
The variable is _____________

Answer: a solution

Explanation:
Substitute the solution x in the inequality.
x = 3
3 ≤ 3
3 is less than or equal to 3.
Thus the variable is a solution.

Question 5.
d < 0.52, d = 0.51
The variable is _____________

Answer: not a solution

Explanation:
Substitute the solution d in the inequality.
-0.51 < -0.52
-0.51 is greater than -0.52
The variable is not the solution.

Question 6.
t ≥ \(\frac{2}{3}\), t = \(\frac{3}{4}\)
The variable is _____________

Answer: a solution

Explanation:
Substitute the solution t in the inequality.
t = \(\frac{3}{4}\)
\(\frac{3}{4}\) ≥ \(\frac{2}{3}\)
\(\frac{3}{4}\) is greater than \(\frac{2}{3}\)
Thus the variable is a solution.

On Your Own

Practice: Copy and Solve Determine whether s = \(\frac{3}{5}\), s = 0, or s = 1.75 are solutions of the inequality.

Question 7.
s > 1
Type below:
_____________

Answer:
s > 1
s = \(\frac{3}{5}\)
\(\frac{3}{5}\) > -1
\(\frac{3}{5}\) is greater than -1.
The variable is the solution.
s = 0
0 > -1
0 is greater than -1
Thus the variable is a solution.
s = 1.75
1.75 > -1
1.75 is greater than -1
s > -1
Thus the variable is a solution.

Question 8.
s ≤ 1 \(\frac{2}{3}\)
Type below:
_____________

Answer:
s ≤ 1 \(\frac{2}{3}\)
s = \(\frac{3}{5}\)
\(\frac{3}{5}\) ≤ 1 \(\frac{2}{3}\)
\(\frac{3}{5}\) is less than but not equal to 1 \(\frac{2}{3}\)
The variable is not the solution.
s ≤ 1 \(\frac{2}{3}\)
s = 0
0 ≤ 1 \(\frac{2}{3}\)
The variable is not the solution.
s = 1.75
1.75 ≤ 1 \(\frac{2}{3}\)
The variable is not the solution.

Question 9.
s < 0.43
Type below:
_____________

Answer:
s < 0.43
\(\frac{3}{5}\) < 0.43
\(\frac{3}{5}\) = 0.6
0.6 is not less than 0.43
Thus the variable is not the solution.
s = 0
0 < 0.43
0 is less than 0.43
Thus the variable is the solution.
s = 1.75
1.75 < 0.43
1.75 is greater than 0.43
Thus the variable is not the solution.

Give two solutions of the inequality.

Question 10.
e < 3
Type below: _____________

Answer:
The solution to the inequality must be whole numbers less than 3.
e = 1 and 2 are the solutions because 1 and 2 are less than 3.
Thus the 2 solutions are 1 and 2.

Question 11.
p > 12
Type below:
_____________

Answer:
The solution to the inequality must be whole numbers greater than -12
p = 0 and -5 are the solutions because 0 and -5 are greater than -12.
Thus the 2 solutions are 0 and -5.

Question 12.
y ≥ 5.8
Type below:
_____________

Answer:
The solution to the inequality must be whole numbers greater than or equal to 5.8
y = 5.8 and 5.9 are the solutions because 5.8 and 5.9 greater than or equal to 5.8
Thus the 2 solutions are 5.8 and 5.9

Question 13.
Connect Symbols and Words A person must be at least 18 years old to vote. The inequality a ≥ 18 represents the possible ages a in years at which a person can vote. Determine whether a = 18, a = 17\(\frac{1}{2}\), and a = 91.5 are solutions of the inequality, and tell what the solutions mean.
Type below:
_____________

Answer:
a ≥ 18
Substitute the values of a in the inequality
a = 18
18 ≥ 18
Thus the variable is the solution.
a = 17\(\frac{1}{2}\)
17\(\frac{1}{2}\) ≥ 18
17\(\frac{1}{2}\) is less than 18.
The variable is not the solution.
a = 91.5
91.5 > 18
The solution is mean.

Problem Solving + Applcations – Page No. 468

The table shows ticket and popcorn prices at five movie theater chains. Use the table for 14–15.
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 10

Question 14.
The inequality p < 4.75 represents the prices p in dollars that Paige is willing to pay for popcorn. The inequality p < 8.00 represents the prices p in dollars that Paige is willing to pay for a movie ticket. At how many theaters would Paige be willing to buy a ticket and popcorn? ______ theater

Answer: 1

Explanation:
The inequality p < 4.75 represents the prices p in dollars that Paige is willing to pay for popcorn. The inequality p < 8.00 represents the prices p in dollars that Paige is willing to pay for a movie ticket.
From the above table, we can see that there is the only theatre with 8.00 and 4.75
So, Paige is willing to buy a ticket and popcorn from 1 theatre.

Question 15.
Sense or Nonsense? Edward says that inequality d ≥ 4.00 represents the popcorn prices in the table, where d is the price of popcorn in dollars. Is Edward’s statement sense or nonsense? Explain. Type below: _____________

Answer: Edward’s statement makes sense because all of the popcorn prices in the table are greater than or equal to $4.00.

Question 16.
Use Math Vocabulary Explain why the statement t > 13 is an inequality.
Type below:
_____________

Answer: The statement is equality because it compares two amounts t and 13 using an inequality symbol.

Question 17.
The minimum wind speed for a storm to be considered a hurricane is 74 miles per hour. The inequality w ≥ 74 represents the possible wind speeds of a hurricane.
Two possible solutions for the inequality w ≥ 74 are _____ and _____.
Two possible solutions for the inequality w ≥ 74 are _____ and _____

Answer: 75 and 80

Explanation:
Given that w is greater than or equal to 74.
The two possible solutions for the inequality w ≥ 74 are 75 and 80.

Solutions of Inequalities – Page No. 469

Determine whether the given value of the variable is a solution of the inequality.

Question 1.
s ≥ 1, s = 1
The variable is _____________

Answer: a solution

Explanation:
The inequality is s ≥ 1
s = 1
1 ≥ 1
1 is a positive number so 1 will be greater than or equal to -1
Thus the variable is a solution.

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 469 Q2

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 469 Q3

Question 4.
u > \(\frac{-1}{2}\), u = 0
The variable is _____________

Answer: a solution

Explanation:
The inequality is u > \(\frac{-1}{2}\)
u = 0
0 > \(\frac{-1}{2}\)
0 is greater than \(\frac{-1}{2}\)
Thus the variable is a solution.

Question 5.
q ≥ 0.6, q = 0.23
The variable is _____________

Answer: not a solution

Explanation:
The inequality is q ≥ 0.6
q = 0.23
0.23 is less than 0.6
Thus the variable is a solution.

Question 6.
b < 2 \(\frac{3}{4}\), b = \(\frac{2}{3}\)
The variable is _____________

Answer: a solution

Explanation:
The inequality is b < 2 \(\frac{3}{4}\)
b = \(\frac{2}{3}\)
\(\frac{2}{3}\) < 2 \(\frac{3}{4}\)
\(\frac{2}{3}\) is less than 2 \(\frac{3}{4}\)
Thus the variable is a solution.

Give two solutions of the inequality.

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 469 Q7

Question 8.
z ≥ 3
Type below:
_____________

Answer:
z = -3 and -2 because -3 and -2 are greater than or equal to -3
Thus the two solutions of the inequality are -3 and -2

Question 9.
f ≤ 5
Type below:
_____________

Answer:
f = -5 and -6 because -5 and -6 are less than or equal to -5
Thus the two solutions of the inequality are -5 and -6.

Problem Solving

Question 10.
The inequality s ≥ 92 represents the score s that Jared must earn on his next test to get an A on his report card. Give two possible scores that Jared could earn to get the A.
Type below:
_____________

Answer: Two possible scores that Jared could earn to get the A are 92 and 100.

Question 11.
The inequality m ≤ $20 represents the amount of money that Sheila is allowed to spend on a new hat. Give two possible money amounts that Sheila could spend on the hat.
Type below:
_____________

Answer: Two possible money amounts that Sheilla could spend on the hat are $15 or $10.

Question 12.
Describe a situation and write an inequality to represent the situation. Give a number that is a solution and another number that is not a solution of the inequality.
Type below:
_____________

Answer:
In the United States, the minimum age required to run for president is 35. This can be represented by the inequality a ≥ 35.
A number that is a solution is 55 and a number that is not a solution is 29.

Lesson Check – Page No. 470

Question 1.
Three of the following are solutions of g < 1\(\frac{1}{2}\). Which one is not a solution?
g = 4     g = 7\(\frac{1}{2}\)   g = 0    g = 2\(\frac{1}{2}\)
Type below:
_____________

Answer: g = 0

Explanation:
g < 1\(\frac{1}{2}\).
g = 4
-4 < 1\(\frac{1}{2}\)
g = 7\(\frac{1}{2}\)
7\(\frac{1}{2}\) < 1\(\frac{1}{2}\).
g = 2\(\frac{1}{2}\)
2\(\frac{1}{2}\) < 1\(\frac{1}{2}\)
g = 0
0 < 1\(\frac{1}{2}\)
Thus 0 is not the solution.

Question 2.
The inequality w ≥ 3.2 represents the weight of each pumpkin, in pounds, that is allowed to be picked to be sold. The weights of pumpkins are listed. How many pumpkins can be sold? Which pumpkins can be sold?
3.18 lb, 4 lb, 3.2 lb, 3.4 lb, 3.15 lb
Type below:
_____________

Answer: 3.2 lb, 3.4 lb

Explanation:
The inequality w ≥ 3.2 represents the weight of each pumpkin, in pounds, that is allowed to be picked to be sold.
Substitute the solutions in the inequality.
w = 3.18
3.18 ≥ 3.2
3.18 is less than 3.2
3.18 < 3.2 lb
w = 4 lb
4 ≥ 3.2
4 is greater than 3.2
4 > 3.2
w = 3.2 lb
3.2 ≥ 3.2
3.2 lb is greater than 3.2 lb
w = 3.4 lb
3.4 ≥ 3.2
3.4 lb is greater than 3.2 lb
w = 3.15 lb
3.15 < 3.2
Thus 3.2 lb, 3.4 lb pumpkins can be sold.

Spiral Review

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 470 Q3

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 470 Q4

Question 5.
Tina bought a t-shirt and sandals. The total cost was $41.50. The t-shirt cost $8.95. The equation 8.95 + c = 41.50 can be used to find the cost c in dollars of the sandals. How much did the sandals cost?
$ _______

Answer: $32.55

Explanation:
Tina bought a t-shirt and sandals. The total cost was $41.50. The t-shirt cost $8.95.
The equation is 8.95 + c = 41.50
c = 41.50 – 8.95
c = $32.55
The cost of the sandal is 32.55

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 470 Q6

Share and Show – Page No. 473

Write an inequality for the word sentence. Tell what type of numbers the variable in the inequality can represent.

Question 1.
The elevation e is greater than or equal to 15 meters.
Type below:
_____________

Answer:
The phrase greater than or equal to represents “≥”
Thus the inequality is e ≥ 15

Question 2.
A passenger’s age a must be more than 4 years.
Type below:
_____________

Answer:
The phrase more than represents the greater than symbol “>”
Thus the inequality is a > 4

Write a word sentence for the inequality.

Question 3.
b < \(\frac{1}{2}\)
Type below:
_____________

Answer:
By seeing the above inequality we can write the word sentence for inequality as,
b is less than \(\frac{1}{2}\)

Question 4.
m ≥ 55
Type below:
_____________

Answer:
By seeing the above inequality we can write the word sentence for inequality as,
m is greater than or equal to 55.

On Your Own

Question 5.
Compare Explain the difference between t ≤ 4 and t < 4.
Type below:
_____________

Answer:
t ≤ 4 is t is less than or equal to 4 which means t is equal to 4 or 3.9.
t < 4 is t is less than 4 which means t is equal to 3, 2, or 1 or 0.

Question 6.
A children’s roller coaster is limited to riders whose height is at least 30 inches and at most 48 inches. Write two inequalities that represent the height h of riders for the roller coaster.
Type below:
_____________

Answer:
h represents the height of riders for the roller coaster.
A children’s roller coaster is limited to riders whose height is at least 30 inches and at most 48 inches.
ar least 30 inches means h must be greater than or equal to 30 inches.
i.e., h ≥ 30 inches
at most 48 inches means h must be less than 48 inches.
i.e., h < 48 inches

Question 7.
Match the inequality with the word sentence it represents.
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 11
Type below:
_____________

Answer:
Go-Math-Grade-6-Answer-Key-Chapter-8-Solutions-of-Equations-img-11

Make Generalizations – Page No. 474

The reading skill make generalizations can help you write inequalities to represent situations. A generalization is a statement that is true about a group of facts.

Sea otters spend almost their entire lives in the ocean. Their thick fur helps them to stay warm in cold water. Sea otters often float together in groups called rafts. A team of biologists weighed the female sea otters in one raft off the coast of Alaska. The chart shows their results.
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 12

Question 8.
Write two inequalities that represent generalizations about the sea otter weights.
Type below:
_____________

Answer:
First, list the weights in pounds in order from least to greatest.
50, 51, 54, 58, 61, 61, 62, 62, 66, 68, 69, 71
Next, write an inequality to describe the weights by using the least weight on the list. Let w represent weights of the otters in the pounds.
The least weight is 50 pounds, so all of the weights are greater than or equal to 50 pounds.
w ≥ 50
Now write an inequality to describe the weights by using the greatest weights in the list.
The greatest weight is 71 pounds, so all of the weights are less than or equal to 71 pounds.
w ≤ 71

Question 9.
Use the chart at the right to write two inequalities that represent generalizations about the number of sea otter pups per raft.
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 13
Type below:
_____________

Answer:
First, list the number of pups in order from least to greatest.
6, 6, 7, 10, 15, 16, 20, 23
Next, write an inequality to describe the number of pups by using the least number of pups on the list. Let n represent the number of pups.
The least weight is 6 pups. So all of the pups will be greater than or equal to 6.
n ≥ 6
Now write an inequality to describe the number of pups by using the greatest weights in the list.
The greatest weight is 23 pups so all of the weights are less than or equal to 23 pups.
n ≤ 23 pups

Write Inequalities – Page No. 475

Write an inequality for the word sentence. Tell what type of numbers the variable in the inequality can represent.

Question 1.
The width w is greater than 4 centimeters.
Type below:
_____________

Answer:
The inequality symbol for “greater than” is >. w > 4, where w is the width in centimeters. w is a positive number.

Question 2.
The score s in a basketball game is greater than or equal to 10 points
Type below:
_____________

Answer:
The inequality symbol for “greater than or equal to” is ≥. s ≥ 10, where s is the score in the basketball game. s is a positive number.

Question 3.
The mass m is less than 5 kilograms
Type below:
_____________

Answer:
The inequality symbol for “less than” is <. m < 5, where m is the mass in kilograms. m is a positive number.

Question 4.
The height h is greater than 2.5 meters
Type below:
_____________

Answer:
The inequality symbol for “greater than” is >. h > 2.5, where h is the height in meters. h is a positive number.

Question 5.
The temperature t is less than or equal to −3°.
Type below:
_____________

Answer:
The inequality symbol for “less than or equal to” is ≤. t ≤  −3° where t is the temperature in degrees. t is a negative number.

Write a word sentence for the inequality.

Question 6.4
k < 7
Type below:
_____________

Answer: The word sentence for the inequality is k is less than -7.

Question 7.
z ≥ 2 \(\frac{3}{5}\)
Type below:
_____________

Answer: The word sentence for the inequality is z is greater than or equal to 2 \(\frac{3}{5}\).

Problem Solving

Question 8.
Tabby’s mom says that she must read for at least 30 minutes each night. If m represents the number of minutes reading, what inequality can represent this situation?
Type below:
_____________

Answer: m ≥ 30

Explanation:
Tabby’s mom says that she must read for at least 30 minutes each night.
m represents the number of minutes of reading.
m is greater than or equal to 30.
Thus the inequality is m ≥ 30.

Question 9.
Phillip has a $25 gift card to his favorite restaurant. He wants to use the gift card to buy lunch. If c represents the cost of his lunch, what inequality can describe all of the possible amounts of money, in dollars, that Phillip can spend on lunch?
Type below:
_____________

Answer: c ≤ 25

Explanation:
Phillip has a $25 gift card to his favorite restaurant.
He wants to use the gift card to buy lunch.
c represents the cost of his lunch
c is less than or equal to 25.
Thus the inequality is c ≤ 25.

Question 10.
Write a short paragraph explaining to a new student how to write an inequality.
Type below:
_____________

Answer:
Inequality is a statement that two quantities are not equal.
To know which direction to shade a graph, I write inequalities with the variable on the left side of the inequality symbol. I know that the symbol has to point to the same number after I rewrite the inequality.
For example, I write 4 < y as y > 4
Now the inequality symbol points in the direction in that I should draw the shaded arrow on my graph.

Lesson Check – Page No. 476

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 476 Q1

Question 2.
Describe the meaning of y ≥ 7.9 in words.
Type below:
_____________

Answer: y ≥ 7.9 means y is greater than or equal to 7.9

Spiral Review

Question 3.
Let y represent Jaron’s age in years. If Dawn were 5 years older, she would be Jaron’s age. Which expression represents Dawn’s age?
Type below:
_____________

Answer: y – 5

Explanation:
Let y represent Jaron’s age in years.
If Dawn were 5 years older, she would be Jaron’s age.
We have to subtract 5 years to know the age of Jaron.
Thus the expression is y – 5.

Question 4.
Simplify the expression 7 × 3g.
Type below:
_____________

Answer: 21g

Question 5.
What is the solution of the equation 8 = 8f?
f = ________

Answer:
8 = 8f
f = 8/8 = 1
f = 1
The solution for the equation 8 = 8f is 1.

Question 6.
Which of the following are solutions of the inequality k ≤ 2?
k = 0   k = 2   k = 4   k = 1   k = 1 \(\frac{1}{2}\)
Type below:
_____________

Answer: k = -2 k = -4

Explanation:
k = 0 in the inequality
k ≤ 2
0 ≤ 2
0 is less than but not equal to -2
Thus 0 is not the solution.
k = 2
k ≤ 2
-2 ≤ 2
Thus -2 is the solution.
k = 4
k ≤ 2
-4 ≤ 2
Thus -4 is the solution.
k = 1
1 ≤ 2
1 ≤ 2
1 is greater than but not equal to -2
Thus 1 is not the solution.
k = 1 \(\frac{1}{2}\)
1 \(\frac{1}{2}\) ≤ 2
1 \(\frac{1}{2}\) ≤ 2
1 \(\frac{1}{2}\) is less than but not equal to -2
Thus 1 \(\frac{1}{2}\) is not the solution.

Share and Show – Page No. 479

Graph the inequality.

Question 1.
m < 15
Type below:
_____________

Answer:

Go Math Grade 6 Answer Key Grap the inequality solution img-1

Question 2.
c ≥ 1.5
Type below:
_____________

Answer:
Go Math Answer Key Grade 6 Chapter 8 Graph the inequalities img-2

Question 3.
b ≤ \(\frac{5}{8}\)
Type below:
_____________

Answer:

Go Math Solution Key for Grade 6 Chapter 8 Graph the inequalities img-3

On Your Own

Practice: Copy and Solve Graph the inequality.

Question 4.
a < \(\frac{2}{3}\)
Type below:
_____________

Answer:
HMH Go Math Grade 6 Chapter 8 Graph the inequalities img-4

Question 5.
x > 4
Type below:
_____________

Answer:
HMH Go Math Answer Key Grade 6 Chapter 8 graph inequalities img-5

Question 6.
k ≥ 0.3
Type below:
_____________

Answer:
Go math grade 6 chapter 8 answer key graph inequalities img-6

Question 7.
t ≤ 6
Type below:
_____________

Answer:
Go math key grade 6 chapter 8 graph inequalities img-7

Write the inequality represented by the graph.

Question 8.
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 14
Type below:
_____________

Answer: m < 6

Question 9.
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 15
Type below:
_____________

Answer: n ≥ -7

Question 10.
Model Mathematics The inequality w ≥ 60 represents the wind speed w in miles per hour of a tornado. Graph the solutions of the inequality on the number line.
Type below:
_____________

Answer:

Go Math Answer Key Grade 6 Chapter 8 Graph inequalities img-8

Question 11.
Graph the solutions of the inequality c < 12 ÷ 3 on the number line
Type below:
_____________

Answer:
c < 12 ÷ 3
c < 4
Go Math Grade 6 Chapter 8 Answer Key Graph inequalities img-9

Problem Solving + Applications – Page No. 480

The table shows the height requirements for rides at an amusement park. Use the table for 12–16
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 16

Question 12.
Write an inequality representing t, the heights in inches of people who can go on Twirl & Whirl.
Type below:
_____________

Answer:
The minimum height of people who can go on Twirl and Whirl is 48 inches.
So, inequality is t ≥ 48.

Question 13.
Graph your inequality from Exercise 12.
Type below:
_____________

Answer:
Draw a full circle at 48 to show that 48 is a solution.
Shade to the right of 48 to show that values greater than or equal to 48 are solutions.

Question 14.
Write an inequality representing r, the heights in inches of people who can go on Race Track.
Type below:
_____________

Answer:
The minimum height of people who can go on Race track is 24 inches.
So, the inequality is r ≥ 42.

Question 15.
Graph your inequality from Exercise 14.
Type below:
_____________

Answer:
Draw a full circle at 42 to show that 42 is a solution.
Shade to the right of 42 to show that values greater than or equal to 48 are solutions.

Question 16.
Write an inequality representing b, the heights in inches of people who can go on both River Rapids and Mighty Mountain. Explain how you determined your answer.
Type below:
_____________

Answer:
You need to be at least 38 inches tall to go on River Rapids and at least 44 inches tall to go on Mighty mountain.
So, you need to be at least 44 inches tall to go on both rides.
The inequality is b ≥ 44.

Question 17.
Alena graphed the inequality c ≤ 25. Darius said that 25 is not part of the solution of the inequality. Do you agree or disagree with Darius? Use numbers and words to support your answer
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 17
Type below:
_____________

Answer: Yes I agree with Darius.
That dark circle and the arrow to the left indicates that c ≤ 25

Graph Inequalities – Page No. 481

Graph the inequality.

Question 1.
h ≥ 3
Type below:
_____________

Answer:
Go Math Answer Key Grade 6 Chapter 8 Graph inequalities image-1

Question 2.
x < \(\frac{-4}{5}\)
Type below:
_____________

Answer:
HMH Go Math Grade 6 Chapter 8 Key Graph Inequalities image-2

Question 3.
y > 2
Type below:
_____________

Answer:
HMH Go Math Solution Key for Grade 6 Chapter 8 Graph inequalities image-3

Question 4.
n ≥ 1 \(\frac{1}{2}\)
Type below:
_____________

Answer:
Go Math Key for Grade 6 Chapter 8 Graoh inequalities image-4

Question 5.
c ≤ 0.4
Type below:
_____________

Answer:
Go Math Grade 6 Answer Key chapter 8 graph inequalities image-5

Write the inequality represented by the graph.

Question 6.
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 18
Type below:
_____________

Answer: n > 3

Question 7.
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 19
Type below:
_____________

Answer: n > -5

Problem Solving

Question 8.
The inequality x ≤ 2 represents the elevation x of a certain object found at a dig site. Graph the solutions of the inequality on the number line.
Type below:
_____________

Answer:
The inequality x ≤ 2 represents the elevation x of a certain object found at a dig site.
Go math answer key grade 6 chapter 8 graph inequalities image-6

Question 9.
The inequality x ≥ 144 represents the possible scores x needed to pass a certain test. Graph the solutions of the inequality on the number line.
Type below:
_____________

Answer:
Go Math Grade 6 Chapter 8 Answer Key Graph inequalities image-7

Question 10.
Write an inequality and graph the solutions on a number line.
Type below:
_____________

Answer:
The inequality is n ≥ -7
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 15

Lesson Check – Page No. 482

Question 1.
Write the inequality that is shown by the graph.
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 20
Type below:
_____________

Answer: x ≥ -2
The number line at right shows the solutions of the inequality x ≥ -2

Question 2.
Describe the graph of g < 0.6.
Type below:
_____________

Answer:
Go Math Answer Key Grade 6 Chapter 8 solution img-5

Spiral Review

Question 3.
Write an expression that shows the product of 5 and the difference of 12 and 9.
Type below:
_____________

Answer:
The equation for the product of 5 and the difference of 12 and 9
5 × 12 – 9
The equation is 5(12 – 9).

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 482 Q4

Question 5.
The equation 12x = 96 gives the number of egg cartons x needed to package 96 eggs. Solve the equation to find the number of cartons needed.
________ cartons

Answer: 8

Explanation:
Given,
The equation 12x = 96 gives the number of egg cartons x needed to package 96 eggs.
12x = 96
x = 96/12 = 8
Thus 8 number of cartons are needed.

Question 6.
The lowest price on an MP3 song is $0.35. Write an inequality that represents the cost c of an MP3 song.
Type below:
_____________

Answer:
Given that,
The lowest price on an MP3 song is $0.35.
c ≥ 0.35
That is an inequality to represent the cost of an MP3 song.

Chapter 8 Review/Test – Page No. 483

Question 1.
For numbers 1a–1c, choose Yes or No to indicate whether the given value of the variable is a solution of the equation.
1a. \(\frac{2}{5}\)v=10; v = 25
1b. n + 5 = 15; n = 5
1c. 5z = 25; z = 5
1a. _____________
1b. _____________
1c. _____________

Answer:
1a. \(\frac{2}{5}\)v=10; v = 25
\(\frac{2}{5}\) × 25=10
2 × 5 = 10
10 = 10
The variable is a solution.
Thus the answer is yes.
1b. n + 5 = 15; n = 5
Substitute n = 5
5 + 5 = 15
10 ≠ 15
The variable is not a solution.
The answer is no.
1c. 5z = 25; z = 5
Substitute z = 5
5 × 5 = 25
25 = 25
The variable is a solution.
Thus the answer is yes.

Question 2.
The distance from third base to home plate is 88.9 feet. Romeo was 22.1 feet away from third base when he was tagged out. The equation 88.9 − t = 22.1 can be used to determine how far he needed to run to get to home plate. Using substitution, the coach determines that Romeo needed to run _____ feet to get to home plate.
Using substitution, the coach determines that Romeo needed to run _____________ feet to get to home plate

Answer: 66.8 feet

Explanation:
The distance from third base to home plate is 88.9 feet.
Romeo was 22.1 feet away from third base when he was tagged out.
The equation is 88.9 − t = 22.1
88.9 − t = 22.1
88.9 – 22.1 = t
t = 66.8 feet
Thus Using substitution, the coach determines that Romeo needed to run 66.8 feet to get to the home plate.

Question 3.
There are 84 grapes in a bag. Four friends are sharing the grapes. Write an equation that can be used to find out how many grapes g each friend will get if each friend gets the same number of grapes.
Type below:
_____________

Answer:
84 = 4g
84 is the total amount of grapes
4 is the number of friends
g = how many grapes each friend will get

Question 4.
Match each scenario with the equation that can be used to solve it.
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 21
Type below:
_____________

Answer:
Go-Math-Grade-6-Answer-Key-Chapter-8-Solutions-of-Equations-img-21

Chapter 8 Review/Test Page No. 484

Question 5.
Frank’s hockey team attempted 15 more goals than Spencer’s team. Frank’s team attempted 23 goals. Write and solve an equation that can be used to find how many goals Spencer’s team attempted.
______ goals

Answer: 8 goals

Explanation:
Frank’s hockey team attempted 15 more goals than Spencer’s team.
Frank’s team attempted 23 goals.
Let x be the Spencer’s team
The phrase more than indicates addition operation.
x + 15 = 23
x = 23 – 15
x = 8 goals

Question 6.
Ryan solved the equation 10 + y = 17 by drawing a model. Use numbers and words to explain how Ryan’s model can be used to find the solution
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 22
Type below:
_____________

Answer: y = 7

Explanation:

  • Draw 11 rectangles on your MathBoard to represent the two sides of the equation.
  • Use algebra tiles to model the equation. Model y + 10 in the left rectangle, and model 17 in the right rectangle.
  • To solve the equation, get the y tile by itself on one side. If you remove a tile from one side, you can keep the two sides equal by removing the same type of tile from the other side.
  • Remove ten 1 tiles on the left side and ten 1 tiles on the right side.
  • The remaining titles will be seven 1 tiles on the right sides.

Thus 10 + y = 17
y = 17 – 10 = 7
y = 7

Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations Page 484 Q7

Question 8.
Select the equations that have the solution m = 17. Mark all that apply.
Options:
a. 3 + m = 21
b. m − 2 = 15
c. 14 = m − 3
d. 2 = m − 15

Answer: B, C, D

Explanation:
a. 3 + m = 21
3 + 17 = 21
20 ≠ 21
b. m − 2 = 15
17 – 2 = 15
15 = 15
c. 14 = m − 3
14 = 17 – 3
14 = 14
d. 2 = m − 15
2 = 17 – 15
2 = 2
Thus the correct answers are B, C and D.

Chapter 8 Review/Test Page No. 485

Question 9.
Describe how you could use algebra tiles to model the equation 4x = 20.
Type below:
_____________

Answer:
4x = 20
x = 20/4 = 5
x = 5
Go Math Grade 6 Solution Key Chapter 8 solution img-3

Question 10.
For numbers 10a–10d, choose Yes or No to indicate whether the equation has the solution x = 12.
10a. \(\frac{3}{4}\)x = 9
10b. 3x = 36
10c. 5x = 70
10d. \(\frac{x}{3}\) = 4
10a. _____________
10b. _____________
10c. _____________
10d. _____________

Answer:
10a. Yes
10b. Yes
10c. No
10d. Yes

Explanation:
10a. \(\frac{3}{4}\)x = 9
\(\frac{3}{4}\) × 12 = 9
3 × 3 = 9
9 = 9
Thus the answer is yes.
10b. 3x = 36
x = 12
3 × 12 = 36
36 = 36
Thus the answer is yes.
10c. 5x = 70
x = 12
5 × 12 = 70
60 ≠ 70
Thus the answer is no.
10d. \(\frac{x}{3}\) = 4
x/3 = 4
x = 4 × 3
x = 12
Thus the answer is yes.

Question 11.
Bryan rides the bus to and from work on the days he works at the library. In one month, he rode the bus 24 times. Solve the equation 2x = 24 to find the number of days Bryan worked at the library. Use a model.
Type below:
_____________

Answer:
2x = 24
x = 24/2 = 12
Thus x = 12
Go Math Grade 6 Key chapter 8 solution img-4

Chapter 8 Review/Test – Page No. 486

Question 12.
Betty needs \(\frac{3}{4}\) of a yard of fabric to make a skirt. She bought 9 yards of fabric.
Part A
Write and solve an equation to find how many skirts x she can make from 9 yards of fabric.
________ skirts

Answer: 12 skirts

Explanation:
Betty needs \(\frac{3}{4}\) of a yard of fabric to make a skirt.
She bought 9 yards of fabric.
x × \(\frac{3}{4}\) = 9
x = 9 × \(\frac{4}{3}\)
x = 3 × 4 = 12
x = 12
she can make 12 skirts from 9 yards of fabric.

Question 12.
Part B
Explain how you determined which operation was needed to write the equation
Type below:
_____________

Answer: Division operation is needed to write the equation to know how many x skirts she can make from 9 yards of fabric.

Question 13.
Karen is working on her math homework. She solves the equation \(\frac{b}{8}\) = 56 and says that the solution is b = 7. Do you agree or disagree with Karen? Use words and numbers to support your answer. If her answer is incorrect, find the correct answer.
Type below:
_____________

Answer:
Karen is working on her math homework.
She solves the equation \(\frac{b}{8}\) = 56 and says that the solution is b = 7.
I Disagree with Karen.
b/8 = 56; multiply both sides by 8 to solve for b, and you get b = 448

Chapter 8 Review/Test Page No. 487

Question 14.
There are 70 historical fiction books in the school library. Historical fiction books make up \(\frac{1}{10}\) of the library’s collection. The equation \(\frac{1}{10}\)b = 70 can be used to find out how many books the library has. Solve the equation to find the total number of books in the library’s collection. Use numbers and words to explain how to solve \(\frac{1}{10}\)b = 70.
Type below:
_____________

Answer:
Given
Number of historical books = 70
The equation used to find the totals number of books in the library collection.
\(\frac{1}{10}\)b = 70
b = 70 × 10
b = 700
Hence there are 700 books in the library collection.

Question 15.
Andy drove 33 miles on Monday morning. This was \(\frac{3}{7}\) of the total number of miles he drove on Monday. Solve the equation \(\frac{3}{7}\)m = 33 to find the total number of miles Andy drove on Monday.
______ miles

Answer: 77 miles

Explanation:
Andy drove 33 miles on Monday morning.
This was \(\frac{3}{7}\) of the total number of miles he drove on Monday.
\(\frac{3}{7}\)m = 33
3 × m = 33 × 7
3 × m = 231
m = 231/3
m = 77 miles
Therefore the total number of miles Andy drove on Monday is 77 miles.

Question 16.
The maximum number of players allowed on a lacrosse team is 23. The inequality t≤23 represents the total number of players t allowed on the team.
Two possible solutions for the inequality are _____ and _____.
Two possible solutions for the inequality are _____ and _____

Answer:
The maximum number of players allowed on a lacrosse team is 23.
t ≤ 23
Thus the two possible solutions for the inequality are 22 and 23.

Question 17.
Mr. Charles needs to have at least 10 students sign up for homework help in order to use the computer lab. The inequality h ≥ 10 represents the number of students h who must sign up. Select possible solutions of the inequality. Mark all that apply.
Options:
a. 7
b. 8
c. 9
d. 10
e. 11
f. 12

Answer: D, E

Explanation:
Mr. Charles needs to have at least 10 students sign up for homework help in order to use the computer lab.
h ≥ 10
The number near to 10 is 10 and 11
Thus the correct answers are options D and E.

Chapter 8 Review/Test Page No. 488

Question 18.
The maximum capacity of the school auditorium is 420 people. Write an inequality for the situation. Tell what type of numbers the variable in the inequality can represent.
Type below:
_____________

Answer:
The maximum capacity of the school auditorium is 420 people
Let x be the maximum people
The inequality is x is less than or equal to 420.
x ≤ 420

Question 19.
Match the inequality to the word sentence it represents
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 23
Type below:
_____________

Answer:
Go-Math-Grade-6-Answer-Key-Chapter-8-Solutions-of-Equations-img-23

Question 20.
Cydney graphed the inequality d ≤ 14.
Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations img 24
Part A
Dylan said that 14 is not a solution of the inequality. Do you agree or disagree with Dylan? Use numbers and words to support your answer
Type below:
_____________

Answer: Agree with Dylan. Because the dark circle shows that it is not the solution.

Question 20.
Part B
Suppose Cydney’s graph had an empty circle at 14. Write the inequality represented by this graph.
Type below:
_____________

Answer: y < 14
HMH Go Math Grade 6 Chapter Key solution img-10

Conclusion:

I believe the information provided in the above article regarding the Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations is satisfactory for all the students. Get all the answer keys of all the chapters on ccssanswers.com For any queries you can post your comments in the below comment section.

Go Math Grade 5 Answer Key Chapter 9 Algebra: Patterns and Graphing

go-math-grade-5-chapter-9-algebra-patterns-and-graphing-answer-key

Redefine your true self with the Go Math Answer Key for Grade 5 curated by subject experts. Score higher grades in your exams and refer to Go Math Grade 5 Answer Key Chapter 9 Algebra: Patterns and Graphing to have strong command over fundamentals. Download the HMH Go Math 5th Grade Solution Key Chapter 9 free of cost and kick start your preparation immediately.

Go Math Grade 5 Answer Key Chapter 9 Algebra: Patterns and Graphing

You will get the necessary skillset needed to draw the line plots and graphs from 5th Grade Go Math Answer Key Ch 9. Access Detailed Solutions to all the problems and learn how to solve related problems when you encounter them during your exams. Seek Homework Help needed by accessing the Go Math Grade 5 Solution Key Chapter 9 Patterns and Graphing. Cross Check the Solutions from our Go Math Grade 5 Answer Key Algebra: Patterns and Graphing and understand the areas you are facing difficulty.

Lesson 1: Line Plot

Line Plot

Lesson 2: Ordered Pairs

Ordered Pairs

Lesson 3: Investigate • Graph Data

Lesson 4: Line Graphs

Line Graphs

Mid-Chapter Checkpoint

Lesson 5: Numerical Patterns

Numerical Patterns

Lesson 6: Problem Solving • Find a Rule

Lesson 7: Graph and Analyze Relationships

Chapter 9 Review/Test

Share and Show – Page No. 371

Use the data to complete the line plot. Then answer the questions.

Lilly needs to buy beads for a necklace. The beads are sold by mass. She sketches a design to determine what beads are needed and then writes down their sizes. The sizes are shown below.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 1
\(\frac{2}{5} g, \frac{2}{5} g, \frac{4}{5} g, \frac{2}{5} g, \frac{1}{5} g, \frac{1}{5} g, \frac{3}{5} g, \frac{4}{5} g, \frac{1}{5} g, \frac{2}{5} g, \frac{3}{5} g, \frac{3}{5} g, \frac{2}{5} g\)
Think: There are ___ Xs above \(\frac{1}{5}\) on the line plot, so the combined mass of the beads is _____ fifths, or _____ gram.

Question 1.
What is the combined mass of the beads with a mass of 1/5 gram?
\(\frac{□}{□}\) grams

Answer: \(\frac{3}{5}\) grams

Explanation:
For first we will count the number of \(\frac{1}{5}\) grams for each amount. Draw an x for the number of times each amount is recorded to complete the line plot.
There are 3 xs above \(\frac{1}{5}\) on the line plot, so the combined mass of the beads is 3 fifths
3 × \(\frac{1}{5}\) = 3/5 gram.

Question 2.
What is the combined mass of all the beads with a mass of \(\frac{2}{5}\) gram?
_____ grams

Answer: 2

Explanation:
For first we will count the number of \(\frac{2}{5}\) grams for each amount. Draw an x for the number of times each amount is recorded to complete the line plot.
There are 5 xs above \(\frac{2}{5}\) on the line plot, so the combined mass of the beads is 5 two fifths.
5 × \(\frac{2}{5}\) = 2 grams

Question 3.
What is the combined mass of all the beads on the necklace?
_____ grams

Answer: 6

Explanation:
Total mass of all the beads on the necklace is \(\frac{3}{5}\) + 2 + \(\frac{8}{5}\) + \(\frac{9}{5}\) = \(\frac{30}{5}\) = 6
Therefore the combined mass of all the beads on the necklace is 6.

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Page 371 Q4

On Your Own

Use the data to complete the line plot. Then answer the questions.

A breakfast chef used different amounts of milk when making pancakes, depending on the number of pancakes ordered. The results are shown below.
\(\frac{1}{2} c, \frac{1}{4} c, \frac{1}{2} c, \frac{3}{4} c, \frac{1}{2} c, \frac{3}{4} c, \frac{1}{2} c, \frac{1}{4} c, \frac{1}{2} c, \frac{1}{2} c\)
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 2

Question 5.
How much milk combined is used in \(\frac{1}{4}\)-cup amounts?
\(\frac{□}{□}\) cup

Answer: \(\frac{1}{2}\) cup

Explanation:
For first we will count the number of \(\frac{1}{4}\) cups for each amount.
2 × \(\frac{1}{4}\) = \(\frac{1}{2}\)

Question 6.
How much milk combined is used in \(\frac{1}{2}\)-cup amounts?
______ cups

Answer: 3 cups

Explanation:
For first we will count the number of \(\frac{1}{2}\) cups for each amount.
There are 6 \(\frac{1}{2}\) cups
6 × \(\frac{1}{2}\) = 3 cups

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Page 371 Q7

Question 8.
How much milk is used in all the orders of pancakes?
_____ cups

Answer: 5 cups

Explanation:
\(\frac{1}{2} c\) + [/latex]\frac{1}{4} c[/latex] + [/latex]\frac{1}{2} c[/latex] + [/latex]\frac{3}{4} c[/latex] + [/latex]\frac{1}{2} c[/latex] + [/latex]\frac{3}{4} c[/latex] + [/latex]\frac{1}{2} c[/latex] +[/latex]\frac{1}{4} c[/latex] + [/latex]\frac{1}{2} c[/latex] + [/latex]\frac{1}{2} c[/latex]
= 3 + [/latex]\frac{1}{4} c[/latex] + [/latex]\frac{3}{4} c[/latex] + [/latex]\frac{3}{4} c[/latex] + [/latex]\frac{1}{4} c[/latex]
= 3 + 1 + 1 = 5cups
Therefore 5 cups of milk is used in all the orders of pancakes.

Question 9.
What is the average amount of milk used for an order of pancakes?
\(\frac{□}{□}\) cup of milk

Answer: \(\frac{1}{2}\) cup of milk

Explanation:
There are 6 \(\frac{1}{2}\) cups of milk.
The average amount of milk used for an order of pancakes is \(\frac{1}{2}\) cup.

Question 10.
Describe an amount you could add to the data that would make the average increase.
Type below:
_________

Answer: \(\frac{3}{4}\) cup
We can add \(\frac{3}{4}\) to the data to increase the average amount of milk.

UNLOCK the Problem – Page No. 372

Question 11.
For 10 straight days, Samantha measured the amount of food that her cat Dewey ate, recording the results, which are shown below. Graph the results on the line plot. What is the average amount of cat food that Dewey ate daily?
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 3
\(\frac{1}{2} c, \frac{3}{8} c, \frac{5}{8} c, \frac{1}{2} c, \frac{5}{8} c, \frac{1}{4} c, \frac{3}{4} c, \frac{1}{4} c, \frac{1}{2} c, \frac{5}{8} c\)
a. What do you need to know?
Type below:
_________

Answer: I need to know the average amount of cat food that Dewey ate daily.

Question 11.
b. How can you use a line plot to organize the information?
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 4
Type below:
_________

Answer:
Go Math Grade 5 Chapter 10 Answer Key solution image-2
We can draw the line plot by using the given information.

Question 11.
c. What steps could you use to find the average amount of food that Dewey ate daily?
Type below:
_________

Answer: \(\frac{1}{2}\) cup

Explanation:
Number of days = 10
1/4 + 1/4 + 3/8 + 1/2 + 1/2 + 1/2 + 5/8 + 5/8 + 5/8 + 3/4 = 1 + 1 + 1/4 + 3/8 + 1/2 + 15/8
2 + 18/8 + 3/4 = 2 + 3 = 5
The average amount of food is 5 ÷ 10 = 5/10 = \(\frac{1}{2}\) cup

Question 11.
d. Fill in the blanks for the totals of each amount measured.
\(\frac{1}{4}\) cup: __________
\(\frac{3}{8}\) cup: __________
\(\frac{1}{2}\) cup: __________
\(\frac{5}{8}\) cup: __________
\(\frac{3}{4}\) cup: __________
Type below:
_________

Answer:
There are 2 xs above \(\frac{1}{4}\) cup: 2
There is 1 x above \(\frac{3}{8}\) cup: 1
There are 3 xs above \(\frac{1}{2}\) cup: 3
There are 3 xs above \(\frac{5}{8}\) cup: 3
There is 1 x above \(\frac{3}{4}\) cup: 1

Question 11.
e. Find the total amount of cat food eaten over 10 days.
_____ + _____ + _____ + _____ + _____ = _____
So, the average amount of food Dewey ate daily was ______.
Type below:
_________

Answer:
Number of days = 10
1/4 + 1/4 + 3/8 + 1/2 + 1/2 + 1/2 + 5/8 + 5/8 + 5/8 + 3/4 = 1 + 1 + 1/4 + 3/8 + 1/2 + 15/8
2 + 18/8 + 3/4 = 2 + 3 = 5 cups

Question 12.
Test Prep How many days did Dewey eat the least amount of cat food?
Options:
a. 1 day
b. 2 day
c. 3 day
d. 4 day

Answer: 1 day
By seeing the above line plot we can say that Dewey eats the least amount of cat food on day 1.
Thus the correct answer is option A.

Share and Show – Page No. 375

Use Coordinate Grid A to write an ordered pair for the given point.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 5

Question 1.
C( _____ , _____ )

Answer: 6, 3

Explanation:
Locate the point for which you want to write an ordered pair.
Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate.
Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate.
So, the ordered pair for C is (6, 3).

Question 2.
D( _____ , _____ )

Answer: 3, 0

Explanation:
Locate the point for which you want to write an ordered pair.
Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate.
Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate.
Thus the ordered pair for D is (3, 0)

Question 3.
E( _____ , _____ )

Answer: 9, 9

Explanation:
Locate the point for which you want to write an ordered pair.
Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate.
Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate.
Thus the ordered pair for E (9, 9)

Question 4.
F( _____ , _____ )

Answer: 10, 5

Explanation:
Locate the point for which you want to write an ordered pair.
Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate.
Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate.
Thus the ordered pair for F is (10, 5)

Plot and label the points on Coordinate Grid A.

Question 5.
M (0, 9)
Type below:
_________

Answer:

Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-5

Question 6.
H (8, 6)
Type below:
_________

Answer:

Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-5-1

Question 7.
K (10, 4)
Type below:
_________

Answer:

Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-5-2

Question 8.
T (4, 5)
Type below:
_________

Answer:

Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-5-3

Question 9.
W (5, 10)
Type below:
_________

Answer:

Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-5-4

Question 10.
R (1, 3)
Type below:
_________

Answer:

Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-5-5

On Your Own

Use Coordinate Grid B to write an ordered pair for the given point.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 6

Question 11.
G( _____ , _____ )

Answer: 6, 4

Explanation:
Locate the point for which you want to write an ordered pair.
Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate.
Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate.
So, the ordered pair for G is (6, 4)

Question 12.
H( _____ , _____ )

Answer: 4, 9

Explanation:
Locate the point for which you want to write an ordered pair.
Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate.
Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate.
So, the ordered pair for H is (4, 9)

Question 13.
I( _____ , _____ )

Answer: 0, 7

Explanation:
Locate the point for which you want to write an ordered pair.
Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate.
Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate.
So, the ordered pair for I is (0, 7)

Question 14.
J( _____ , _____ )

Answer: 9, 5

Explanation:
Locate the point for which you want to write an ordered pair.
Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate.
Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate.
So, the ordered pair for J is (9, 5)

Question 15.
K( _____ , _____ )

Answer: 3, 3

Explanation:
Locate the point for which you want to write an ordered pair.
Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate.
Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate.
So, the ordered pair for K is (3, 3)

Question 16.
L( _____ , _____ )

Answer: 5, 2

Explanation:
Locate the point for which you want to write an ordered pair.
Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate.
Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate.
So, the ordered pair for L is (5, 2)

Question 17.
M( _____ , _____ )

Answer: 1, 1

Explanation:
Locate the point for which you want to write an ordered pair.
Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate.
Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate.
So, the ordered pair for M is (1, 1)

Question 18.
N( _____ , _____ )

Answer: 2, 5

Explanation:
Locate the point for which you want to write an ordered pair.
Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate.
Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate.
So, the ordered pair for N is (2, 5)

Question 19.
O( _____ , _____ )

Answer: 7, 8

Explanation:
Locate the point for which you want to write an ordered pair.
Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate.
Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate.
So, the ordered pair for O is (7, 8)

Question 20.
P( _____ , _____ )

Answer: 10, 10

Explanation:
Locate the point for which you want to write an ordered pair.
Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate.
Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate.
So, the ordered pair for P is (10, 10)

Plot and label the points on Coordinate Grid B.

Question 21.
W (8, 2)

Answer:

Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-6-1

Question 22.
E (0, 4)

Answer:

Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-6-2

Question 23.
X (2, 9)

Answer:

Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-6-3

Question 24.
B (3, 4)

Answer:

Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-6-4

Question 25.
R (4, 0)

Answer:

Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-6-5

Question 26.
F (7, 6)

Answer:

Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-6-6

Question 27.
T (5, 7)

Answer:

Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-6-7

Question 28.
A (7, 1)

Answer:

Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-6-8

Question 29.
S (10, 8)

Answer:

Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-6-9

Question 30.
Y (1, 6)

Answer:

Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-6-10

Question 31.
Q (3, 8)

Answer:

Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-6-11

Question 32.
V (3, 1)

Answer:

Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-6-12

Problem Solving – Page No. 376

Nathan and his friends are planning a trip to New York City. Use the map for 33–38. Each unit represents 1 city block.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 7

Question 33.
What ordered pair gives the location of Bryant Park?
( _____ , _____ )

Answer: 4, 8

Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-7

Question 34.
What’s the Error? Nathan says that Madison Square Garden is located at (0, 3) on the map. Is his ordered pair correct? Explain.
Type below:
__________

Answer: He needs to put point 3 on Y-axis but he placed on X-Axis.

Question 35.
The Empire State Building is located 5 blocks right and 1 block up from (0, 0). Write the ordered pair for this location. Plot and label a point for the Empire State Building.
Type below:
__________

Answer: 5, 1

Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-7-1

Question 36.
Paulo walks from point B to Bryant Park. Raul walks from point B to Madison Square Garden. If they only walk along the grid lines, who walks farther? Explain.
__________

Answer: Paulo
By seeing the above graph we can say that Paulo walks farther along the grid lines.

Question 37.
Explain how to find the distance between Bryant Park and a hot dog stand at the point (4, 2).
_____ city blocks

Answer: 6

Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-7-2

Question 38.
Test Prep Use the map above. Suppose a pizzeria is located at point B. What ordered pair describes this point?
Options:
a. (4,2)
b. (3,4)
c. (2,4)
d. (4,4)

Answer: (2,4)

Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-7-3

Share and Show – Page No. 379

Graph the data on the coordinate grid.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 8
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 9

Question 1.
a. Write the ordered pairs for each point.
Type below:
__________

Answer: A(1, 30), B (2, 35), C (3, 38), D (4, 41), E (5, 44)

Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-9

Question 1.
b. What does the ordered pair (3, 38) tell you about Ryan’s age and height?
Type below:
__________

Answer: The ordered pair tells that their age of Ryan is 3 and their height is 38 inches.

Question 1.
c. Why would points (6, 42) be nonsense?
Type below:
__________

Answer: The point (6, 42) be nonsense because the height will be increased. In the above-ordered pair the height is decreased. So, the statement is nonsense.

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 10
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 11

Question 2.
a. Write the ordered pairs for each point.
Type below:
__________

Answer: We can write the ordered pairs by using the above table Day is the x-axis and height is the y-axis. The coordinates are A (5,1), B (10,3), C (15, 8), D (20,12), E (25,16), F(30,19).

Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-11

Question 2.
b. How would the ordered pairs be different if the heights of the plants were measured every 6 days for 30 days instead of every 5 days?

Answer:
If the heights of the plants were measured every 6 days for 30 days instead of every 5 days the coordinates will be A (6,1), B (12,3), C (18, 8), D (24,12), E (30,16)

Problem Solving – Page No. 380

What’s the Error?

Question 3.
Mary places a miniature car onto a track with launchers. The speed of the car is recorded every foot. Some of the data is shown in the table. Mary graphs the data on the coordinate grid below.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 12
Look at Mary’s graphed data.
Find her error.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 14
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 13
Graph the data and correct
the error.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 15
• Describe the error Mary made.
Type below:
__________

Answer:
Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-14-1

Graph the data and correct the error
Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-15

Share and Show – Page No. 383

Use the table at the right for 1–3.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 16

Question 1.
What scale and intervals would be appropriate to make a graph of the data?
Type below:
__________

Answer:
Scale is 1 cm = 10°F
Months will be on the x-axis.
The temperature will be on the y-axis.

Question 2.
Write the related pairs as ordered pairs.
Type below:
__________

Answer: The related pairs are A (Jan, 40), B (Feb, 44), C (Mar, 54), D (Apr, 62), E (May, 70)

Question 3.
Make a line graph of the data.
Type below:
__________

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Page 383 Q4

On Your Own

Use the table at the right for 5–7.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 17

Question 5.
Write the related number pairs for the plant height as ordered pairs.
Type below:
__________

Answer: The related number pairs of the above table are A (1, 20), B(2, 25), C (3, 29), D (4, 32)

Question 6.
What scale and intervals would be appropriate to make a graph of the data?
Type below:
__________

Answer: The above table says that the X-Axis is Month and Y-Axis is Height in inches.
Scale is 1 cm = 5 inches.

Explanation:
The horizontal axis could represent months from 1 to 4. In this case, the scale interval is one month.
The vertical axis could represent height from 20 inches to 32 inches but we can show a break in the scale between 1 inch and 16 inches since there are no heights between 0 inches and 20 inches, the scale interval is 1 inch.

Question 7.
Make a line graph of the data.
Type below:
__________

Question 8.
Use the graph to find the difference in height between Month 1 and Month 2.
Type below:
__________

Answer: By observing the above graph we can say that the difference between months 1 and 2 is 5 inches.
25 – 20 = 5 inches
From the graph we can see that the plant grew the most between 1 and 2 months (about 5 inches), the least change is between 3 and 4 months (about 3 inches).

Question 9.
Use the graph to estimate the height at 1 \(\frac{1}{2}\) months.
_____ in.

Answer: The estimated height at 1 \(\frac{1}{2}\) months is 22.5 inches.
The average of month 1 and month 2 is (20 + 25) ÷ 2 = 45/2 = 22.5 inches.

Connect to science – Page No. 384

Evaporation changes water on Earth’s surface into water vapor. Water vapor condenses in the atmosphere and returns to the surface as precipitation. This process is called the water cycle. The ocean is an important part of this cycle. It influences the average temperature and precipitation of a place.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 18
The overlay graph below uses two vertical scales to show monthly average precipitation and temperatures for Redding, California.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 19

Use the graph for 10–13.

Question 10.
About how much precipitation falls in Redding, California, in February?
_____ inches

Answer: From the graph, we can see that the precipitation in February is 4.2 inches.

Question 11.
What is the average temperature for Redding, California, in February?
_____ °F

Answer: From the graph, we can see that the temperature in February is 50°F.

Question 12.
Explain how the overlay graph helps you relate precipitation and temperature for each month.
Type below:
__________

Answer: The average temperature for each month is plotted on the graph with the blue line and the red bar graph represents the precipitation. As the temperature increases the precipitation decreases.

Question 13.
Describe how the average temperature changes in the first 5 months of the year.
Type below:
__________

Answer: From the graph, we can see that the temperature in the first 5 months of the year but the amount of precipitation is decreasing. It’s logical because when the temperature is increasing the amount of precipitation is decreasing.

Question 14.
Test Prep Which day had an increase of 3 feet of snow from the previous day?
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 20
Options:
a. Day 2
b. Day 3
c. Day 5
d. Day 6

Answer: Day 5

Explanation:
By seeing the above graph we can say that the snow level has increased 3 feet from day 4 to Day 5.
Thus the correct answer is option C.

Mid-Chapter Checkpoint – Vocabulary – Page No. 385

Choose the best term from the box.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Mid-Chapter Checkpoint img 21

Question 1.
The ______ is the horizontal number line on the coordinate grid.
__________

Answer: X-Axis
The X-Axis is the horizontal number line on the coordinate grid.

Question 2.
A ______ is a graph that uses line segments to show how data changes over time.
__________

Answer: Line graph
A Line graph is a graph that uses line segments to show how data changes over time.

Concepts and Skills

Use the line plot at the right for 3–5.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Mid-Chapter Checkpoint img 22

Question 3.
How many kittens weigh at least \(\frac{3}{8}\) of a pound?
______ kittens

Answer: 9

Explanation:
The line plot shows that there are 4 xs above \(\frac{3}{8}\), 3 xs above \(\frac{1}{2}\) and 2 xs on \(\frac{5}{8}\).
To find the kittens weigh at least \(\frac{3}{8}\) we need to add all above \(\frac{3}{8}\)
= 4 + 3 + 2 = 9

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Page 385 Q4

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Page 385 Q5

Use the coordinate grid at the right for 6–13.

Write an ordered pair for the given point.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Mid-Chapter Checkpoint img 23

Question 6.
A( ______ , ______ )

Answer: 1, 6
The ordered pair for A is (1,6)

Question 7.
B( ______ , ______ )

Answer: 2, 2
The ordered pair for B is (2, 2)

Question 8.
C( ______ , ______ )

Answer: 4, 4
The ordered pair for C is (4, 4)

Question 9.
D( ______ , ______ )

Answer: 0, 3
The ordered pair for D is (0, 3)

Plot and label the point on the coordinate grid.

Question 10.
E(6, 2)
Type below:
__________

Answer:
Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-23-1

Question 11.
F(5, 0)
Type below:
__________

Answer:
Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-23-2

Question 12.
G(3, 4)
Type below:
__________

Answer:
Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-23-3

Question 13.
H(3, 1)
Type below:
__________

Answer:
Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-23-4

Mid-Chapter Checkpoint – Page No. 386

Question 14.
Jane drew a point that was 1 unit to the right of the y-axis and 7 units above the x-axis. What is the ordered pair for this location?
( ______ , ______ )

Answer: (1, 7)
The ordered pair for the location is (1, 7).

Question 15.
The graph below shows the amount of snowfall in a 6-hour period.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Mid-Chapter Checkpoint img 24
Between which hours did the least amount of snowfall?
between hour ______ and hour ______

Answer: From the graph, we can see that the least amount of snowfall between 2 hours and 4 hours, 0 inches.
Go Math Grade 5 Answer Key Chapter 9 mid chapter solution

Question 16.
Joy recorded the distances she walked each day for five days. How far did she walk in 5 days?
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Mid-Chapter Checkpoint img 25
______ \(\frac{□}{□}\) miles

Answer: 2 \(\frac{1}{6}\) miles

Explanation:
There are 3 xs above \(\frac{1}{3}\) = 3 × \(\frac{1}{3}\) = 1
There are 1 x above \(\frac{1}{2}\) = 1 × \(\frac{1}{2}\) = \(\frac{1}{2}\)
There is 1 x above \(\frac{2}{3}\) = 1 × \(\frac{2}{3}\) = \(\frac{2}{3}\)
1 + \(\frac{2}{3}\) + \(\frac{1}{2}\) = (6 + 3 + 4)/6 = 13/6
The mixed fraction of 13/6 is 2 \(\frac{1}{6}\) miles
Thus she walked 2 \(\frac{1}{6}\) miles in 5 days.

Share and Show – Page No. 389

Use the given rules to complete each sequence. Then, complete the rule that describes how nickels are related to dimes.

Question 1.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 26
Type below:
__________

Answer: The number of Dimes is 2 times the number of Nickels.
We need to add 5 to Nickels = 5 + 5 + 5 + 5 + 5 = 25
We need to add 10 to Dimes = 10 + 10 + 10 + 10 + 10 = 50

Complete the rule that describes how one sequence is related to the other. Use the rule to find the unknown term.

Question 2.
Multiply the number of books by ______ to find the amount spent.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 27
______
Explain:
__________

Answer: The amount spent is 4 times the number of books so we multiply the number of books by 4 to find the amount spent.
Multiply 4 to the amount spent = 24 × 4 = 96

Question 3.
Divide the weight of the bag by _____ to find the number of marbles.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 28
______
Explain:
__________

Answer: The weight of Bag is 3 times the number of marbles So, we divide the weight of Bag by 3 to find the number of marbles.
Divide 360 by 3
360/3 = 120

On Your Own

Complete the rule that describes how one sequence is related to the other. Use the rule to find the unknown term.

Question 4.
Multiply the number of eggs by _______ to find the number of muffins.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 29
Type below:
__________

Answer: The muffins is 6 times the number of eggs so we multiply the number of eggs by 6 to find the muffins.
The unknown term in the table we will find when multiply 18 by 6.
18 × 6 = 108
The unknown term is 108.

Question 5.
Divide the number of meters by _______ to find the number of laps.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 30
Type below:
__________

Answer: The number of meters is 400 times the number of laps so we divide the number of meters by 400 to find the number of laps.
The unknown term in the table we will find when divide 6400 by 400.
6400 ÷ 400 = 16
The unknown term is 16.

Question 6.
Suppose the number of eggs used in Exercise 4 is changed to 3 eggs for each batch of 12 muffins, and 48 eggs are used. How many batches and how many muffins will be made?
______ batches
______ muffins

Answer: 16 batches 192 muffins will be made.

Explanation:
If we change to 3 eggs for each batch of 12 muffins and 48 eggs are used we will have 16 batches.
16 × 3 = 48
The muffins are 4 times the number of eggs so we multiply the number of eggs by 4 to fins the number of muffins.
If the number of batches is 16 and there are 48 eggs to find the number of muffins we will multiply the number of eggs 48 with 4:
48 × 4 = 192
192 muffins will be made.

Problem Solving – Page No. 390

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Page 390 Q7

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Page 390 Q8Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Page 390 Q8.1

Question 9.
In the cafeteria, tables are arranged in groups of 4, with each table seating 8 students. How many students can sit at 10 groups of tables? Write the rule you used to find the number of students.
______ students

Answer: 320 students

Explanation:
Tables are arranged in groups of 4, with each table seating 8 students, so in one group sit
4 × 8 = 32 students
To find how many students can sit at 10 groups of tables, we will find when multiplying 32 students with 10.
32 × 10 = 320
Finally, 320 students can sit at 10 groups of tables. The rule which we used to find the number of students is to multiply by 32 which is marked is a solution.

Question 10.
Test Prep What is the unknown number in Sequence 2 in the chart? What rule could you write that relates Sequence 1 to Sequence 2?
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 31
Options:
a. 70; Multiply by 2.
b. 100; Add 25.
c. 105; Multiply by 3.
d. 150; Add 150.

Answer: 105; Multiply by 3.

Explanation:
The unknown number in Sequence number 7 we will get when multiply 35 with 3 because the rule that releases the number of miles to the number of runners is multiplying by 3.
The unknown number is: 35 × 3 = 105
Thus the correct answer is option C.

Share and Show – Page No. 393

Question 1.
Max builds rail fences. For one style of fence, each section uses 3 vertical fence posts and 6 horizontal rails. How many posts and rails does he need for a fence that will be 9 sections long?
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 32
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 33
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 34
First, think about what the problem is asking and what you know. As each section of fence is added, how does the number of posts and the number of rails change?

Next, make a table and look for a pattern. Use what you know about 1, 2, and 3 sections. Write a rule for the number of posts and rails needed for 9 sections
of fence.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 35
Possible rule for posts: _____________
Possible rule for rails: ______________
Finally, use the rule to solve the problem.
Type below:
__________

Answer:
Possible rule for posts: 27
Possible rule for rails: 54

Explanation:
The number of posts is 3 times the number of sections. So, we multiply the number of posts by 3.
With using the rule the unknown number is 9 × 3 = 27
Thus the possible rule for posts is 27.
Now multiply the number of rails by 2.
With using the rule the unknown number is 27 × 2 = 54
Thus the possible rule for rails is 54.

Question 2.
What if another style of rail fencing has 6 rails between each pair of posts? How many rails are needed for 9 sections of this fence?
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 36
Possible rule for rails: ____________________
______ rails

Answer: 108 rails

Explanation:
The number of posts is 3 times the number of sections. So, we multiply the number of posts by 3.
Using the rule the unknown number is 9 × 3 = 27
Thus the possible rule for posts is 27.
Now multiply the number of rails by 4.
Using the rule the unknown number is 27 × 4 = 108
Thus the possible rule for rails is 108.

Question 3.
Leslie is buying a coat on layaway for $135. She will pay $15 each week until the coat is paid for. How much will she have left to pay after 8 weeks?
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 37
$ ______

Answer: $15

Explanation:
Leslie is buying a coat on layaway for $135. She will pay $15 each week until the coat is paid for.
Multiply the number of weeks by 15.
15 × 8 = $120
Now subtract $120 from $135
= $135 – $120 = $15

On Your Own – Page No. 394

Question 4.
Jane works as a limousine driver. She earns $50 for every 2 hours that she works. How much does Jane earn in one week if she works 40 hours per week? Write a rule and complete the table.
Possible rule: _____________
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 38
$ ______

Answer: 1000

Explanation:
The possible rule for Hour Worked: We can see that the difference between terms is 2.
So, the rule which describes this is Add 2.
The possible rule for Jane’s Pay: We can see that the difference between terms is 50.
So, the rule which describes this is Add 50.
Jane’s Pay is 25 times the hours worked so, we will multiply the hours worked by 25 to find Jane’s Pay.
The unknown number Jane’s Pay we will find when multiplying 40 with 25:
40 × 25 = 1000
She earns 1000 dollars.

Question 5.
Rosa joins a paperback book club. Members pay $8 to buy 2 tokens, and can trade 2 tokens for 4 paperback books. Rosa buys 30 tokens and trades them for 60 paperback books. How much money does she spend? Write a rule and complete the table.
Possible rule: _______________
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 39
$ ______

Answer: 120

Explanation:
Possible rule for Tokens: We can see that the difference between terms is 8.
So, the rule which describes this is Add 8.
Possible rule for Games: We can see that the difference between terms is 4.
So, the rule which describes this is Add 4.
Tokens are 2 times the games so, we will divide the tokens by 2 to find how many games can she3 play.
The unknown number of games we will find when dividing 120 by 2:
120 ÷ 2 = 60
She can play 60 games for 120 tokens.

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Page 394 Q6

Question 7.
Test Prep Which expression could describe the next figure in the pattern, Figure 4?
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 40
Options:
a. 2 × 5
b. 2 + 4 + 4
c. 2 + 4 + 4 + 4
d. 16

Answer: 2 + 4 + 4 + 4

Explanation:
We can see that the difference between two consecutive figures is 4 squares.
So, the rule which describes this is Add 4.
Thus figure 4 has 14 squares.
Thus the correct answer is option C.

Share and Show – Page No. 397

Graph and label the related number pairs as ordered pairs.
Then complete and use the rule to find the unknown term.

Question 1.
Multiply the number of tablespoons by ___ to find its weight in ounces.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 41
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 42
Type below:
_________

Answer: Multiply the number of tablespoons by 2 to find its weight in ounces.
5 × 2 = 10
Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-42-1

Question 2.
Multiply the number of hours by ____ to find the distance in miles.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 43
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 44
Type below:
_________

Answer: Multiply the number of hours by 3 to find the distance in miles.
4 × 3 = 12 miles
Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-42-2

On Your Own

Graph and label the related number pairs as ordered pairs.
Then complete and use the rule to find the unknown term.

Question 3.
Multiply the number of inches by ____ to find the distance in miles.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 45
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 46
Type below:
_________

Answer: Multiply the number of inches by 5 to find the distance in miles.
10 × 5 = 50
Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-46-1

Question 4.
Multiply the number of centiliters by ____ to find the equivalent number of milliliters.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 47
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 48

Answer:
Multiply the number of centiliters by 10 to find the equivalent number of milliliters.
5 × 10 = 50 milliliters
Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-46-2

Problem Solving – Page No. 398

Sense or Nonsense?

Question 5.
Elsa solved the following problem.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 49
Lou and George are making chili for the Annual Firefighter’s Ball. Lou uses 2 teaspoons of hot sauce for every 2 cups of chili that he makes, and George uses 3 teaspoons of the same hot sauce for every cup of chili in his recipe. Who has the hotter chili, George or Lou?

Write the related number pairs as ordered pairs and then graph them. Use the graph to compare who has the hotter chili, George or Lou.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 50
Lou’s chili: (2, 2), (4, 4), (6, 6), (8, 8)
George’s chili: (1, 3), (2, 6), (3, 9), (4, 12)

Elsa said that George’s chili was hotter than Lou’s because the graph showed that the amount of hot sauce in George’s chili was always 3 times as great as the amount of hot sauce in Lou’s chili.

Does Elsa’s answer make sense, or is it nonsense?
Explain.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 51

Answer: Elsa’s Answer makes sense.

Explanation:
Elsa’s answer makes sense because the amount of hot sauce in George’s chili was always 3 times as great as the amount of hot sauce in Lou’s chili. To prove this we will take two points from the graph which has an equal amount of cups of chili and compares the amount of hot sauce in George’s chili with the amount of hot sauce in Lou’s chili.
If we take 4 cups of George’s chili and Lou’s chili the amount of hot sauce in George’s chili is 12 teaspoons and the amount of hot sauce in Lou’s chili is 4 teaspoons.
12 is 3 times greater than 4 so Elsa’s answer makes sense.

Chapter Review/Test – Vocabulary – Page No. 399

Choose the best term from the box.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Chapter Review/Test img 52

Question 1.
The __________ is the point where the x-axis and y-axis meet. Its __________ is 0, and its __________ is 0.
The ________ is the point where the x-axis and y-axis meet.
Its ________ is 0,
and its ________ is 0.

Answer:
The Origin is the point where the x-axis and y-axis meet.
Its x-coordinate is 0,
and its y-coordinate is 0.

Question 2.
A __________ uses line segments to show how data changes over time.

Answer: A line graph uses line segments to show how data changes over time.

Check Concepts

Use the table for 3–4.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Chapter Review/Test img 53

Question 3.
Write related number pairs of data as ordered pairs.
Type below:
__________

Answer:
The ordered pair for week 1 is (1, 2)
The ordered pair for week 2 is (2, 6)
The ordered pair for week 3 is (3, 14)
The ordered pair for week 4 is (4, 16)

Question 4.
Make a line graph of the data.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Chapter Review/Test img 54
Type below:
__________

Answer:
Go-Math-Grade-5-Answer-Key-Chapter-9-Algebra-Patterns-and-Graphing-img-54

The ordered pair for week 1 is (1, 2)
The ordered pair for week 2 is (2, 6)
The ordered pair for week 3 is (3, 14)
The ordered pair for week 4 is (4, 16)

Complete the rule that describes how one sequence is related to the other. Use the rule to find the unknown term.

Question 5.
Multiply the number of eggs by ________ to find the number of cupcakes.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Chapter Review/Test img 55
_______

Answer:
Multiply the number of eggs by 6 to find the number of cupcakes.
The unknown number in batches 6 we will get when multiply 18 with 6 because the rule that releases the number of eggs to the number of cupcakes is multiplying by 6.
The number of eggs is multiple of 3 and the number of cupcakes is multiple of 6.

Chapter Review/Test – Page No. 400

Fill in the bubble completely to show your answer.

Question 6.
The letters on the coordinate grid represent the locations of the first four holes on a golf course.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Chapter Review/Test img 56
Which ordered pair describes the location of the hole labeled T?
Options:
a. (0, 7)
b. (1, 7)
c. (7, 0)
d. (7, 1)

Answer: (0, 7)
By seeing the above graph we can find the location of the hole label T i.e., (0, 7)

Use the line plot at the right for 7–8.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Chapter Review/Test img 57

Question 7.
What is the average of the data in the line plot?
Options:
a. \(\frac{1}{2}\) pound
b. 1 pound
c. 6 pounds
d. 6 \(\frac{3}{4}\) pounds

Answer: 6 pounds

Explanation:
There are 3 xs above \(\frac{1}{2}\) pound = 3 × \(\frac{1}{2}\) = 3/2
There are 4 xs above \(\frac{2}{3}\) pound = 4 × \(\frac{2}{3}\) = 8/3
There is 1 x above \(\frac{5}{6}\) pound = 5/6
There are 2 xs above \(\frac{1}{6}\) = 2/6
There are 2 xs above \(\frac{1}{3}\) = 2/3
3/2 + 8/3 + 5/6 + 2/6 + 2/3 = 6 pounds
Thus the correct answer is option C.

Question 8.
How many bags of rice weigh at least \(\frac{1}{2}\) pound?
Options:
a. 2
b. 3
c. 5
d. 8

Answer: 8

Explanation:
By seeing the above line plot we can find the number of bags of rice weigh at least \(\frac{1}{2}\) pound
There are 3 xs above \(\frac{1}{2}\) pound = 3 × \(\frac{1}{2}\) = 3/2
There are 4 xs above \(\frac{2}{3}\) pound = 4 × \(\frac{2}{3}\) = 8/3
There is 1 x above \(\frac{5}{6}\) pound = 5/6
Total number of bags of rice weigh at least \(\frac{1}{2}\) pound = 3 + 4 + 1 = 8
Thus the correct answer is option D.

Chapter Review/Test – Page No. 401

Fill in the bubble completely to show your answer.

Use the table for 9–10.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Chapter Review/Test img 58

Question 9.
Compare Tori’s and Martin’s savings. Which of the following statements is true?
Options:
a. Tori saves 4 times as much per week as Martin.
b. Tori will always have exactly $15 more in savings than Martin has.
c. Tori will save 15 times as much as Martin will.
d. On week 5, Martin will have $30 and Tori will have $90.

Answer: Tori saves 4 times as much per week as Martin.

Explanation:
By seeing the above table we can say that Tori saves 4 times as much per week as Martin.
Thus the correct answer is option A.

Question 10.
What rule could you use to find Tori’s savings after 10 weeks?
Options:
a. Add 10 from one week to the next.
b. Multiply the week by 2.
c. Multiply Martin’s savings by 4.
d. Divide Martin’s savings by 4.

Answer: Multiply Martin’s savings by 4.

Explanation:
We can find the savings of Tori by multiplying the savings of Martins by 4.
Thus the suitable statement is Multiply Martin’s savings by 4.
Therefore the correct answer is option C.

Question 11.
In an ordered pair, the x-coordinate represents the number of hexagons and the y-coordinate represents the total number of sides. If the x-coordinate is 7, what is the y-coordinate?
Options:
a. 6
b. 7
c. 13
d. 42

Answer: 6

Explanation:
Given that x-coordinate represents the number of hexagons.
Thus x-coordinate is 6.
And also given that the y-coordinate represents the number of sides.
The figure hexagon contains 6 sides.
So, the y-coordinate is 6.
Thus the ordered pair is (7, 6)
Therefore the correct answer is option A.

Question 12.
Point A is 2 units to the right and 4 units up from the origin. What ordered pair describes point A?
Options:
a. (2, 0)
b. (2, 4)
c. (4, 2)
d. (0, 4)

Answer: (2, 4)

Explanation:
Point A is 2 units to the right and 4 units up from the origin.
2 units will be located on the x-axis and 4 units will be on the y-axis.
Thus the ordered pair for point A is (2, 4)
Therefore the correct answer is option B.

Chapter Review/Test – Page No. 402

Constructed Response

Question 13.
Mr. Stevens drives 110 miles in 2 hours, 165 miles in 3 hours, and 220 miles in 4 hours. How many miles will he drive in 5 hours?
Explain how the number of hours he drives is related to the number of miles he drives.
_____ miles

Answer: 275 miles

Explanation:
Given that, Mr. Stevens drives 110 miles in 2 hours, 165 miles in 3 hours, and 220 miles in 4 hours.
We have to divide the number of miles by number of hours
That means, 110/2, 165/3, 220/4
the distance gone in 5 hours can be found with this equation
110/2 x ?/5
multiply 110 by 5 then divide the product by 2
110 × 5= 550
550/2 =275
Thus the answer is Mr. Stevens goes 275 miles in 5 hr.

Performance Task

Question 14.
Tim opens the freezer door and measures the temperature of the air inside. He continues to measure the temperature every 2 minutes, as the door stays open, and records the data in the table.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Chapter Review/Test img 59
A). On the grid below, make a line graph showing the data in the table.
Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Chapter Review/Test img 60
Type below:
__________

Question 14.
B). Use the graph to estimate the temperature at 7 minutes.
Estimate: _____ °F

Answer: By seeing the above graph we can say that the estimated temperature at 7 minutes is 15°F.

Question 14.
C). Write a question that can be answered by making a prediction. Then answer your question and explain how you made your prediction.
Type below:
__________

Question: Estimate the temperature at 5 minutes by using the graph.
Answer: By seeing the above table we can say that the estimated temperature at 5 minutes is 13°F

Conclusion

Fall in love with Maths by utilizing the Go Math 5th Standard 5 Answer Key. Make use of the Go Math Grade 5 Answer Key Chapter 9 Algebra: Patterns and Graphing as a reference for all your queries. Keep in touch with our site to avail updates on Class Specific Go Math Answer Key at your fingertips.

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms

go-math-grade-6-chapter-10-area-of-parallelograms-answer-key

Get Chapter 10 Area of Parallelograms Go Math Grade 6 Answer Key from this page. Here you can know the formulas of the area of a parallelogram. In order to solve the problems first, you have to know what is parallelogram and how to calculate the area of a parallelogram. Download HMH Go Math Grade 6 Solution Key Area of Parallelograms pdf here.

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms

Check out the topics covered in Chapter 10 Area of Parallelograms before you start practicing the problems. Area of Parallelograms includes topics like the area of triangles, Area of Trapezoids, Area of Regular Polygons, Composite Figures, etc. Practice the problems a number of times and enhance your math skills. After that solve the questions given in the mid-chapter checkpoint and review test. We have also provided the solutions of mid-chapter and review test here.

Lesson 1: Algebra • Area of Parallelograms

Area of Parallelograms

Lesson 2: Investigate • Explore Area of Triangles

Explore Area of Triangles

Lesson 3: Algebra • Area of Triangles

Lesson 4: Investigate • Explore Area of Trapezoids

Area of Trapezoids

Lesson 5: Algebra • Area of Trapezoids

Mid-Chapter Checkpoint

Lesson 6: Area of Regular Polygons

Area of Regular Polygons

Lesson 7: Composite Figures

Lesson 8: Problem Solving • Changing Dimensions

Lesson 9: Figures on the Coordinate Plane

Chapter 10 Review/Test

Share and Show – Page No. 535

Find the area of the parallelogram or square.

Question 1.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 1
_______ m2

Answer: 9.96

Explanation:
Given that
Base = 8.3 m
Height = 1.2 m
We know that the area of the parallelogram is base × height
A = bh
A = 8.3 m × 1.2 m
A = 9.96 square meters
Thus the area of the parallelogram for the above figure is 9.96 m²

Question 2.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 2
_______ ft2

Answer: 90

Explanation:
Given,
Base = 15 ft
Height = 6 ft
Area = ?
We know that,
Area of the parallelogram = bh
A = 15 ft × 6 ft
A = 90 square feet
Thus the area of the parallelogram for the above figure is 90 ft²

Question 3.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 3
_______ mm2

Answer: 6.25

Explanation:
The above figure is a square
The side of the square is a × a
A = 2.5 mm × 2.5 mm
A = 6.25 square mm
Thus the area of the square is 6.25 mm²

Question 4.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 4
\(\frac{□}{□}\) ft2

Answer: 1/2

Explanation:
Given
Base = 3/4 ft
Height = 2/3 ft
Area of the parallelogram is base × height
A = bh
A = 3/4 × 2/3
A = 1/2
Thus the area of the above parallelogram is 1/2 ft²

Find the unknown measurement for the parallelogram.

Question 5.
Area = 11 yd2
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 5
_______ yd

Answer: 2

Explanation:
Given,
A = 11 yd²
B = 5 1/2 yd
We know that
A = bh
11 = 5 1/2 × h
11 = 11/2 × h
22 = 11 × h
H = 2 yd
Thus the height of the above figure is 2 yards.

Question 6.
Area = 32 yd2
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 6
_______ yd

Answer: 8 yd

Explanation:
Given
Area = 32 yd2
Base = 4 yd
Height = ?
We know that
A = b × h
32 = 4 yd × h
H = 32/4
H = 8 yd
Therefore the height of the above figure is 8 yards.

On Your Own

Find the area of the parallelogram.

Question 7.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 7
_______ m2

Answer: 58.24

Explanation:
Given
Base = 9.1 m
Height = 6.4 m
A = b × h
A = 9.1 m × 6.4 m
A = 58.24 square meters
Thus the area of the parallelogram for the above figure is 58.24 m²

Question 8.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 8
_______ ft2

Answer: 168

Explanation:
Given
Base = 21 ft
Height = 8ft
We know that the area of the parallelogram is  base × height
A = 21 ft × 8ft
A = 168 square feet
Therefore the area of the above figure is 168 ft²

Find the unknown measurement for the figure.

Question 9.
square
A = ?
s = 15 ft
A = _______ ft

Answer: 225

Explanation:
Given,
S = 15 ft
The area of the square is s × s
A = 15 ft × 15 ft
A = 225 ft²
Thus the area of the square is 225 square feet.

Question 10.
parallelogram
A = 32 m2
b = ?
h = 8 m
b = _______ m

Answer: 4

Explanation:
Given
A = 32 m²
H = 8m
B = ?
To find the base we have to use the area of parallelogram formula
A = bh
32 m² = b × 8 m
B = 32/8
B = 4 m
Thus the base is 4 meters

Question 11.
parallelogram
A = 51 \(\frac{1}{4}\) in.2
b = 8 \(\frac{1}{5}\) in.
h = ?
________ \(\frac{□}{□}\) in.

Answer: 6 \(\frac{1}{4}\) in.

Explanation:
Given,
A = 51 \(\frac{1}{4}\) in.2
b = 8 \(\frac{1}{5}\) in.
H = ?
We know that the area of the parallelogram is  base × height
A = bh
51 \(\frac{1}{4}\) = h × 8 \(\frac{1}{5}\) in.
h = 51 \(\frac{1}{4}\) ÷ 8 \(\frac{1}{5}\) in.
h = 205/4 ÷ 41/5
h = 1025/164
h = 6 \(\frac{1}{4}\) in.
Thus the height of the parallelogram is 6 \(\frac{1}{4}\) in.

Question 12.
parallelogram
A = 121 mm2
b = 11 mm
h = ?
________ mm

Answer: 11 mm

Explanation:
Given
A = 121 mm²
B = 11 mm
H = ?
We know that
A = b × h
121 mm² = 11 mm × h
H = 121/11
H = 11 mm
Thus the height is 11 mm.

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 535 Q13

Problem Solving + Applications – Page No. 536

Question 14.
Jane’s backyard is shaped like a parallelogram. The base of the parallelogram is 90 feet, and the height is 25 feet. What is the area of Jane’s backyard?
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 9
________ ft2

Answer: 2250

Explanation:
Jane’s backyard is shaped like a parallelogram.
The base of the parallelogram is 90 feet, and the height is 25 feet.
A = bh
A = 90 ft × 25 ft
A = 2250 square feet
Therefore the area of the parallelogram for the above figure is 2250 ft2

Question 15.
Jack made a parallelogram by putting together two congruent triangles and a square, like the figures shown at the right. The triangles have the same height as the square. What is the area of Jack’s parallelogram?
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 10
________ cm2

Answer: 104

Explanation:
Jack made a parallelogram by putting together two congruent triangles and a square, like the figures shown at the right.
The triangles have the same height as the square.
Base = 8 cm + 5 cm = 13 cm
Height = 8 cm
Area = bh
A = 13 cm × 5 cm
A = 104 square cm
Thus the area of the parallelogram is 104 cm2

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 536 Q16

Question 17.
Verify the Reasoning of Others Li Ping says that a square with 3-inch sides has a greater area than a parallelogram that is not a square but has sides that have the same length. Does Li Ping’s statement make sense? Explain.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 11
Type below:
_______________

Answer: 9

Explanation:
Base = 3 in
Height = 3 in
A = bh
A = 3 in × 3 in
A = 9 square inches
Therefore the area of the above figure is 9 in²

Question 18.
Find the area of the parallelogram.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 12
________ in.2

Answer: 60

Explanation:
Base = 12 in
H = 5 in
A = bh
A = 12 in × 5 in
A = 60 square inches
A = 60 in²

Area of Parallelograms – Page No. 537

Find the area of the figure.

Question 1.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 13
________ ft2

Answer: 126

Explanation:
The base of the figure is 18 ft
Height = 7 ft
The area of the parallelogram is bh
A = 18 ft × 7 ft
A = 126 square feet
Thus the area of the parallelogram is 126 ft2

Question 2.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 14
________ cm2

Answer: 35

Explanation:
Base = 7 cm
Height = 5 cm
A = bh
A = 7 cm × 5 cm
A = 35 square cm
A = 35 cm2

Find the unknown measurement for the figure.

Question 3.
parallelogram
A = 9.18 m2
b = 2.7 m
h = ?
h = ________ m

Answer: 3.4

Explanation:
A = 9.18 m2
b = 2.7 m
h = ?
A = bh
9.18 m2 = 2.7 m × h
h = 9.18/2.7
A = 3.4 m

Question 4.
parallelogram
A = ?
b = 4 \(\frac{3}{10}\) m
h = 2 \(\frac{1}{10}\) m
A = ________ \(\frac{□}{□}\) m2

Explanation:
b = 4 \(\frac{3}{10}\) m
h = 2 \(\frac{1}{10}\) m
A = ?
A = bh
A = 4 \(\frac{3}{10}\) m × 2 \(\frac{1}{10}\) m
A = \(\frac{43}{10}\) m × \(\frac{21}{10}\) m
A = \(\frac{903}{100}\) m²
A = 9 \(\frac{3}{100}\) m²

Question 5.
square
A = ?
s = 35 cm
A = ________ cm2

Answer: 1225

Explanation:
s = 35 cm
A = s × s
A = 35 cm × 35 cm
A = 1225 cm2
Area of the parallelogram is 1225 cm2

Question 6.
parallelogram
A = 6.3 mm2
b = ?
h = 0.9 mm
b = ________ mm

Answer: 7

Explanation:
A = 6.3 mm2
b = ?
h = 0.9 mm
A = bh
6.3 mm2 = b × 0.9 mm
b = 6.3/0.9
b = 7 mm
Thus the base of the parallelogram is 7 mm.

Problem Solving

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 537 Q7

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 537 Q8

Question 9.
Copy the two triangles and the square in Exercise 15 on page 536. Show how you found the area of each piece. Draw the parallelogram formed when the three figures are put together. Calculate its area using the formula for the area of a parallelogram.
Type below:
_______________

Answer:
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 10
First, we need to add the base of the triangle and square
So, base = 8 cm + 5 cm
base = 13 cm
The height of the triangle and square are the same.
So, h = 8 cm
Area of the parallelogram is base × height
A = bh
A = 13 cm × 5 cm
A = 104 square cm
Thus the area of the parallelogram is 104 cm2

Lesson Check – Page No. 538

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 538 Q1
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 538 Q1.1

Question 2.
Square County is a square-shaped county divided into 16 equal-sized square districts. If the side length of each district is 4 miles, what is the area of Square County?
________ square miles

Answer: 256 square miles

Explanation:
Square County is a square-shaped county divided into 16 equal-sized square districts.
If the side length of each district is 4 miles
4 × 4 = 16
A = 16 × 16 = 256 square miles

Spiral Review

Question 3.
Which of the following values of y make the inequality y < 4 true?
y = 4     y = 6      y = 0    y = 8    y = 2
Type below:
_______________

Answer: y = -6

Question 4.
On a winter’s day, 9°F is the highest temperature recorded. Write an inequality that represents the temperature t in degrees Fahrenheit at any time on this day.
Type below:
_______________

Answer: t ≤ 9

Explanation:
On a winter’s day, 9°F is the highest temperature recorded.
t will be less than or equal to 9.
The inequality is t ≤ 9

Question 5.
In 2 seconds, an elevator travels 40 feet. In 3 seconds, the elevator travels 60 feet. In 4 seconds, the elevator travels 80 feet. Write an equation that gives the relationship between the number of seconds x and the distance y the elevator travels.
Type below:
_______________

Answer: y = 20x

Explanation:
x represents the number of seconds
y represents the distance the elevator travels.
The elevator travels 20 feet per second.
Thus the equation is y = 20x

Question 6.
The linear equation y = 4x represents the number of bracelets y that Jolene can make in x hours. Which ordered pair lies on the graph of the equation?
Type below:
_______________

Answer: (4, 16)

Explanation:
y = 4x
If x = 4
Then y = 4(4)
y = 16
Thus the ordered pairs are (4, 16)

Share and Show – Page No. 541

Question 1.
Trace the parallelogram, and cut it into two congruent triangles. Find the areas of the parallelogram and one triangle, using square units.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 15
Type below:
_______________

Answer:
Base = 9 units
Height = 4 units
Area of the parallelogram = base × height
A = 9 × 4
A = 36 sq. units
Area of the triangle = ab/2
A = (9 × 4)/2
A = 18 sq. units
Area of another triangle = ab/2
A = (9 × 4)/2
A = 18 sq. units

Find the area of each triangle.

Question 2.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 16
_______ in.2

Answer: 40

Explanation:
The area of the right triangle is bh/2
A = (8 × 10)/2
A = 80/2
A = 40 in.2
Thus the area of the triangle for the above figure is 40 in.2

Question 3.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 17
_______ ft2

Answer: 180

Explanation:
The area of the right triangle is bh/2
A = (18 × 20)/2
A = 360/2
A = 180 ft2

Question 4.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 18
_______ yd2

Answer: 22

Explanation:
The area of the right triangle is bh/2
A = (4 × 11)/2
A = 44/2
A = 22
A = 22 yd2
Thus the area of the triangle is 22 yd2

Question 5.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 19
_______ mm2

Answer: 495

Explanation:
The area of the right triangle is bh/2
A = (30 × 33)/2
A = 990/2
A = 495 mm2
Thus the area of the triangle is 495 mm2

Question 6.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 20
_______ in.2

Answer: 190

Explanation:
The area of the right triangle is bh/2
A = (19 × 20)/2
A = 380/2
A = 190 in.2
Thus the area of the triangle is 190 in.2

Question 7.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 21
_______ cm2

Answer: 96

Explanation:
The area of the right triangle is bh/2
A = (16 × 12)/2
A = 192/2
A = 96 Sq. cm
Thus the area of the triangle is 96 Sq. cm

Problem Solving + Applications

Question 8.
Communicate Describe how you can use two triangles of the same shape and size to form a parallelogram.
Type below:
_______________

Answer: Put them together like a puzzle. if the sides are parallel then it would be a parallelogram.

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 541 Q9

Sense or Nonsense? – Page No. 542

Question 10.
Cyndi and Tyson drew the models below. Each said his or her drawing represents a triangle with an area of 600 square inches. Whose statement makes sense? Whose statement is nonsense? Explain your reasoning.
Tyson’s Model:
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 22

Cyndi’s Model:
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 23
Type below:
_______________

Answer: Tyson’s Model makes sense.
The base of the figure is 30 in.
The height of the figure is 40 in
Area of the triangle = bh/2
A = (30 × 40)/2
A = 1200/2 = 600 sq. in
Cyndi’s Model doesn’t make sense because there is no base for the triangle.

Question 11.
A flag is separated into two different colors. Find the area of the white region. Show your work.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 24
_______ ft.2

Answer: 7.5 ft.2

Explanation:
A flag is separated into two different colors.
B = 5 ft
H = 3 ft
Area of the triangle = bh/2
A = (3 × 5)/2
A = 15/2
A = 7.5 sq. ft

Explore Area of Triangles – Page No. 543

Find the area of each triangle.

Question 1.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 25
_______ ft2

Answer: 30

Explanation:
Given,
Base = 6 ft
Height = 10 ft
Area of the triangle = bh/2
A = (6 ft × 10 ft)/2
A = 60 sq. ft/2
A = 30 ft2
Thus the area of the triangle for the above figure is 0 ft2

Question 2.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 26
_______ cm2

Answer: 925

Explanation:
Given,
Base = 50 cm
Height = 37 cm
Area of the triangle = bh/2
A = (50 × 37)/2
A = 1850/2
A = 925 sq. cm
Therefore the area of the above figure is 925 cm2

Question 3.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 27
_______ mm2

Answer: 400

Explanation:
Given,
Base = 40 mm
Height = 20 mm
Area of the triangle = bh/2
A = (40 × 20)/2
A = 800/2
A = 400 mm2
Therefore the area of the above figure is 400 mm2

Question 4.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 28
_______ in.2

Answer: 180

Explanation:
Given,
Base = 12 in.
Height = 30 in.
Area of the triangle = bh/2
A = (12 × 30)/2
A = 360/2
A = 180 in.2
Therefore the area of the above figure is 180 in.2

Question 5.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 29
_______ cm2

Answer: 225

Explanation:
Given,
Base = 15 cm
Height = 30 cm
Area of the triangle = bh/2
A = (15 × 30)/2
A = 450/2
A = 225 cm2
Therefore the area of the above figure is 225 cm2

Question 6.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 30
_______ cm2

Answer: 450

Explanation:
Given,
Base = 20 cm
Height = 45 cm
Area of the triangle = bh/2
A = (20 × 45)/2
A = 900/2
A = 450 cm2
Therefore the area of the above figure is 450 cm2

Problem Solving

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 543 Q7

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 543 Q8

Question 9.
Draw 3 triangles on grid paper. Draw appropriate parallelograms to support the formula for the area of the triangle. Tape your drawings to this page.
Type below:
_______________

Lesson Check – Page No. 544

Question 1.
What is the area of a triangle with a height of 14 feet and a base of 10 feet?
_______ ft2

Answer: 70

Explanation:
Given,
Base = 10 feet
Height = 14 feet
Area of the triangle = bh/2
A = (14 × 10)/2
A = 140/2
A = 70 ft2
Therefore the area of the triangle is 70 ft2

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 544 Q2

Spiral Review

Question 3.
Jack bought 3 protein bars for a total of $4.26. Which equation could be used to find the cost c in dollars of each protein bar?
Type below:
_______________

Answer: 3c = 4.26

Explanation:
Jack bought 3 protein bars for a total of $4.26.
c represents the cost of each protein bar
3c = 4.26

Question 4.
Coach Herrera is buying tennis balls for his team. He can solve the equation 4c = 92 to find how many cans c of balls he needs. How many cans does he need?
_______ cans

Answer: 23

Explanation:
Coach Herrera is buying tennis balls for his team.
4c = 92
c = 92/4
c = 23
Therefore he need 23 cans.

Question 5.
Sketch the graph of y ≤ 7 on a number line.
Type below:
_______________

Answer:
Go Math Grade 6 Answer Key Chapter 10 solution img-1

Question 6.
A square photograph has a perimeter of 20 inches. What is the area of the photograph?
_______ in.2

Answer: 25

Explanation:
A square photograph has a perimeter of 20 inches.
p = 4s
20 = 4s
s = 20/4
s = 5 in.
Area of the square is s × s
A = 5 × 5 = 25
Thus the area of square photograph = 25 in.2

Share and Show – Page No. 547

Question 1.
Find the area of the triangle.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 31
A = _______ cm2

Answer: 56

Explanation:
B = 14 cm
H = 8 cm
Area of the triangle = bh/2
A = (14 × 8)/2
A = 14 × 4
A = 56 sq. cm
Thus the area of the above figure is 56 cm2

Question 2.
The area of the triangle is 132 in.2. Find the height of the triangle
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 32
h = _______ in.

Answer: 12

Explanation:
B = 22 in.
H = ?
A = 132 in.2
Area of the triangle = bh/2
132 sq. in  = 22 in × h
h = 132 sq. in/22 in
h = 12 in
Thus the height of the above figure is 12 in.

Find the area of the triangle.

Question 3.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 33
A = _______ mm2

Answer: 540

Explanation:
B = 27 mm
H = 40 mm
Area of the triangle = bh/2
A = (27 × 40)/2
A = 27 × 20 = 540
A = 540 mm2
Therefore the area of the above figure is 540 mm2

Question 4.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 34
A = _______ mm2

Answer: 11

Explanation:
B = 5.5 mm
H = 4 mm
Area of the triangle = bh/2
A = (5.5 mm × 4 mm)/2
A = 5.5 mm × 2 mm
A = 11 mm2
Therefore the area of the above figure is 11 mm2

On Your Own

Find the unknown measurement for the figure.

Question 5.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 35
h = _______ in.

Answer: 21

Explanation:
B = 5 in
H =?
A = 52.5 sq. in
Area of the triangle = bh/2
52.5 sq. in = (5 × h)/2
52.5 sq. in × 2 = 5h
h = 21 in
Thus the height of the above figure is 21 in

Question 6.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 36
h = _______ cm

Answer: 4.3

Explanation:
B = 80 mm = 8 cm
H = ?
A = 17.2 sq. cm
Area of the triangle = bh/2
17.2 sq. cm = (8 cm × h)/2
17.2 × 2 = 8 × h
h = 4.3 cm
Thus the height of the above figure is 4.3 cm

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 547 Q7

Unlock the Problem – Page No. 548

Question 8.
Alani is building a set of 4 shelves. Each shelf will have 2 supports in the shape of right isosceles triangles. Each shelf is 14 inches deep. How many square inches of wood will she need to make all of the supports?
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 37
a. What are the base and height of each triangle?
Base: ___________ in.
Height: ___________ in.

Answer:
Base: 14 in
Height: 14 in

Explanation:
Given that,
Each shelf is 14 inches deep.
Height = 14 inches
By seeing the above figure we can say that the base of the shelves is 14 inches
Base = 14 inches

Question 8.
b. What formula can you use to find the area of a triangle?
Type below:
_______________

Answer: The formula to find the Area of the triangle = bh/2

Question 8.
c. Explain how you can find the area of one triangular support.
Type below:
_______________

Answer:
We can find the area of one triangle support by substituting the base and height in the formula.
A = (14 × 14)/2
A = 98 sq. in

Question 8.
d. How many triangular supports are needed to build 4 shelves?
_______ supports

Answer: 8
By seeing the above figure we can say that 8 triangular supports are needed to build 4 shelves.

Question 8.
e. How many square inches of wood will Alani need to make all the supports?
_______ in.2

Answer: 784

Explanation:
The depth of each shelf made by Alamo is 14 inches.
So the base of the right isosceles triangular supporter is 14 inches.
So one equal side is 14 cm. Now by using the Pythagoras theorem we can calculate the other side of the supporter = = 19.8 inches.
The area of the right isosceles triangle is given by × base ×height. Here the base and height are equal to 14 inches.
Therefore the area of each right isosceles triangular supporter is
A = (14 × 14)/2
A = 98 sq. in
Each shelf would require two such supporters and there are 4 such shelves. Thus the total number of supporters required is 8.
Square inches of wood necessary for 8 right isosceles triangular supporters = 98 × 8 = 784 square inches.

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 548 Q9

Question 10.
The area of a triangle is 30 ft2.
For numbers, 10a–10d, select Yes or No to tell if the dimensions given could be the height and base of the triangle.
10a. h = 3, b = 10
10b. h = 3, b = 20
10c. h = 5, b = 12
10d. h = 5, b = 24
10a. ___________
10b. ___________
10c. ___________
10d. ___________

Answer:
10a. No
10b. yes
10c. Yes
10d. No

Explanation:
The area of a triangle is 30 ft2.
10a. h = 3, b = 10
Area of the triangle = bh/2
A = (3 × 10)/2
A = 15 ft2.
Thus the answer is no.
10b. h = 3, b = 20
Area of the triangle = bh/2
A = (3 × 20)/2
A = 30 ft2.
Thus the answer is yes.
10c. h = 5, b = 12
Area of the triangle = bh/2
A = (5 × 12)/2
A = 30 ft2.
Thus the answer is yes.
10d. h = 5, b = 24
Area of the triangle = bh/2
A = (5 × 24)/2
A = 60 ft2.
Thus the answer is no.

Area of Triangles – Page No. 549

Find the area.

Question 1.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 38
_______ in.2

Answer: 45

Explanation:
Given,
Base = 15 in.
Height = 6 in.
Area of the triangle = bh/2
A = (15 × 6)/2
A = 90/2
A = 45 in.2

Question 2.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 39
_______ m2

Answer: 0.36

Explanation:
Given,
Base = 1.2 m
Height = 0.6 m
Area of the triangle = bh/2
A = (1.2 × 0.6)/2
A = 0.72/2
A = 0.36 m2

Question 3.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 40
_______ ft2

Answer: 6

Explanation:
Given,
Base = 4 1/2 ft
Height = 2 2/3 ft
Area of the triangle = bh/2
A = (4 1/2 × 2 2/3)/2
A = 12/2
A = 6 ft2

Find the unknown measurement for the triangle.

Question 4.
A = 0.225 mi2
b = 0.6 mi
h = ?
h = _______ mi

Answer: 0.75

Explanation:
Given,
A = 0.225 mi2
b = 0.6 mi
h = ?
Area of the triangle = bh/2
0.225 = (0.6 × h)/2
0.450 = 0.6 × h
h = 0.450/0.6
h = 0.75 mi

Question 5.
A = 4.86 yd2
b = ?
h = 1.8 yd
b = _______ yd

Answer: 5.4 yd

Explanation:
Given,
A = 4.86 yd2
b = ?
h = 1.8 yd
Area of the triangle = bh/2
4.86 yd2 = (b × 1.8 yd)/2
4.86 × 2 = b × 1.8
9.72 = b × 1.8
b = 9.72/1.8
b = 5.4 yd

Question 6.
A = 63 m2
b = ?
h = 12 m
b = _______ m

Answer: 10.5

Explanation:
Given,
A = 63 m2
b = ?
h = 12 m
Area of the triangle = bh/2
63 = (b × 12)/2
63 = b × 6
b = 63/6
b = 10.5 m

Question 7.
A = 2.5 km2
b = 5 km
h = ?
h = _______ km

Answer: 1

Explanation:
Given,
A = 2.5 km2
b = 5 km
h = ?
Area of the triangle = bh/2
2.5 = (5 km × h)/2
2.5 km2 = 2.5 km × h
h = 2.5/2.5
h = 1 km

Problem Solving

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 549 Q8

Question 9.
Alicia is making a triangular sign for the school play. The area of the sign is 558 in.2. The base of the triangle is 36 in. What is the height of the triangle?
_______ in.

Answer: 31

Explanation:
Given,
Alicia is making a triangular sign for the school play.
The area of the sign is 558 in.2
The base of the triangle is 36 in.
Area of the triangle = bh/2
558 = (36 × h)/2
558 = 18 × h
h = 558/18
h = 31 inches

Question 10.
Describe how you would find how much grass seed is needed to cover a triangular plot of land.
Type below:
_______________

Answer:

You will need to find the area
A=height multiplied by the base divided by 2
Area of the triangle = bh/2

Lesson Check – Page No. 550

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 550 Q1

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 550 Q2

Spiral Review

Question 3.
Tina bought a t-shirt and sandals. The total cost was $41.50. The t-shirt cost $8.95. The equation 8.95 + c = 41.50 can be used to find the cost c in dollars of the sandals. How much did the sandals cost?
$ _______

Answer: $32.55

Explanation:
Tina bought a t-shirt and sandals.
The total cost was $41.50.
The t-shirt cost $8.95.
8.95 + c = 41.50
c = 41.50 – 8.95
c = $32.55

Question 4.
There are 37 paper clips in a box. Carmen places more paper clips in the box. Write an equation to show the total number of paper clips p in the box after Carmen places n more paper clips in the box.
Type below:
_______________

Answer: 37 + n = p

Explanation:
There are 37 paper clips in a box. Carmen places more paper clips in the box.
n represents number of paper clips in the box
The equation is 37 + n = p

Question 5.
Name another ordered pair that is on the graph of the equation represented by the table.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 41
Type below:
_______________

Answer: The ordered pairs are (1, 6), (2, 12), (3, 18), (4, 16)

Question 6.
Find the area of the triangle that divides the parallelogram in half.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 42
_______ cm2

Answer: 58.5

Explanation:
Given,
b = 13 cm
h = 9 cm
Area of the triangle = bh/2
A = (13 × 9)/2
A = 117/2
A = 58.5 cm2

Share and Show – Page No. 553

Question 1.
Trace and cut out two copies of the trapezoid. Arrange the trapezoids to form a parallelogram. Find the areas of the parallelogram and one trapezoid using square units
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 43
Type below:
_______________

Answer:
Figure 1:
Base 1 = 3 units
Base 2= 7 units
Height = 4 units
Area of the trapezium = (b1 + b2)h/2
A = (3 + 7)4/2
A = 10 × 2
A = 20 sq. units
Figure 2:
Base 1 = 7 units
Base 2= 3 units
Height = 4 units
Area of the trapezium = (b1 + b2)h/2
A = (7 + 3)4/2
A = 10 × 2
A = 20 sq. units

Find the area of the trapezoid.

Question 2.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 44
_______ cm2

Answer: 40

Explanation:
Base 1 = 6 cm
Base 2 = 10 cm
Height = 5 cm
We know that the Area of the trapezium is the sum of bases into height divided by 2.
Area of the trapezium = (b1 + b2)h/2
A = (6 cm + 10 cm)5 /2
A = (16 × 5)/2
A = 40 sq. cm

Question 3.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 45
_______ in.2

Answer: 48

Explanation:
b1 = 3 in
b2 = 9 in.
h = 8 in.
We know that the Area of the trapezium is the sum of bases into height divided by 2.
Area of the trapezium = (b1 + b2)h/2
A = (3 + 9)8/2
A = 12 × 4
A = 48 sq. in

Question 4.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 46
_______ ft2

Answer: 64

Explanation:
b1 = 11 ft
b2 = 5 ft
h = 8 ft
We know that the Area of the trapezium is the sum of bases into height divided by 2.
Area of the trapezium = (b1 + b2)h/2
A = (11 + 5)8/2
A = 16 × 4
A = 64 sq. ft

Question 5.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 47
_______ cm2

Answer: 266

Explanation:
b1 = 16 cm
b2 = 22 cm
h = 14 cm
We know that the Area of the trapezium is the sum of bases into height divided by 2.
Area of the trapezium = (b1 + b2)h/2
A = (16 + 22)14/2
A = 38 × 7
A = 266 sq. cm

Question 6.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 48
_______ mm2

Answer: 71.5

Explanation:
b1 = 8 mm
b2 = 14 mm
h = 6.5 mm
We know that the Area of the trapezium is the sum of bases into height divided by 2.
Area of the trapezium = (b1 + b2)h/2
A = (8 + 14)6.5/2
A = 11 × 6.5
A = 71.5 sq. mm

Question 7.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 49
_______ in.2

Answer: 31.5

Explanation:
b1 = 3 1/2 in.
b2 = 8 1/2 in.
h = 5 1/4 in.
We know that the Area of the trapezium is the sum of bases into height divided by 2.
Area of the trapezium = (b1 + b2)h/2
b = 3 1/2 + 8 1/2
b = 12
A = 5 1/4 × 12/2
A = 5 1/4 × 6
A = 31.5 sq. in

Problem Solving + Applications

Question 8.
Describe a Method Explain one way to find the height of a trapezoid if you know the area of the trapezoid and the length of both bases.
Type below:
_______________

Answer:
1) Add the length of both bases: [Total Length = Length 1 + Length 2]
2) Divide the length that you found by 2. [Average Length = Total Length ÷ 2]
3) Divide the Area by the length found [Height = Area ÷ average length]

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 553 Q9

What’s the Error? – Page No. 554

Question 10.
Except for a small region near its southeast corner, the state of Nevada is shaped like a trapezoid. The map at the right shows the approximate dimensions of the trapezoid. Sabrina used the map to estimate the area of Nevada.
Look at how Sabrina solved the problem. Find her error.
Two copies of the trapezoid can be put together to form a rectangle.
length of rectangle: 200 + 480 = 680 mi
width of rectangle: 300 mi
A = lw
A = 680 × 300
A = 204,000
The area of Nevada is about 204,000 square miles.
Describe the error. Find the area of the trapezoid to estimate the area of Nevada.
Type below:
_______________

Answer:
The area of Nevada is she didn’t divide by 2.
Area of the trapezium = (b1 + b2)h/2
A = (200 + 480)300/2
A = 680 × 150
A = 102000 sq. miles

Question 11.
A photo was cut in half at an angle. What is the area of one of the cut pieces?
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 50
_______ in.2

Answer: 30

Explanation:
b1= 3 in
b2 = 7 in
h = 6 in.
Area of the trapezium = (b1 + b2)h/2
A = (3 + 7)6/2
A = 10 × 3
A = 30 sq. in
Thus the area of the trapezium is 30 in.2

Explore Area of Trapezoids – Page No. 555

Question 1.
Trace and cut out two copies of the trapezoid. Arrange the trapezoids to form a parallelogram. Find the areas of the parallelogram and the trapezoids using square units.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 51
Type below:
_______________

Answer:
Figure 1:
b1 = 2 units
b2 = 6 units
h = 3 units
Area of the trapezium = (b1 + b2)h/2
A = (2 + 6)3/2
A = (8)(3)/2
A = 24/2 = 12
A = 12 sq. units
Figure 2:
b1 = 6 units
b2 = 2 units
h = 3 units
Area of the trapezium = (b1 + b2)h/2
A = (6 + 2)3/2
A = (8)(3)/2
A = 24/2 = 12
The area of figure 2 is 12 sq. units

Find the area of the trapezoid.

Question 2.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 52
_______ in.2

Answer: 38.5

Explanation:
Given,
b1 = 9 in
b2 = 2 in
h = 7 in
Area of the trapezium = (b1 + b2)h/2
A = (9 + 2)7/2
A = (11 × 7)/2
A = 77/2 = 38.5 in.2

Question 3.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 53
_______ yd2

Answer: 3600

Explanation:
Given,
b1 = 24 yd
b2 = 48 yd
h = 100 yd
Area of the trapezium = (b1 + b2)h/2
A = (24 + 48)100/2
A = 72 × 50
A = 3600 yd2

Question 4.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 54
_______ ft2

Answer: 64

Explanation:
Given,
b1 = 4.5 ft
b2 = 11.5 ft
h = 8 ft
Area of the trapezium = (b1 + b2)h/2
A = (4.5 + 11.5)8/2
A = 16 × 4
A = 64 sq. ft

Problem Solving

Question 5.
A cake is made out of two identical trapezoids. Each trapezoid has a height of 11 inches and bases of 9 inches and 14 inches. What is the area of one of the trapezoid pieces?
_______ in.2

Answer: 126.5

Explanation:
Given,
A cake is made out of two identical trapezoids.
Each trapezoid has a height of 11 inches and bases of 9 inches and 14 inches.
Area of the trapezium = (b1 + b2)h/2
A = (9 + 14)11/2
A = 23 × 11/2
A = 126.5 in.2

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 555 Q6

Question 7.
Find the area of a trapezoid that has bases that are 15 inches and 20 inches and a height of 9 inches.
_______ in.2

Answer: 157.5

Explanation:
b1 = 15 inches
b2 = 20 inches
h = 9 inches
Area of the trapezium = (b1 + b2)h/2
A = (15 + 20)9/2
A = (35 × 9)/2
A = 157.5 sq. in

Lesson Check – Page No. 556

Question 8.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 55
_______ yd2

Answer: 84

Explanation:
b1 = 9 yd
b2 = 15 yd
h = 7 yd
Area of the trapezium = (b1 + b2)h/2
A = (9 + 15)7/2
A = 24 × 3.5
A = 84 sq. yd

Question 2.
Maggie colors a figure in the shape of a trapezoid. The trapezoid is 6 inches tall. The bases are 4.5 inches and 8 inches. What is the area of the figure that Maggie colored?
_______ in.2

Answer: 37.5

Explanation:
Maggie colors a figure in the shape of a trapezoid.
The trapezoid is 6 inches tall.
The bases are 4.5 inches and 8 inches.
b1 = 4.5 in
b2 = 8 in
h = 6 in
Area of the trapezium = (b1 + b2)h/2
A = (4.5 in + 8 in)6/2
A = 12.5 in × 3
A = 37.5 sq. in

Spiral Review

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 556 Q3

Question 4.
Ginger makes pies and sells them for $14 each. Write an equation that represents the situation, if y represents the money that Ginger earns and x represents the number of pies sold.
Type below:
_______________

Answer: y = 14x

Explanation:
Ginger makes pies and sells them for $14 each.
y represents the money that Ginger earns
x represents the number of pies sold
The equation is y = 14x

Question 5.
What is the equation for the graph shown below?
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 56
Type below:
_______________

Answer: y = 2x
By seeing the graph we can say that y = 2x

Question 6.
Cesar made a rectangular banner that is 4 feet by 3 feet. He wants to make a triangular banner with the same area as the other. The triangular banner will have a base of 4 feet. What should its height be?
_______ feet

Answer: 6

Explanation:
6 Because 4×3=12 and (4× 6)/2=12

Share and Show – Page No. 559

Question 1.
Find the area of the trapezoid.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 57
A = _______ cm2

Answer: 18

Explanation:
Given,
b1 = 6 cm
b2 = 3 cm
h = 4 cm
We know that,
Area of the trapezium = (b1 + b2)h/2
A = (6 cm + 3 cm)4 cm/2
A = 9 cm × 2 cm
A = 18 sq. cm
Therefore the area of the trapezoid is 18 cm2

Question 2.
The area of the trapezoid is 45 ft2. Find the height of the trapezoid.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 58
h = _______ ft

Answer: 5

Explanation:
b1 = 10 ft
b2 = 8 ft
The area of the trapezoid is 45 ft2
We know that,
Area of the trapezium = (b1 + b2)h/2
45 ft2 = (10 ft + 8 ft)h/2
90 = 18 × h
h = 90/18
h = 5 ft
Thus the height of the above figure is 5 ft.

Question 3.
Find the area of the trapezoid.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 59
_______ mm2

Answer: 540

Explanation:
b1 = 17 mm
b2 = 43 mm
h = 18 mm
We know that,
Area of the trapezium = (b1 + b2)h/2
A = (17 + 43)18/2
A = 60 mm × 9 mm
A = 540 sq. mm
Thus the area of the trapezoid is 540 mm2

On Your Own

Find the area of the trapezoid.

Question 4.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 60
A = _______ in.2

Answer: 266

Explanation:
Given,
b1 = 17 in
b2 = 21 in
h = 14 in
We know that,
Area of the trapezium = (b1 + b2)h/2
A = (17 in + 21 in)14/2
A = 38 in × 7 in
A = 266 sq. in
Therefore Area of the trapezium is 266 in.2

Question 5.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 61
A = _______ m2

Answer: 25.2 m2

Explanation:
Given,
b1 = 9.2 m
b2 = 2.8 m
h = 4.2 m
We know that,
Area of the trapezium = (b1 + b2)h/2
A = (9.2 + 2.8)4.2/2
A = 12 × 2.1
A = 25.2 sq. m
Therefore the area of the trapezium is 25.2 m2

Find the height of the trapezoid.

Question 6.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 62
h = _______ in.

Answer: 25

Explanation:
Given,
b1 = 27.5 in
b2 = 12.5 in
h = ?
A = 500 sq. in
We know that,
Area of the trapezium = (b1 + b2)h/2
500 sq. in = (27.5 in + 12.5 in)h/2
500 sq. in = 40 × h/2
500 sq. in = 20h
h = 500/20
h = 25 inches
Thus the height of the above figure is 25 inches.

Question 7.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 63
h = _______ cm

Answer: 15

Explanation:
A = 99 sq. cm
b1 = 3.2 cm
b2 = 10 cm
h = ?
We know that,
Area of the trapezium = (b1 + b2)h/2
99 sq. cm = (3.2 cm+ 10 cm)h/2
99 sq. cm = (13.2 cm)h/2
99 sq. cm = 6.6 × h
h = 99 sq. cm/6.6 cm
h = 15 cm

Problem Solving + Applications – Page No. 560

Use the diagram for 8–9.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 64

Question 8.
A baseball home plate can be divided into two trapezoids with the dimensions shown in the drawing. Find the area of home plate.
_______ in.2

Answer: 21.75

Explanation:
The bases of the trapezoid area are 8.5 in and 17 in and the height is 8.5 in.
We know that,
Area of the trapezium = (b1 + b2)h/2
A = 1/2 (8.5 + 17)8.5
A = (25.5)(8.5)/2
A = 1/2 × 216.75
The area of the home plate is double the area of a trapezoid.
So, the area of the home plate is 216.75 sq. in.

Question 9.
Suppose you cut the home plate along the dotted line and rearranged the pieces to form a rectangle. What would the dimensions and the area of the rectangle be?
Type below:
_______________

Answer:
The dimensions of the rectangle would be 25.5 in by 8.5 in.
The area would be 216.75 sq. in.

Question 10.
A pattern used for tile floors is shown. A side of the inner square measures 10 cm, and a side of the outer square measures 30 cm. What is the area of one of the yellow trapezoid tiles?
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 65
_______ cm2

Answer: 200 sq. cm

Explanation:
A side of the inner square measures 10 cm, and a side of the outer square measures 30 cm.
The bases of the trapezoid are 10 cm and 30 cm and the height of the trapezoid is 10 cm.
We know that,
Area of the trapezium = (b1 + b2)h/2
A = (10 + 30)10/2
A = 40 cm × 5 cm
A = 200 sq. cm
So, the area of one of the yellow trapezoid tiles is 200 sq. cm

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 560 Q11

Question 12.
Which expression can be used to find the area of the trapezoid? Mark all that apply.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 66
Options:
a. \(\frac{1}{2}\) × (4 + 1.5) × 3.5
b. \(\frac{1}{2}\) × (1.5 + 3.5) × 4
c. \(\frac{1}{2}\) × (4 + 3.5) × 1.5
d. \(\frac{1}{2}\) × (5) × 4

Answer: \(\frac{1}{2}\) × (1.5 + 3.5) × 4

Explanation:
b1 = 3.5 ft
b2 = 1.5 ft
h = 4 ft
We know that,
Area of the trapezium = (b1 + b2)h/2
A = (3.5 ft + 1.5 ft)4ft/2
A = \(\frac{1}{2}\) × (1.5 + 3.5) × 4
Thus the correct answer is option B.

Area of Trapezoids – Page No. 561

Find the area of the trapezoid.

Question 1.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 67
_______ cm2

Answer: 252 cm2

Explanation:
Given that,
long base b1 = 17 cm
short base b2 = 11 cm
h = 18 cm
We know that,
The Area of the trapezium = (b1 + b2)h/2
A = (17 cm + 11 cm)18 cm/2
A = 28 cm × 9 cm
A = 252 cm2
Thus the area of the trapezium for the above figure is 252 cm2

Question 2.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 68
_______ ft2

Answer: 30 ft2

Explanation:
Given,
b1 = 6.5 ft
b2 = 5.5 ft
h = 5 ft
We know that,
The Area of the trapezium = (b1 + b2)h/2
A = (6.5 + 5.5)5/2
A = 12 ft × 2.5 ft
A = 30 sq. ft
Therefore the area of the trapezium is 30 ft2

Question 3.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 69
_______ cm2

Answer: 0.08 cm2

Explanation:
Given,
b1 = 0.6 cm
b2 = 0.2 cm
h = 0.2 cm
We know that,
The Area of the trapezium = (b1 + b2)h/2
A = (0.6 cm + 0.2 cm)0.2 cm/2
A = 0.8 cm × 0.1 cm
A = 0.08 sq. cm
Thus the area of the trapezium is 0.08 sq. cm

Question 4.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 70
_______ in.2

Answer: 37.5 in.2

Explanation:
Given,
b1 = 5 in
b2 = 2 1/2
h = 10 in
We know that,
The Area of the trapezium = (b1 + b2)h/2
A = (5 in + 2 1/2 in)10/2
A = 7 1/2 × 5
A = 37.5 sq. in
Thus the area of the trapezium is 37.5 in.2

Problem Solving

Question 5.
Sonia makes a wooden frame around a square picture. The frame is made of 4 congruent trapezoids. The shorter base is 9 in., the longer base is 12 in., and the height is 1.5 in. What is the area of the picture frame?
_______ in.2

Answer: 63

Explanation:
Given,
Sonia makes a wooden frame around a square picture.
The frame is made of 4 congruent trapezoids.
The shorter base is 9 in., the longer base is 12 in., and the height is 1.5 in.
We know that,
The Area of the trapezium = (b1 + b2)h/2
A = (9 in + 12 in)1.5/2
A = 21 in × 1.5 in/2
A = 63 sq. in
Thus the area of the trapezium is 63 in.2

Question 6.
Bryan cuts a piece of cardboard in the shape of a trapezoid. The area of the cutout is 43.5 square centimeters. If the bases are 6 centimeters and 8.5 centimeters long, what is the height of the trapezoid?
_______ cm

Answer: 6 cm

Explanation:
Given,
Bryan cuts a piece of cardboard in the shape of a trapezoid.
The area of the cutout is 43.5 square centimeters.
If the bases are 6 centimeters and 8.5 centimeters long.
We know that,
The Area of the trapezium = (b1 + b2)h/2
43.5 sq. cm = (6 + 8.5)h/2
43.5 × 2 = 14.5 × h
h = 6 cm
Therefore the height of the trapezoid is 6 cm.

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 561 Q7

Lesson Check – Page No. 562

Question 1.
Dominic is building a bench with a seat in the shape of a trapezoid. One base is 5 feet. The other base is 4 feet. The perpendicular distance between the bases is 2.5 feet. What is the area of the seat?
_______ ft2

Answer: 11.25 sq. ft

Explanation:
Given,
Dominic is building a bench with a seat in the shape of a trapezoid.
One base is 5 feet. The other base is 4 feet.
The perpendicular distance between the bases is 2.5 feet.
We know that,
The Area of the trapezium = (b1 + b2)h/2
A = (5 ft + 4 ft)2.5/2
A = 4.5 ft × 2.5 ft
A = 11.25 sq. ft
Thus the area of the seat is 11.25 sq. ft

Question 2.
Molly is making a sign in the shape of a trapezoid. One base is 18 inches and the other is 30 inches. How high must she make the sign so its area is 504 square inches?
_______ in.

Answer: 21 in.

Explanation:
Given,
Molly is making a sign in the shape of a trapezoid.
One base is 18 inches and the other is 30 inches.
A = 504 sq. in
We know that,
The Area of the trapezium = (b1 + b2)h/2
504 sq. in = (18 + 30)h/2
504 sq. in = 24 × h
h = 504 sq. in÷ 24 in
h = 21 inches
Thus the height of the trapezoid is 21 inches.

Spiral Review

Question 3.
Write these numbers in order from least to greatest.
3 \(\frac{3}{10}\)     3.1       3 \(\frac{1}{4}\)
Type below:
_______________

Explanation:
First, convert the fraction into a decimal.
3 \(\frac{3}{10}\) = 3.3
3 \(\frac{1}{4}\) = 3.25
Now write the numbers from least to greatest.
3.1 3.25 3.3

Question 4.
Write these lengths in order from least to greatest.
2 yards       5.5 feet        70 inches
Type below:
_______________

Answer: 5.5 feet, 70 inches, 2 yards

Explanation:
First, convert from inches to feet.
1 feet = 12 inches
70 inches = 5.8 ft
1 yard = 3 feet
2 yards = 2 × 3 ft
2 yards = 6 feet
Now write the numbers from least to greatest.
5.5 ft 5.8 ft 6 ft

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 562 Q5

Question 6.
Brian frosted a cake top shaped like a parallelogram with a base of 13 inches and a height of 9 inches. Nancy frosted a triangular cake top with a base of 15 inches and a height of 12 inches. Which cake’s top had the greater area? How much greater was it?
Type below:
_______________

Explanation:
Parallelogram Formula = Base × Height
A=bh
A=13 × 9=117 in
Triangle Formula=
A=1/2bh
A=1/2 × 15 × 12 = 90 in
Brian’s cake top has a greater area, and by 27 inches.

Mid-Chapter Checkpoint – Vocabulary – Page No. 563

Choose the best term from the box to complete the sentence.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 71

Question 1.
A _____ is a quadrilateral that always has two pairs of parallel sides.
Type below:
_______________

Answer: A parallelogram is a quadrilateral that always has two pairs of parallel sides.

Question 2.
The measure of the number of unit squares needed to cover a surface without any gaps or overlaps is called the _____.
Type below:
_______________

Answer: The measure of the number of unit squares needed to cover a surface without any gaps or overlaps is called the Area.

Question 3.
Figures with the same size and shape are _____.
Type below:
_______________

Answer: Figures with the same size and shape are Congruent.

Concepts and Skills

Find the area.

Question 4.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 72
_______ cm2

Answer: 19.38

Explanation:
b = 5.7 cm
h = 3.4 cm
Area of parallelogram = bh
A = 5.7 cm × 3.4 cm
A = 19.38 cm2
Thus the area of the parallelogram is 19.38 cm2

Question 5.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 73
_______ \(\frac{□}{□}\) in.2

Answer: 42 \(\frac{1}{4}\) in.2

Explanation:
b = 6 \(\frac{1}{2}\)
h = 6 \(\frac{1}{2}\)
Area of parallelogram = bh
A = 6 \(\frac{1}{2}\) × 6 \(\frac{1}{2}\)
A = 42 \(\frac{1}{4}\) in.2
Thus the area of the parallelogram is 42 \(\frac{1}{4}\) in.2

Question 6.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 74
_______ mm2

Answer: 57.4

Explanation:
b = 14 mm
h = 8.2 mm
A = bh/2
A = (14 mm × 8.2 mm)/2
A = 57.4 mm2

Question 7.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 75

Answer: 139.5

Explanation:
b1 = 13 cm
b2= 18 cm
h = 9 cm
Area of the trapezium = (b1 + b2)h/2
A = (13 + 18)9/2
A = 31 × 4.5
A = 139.5 sq. cm

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 563 Q8

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 563 Q9

Page No. 564

Question 10.
The height of a parallelogram is 3 times the base. The base measures 4.5 cm. What is the area of the parallelogram?
_______ cm2

Answer: 60.75

Explanation:
The height of a parallelogram is 3 times the base. The base measures 4.5 cm.
A = bh
h = 3 × 4.5
h = 13.5 cm
b = 4.5 cm
A = 13.5 cm × 4.5 cm
A = 60.75 cm2

Question 11.
A triangular window pane has a base of 30 inches and a height of 24 inches. What is the area of the window pane?
_______ in.2

Answer: 360

Explanation:
A triangular window pane has a base of 30 inches and a height of 24 inches.
b = 30 in
h = 24 in
A = bh/2
A = (30 × 24)/2
A = 30 × 12
A = 360 in.2

Question 12.
The courtyard behind Jennie’s house is shaped like a trapezoid. The bases measure 8 meters and 11 meters. The height of the trapezoid is 12 meters. What is the area of the courtyard?
_______ m2

Answer: 114

Explanation:
Given,
The courtyard behind Jennie’s house is shaped like a trapezoid.
The bases measure 8 meters and 11 meters.
The height of the trapezoid is 12 meters.
Area of the trapezium = (b1 + b2)h/2
A = (8 + 11)12/2
A = 19 × 6
A = 114 m2

Question 13.
Rugs sell for $8 per square foot. Beth bought a 9-foot-long rectangular rug for $432. How wide was the rug?
_______ feet

Answer: 6 feet

Explanation:
If you know the rugs sell for 8$ per square foot and the total spend was $432.
You divide 432 by 8 to find the total number of square feet of the rug.
To find the total square foot you find the area.
So the area of a rectangle is L × W. So 54 = 9 × width.
So just divide 54 by 9 and you get the width of the rug.
The width is 6 feet.
Now you check. A nine by 6 rug square foot is 54. and then times by 8 and you get 432 total.

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 564 Q14

Share and Show – Page No. 567

Find the area of the regular polygon.

Question 1.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 76
_______ cm2

Answer: 120

Explanation:
b = 5 cm
h = 6 cm
Number of congruent figures inside the figure: 8
Area of each triangle = bh/2
A = (5 cm)(6 cm)/2
A = 15 sq. cm
Now to find the area of the regular polygon we have to multiply the area of each triangle and number of congruent figures.
Area of regular octagon = 8 × 15 sq. cm
A = 120 sq. cm
Therefore the area of the regular octagon for the above figure = 120 sq. cm

Question 2.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 77
_______ m2

Answer: 60

Explanation:
Given,
b = 6 m
h = 4 m
Number of congruent figures inside the figure: 5
Area of each triangle = bh/2
A = (6 m)(4 m)/2
A = 12 sq. m
Now to find the area of the regular polygon we have to multiply the area of each triangle and number of congruent figures.
Area of regular pentagon = 5 × 12 sq. m
A = 60 sq. m
Therefore the area of the above figure is 60 sq. m.

Question 3.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 78
_______ mm2

Answer: 480

Explanation:
Given,
b = 8 mm
h = 12 mm
Number of congruent figures inside the figure: 10
Area of each triangle = bh/2
A = (12 mm)(8 mm)/2
A = 48 sq. mm
Now to find the area of the regular polygon we have to multiply the area of each triangle and number of congruent figures.
Area of regular polygon = 10 × 48 sq. mm
A = 480 sq. mm
Therefore, the area of the regular polygon is 480 sq. mm

On Your Own

Find the area of the regular polygon.

Question 4.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 79
_______ cm2

Answer: 168

Explanation:
Given,
b = 8 cm
h = 7 cm
Number of congruent figures inside the figure: 6
Area of each triangle = bh/2
A = (8 cm)(7 cm)/2
A = 28 sq. cm
Now to find the area of the regular polygon we have to multiply the area of each triangle and number of congruent figures.
Area of regular hexagon = 6 × 28 sq. cm
A = 168 sq. cm
Thus the area of the above figure is 168 sq. cm

Question 5.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 80
_______ in.2

Answer: 6020

Explanation:
Given,
b = 28 in
h = 43 in
Number of congruent figures inside the figure: 10
Area of each triangle = bh/2
A = (28 in)(43 in)/2
A = 602 sq. in
Now to find the area of the regular polygon we have to multiply the area of each triangle and number of congruent figures.
Area of regular polygon = 10 × Area of each triangle
A = 10 × 602 sq. in
A = 6020 sq. in
Therefore the area of the regular polygon is 6020 sq. in

Question 6.
Explain A regular pentagon is divided into congruent triangles by drawing a line segment from each vertex to the center. Each triangle has an area of 24 cm2. Explain how to find the area of the pentagon
Type below:
_______________

Answer: 120

Explanation:
Given,
Each triangle has an area of 24 cm2.
Pentagon has 5 sides. The number of congruent figures is 5.
Now to find the area of the regular polygon we have to multiply the area of each triangle and number of congruent figures.
Area of regular pentagon = 5 × 24 sq. cm
A = 120 sq. cm
Therefore the area of the pentagon is 120 sq. cm

Page No. 568

Question 7.
Name the polygon and find its area. Show your work.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 81
_______ in.2

Answer: 76.8 sq. in

Explanation:
b = 4 in
h = 4.8 in
Number of configured figures of the regular polygon: 8
Area of the triangle = bh/2
A = (4)(4.8)/2
A = 9.6 sq. in.
Now to find the area of the regular polygon we have to multiply the area of each triangle and number of congruent figures.
Area of regular polygon = 8 × area of the triangle
A = 8 × 9.6 sq. in.
A = 76.8 sq. in
Thus the area of the regular polygon is 76.8 sq. in.

Regular polygons are common in nature

One of the bestknown examples of regular polygons in nature is the small hexagonal cells in honeycombs constructed by honeybees. The cells are where bee larvae grow. Honeybees store honey and pollen in the hexagonal cells. Scientists can measure the health of a bee population by the size of the cells.

Question 8.
Cells in a honeycomb vary in width. To find the average width of a cell, scientists measure the combined width of 10 cells, and then divide by 10.
The figure shows a typical 10-cell line of worker bee cells. What is the width of each cell?
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 82
_______ cm

Answer: 0.52 cm

Explanation:
Since the combined width of 10 cells is 5.2 cm, the width of each cell is 5.2 ÷ 10 = 0.52 cm.

Question 9.
The diagram shows one honeycomb cell. Use your answer to Exercise 8 to find h, the height of the triangle. Then find the area of the hexagonal cell.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 83
Type below:
_______________

Answer: 0.234 sq. cm

Explanation:
The length of the h, the height of the triangle, is half of the width of each cell.
Since the width of each cell is 0.52 cm
h = 0.52 ÷ 2 = 0.26 cm
Area of the triangle = bh/2
A = (0.3)(0.26)/2
A = 0.078/2
A = 0.039
The area of the hexagon is:
6 × 0.039 = 0.234 sq. cm.

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 568 Q10

Area of Regular Polygons – Page No. 569

Find the area of the regular polygon.

Question 1.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 84
_______ mm2

Answer: 168

Explanation:
Given,
b = 8 mm
h = 7 mm
Number of congruent figures inside the figure: 6
Area of each triangle = bh/2
A = (8)(7)/2
A = 28 sq. mm
Now to find the area of regular polygon we have to multiply the area of each triangle and number of congruent figures.
Area of regular polygon = 6 × 28 sq. mm
A = 168 sq. mm
Therefore the area of the regular polygon for the above figure is 168 sq. mm

Question 2.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 85
_______ yd2

Answer: 139.5

Explanation:
Given,
b = 9 yd
h = 6.2 yd
Number of congruent figures inside the figure: 5
Area of each triangle = bh/2
A = (9 yd) (6.2 yd)/2
A = 9 yd × 3.1 yd
A = 27.9 sq. yd
Now to find the area of the regular polygon we have to multiply the area of each triangle and number of congruent figures.
Area of regular polygon = 5 × 27.9 sq. yd
A = 139.5 sq. yd
Thus the area of the regular polygon for the above figure is 139.5 sq. yd.

Question 3.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 86
_______ in.2

Answer: 52.8

Explanation:
Given,
b = 3.3 in
h = 4 in
Number of congruent figures inside the figure: 8
Area of each triangle = bh/2
A = (3.3 in)(4 in)/2
A = 6.6 sq. in
Now to find the area of the regular polygon we have to multiply the area of each triangle and number of congruent figures.
Area of regular polygon = 8 × 6.6 sq. in
A = 52.8 sq. in
The area of the regular polygon is 52.8 sq. in

Problem Solving

Question 4.
Stu is making a stained glass window in the shape of a regular pentagon. The pentagon can be divided into congruent triangles, each with a base of 8.7 inches and a height of 6 inches. What is the area of the window?
_______ in.2

Answer: 130.5

Explanation:
Stu is making a stained glass window in the shape of a regular pentagon.
The pentagon can be divided into congruent triangles, each with a base of 8.7 inches and a height of 6 inches.
Number of congruent figures inside the figure: 5
Area of each triangle = bh/2
A = (8.7 in)(6 in)/2
A = 8.7 in × 3 in
A = 26.1 sq. in.
Now to find the area of the regular polygon we have to multiply the area of each triangle and number of congruent figures.
Area of regular polygon = 5 × 26.1 sq. in
A = 130.5 sq. in
Thus the area of the window is 130.5 sq. in

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 569 Q5

Question 6.
A square has sides that measure 6 inches. Explain how to use the method in this lesson to find the area of the square.
Type below:
_______________

Answer: 36 sq. in

Explanation:
A square has sides that measure 6 inches.
s = 6 in
We know that,
Area of the square = s × s
A = 6 in × 6 in
A = 36 sq. in
Thus the area of the square is 36 sq. in

Lesson Check – Page No. 570

Question 1.
What is the area of the regular hexagon?
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 87
________ \(\frac{□}{□}\) m2

Answer: 30 \(\frac{3}{5}\) m2

Explanation:
Given,
b = 3 \(\frac{2}{5}\) m
h = 3 m
Area of each triangle = bh/2
A = 3 \(\frac{2}{5}\) m × 3/2 m
A = 5.1 sq. m
Now to find the area of the regular polygon we have to multiply the area of each triangle and number of congruent figures.
Area of the regular hexagon = 6 × 5.1 = 30.6
= 30 \(\frac{6}{10}\) m2
= 30 \(\frac{3}{5}\) m2
Therefore the area of the regular hexagon is 30 \(\frac{3}{5}\) m2

Question 2.
A regular 7-sided figure is divided into 7 congruent triangles, each with a base of 12 inches and a height of 12.5 inches. What is the area of the 7-sided figure?
________ in.2

Answer: 525 sq. in

Explanation:
A regular 7-sided figure is divided into 7 congruent triangles, each with a base of 12 inches and a height of 12.5 inches.
Area of each triangle = bh/2
A = (12 in)(12.5 in)/2
A = 12.5 in × 6 in
A = 75 sq. inches
Thus the area of each triangle = 75 sq. in
Now to find the area of the regular polygon we have to multiply the area of each triangle and number of congruent figures.
Area of regular polygon = 7 × 75 sq. in
A = 525 sq. in
Thus the area of the 7-sided figure is 525 sq. in

Spiral Review

Question 3.
Which inequalities have b = 4 as one of its solutions?
2 + b ≥ 2      3b ≤ 14
8 − b ≤ 15     b − 3 ≥ 5
Type below:
_______________

Answer: b − 3 ≥ 5

Explanation:
Substitute b = 4 in the inequality
i. 2 + b ≥ 2
2 + 4 ≥ 2
6 ≥ 2
ii. 3b ≤ 14
3(4) ≤ 14
12 ≤ 14
iii. 8 − b ≤ 15
8 – 4 ≤ 15
4 ≤ 15
iv. b − 3 ≥ 5
4 – 3 ≥ 5
1 ≥ 5
1 is not greater than or equal to 5.

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 570 Q4

Question 5.
What is the area of triangle ABC?
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 88
________ ft2

Answer: 30 ft2

Explanation:
b = 6 ft
h = 10 ft
We know that,
Area of each triangle = bh/2
A = (6 ft)(10 ft)/2
A = 60 sq. ft/2
A = 30 sq. ft
Therefore the area of triangle ABC is 30 sq. ft

Question 6.
Marcia cut a trapezoid out of a large piece of felt. The trapezoid has a height of 9 cm and bases of 6 cm and 11 cm. What is the area of Marcia’s felt trapezoid?
________ cm2

Answer: 76.5 cm2

Explanation:
Marcia cut a trapezoid out of a large piece of felt.
The trapezoid has a height of 9 cm and bases of 6 cm and 11 cm.
Area of the trapezium = (b1 + b2)h/2
A = (6 + 11)9/2
A = 17 cm × 4.5 cm
A = 76.5 sq. cm
Therefore the area of Marcia’s felt trapezoid is 76.5 cm2

Share and Show – Page No. 573

Question 1.
Find the area of the figure.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 89
________ ft2

Answer: 126 sq. ft

Explanation:
Figure 1:
l = 10 ft
w = 5 ft
A = lw
A = 10 ft × 5 ft
A = 50 sq. ft
Figure 2:
l = 10 ft
w = 5 ft
A = lw
A = 10 ft × 5 ft
A = 50 sq. ft
Figure 3:
b = 5 ft + 5 ft + 3 ft
b = 13 ft
h = 4 ft
Area of triangle = bh/2
A = 13 ft × 4 ft/2
A = 13 ft × 2 ft
A = 26 sq. ft
Add the areas of all the figures = 50 sq. ft + 50 sq. ft + 26 sq. ft
Thus the Area of the composite figure is 126 sq. ft.

Find the area of the figure.

Question 2.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 90
________ mm2

Answer: 128.2 sq. mm

Explanation:
Figure 1:
b1 = 11 mm
b2 = 11 mm
h = 8.2 mm
Area of the trapezoid = (b1 + b2)h/2
A = (11 mm + 11 mm)8.2 mm/2
A = 22 mm × 4.1 mm
A = 90.2 sq. mm
Figure 2:
b1 = 11mm
b2 = 8mm
h = 4mm
Area of the trapezoid = (b1 + b2)h/2
A = (11mm + 8mm)4mm/2
A = 19mm × 2mm
A = 38 sq. mm
Add the areas of both figures = 90.2 sq. mm + 38 sq. mm
Thus the area of the figure is 128.2 sq. mm

Question 3.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 91
________ m2

Answer: 144 sq. m

Explanation:
Figure 1:
l = 12 m
w = 7 m
Area of Rectangle = lw
A = 12m × 7m
A = 84 sq. m
Figure 2:
Area of right triangle = ab/2
a = 5m
b = 12m
A = (5m)(12m)/2
A = 30 sq. m
Figure 3:
Area of right triangle = ab/2
a = 5m
b = 12m
A = (5m)(12m)/2
A = 30 sq. m
Area of all figures = 84 sq. m + 30 sq. m + 30 sq. m = 144 sq. m.
Therefore the area of the figure is 144 sq. m

On Your Own

Question 4.
Find the area of the figure.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 92
________ in.2

Answer: 184 sq. in

Explanation:
Figure 1:
b = 8 in
h = 6 in
Area of right triangle = ab/2
A = 8 in × 6 in/2
A = 24 sq. in
Figure 2:
Area of Rectangle = lw
A = 16 in × 6 in
A = 96 sq. in
Figure 3:
Area of right triangle = ab/2
b = 8 in
h = 8 in
A = 8 in × 8 in/2
A = 32 sq. in
Figure 4:
Area of right triangle = ab/2
b = 8 in
h = 8 in
A = 8 in × 8 in/2
A = 32 sq. in
Area of all figures = 24 sq. in + 96 sq. in + 32 sq. in + 32 sq. in = 184 sq. in
Thus the area of the figure = 184 sq. in.

Question 5.
Attend to Precision Find the area of the shaded region.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 93
________ m2

Answer: 96.05 sq. m

Explanation:
Figure 1:
Area of Rectangle = lw
A = 12.75 m × 8.8 m
A = 112.2 sq. m
Figure 2:
Area of Rectangle = lw
l = 4.25 m
w = 3.3 m
A = 4.25 m × 3.3 m
A = 16.15 sq. m
Area of all the figures = 112.2 sq. m + 16.15 sq. m = 90.05 sq. m
Therefore the area of the figure = 90.05 sq. m

Unlock the Problem – Page No. 574

Question 6.
Marco made the banner shown at the right. What is the area of the yellow shape?
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 99
a. Explain how you could find the area of the yellow shape if you knew the areas of the green and red shapes and the area of the entire banner.
Type below:
_______________

Answer: I can find the area of the yellow shape by subtracting the areas of the green and red shapes from the area of the entire banner.

Question 6.
b. What is the area of the entire banner? Could you explain how you found it?
The area of the banner is ________ in.2

Answer: 1440 sq. in

Explanation:
The banner is a rectangle with a width of 48 inches and a length of 30 inches.
A = lw
A = 48 in × 30 in
A = 1440 sq. in
Therefore, the area of the banner is 1440 sq. in.

Question 6.
c. What is the area of the red shape? What is the area of each green shape?
The area of the red shape is ________ in.2
The area of each green shape is ________ in.2

Answer:
The area of the red shape is 360 in.2
The area of each green shape is 360 in.2

Explanation:
The red shape is a triangle with a base of 30 inches and a height of 24 inches.
A = bh/2
A = (30)(24)/2
A = 360 sq. in.
The area of the red triangle is 360 sq. in.
Each green shape is a triangle with a base of 15 inches and a height of 48 inches.
A = bh/2
A = 1/2 × 15 × 48
A = 720/2
A = 360 sq. in
Therefore the area of each green triangle is 360 sq. in.

Question 6.
d. What equation can you write to find A, the area of the yellow shape?
Type below:
_______________

Answer: A = 1440 – (360 + 360 + 360)

Question 6.
e. What is the area of the yellow shape?
The area of the yellow shape is ________ in.2

Answer: 360 sq. in

Explanation:
A = bh/2
A = 1/2 × 15 × 48
A = 720/2
A = 360 sq. in
Therefore the area of the yellow shape is 360 sq. in

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 574 Q7

Question 8.
Sabrina wants to replace the carpet in a few rooms of her house. Select the expression she can use to find the total area of the floor that will be covered. Mark all that apply.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 95
Options:
a. 8 × 22 + 130 + \(\frac{1}{2}\) × 10 × 9
b. 18 × 22 − \(\frac{1}{2}\) × 10 × 9
c. 18 × 13 + \(\frac{1}{2}\) × 10 × 9
d. \(\frac{1}{2}\) × (18 + 8) × 22

Answer: 8 × 22 + 130 + \(\frac{1}{2}\) × 10 × 9

Explanation:
Figure 1:
l = 13 ft
w = 10 ft
Area of the rectangle = lw
A = 13 ft × 10 ft = 130
Figure 2:
b = 9 ft
h = 10 ft
Area of the triangle = bh/2
A = (9)(10)/2
A = 45 sq. ft
Figure 3:
Area of the rectangle = lw
l = 22 ft
w = 8 ft
The area of the composite figure is 8 × 22 + 130 + \(\frac{1}{2}\) × 10 × 9
Thus the correct answer is option A.

Composite Figures – Page No. 575

Find the area of the figure

Question 1.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 96
________ cm2

Answer: 37 cm2

Explanation:
Area of square = s × s
A = 3 × 3 = 9 sq. cm
Area of Triangle = bh/2
A = 2 × 8/2 = 8 sq. cm
Area of the trapezoid = (b1 + b2)h/2
A = (5 + 3)5/2
A = 4 × 5 = 20 sq. in
Area of composite figure = 9 sq. cm + 8 sq. cm + 20 sq. in
A = 37 cm2

Question 2.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 97
________ ft2

Answer:

Explanation:
Figure 1:
b = 9 ft
h = 6 ft
Area of Triangle = bh/2
A = (9ft)(6ft)/2
A = 27 sq. ft
Figure 2:
l = 12 ft
w = 9 ft
Area of the rectangle = lw
A = (12ft)(9ft)/2
A = 12 ft × 9 ft
A = 108 sq. ft
Figure 3:
Area of Triangle = bh/2
b = 9 ft
h = 10 ft
A = (10ft)(9ft)/2
A = 45 sq. ft
Area of the composite figure = 27 sq. ft + 108 sq. ft + 45 sq. ft = 180 sq. ft

Question 3.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 98
________ yd2

Answer: 128 yd2

Explanation:
Figure 1:
b1 = 7 yd
b2 = 14 yd
h = 8 yd
Area of the trapezoid = (b1 + b2)h/2
A = (7yd + 14yd)8yd/2
A = 21 yd × 4 yd
A = 84 sq. yd
Figure 2:
b = 11 yd
h = 4 yd
Area of the parallelogram = bh
A = 11yd × 4yd = 44 sq. yd
Area of the composite figure = 84 sq. yd + 44 sq. yd = 128 sq. yd

Problem Solving

Question 4.
Janelle is making a poster. She cuts a triangle out of poster board. What is the area of the poster board that she has left?
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 99
________ in.2

Answer: 155 sq. in

Explanation:
The poster is a parallelogram, and it’s area is:
A = bh
A = 20 x 10
A = 200 sq. in
The area of the triangle that Janelle cut out of the poster board is:
A = 1/2bh
A = 1/2 x 10 x 9
A = 90/2
A = 45 sq. in
The area of the poster board that she has left is 200 sq. in – 45 sq. in = 155 sq. in

Question 5.
Michael wants to place grass on the sides of his lap pool. Find the area of the shaded regions that he wants to cover with grass.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 100
________ yd2

Answer: 204 yd2

Explanation:
The area of the shaded region can be found by finding the total area and subtracting the area of the lap pool.
Total area = Area of the trapezium = 1/2 × (Sum of parallel sides) × distance between them
Sum of parallel sides = 25 yd + (3 + 12) = 40 yd
Distance between them = 12 yd
Total area = 1/2 × 40 × 12 = 240 yd²
Find the area of the lap pool.
Area = length × width = 12 × 3 = 36 yd²
Find the area of the shaded region
Area to be covered with grass = 240 – 36 = 204 yd²

Question 6.
Describe one or more situations in which you need to subtract to find the area of a composite figure.
Type below:
_______________

Answer:
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 93
Figure 1:
Area of Rectangle = lw
A = 12.75 m × 8.8 m
A = 112.2 sq. m
Figure 2:
Area of Rectangle = lw
l = 4.25 m
w = 3.3 m
A = 4.25 m × 3.3 m
A = 16.15 sq. m
Area of all the figures = 112.2 sq. m + 16.15 sq. m = 90.05 sq. m
Therefore the area of the figure = 90.05 sq. m

Lesson Check – Page No. 576

Question 1.
What is the area of the composite figure?
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 101
________ m2

Answer: 227 m2

Explanation:
Figure 1:
b = 7 m
h = 7 m
Area of the triangle = bh/2
A = (7m)(7m)/2
A = 24.5 sq. m
Figure 2:
b1 = 7m
b2 = 10m
h = 9m
Area of the trapezoid = (b1 + b2)h/2
A = (7m + 10m)9m/2
A = 17m × 4.5 m
A = 76.5 sq. m
Area of the rectangle = lw
A = 18m × 7m
A = 126 sq. m
Area of the figures = 24.5 sq. m + 76.5 sq. m + 126 sq. m = 227 sq. m
Thus the area of the figure is 227 sq. m

Question 2.
What is the area of the shaded region?
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 102
________ in.2

Answer: 251.5 in.2

Explanation:
Figure 1:
l = 21 in
w = 15 in
Area of triangle = bh/2
A = 21 in × 15 in/2
A = 157.5 sq. in
Figure 2:
b1 = 12 in
b2 = 15 in
h = 11 in
Area of the trapezoid = (b1 + b2)h/2
A = (12 in + 15 in)11 in/2
A = 27 in × 5.5 in
A = 148.5 sq. in
Figure 3:
b = 13 in
h = 14.4 in
Area of trinagle = bh/2
A = 13 × 14.4in/2
A = 13in × 7.2 in
A = 94 sq. in
The area of the shaded region is 94 sq. in + 157.5 sq. in = 251.5 in.2

Spiral Review

Question 3.
In Maritza’s family, everyone’s height is greater than 60 inches. Write an inequality that represents the height h, in inches, of any member of Maritza’s family.
Type below:
_______________

Answer: h > 60

Explanation:
Given, Maritza’s family, everyone’s height is greater than 60 inches.
The inequality is h > 60

Question 4.
The linear equation y = 2x represents the cost y for x pounds of apples. Which ordered pair lies on the graph of the equation?
Type below:
_______________

Answer: (2, 4)

Explanation:
y = 2x
put x = 2
y = 2(2)
y = 4
The ordered pair is (2,4)

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 576 Q5

Question 6.
A regular hexagon has sides measuring 7 inches. If the hexagon is divided into 6 congruent triangles, each has a height of about 6 inches. What is the approximate area of the hexagon?
________ in.2

Answer: 126 in.2

Explanation:
b = 7 in
h = 6 in
Number of congruent figures: 6
Area of the triangle = bh/2
A = (7in)(6in)/2
A = 21 sq. in
Area of regular hexagon = 6 × area of each triangle
A = 6 × 21 sq. in
A = 126 sq. in
Thus the approximate area of the hexagon is 126 sq. in.

Share and Show – Page No. 579

Question 1.
The dimensions of a 2-cm by 6-cm rectangle are multiplied by 5. How is the area of the rectangle affected?
Type below:
_______________

Answer: 25

Explanation:
The dimensions of a 2-cm by 6-cm rectangle are multiplied by 5.
Original Area:
Area of rectangle = lw
A = 2cm × 6cm = 12 sq. cm
New dimensions:
l = 6 × 5 = 30 cm
w = 2 × 5 = 10 cm
The new area is:
A = 10 cm × 30 cm = 300 sq. cm
New Area/ Original Area = 300/12 = 25
So, the new area is 25 times the original area.

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 579 Q2

Question 3.
Evan bought two square rugs. The larger one measured 12 ft square. The smaller one had an area equal to \(\frac{1}{4}\) the area of the larger one. What fraction of the side lengths of the larger rug were the side lengths of the smaller one?
Type below:
_______________

Answer:
Since the area of the smaller rug is \(\frac{1}{4}\) times the area of the larger rug, the side lengths of the smaller rug are \(\frac{1}{2}\) of the side lengths of the larger one.

Question 4.
On Silver Island, a palm tree, a giant rock, and a buried treasure form a triangle with a base of 100 yd and a height of 50 yd. On a map of the island, the three landmarks form a triangle with a base of 2 ft and a height of 1 ft. How many times the area of the triangle on the map is the area of the actual triangle?
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 103
Type below:
_______________

Answer: 45,000

Explanation:
Area of triangle= (1/2) (base x height)
1 yard = 3 foot
Base of the actual triangle= 100 yards= 300ft
Height of the actual triangle= 50 yards= 150ft.
Area of the actual triangle= (1/2) (300 x 150) = 45000 square ft
The base of the triangle on the map = 2ft
Height of the triangle on the map= 1ft
Area of the triangle on the map= (1/2) (2 x 1) = 1 square ft.
The actual area is 45000 time the area of the map

On Your Own – Page No. 580

Question 5.
A square game board is divided into smaller squares, each with sides one-ninth the length of the sides of the board. Into how many squares is the game board divided?
________ small squares

Answer: 81 small squares

Explanation:
Each side of the game board is divided into 9 lengths.
The game board is divided into 9 × 9 = 81 small squares.
Thus, the board is divided into 81 small squares.

Question 6.
Flynn County is a rectangle measuring 9 mi by 12 mi. Gibson County is a rectangle with an area 6 times the area of Flynn County and a width of 16 mi. What is the length of Gibson County?
________ mi

Answer: 40.5 mi.

Explanation:
Flynn County is a rectangle measuring 9 mi by 12 mi.
Gibson County is a rectangle with an area 6 times the area of Flynn County and a width of 16 mi.
The area of Flynn Country is
A = 9 × 12 = 108 sq. mi
The area of Gibson Country is
A = 6 × 108 = 648 sq. mi
A = lw
648 = 16 × l
l = 648/16
l = 40.5 mi
Therefore the length of Gibson Country is 40.5 miles.

Question 7.
Use Diagrams Carmen left her house and drove 10 mi north, 15 mi east, 13 mi south, 11 mi west, and 3 mi north. How far was she from home?
________ miles

Answer:
15 mi – 11 mi = 4 miles
Thus Carmen is 4 miles from home.

Question 8.
Bernie drove from his house to his cousin’s house in 6 hours at an average rate of 52 mi per hr. He drove home at an average rate of 60 mi per hr. How long did it take him to drive home?
________ hours

Answer: 5.2 hours

Explanation:
Given,
Bernie drove from his house to his cousin’s house in 6 hours at an average rate of 52 mi per hr. He drove home at an average rate of 60 mi per hr.
The distance from Bernie’s house to his cousin’s house is
52 mi/hr × 6hr = 52 × 6mi = 312 miles
On the way back, he drove for
312mi ÷ 60mi/hr = 5.2 hours
Therefore it takes 5.2 hours for Bernie to drive home.

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 580 Q9

Problem Solving Changing Dimensions – Page No. 581

Read each problem and solve.

Question 1.
The dimensions of a 5-in. by 3-in. rectangle are multiplied by 6. How is the area affected?
Type below:
_______________

Answer: 36

Explanation:
Original area: A = 5 × 3 = 15 sq. in
new dimensions:
l = 6 × 5 = 30 in
w = 6 × 3 = 18 in
New Area = l × w
A = 30 in × 18 in
A = 540 sq. in
Thus new area = 540 sq. in
new area/original area = 540/15 = 36
Thus the area was multiplied by 36.

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 581 Q2

Question 3.
The dimensions of a 3-ft by 6-ft rectangle are multiplied by \(\frac{1}{3}\). How is the area affected?
Type below:
_______________

Answer: 1/9

Explanation:
Original area: A = 3 ft × 6 ft = 18 sq. ft
new dimensions:
l = 3 ft × \(\frac{1}{3}\) = 1 ft
w = 6 ft × \(\frac{1}{3}\) = 2 ft
New area: A = 1 ft × 2 ft = 2 sq. ft
new area/original area = 2/18 = 1/9
The area was multiplied by 1/9.

Question 4.
The dimensions of a triangle with base 10 in. and height 4.8 in. are multiplied by 4. How is the area affected?
Type below:
_______________

Answer: 16

Explanation:
original area: A = 10 in × 4.8 in = 48 sq. in
new dimensions:
l = 10 in × 4 = 40 in
w = 4.8 in × 4 = 19.2 in
new area = l × w
A = 40 in × 19.2 in
A = 768 sq. in
new area/original area = 768/48
Thus the area was multiplied by 16.

Question 5.
The dimensions of a 1-yd by 9-yd rectangle are multiplied by 5. How is the area affected?
Type below:
_______________

Answer: 25

Explanation:
original area: A = 1 yd × 9 yd = 9 sq. yd
new dimensions:
l = 1 yd × 5 = 5 yd
w = 9 yd × 5 = 45 yd
new area = 5 yd × 45 yd = 225 sq. yd
new area/original area = 225 sq. yd/9 sq. yd
Thus the area was multiplied by 25.

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 581 Q6

Question 7.
The dimensions of a triangle are multiplied by \(\frac{1}{4}\). The area of the smaller triangle can be found by multiplying the area of the original triangle by what number?
Type below:
_______________

Answer: 1/16

Explanation:
We can find the area of the original triangle by multiplying with \(\frac{1}{4}\)
\(\frac{1}{4}\) × \(\frac{1}{4}\) = \(\frac{1}{16}\)
Thus the area was multiplied by \(\frac{1}{16}\)

Question 8.
Write and solve a word problem that involves changing the dimensions of a figure and finding its area.
Type below:
_______________

Answer:
The dimensions of a triangle with a base 1.5 m and height 6 m are multiplied by 2. How is the area affected?
Original area:
Area of triangle = bh/2
A = (1.5m)(6m)/2
A = 4.5 sq. m
new dimensions:
b = 1.5m × 2 = 3 m
h = 6 m × 2 = 12 m
Area of triangle = bh/2
A = (12m × 3m)/2
A = 6m × 3m
A = 18 sq. m
new area/original area = 18 sq. m/4.5 sq. m
The area was multiplied by 4.

Lesson Check – Page No. 582

Question 1.
The dimensions of Rectangle A are 6 times the dimensions of Rectangle B. How do the areas of the rectangles compare?
Type below:
_______________

Answer: Area of Rectangle A = 36 × Area of Rectangle B

Explanation:
The area of Rectangle A will always be 36 times the area of Rectangle B.
If Rectangle B has length 1 and width 2, Rectangle A will have length 6 and width 12. By multiplying, Rectangle A will have an area of 72 and B 2. Divide the two numbers and you will have 36.

Question 2.
A model of a triangular piece of jewelry has an area that is \(\frac{1}{4}\) the area of the jewelry. How do the dimensions of the triangles compare?
Type below:
_______________

Answer: Model dimensions = 1/2 jewelry dimensions

Explanation:
The dimensions of the model area
1/4 ÷ 2 = 1/2 times the dimensions of the piece of jewelry.

Spiral Review

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 582 Q3

Question 4.
Graph y > 3 on a number line.
Type below:
_______________

Answer:
HMH Go Math Grade 6 Chapter 10 Answer Key img-1

Question 5.
The parallelogram below is made from two congruent trapezoids. What is the area of the shaded trapezoid?
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 104
________ mm2

Answer: 1312.5 sq. mm

Explanation:
Given,
b1 = 25mm
b2 = 50mm
h = 35mm
Area of the trapezoid = (b1 + b2)h/2
A = (25mm + 50mm)35mm/2
A = 75mm × 35mm/2
A = 1312.5 sq. mm
Thus the area of the shaded region is 1312.5 sq. mm

Question 6.
A rectangle has a length of 24 inches and a width of 36 inches. A square with side length 5 inches is cut from the middle and removed. What is the area of the figure that remains?
________ in.2

Answer: 839 sq. in

Explanation:
Area of rectangle = lw
A = 24 in × 36 in
A = 864 sq. in
Area of square = s × s
s = 5 in
A = 5 in × 5 in
A = 25 sq. in
Area of the figure that remains = 864 sq. in – 25 sq. in
A = 839 sq. in

Share and Show – Page No. 585

Question 1.
The vertices of triangle ABC are A(−1, 3), B(−4, −2), and C(2, −2). Graph the triangle and find the length of side \(\overline { BC } \).
________ units

Answer: 6 units
Go Math Grade 6 chapter 10 img-5

Give the coordinates of the unknown vertex of rectangle JKLM, and graph.

Question 2.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 105
Type below:
_______________

Answer:
Go-Math-Grade-6-Answer-Key-Chapter-10-Area-of-Parallelograms-img-105

Question 3.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 106
Type below:
_______________

Answer:
Go-Math-Grade-6-Answer-Key-Chapter-10-Area-of-Parallelograms-img-106

On Your Own

Question 4.
Give the coordinates of the unknown vertex of rectangle PQRS, and graph.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 107
Type below:
_______________

Answer:
Go-Math-Grade-6-Answer-Key-Chapter-10-Area-of-Parallelograms-img-107

Question 5.
The vertices of pentagon PQRST are P(9, 7), Q(9, 3), R(3, 3), S(3, 7), and T(6, 9). Graph the pentagon and find the length of side \(\overline { PQ } \).
________ units

Answer: 4 units
Go Math Grade 6 chapter 10 img-6

Problem Solving + Applications – Page No. 586

The map shows the location of some city landmarks. Use the map for 6–7.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 108

Question 6.
A city planner wants to locate a park where two new roads meet. One of the new roads will go to the mall and be parallel to Lincoln Street which is shown in red. The other new road will go to City Hall and be parallel to Elm Street which is also shown in red. Give the coordinates for the location of the park.
Type below:
_______________

Answer:
Go-Math-Grade-6-Answer-Key-Chapter-10-Area-of-Parallelograms-img-108
By seeing we can say that the coordinates for the location of the park is (1,1)

Question 7.
Each unit of the coordinate plane represents 2 miles. How far will the park be from City Hall?
________ miles

Answer: 8 units

Explanation:
The distance from City Hall to Park is 4 units.
Each unit = 2 miles
So, 2 miles × 4 = 8 miles
The distance from City Hall to Park is 8 miles.

Question 8.
\(\overline { PQ } \) is one side of right triangle PQR. In the triangle, ∠P is the right angle, and the length of side \(\overline { PR } \) is 3 units. Give all the possible coordinates for vertex R.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 109
Type below:
_______________

Answer:
Go-Math-Grade-6-Answer-Key-Chapter-10-Area-of-Parallelograms-img-109
The coordinates of S are (-2,-2)
The coordinates of R are (3,-2)

Question 9.
Use Math Vocabulary Quadrilateral WXYZ has vertices with coordinates W(−4, 0), X(−2, 3), Y(2, 3), and Z(2, 0). Classify the quadrilateral using the most exact name possible and explain your answer.
Type below:
_______________

Answer: Trapezoid
Go Math Grade 6 chapter 11 img
By seeing the above graph we can say that a suitable quadrilateral is a trapezoid.

Question 10.
Kareem is drawing parallelogram ABCD on the coordinate plane. Find and label the coordinates of the fourth vertex, D, of the parallelogram. Draw the parallelogram. What is the length of side CD? How do you know?
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 110
Type below:
_______________

Answer:
Go-Math-Grade-6-Answer-Key-Chapter-10-Area-of-Parallelograms-img-110

Figures on the Coordinate Plane – Page No. 587

Question 1.
The vertices of triangle DEF are D(−2, 3), E(3, −2), and F(−2, −2). Graph the triangle, and find the length of side \(\overline { DF } \).
________ units

Answer: 5 units

Explanation:
Vertical distance of D from 0: |3| = 3 units
Vertical Distance of F from 0: |-2| = 2 units
The points are in different quadrants, so add to find the distance from D to F: 3 + 2 = 5

Graph the figure and find the length of side \(\overline { BC } \).

Question 2.
A(1, 4), B(1, −2), C(−3, −2), D(−3, 3)
________ units

Answer: 4 units
Go Math Grade 6 chapter 10 img-1

Question 3.
A(−1, 4), B(5, 4), C(5, 1), D(−1, 1)
________ units

Answer: 3 units
Go Math Grade 6 chapter 10 img-2

Problem Solving

Question 4.
On a map, a city block is a square with three of its vertices at (−4, 1), (1, 1), and (1, −4). What are the coordinates of the remaining vertex?
Type below:
_______________

Answer: (-4, -4)
Go Math Grade 6 chapter 10 img-3

Question 5.
A carpenter is making a shelf in the shape of a parallelogram. She begins by drawing parallelogram RSTU on a coordinate plane with vertices R(1, 0), S(−3, 0), and T(−2, 3). What are the coordinates of vertex U?
Type below:
_______________

Answer: (2, 3)
Go Math Grade 6 chapter 10 img-4

Question 6.
Explain how you would find the fourth vertex of a rectangle with vertices at (2, 6), (−1, 4), and (−1, 6).
Type below:
_______________

Answer:

Explanation:
Midpoint of AC = (2 + (-1))/2 = 1/2; (6 + 6)/2 = 6
Midpoint of AC = (1/2, 6)
Midpoint of BD = (-1 + a)/2 = (-1 + a)/2; (b + 4)/2
(-1 + a)/2 = 1/2
-1 + a = 1
a = 2
(b + 4)/2 = 6
b + 4 = 12
b = 12 – 4
b = 8
So, the fouth vertex D is (2, 8)

Lesson Check – Page No. 588

Question 1.
The coordinates of points M, N, and P are M(–2, 3), N(4, 3), and P(5, –1). What coordinates for point Q make MNPQ a parallelogram?
Type below:
_______________

Answer: Q (-1, -1)

Question 2.
Dirk draws quadrilateral RSTU with vertices R(–1, 2), S(4, 2), T(5, –1), and U( 2, –1). Which is the best way to classify the quadrilateral?
Type below:
_______________

Answer:
The bases and height are not equal.
So, the best way to classify the quadrilateral is Trapezoid.

Spiral Review

Question 3.
Marcus needs to cut a 5-yard length of yarn into equal pieces for his art project. Write an equation that models the length l in yards of each piece of yarn if Marcus cuts it into p pieces.
Type below:
_______________

Answer:
Given,
Marcus needs to cut a 5-yard length of yarn into equal pieces for his art project.
To find the length we have to divide 5 by p.
Thus the equation is l = 5 ÷ p

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 588 Q4

Question 5.
A trapezoid is 6 \(\frac{1}{2}\) feet tall. Its bases are 9.2 feet and 8 feet long. What is the area of the trapezoid?
________ ft2

Answer: 55.9

Explanation:
Given that,
A trapezoid is 6 \(\frac{1}{2}\) feet tall. Its bases are 9.2 feet and 8 feet long.
We know that
Area of trapezoid = (b1 + b2)h/2
A = (9.2 + 8)6.5/2
A = (17.2 × 6.5)/2
A = 55.9 ft2

Question 6.
The dimensions of the rectangle below will be multiplied by 3. How will the area be affected?
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 111
Type below:
_______________

Answer:
3 × 3 = 9
the area will be multiplied by 9.

Chapter 10 Review/Test – Page No. 589

Question 1.
Find the area of the parallelogram.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 112
________ in.2

Answer: 67.5

Explanation:
b = 9 in
h = 7.5 in
Area of the parallelogram is bh
A = 9 in × 7.5 in
A = 67.5 sq. in
Thus the area of the parallelogram is 67.5 in.2

Question 2.
A wall tile is two different colors. What is the area of the white part of the tile? Explain how you found your answer.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 113
________ in.2

Answer: 11 in.2

Explanation:
b = 5.5 in
h = 4 in
We know that
The area of the triangle is bh/2
A = (5.5 in × 4 in)/2
A = 22/2 sq. in
A = 11 sq. in
Thus the area of one triangle is 11 in.2

Question 3.
The area of a triangle is 36 ft2. For numbers 3a–3d, select Yes or No to tell if the dimensions could be the height and base of the triangle.
3a. h = 3 ft, b = 12 ft
3b. h = 3 ft, b = 24 ft
3c. h = 4 ft, b = 18 ft
3d. h = 4 ft, b = 9 ft
3a. ____________
3b. ____________
3c. ____________
3d. ____________

Answer:
3a. No
3b. Yes
3c. Yes
3d. No

Explanation:
The area of a triangle is 36 ft2.
3a. h = 3 ft, b = 12 ft
The area of the triangle is bh/2
A = (12 × 3)/2
A = 6 × 3 = 18
A = 18 sq. ft
Thus the answer is no.
3b. h = 3 ft, b = 24 ft
The area of the triangle is bh/2
A = (3 × 24)/2
A = 3 × 12
A = 36 sq. ft
Thus the answer is yes.
3c. h = 4 ft, b = 18 ft
The area of the triangle is bh/2
A = (4 × 18)/2
A = 4 × 9
A = 36 sq. ft
Thus the answer is yes.
3d. h = 4 ft, b = 9 ft
The area of the triangle is bh/2
A = (4 × 9)/2
A = 2 ft × 9 ft
A = 18 sq. ft
Thus the answer is no.

Question 4.
Mario traced this trapezoid. Then he cut it out and arranged the trapezoids to form a rectangle. What is the area of the rectangle?
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 114
________ in.2

Answer: 112

Explanation:
b1 = 10 in
b2 = 4 in
h = 8 in
We know that
Area of trapezoid = (b1 + b2)h/2
A = (10 in + 4 in)8 in/2
A = 14 in × 4 in
A = 56 sq. in
Thus the area of the trapezoid for the above figure is 56 sq. in

Chapter 10 Review/Test Page No. 590

Question 5.
The area of the triangle is 24 ft2. Use the numbers to label the height and base of the triangle.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 115
Type below:
_______________

Answer: 6, 8

Explanation:
Go-Math-Grade-6-Answer-Key-Chapter-10-Area-of-Parallelograms-img-115
Area of the triangle = bh/2
A = (6 ft × 8 ft)/2
A = 6 ft × 4 ft
A = 24 ft2

Question 6.
A rectangle has an area of 50 cm2. The dimensions of the rectangle are multiplied to form a new rectangle with an area of 200 cm2. By what number were the dimensions multiplied?
Type below:
_______________

Answer: 2

Explanation:
Let A₁ = the original area a
and A₂ = the new area
and n = the number by which the dimensions were multiplied
A₁ = lw
A₂ = nl × nw = n²lw
A₂/A₁ = (n²lw)/(lw) = 200/50
n² = 4
n = 2

Question 7.
Sami put two trapezoids with the same dimensions together to make a parallelogram.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 116
The formula for the area of a trapezoid is \(\frac{1}{2}\)(b1 + b2)h. Explain why the bases of a trapezoid need to be added in the formula.
Type below:
_______________

Answer:
A trapezoid is a 4-sided figure with one pair of parallel sides. To find the area of a trapezoid, take the sum of its bases, multiply the sum by the height
sum by the height of the trapezoid, and then divide the result by 2.

Question 8.
A rectangular plastic bookmark has a triangle cut out of it. Use the diagram of the bookmark to complete the table.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 117
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 118
Type below:
_______________

Answer: 10 – 0.5 = 9.5
Go-Math-Grade-6-Answer-Key-Chapter-10-Area-of-Parallelograms-img-118

Chapter 10 Review/Test Page No. 591

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 591 Q9

Question 10.
A pillow is in the shape of a regular pentagon. The front of the pillow is made from 5 pieces of fabric that are congruent triangles. Each triangle has an area of 22 in.2. What is the area of the front of the pillow?
________ in.2

Answer: 110 in.2

Explanation:
Given,
Each triangle has an area of 22 in.2
The front of the pillow is made from 5 pieces of fabric that are congruent triangles.
Area of front pillow = 5 × 22 in.2 = 110 in.2

Question 11.
Which expressions can be used to find the area of the trapezoid? Mark all that apply.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 119
Options:
a. \(\frac{1}{2}\) × (5 + 2) × 4.5
b. \(\frac{1}{2}\) × (2 + 4.5) × 5
c. \(\frac{1}{2}\) × (5 + 4.5) × 2
d. \(\frac{1}{2}\) × (6.5) × 5

Answer: \(\frac{1}{2}\) × (2 + 4.5) × 5

Explanation:
b1 = 4.5 in
b2 = 2
h = 5 in
We know that,
Area of trapezoid = (b1 + b2)h/2
A = \(\frac{1}{2}\) × (2 + 4.5) × 5
Thus the correct answer is option B.

Question 12.
Name the polygon and find its area. Show your work.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 120
Type below:
_______________

Answer: 31 sq. in.

Explanation:
b = 5 in
h = 6.2 in
The area of the triangle is bh/2
A = (5 × 6.2)/2
A = 31/2
A = 15.5 sq. in
There are 2 triangles.
To find the area of the regular polygon we have to multiply the area of the triangle and number of triangles.
A = 15.5 × 2 = 31

Chapter 10 Review/Test Page No. 592

Question 13.
A carpenter needs to replace some flooring in a house.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 121
Select the expression that can be used to find the total area of the flooring to be replaced. Mark all that apply.
Options:
a. 19 × 14
b. 168 + 12 × 14 + 60
c. 19 × 24 − \(\frac{1}{2}\) × 10 × 12
d. 7 × 24 + 12 × 14 + \(\frac{1}{2}\) × 10 × 12

Answer: B, C, D

Explanation:
Go-Math-Grade-6-Answer-Key-Chapter-10-Area-of-Parallelograms-img-121

Here we have to use the Area of the parallelogram, Area of the rectangle, and area of triangle formulas.
Thus the suitable answers are 168 + 12 × 14 + 60, 19 × 24 − \(\frac{1}{2}\) × 10 × 12 and 7 × 24 + 12 × 14 + \(\frac{1}{2}\) × 10 × 12.

Question 14.
Ava wants to draw a parallelogram on the coordinate plane. She plots these 3 points.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 122
Part A
Find and label the coordinates of the fourth vertex, K, of the parallelogram. Draw the parallelogram
Type below:
_______________

Answer: K (2, 1)
Go-Math-Grade-6-Answer-Key-Chapter-10-Area-of-Parallelograms-img-122

Question 14.
Part B
What is the length of side JK? How do you know?
Type below:
_______________

Answer:
By using the above graph we can find the length of JK.
The length of the JK is 2 units.

Chapter 10 Review/Test Page No. 593

Question 15.
Joan wants to reduce the area of her posters by one-third. Draw lines to match the original dimensions in the left column with the correct new area in the right column. Not all dimensions will have a match.
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 123
Type below:
_______________

Answer:
Go-Math-Grade-6-Answer-Key-Chapter-10-Area-of-Parallelograms-img-123

Question 16.
Alex wants to enlarge a 4-ft by 6-ft vegetable garden by multiplying the dimensions of the garden by 2.
Part A
Find each area.
Area of original garden : ________ ft2
Area of enlarged garden : ________ ft2

Answer:
B = 4 ft
w = 6 ft
Area of original garden = 4 ft × 6 ft
A = 24 sq. ft
Now multiply 2 to base and width
b = 4 × 2 = 8 ft
w = 6 × 2 = 12 ft
Area of original garden = bw
A = 8 ft × 12 ft
A = 96 sq. ft

Question 16.
Suppose the point (3, 2) is changed to (3, 1) on this rectangle. What other point must change so the figure remains a rectangle? What is the area of the new rectangle?
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 124
Type below:
_______________

Answer:
Point: (-2, 2) would change to (-2, 1)
Rectangle:
B = 5 units
W = 4 units
Area of the rectangle = b × w
A = 5 × 4 = 20
A = 20 sq. units

Chapter 10 Review/Test Page No. 594

Question 18.
Look at the figure below. The area of the parallelogram and the areas of the two congruent triangles formed by a diagonal are related. If you know the area of the parallelogram, how can you find the area of one of the triangles?
Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms img 125
Type below:
_______________

Answer:
Each of the diagonals of a parallelogram divides it into two congruent triangles, as we saw when we proved properties like that the opposite sides are equal to each other or that the two pairs of opposite angles are congruent. Since those two triangles are congruent, their areas are equal.
We also saw that the diagonals of the parallelogram bisect each other, and so create two additional pairs of congruent triangles.
When comparing the ratio of areas of triangles, we often look for an equal base or an equal height.

Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms Page 594 Q19

Question 20.
Eliana is drawing a figure on the coordinate grid. For numbers 20a–20d, select True or False for each statement.
20a. The point (−1, 1) would be the fourth vertex of a square.
20b. The point (1, 1) would be the fourth vertex of a trapezoid.
20c. The point (2, -1) would be the fourth vertex of a trapezoid.
20d. The point (−1, -1) would be the fourth vertex of a square.
20a. ____________
20b. ____________
20c. ____________
20d. ____________

Answer:
20a. False
20b. False
20c. True
20d. True

Conclusion:

With the help of the above-provided links you can complete the homework within time without any mistakes. Test your knowledge by solving the problems mentioned in our website. Stay with us to get the solution keys of all Go Math Grade 6 Chapters from 1 to 13.

Go Math Grade 6 Answer Key Chapter 7 Exponents

go-math-grade-6-chapter-7-exponents-answer-key

The solutions of Grade 6 Go Math Answer Key for Chapter 7 Exponents are available in simple PDFs here. With the help off the HMH Go Math Grade 6 Chapter 7 Exponents Answer Ley can be easily downloaded by the students by using the provided links. You can understand the concept of the standard form in this article. So, Download a free pdf of Go Math Grade 6 Answer Key Chapter 7 Exponents.

Go Math Grade 6 Answer Key Chapter 7 Exponents

Our main aim is to provide a brief explanation of all the questions. We have provided the table of contents of chapter 7 Exponents in the below section. So, once go through the topics before you start your preparation. This will help you to know in which topic you are lagging. Hence make use of the resources provided on this page and try to score good marks in the exams. After your preparation we suggest the students to test your skills by solving the questions in the mid-chapter checkpoint and review test.

Lesson 1: Exponents

Lesson 2: Evaluate Expressions Involving Exponents

Lesson 3: Write Algebraic Expressions

Lesson 4: Identify Parts of Expressions

Lesson 5: Evaluate Algebraic Expressions and Formulas

Mid-Chapter Checkpoint

Lesson 6: Use Algebraic Expressions

Lesson 7: Problem Solving • Combine Like Terms

Lesson 8: Generate Equivalent Expressions

Lesson 9: Identify Equivalent Expressions

Chapter 7 Review/Test

Share and Show – Page No. 359

Question 1.
Write 24 by using repeated multiplication. Then find the value of 24.
___________

The repeated multiplication of 24

Answer: 16

Explanation:
The repeated factor is 2
The number 2 is repeated 4 times.
The repeated multiplication of 24 is 2 × 2 × 2 × 2 = 16
Thus the value of 24 is 16.

Use one or more exponents to write the expression.

Question 2.
7 × 7 × 7 × 7
Type below:
_____________

Answer: 74

Explanation:
The repeated factor is 7.
7 is repeated four times.
The exponent of the repeated multiplication 7 × 7 × 7 × 7 is 74

Question 3.
5 × 5 × 5 × 5 × 5
Type below:
_____________

Answer: 55

Explanation:
The repeated factor is 5. The number 5 is repeated five times.
The exponent of the repeated multiplication 5 × 5 × 5 × 5 × 5 is 55

Go Math Grade 6 Answer Key Chapter 7 Exponents Page 359 Q4

On Your Own

Find the value.

Question 5.
202
______

Answer: 20 × 20 = 400

Explanation:
The repeated factor is 20
Write the factor 2 times.
20 × 20 = 400
The value of 202 = 400

Question 6.
821
______

Answer: 82

Explanation:
The repeated factor is 82
Write the factor 1 time.
The value of 821 is 82

Go Math Grade 6 Answer Key Chapter 7 Exponents Page 359 Q7

Go Math Grade 6 Answer Key Chapter 7 Exponents Page 359 Q8

Complete the statement with the correct exponent.

Question 9.
5? = 125
______

Answer: 53

Explanation:
The exponential form of 125 is 5 × 5 × 5 = 53
5? = 125
5? = 53
When bases are equal powers should be equated.
Thus the exponent is 3

Question 10.
16? = 16
______

Answer: 1

Explanation:
The exponential form of 16 is 161
16? = 161
When bases are equal powers should be equated.
Thus the exponent is 1.

Go Math Grade 6 Answer Key Chapter 7 Exponents Page 359 Q11

Go Math Grade 6 Answer Key Chapter 7 Exponents Page 359 Q12

Question 13.
Select the expressions that are equivalent to 32. Mark all that apply.
Options:
a. 25
b. 84
c. 23 × 4
d. 2 × 4 × 4

Answer: 25

Explanation:
The exponent of 32 by using the base 2 is 2 × 2 × 2 × 2 × 2  = 25
32 = 25
Thus the correct answer is option A.

Bacterial Growth – Page No. 360

Bacteria are tiny, one-celled organisms that live almost everywhere on Earth. Although some bacteria cause disease, other bacteria are helpful to humans, other animals, and plants. For example, bacteria are needed to make yogurt and many types of cheese.

Under ideal conditions, a certain type of bacterium cell grows larger and then splits into 2 “daughter” cells. After 20 minutes, the daughter cells split, resulting in 4 cells. This splitting can happen again and again as long as conditions remain ideal.

Complete the table.
Go Math Grade 6 Answer Key Chapter 7 Exponents img 1
Extend the pattern in the table above to answer 14 and 15.

Question 14.
What power of 2 shows the number of cells after 3 hours? How many cells are there after 3 hours?
Type below:
_____________

Answer: 29

Explanation:
So, each cell doubles every 20 mins. After 20 minutes, you have 1(2) = 2 cells. After 40 minutes, you have 2(2) = 4 cells, etc.
1 hour = 60 minutes
3 hours = 3 × 60 minutes = 180 minutes
180/20 = 9 divisions
Thus 29 cells are there after 3 hours.

Question 15.
How many minutes would it take to have a total of 4,096 cells?
_______ minutes

Answer: 240 minutes

Explanation:
First, convert the cells into the exponential form.
The exponential form of 4096 is 2 × 2 × 2 × 2 × 2 × 2× 2 × 2× 2 × 2× 2 × 2 = 212
Multiply the power with 20
12 × 20 = 240
Thus it would take 240 minutes to have a total of 4,096 cells

Exponents – Page No. 361

Exponents

Use one or more exponents to write the expression.

Question 1.
6 × 6
Type below:
_____________

Answer:
The number 6 is used as a repeated factor.
6 is used as a factor 2 times.
Now write the base and exponent for 6 × 6 = 62

Go Math Grade 6 Answer Key Chapter 7 Exponents Page 361 Q2

Go Math Grade 6 Answer Key Chapter 7 Exponents Page 361 Q3

Question 4.
64
_______

Answer:
The repeated factor is 6.
Write the factor 4 times.
The value of 64 is 6 × 6 × 6 × 6 = 1296

Question 5.
16
_______

Answer:
The repeated factor is 1.
Write the factor 6 times.
The value of 16 is 1 × 1 × 1 × 1 × 1 × 1 = 1

Question 6.
105
_______

Answer:
The repeated factor is 10.
Write the factor 5 times.
The value of 105 is 10 × 10 × 10 × 10 × 10 = 1,00,000

Question 7.
Write 144 with an exponent by using 12 as the base.
Type below:
_____________

Answer: 12 × 12 = 122
The exponential form of 144 is 12 × 12 = 122

Question 8.
Write 343 with an exponent by using 7 as the base.
Type below:
_____________

Answer: The exponential form of 343 is 7 × 7 × 7 = 73

Go Math Grade 6 Answer Key Chapter 7 Exponents Page 361 Q9

Go Math Grade 6 Answer Key Chapter 7 Exponents Page 361 Q10

Question 11.
Explain what the expression 45 means and how to find its value.
Type below:
_____________

Answer:
The repeated factor is 4.
Write the factor 5 times.
The value of 45 is 4 × 4 × 4 × 4 × 4 = 1024

Lesson Check – Page No. 362

Question 1.
The number of games in the first round of a chess tournament is equal to 2 × 2 × 2 × 2 × 2 × 2. Write the number of games using an exponent.
Type below:
_____________

Answer: 26

Explanation:

The number 2 is the repeated factor.
2 is repeated 6 times.
2 × 2 × 2 × 2 × 2 × 2 = 26

Go Math Grade 6 Answer Key Chapter 7 Exponents Page 362 Q2

Spiral Review

Question 3.
The table shows the amounts of strawberry juice and lemonade needed to make different amounts of strawberry lemonade. Name another ratio of strawberry juice to lemonade that is equivalent to the ratios in the table.
Go Math Grade 6 Answer Key Chapter 7 Exponents img 2
Type below:
_____________

Answer: 5 : 15

Explanation:
By using the above table we can find the ratio of strawberry juice to lemonade.
2 : 6 = 1 : 3
The ratio of strawberry juice to lemonade next to 4 : 12 is 5 : 15

Question 4.
Which percent is equivalent to the fraction \(\frac{37}{50}\)?
_______ %

Answer: 74%

Explanation:
\(\frac{37}{50}\) × 100
0.74 × 100 = 74
Thus 74% is equivalent to the fraction \(\frac{37}{50}\)

Go Math Grade 6 Answer Key Chapter 7 Exponents Page 362 Q5

Question 6.
Use the formula d = rt to find the distance traveled by car driving at an average speed of 50 miles per hour for 4.5 hours.
_______ miles

Answer: 225 miles

Explanation:
Given,
r = 50 miles/hour
t = 4.5 hours
Use the formula d = rt
d = 50 × 4.5 = 225 miles
Thus the distance traveled by car driving at an average speed of 50 miles per hour for 4.5 hours is 225 miles.

Share and Show – Page No. 365

Question 1.
Evaluate the expression 9 + (52 − 10)
_______

Answer: 24

Explanation:
First write the square for 52
52 is 25
Now simplify the expression 9 + (25 – 10)
9 + 15 = 24
So, 9 + (52 − 10) = 24

Evaluate the expression.

Go Math Grade 6 Answer Key Chapter 7 Exponents Page 365 Q2

Go Math Grade 6 Answer Key Chapter 7 Exponents Page 365 Q3

Question 4.
(8 + 92) − 4 × 10
_______

Answer: 49

Explanation:
First multiply 9 × 9 = 81
(8 + 81) – (4 × 10)
Multiply 4 and 10.
4 × 10 = 40
(8 + 81) – (40)
89 – 40 = 49
(8 + 92) − 4 × 10 = 49

On Your Own

Evaluate the expression

Question 5.
10 + 62 × 2 ÷ 9
_______

Answer: 18

Explanation:
10 + (62 × 2) ÷ 9
Multiply 6 × 6 = 36
10 + (36 × 2) ÷ 9
Multiply 36 and 2 and then divide by 9.
10 + (72 ÷ 9)
10 + 8 = 18
So, 10 + 62 × 2 ÷ 9 = 18

Question 6.
62 − (23 + 5)
_______

Answer: 23

Explanation:

The value of 62 is 6 × 6 = 36
The value of 23 is 2 × 2 × 2 = 8
36 – (8 + 5)
36 – 13 = 23
Thus the answer for the expression for 62 − (23 + 5) is 23.

Question 7.
16 + 18 ÷ 9 + 34
_______

Answer: 99

Explanation:
16 + (18 ÷ 9) + 34
First divide 18 by 9
16 + 2 + 34
18 + 34
The value of 34 is 3 × 3 × 3 × 3 = 81
18 + 81 = 99
Thus the answer for the expression 16 + (18 ÷ 9) + 34 is 99.

Place parentheses in the expression so that it equals the given value.

Question 8.
102 − 50 ÷ 5
value: 10
Type below:
_____________

Answer: 10

Explanation:
102 − 50 ÷ 5
The factor of 102 is 10 × 10 = 100
(102 − 50) ÷ 5
50 ÷ 5 = 10
102 − 50 ÷ 5 = 10
The value of 102 − 50 ÷ 5 = 10

Question 9.
20 + 2 × 5 + 41
value: 38
Type below:
_____________

Answer: 38

Explanation:
20 + 2 × 5 + 41
The value of 41 is 4.
20 + 2 × (5 + 4)
20 + 2 × 9
Now multiply 2 and 9.
20 + 18 = 38
The value of 20 + 2 × 5 + 41 = 38

Question 10.
28 ÷ 22 + 3
value: 4
Type below:
_____________

Answer: 4

Explanation:
28 ÷ 22 + 3
28 ÷ (22 + 3)
The value of 22 is 4
28 ÷ (4 + 3)
28 ÷ 7 = 4
The value of 28 ÷ 22 + 3 is 4.

Problem Solving + Applications – Page No. 366

Use the table for 11–13.
Go Math Grade 6 Answer Key Chapter 7 Exponents img 3

Question 11.
Write an Expression To find the cost of a window, multiply its area in square feet by the price per square foot. Write and evaluate an expression to find the cost of a knot window
$ _______

Answer: 108

Explanation:
To find the cost of the knot window multiply the area with the price per square foot.
Area per square feet is 22
Price per square foot is $27
Cost = 22 × 27 = 4 × 27 = 108
Thus the cost of a knot window is $108

Go Math Grade 6 Answer Key Chapter 7 Exponents Page 366 Q12
Go Math Grade 6 Answer Key Chapter 7 Exponents Page 366 Q12.1

Question 13.
DeShawn bought a tulip window. Emma bought a rose window. Write and evaluate an expression to determine how much more DeShawn paid for his window than Emma paid for hers.
$ _______

Answer: 258

Explanation:
Given that, DeShawn bought a tulip window.
DeShawn bought it for 42 × $33 = 16 × $33 = 528
Emma bought a rose window
Emma bought it for 32 × 30 = 9 × 30 = 270
$528 – $270 = $258
DeShawn paid $258 for his window and Emma paid for hers.

Question 14.
What’s the Error? Darius wrote 17 − 22 = 225. Explain his error.
Type below:
_____________

Answer: 17 – 4 is actually 13 but not 225.

Question 15.
Ms. Hall wrote the expression 2 × (3 + 5)2÷ 4 on the board. Shyann said the first step is to evaluate 52. Explain Shyann’s mistake. Then evaluate the expression
_______

Answer: 32

Explanation:
2 × (3 + 5)2÷ 4
First, add 3 and 5.
2 × (8)2÷ 4
The square of 8 × 8 is 64.
2 × (64 ÷ 4) = 2 × 16 = 32

Evaluate Expressions Involving Exponents – Page No. 367

Evaluate Expressions Involving Exponents

Evaluate the expression.

Question 1.
5 + 17 − 102 ÷ 5
_______

Answer: 2

Explanation:
5 + 17 – (100 ÷ 5)
Divide 100 by 5
(5 + 17) – 20
22 – 20 = 2
So, the value for the expression 5 + 17 − 102 ÷ 5 = 2

Question 2.
72 − 32 × 4
_______

Answer: 13

Explanation:
72 − 32 × 4
72 − (32 × 4)
72 − (9 × 4)
49 – 36 = 13
Thus, 72 − 32 × 4 = 13

Question 3.
24 ÷ (7 − 5)
_______

Answer: 8

Explanation:
24 ÷ (7 − 5)
24 ÷ 2
24 = 2 × 2 × 2 × 2 = 16
16 ÷ 2 = 8
24 ÷ (7 − 5) = 8

Go Math Grade 6 Answer Key Chapter 7 Exponents Page 367 Q4

Go Math Grade 6 Answer Key Chapter 7 Exponents Page 367 Q5

Question 6.
(12 − 8)3 − 24 × 2
_______

Answer: 16

Explanation:
(12 − 8)3 − 24 × 2 = (4)3 − 24 × 2
64 – (24 × 2)
= 64 – 48 = 16
(12 − 8)3 − 24 × 2 = 16

Place parentheses in the expression so that it equals the given value.

Question 7.
12 × 2 + 23
value: 120
Type below:
_____________

Answer:
12 × (2 + 23)
12 × (2 + 8)
12 × 10 = 120
12 × 2 + 23 = 120

Question 8.
72 + 1 − 5 × 3
value: 135
Type below:
_____________

Answer:
(72 + 1 − 5) × 3
(49 + 1 – 5) × 3
(50 – 5) × 3
45 × 3 = 135
72 + 1 − 5 × 3 = 135

Problem Solving

Question 9.
Hugo is saving for a new baseball glove. He saves $10 the first week, and $6 each week for the next 6 weeks. The expression 10 + 62 represents the total amount in dollars he has saved. What is the total amount Hugo has saved?
$ _______

Answer: $46

Explanation:
Hugo is saving for a new baseball glove.
He saves $10 the first week, and $6 each week for the next 6 weeks.
The expression 10 + 62 represents the total amount in dollars he has saved.
10 + 62 = 10 + 36 = 46
The total amount Hugo has saved is $46

Question 10.
A scientist placed 5 fish eggs in a tank. Each day, twice the number of eggs from the previous day hatch. The expression 5 × 26 represents the number of eggs that hatch on the seventh day. How many eggs hatch on the seventh day?
_______ eggs

Answer: 320 eggs

Explanation:
A scientist placed 5 fish eggs in a tank.
Each day, twice the number of eggs from the previous day hatch.
The expression 5 × 26 represents the number of eggs that hatch on the seventh day.
5 × 26 = 5 × 64 = 320 eggs
Therefore 320 eggs hatch on the seventh day.

Question 11.
Explain how you could determine whether a calculator correctly performs the order of operations.
Type below:
_____________

Answer: Create a problem that must use the order of operations and isn’t solved by just left to right. Solve it going left to right. Then solve it using the order of operations. Solve it on the calculator. Your answer on the calculator will match the one using the order of operations.

Lesson Check – Page No. 368

Question 1.
Ritchie wants to paint his bedroom ceiling and four walls. The ceiling and each of the walls are 8 feet by 8 feet. A gallon of paint covers 40 square feet. Write an expression that can be used to find the number of gallons of paint Ritchie needs to buy.

Ritchie wants to paint his bedroom ceiling and four walls
Type below:
_____________

Answer:
Ritchie wants to paint his bedroom ceiling and four walls.
The ceiling and each of the walls are 8 feet by 8 feet.
A gallon of paint covers 40 square feet.
8 × 8 × (4 + 1) ÷ 40
82 (4 + 1) ÷ 40
Thus the expression that can be used to find the number of gallons of paint Ritchie needs to buy is 82 (4 + 1) ÷ 40

Question 2.
A Chinese restaurant uses about 225 pairs of chopsticks each day. The manager wants to order a 30-day supply of chopsticks. The chopsticks come in boxes of 750 pairs. How many boxes should the manager order?
_______ boxes

Answer: 9 boxes

Explanation:
A Chinese restaurant uses about 225 pairs of chopsticks each day.
The manager wants to order a 30-day supply of chopsticks.
Multiply the number of pairs with the number of days
225 × 30 = 6750
The chopsticks come in boxes of 750 pairs.
Now divide the number of chopsticks by the number of pairs.
6750 ÷ 750 = 9 boxes.

Spiral Review

Question 3.
Annabelle spent $5 to buy 4 raffle tickets. How many tickets can she buy for $20?
_______ tickets

Answer: 16 tickets

Explanation:
Annabelle spent $5 to buy 4 raffle tickets.
To find the number of tickets she can buy for $20.
($20 ÷ $5) × 4
4 × 4 = 16 tickets
That means she can buy 16 tickets for $20.

Go Math Grade 6 Answer Key Chapter 7 Exponents Page 368 Q4

Question 5.
How many pounds are equivalent to 40 ounces?
_______ pounds

Answer: 2.5 pounds

Explanation:
Convert from ounces to pounds.
1 pound = 16 ounces
1 ounce = 1/16 pound
40 ounces = 40 × 1/16 pound
40 ounces = 2.5 pounds
Thus, 2.5 pounds are equivalent to 40 ounces

Question 6.
List the expressions in order from least to greatest.
15 33 42 81
Type below:
_____________

Answer:
15 33 42 81
15 = 1 × 1 × 1 × 1 × 1 = 1
33 = 3 × 3 × 3 = 27
42 = 4 × 4 = 16
81 = 8
Thus the order from least to greatest.
15 81 42 33

Share and Show – Page No. 371

Question 1.
Write an algebraic expression for the product of 6 and p.
What operation does the word “product” indicate?
Type below:
_____________

Answer: 6 × p
Explanation:
The word product indicates multiplication.
Multiply 6 with p.
The algebraic expression for the product of 6 and p is 6 × p.

Write an algebraic expression for the word expression.

Question 2.
11 more than e
Type below:
_____________

Answer: 11 + e

Explanation:
The word more than indicates addition operation.
So, the algebraic expression is 11 + e

Question 3.
9 less than the quotient of n and 5
Type below:
_____________

Answer: 9 – (n ÷ 5)

Explanation:
The word “less than” indicates subtraction and the “quotient” indicates division.
So, the expression is 9 – (n ÷ 5)

On Your Own

Write an algebraic expression for the word expression.

Question 4.
20 divided by c
Type below:
_____________

Answer: 20 ÷ c

Explanation:
Here we have to divide 20 by c.
The expression is 20 ÷ c

Question 5.
8 times the product of 5 and t
Type below:
_____________

Answer: 8 × (5t)

Explanation:
The word times indicate multiplication and the product indicates multiplication.
Here we have to multiply 8 with 5 and t.
Thus the expression is 8 × 5 × t = 8 × 5t

Go Math Grade 6 Answer Key Chapter 7 Exponents Page 371 Q6

Question 7.
A state park charges a $6.00 entry fee plus $7.50 per night of camping. Write an algebraic expression for the cost in dollars of entering the park and camping for n nights.
Type below:
_____________

Answer: $6.00 + $7.50 n

Explanation:
Given that, A state park charges a $6.00 entry fee plus $7.50 per night of camping.
Find the camping for n nights. The product of $7.50 camping for n nights.
$7.50 × n
Now add park charges to the camping nights.
$6.00 + $7.50 n
Thus the algebraic expression for the cost in dollars of entering the park and camping for n nights is $6.00 + $7.50 n

Question 8.
Look for Structure At a bookstore, the expression 2c + 8g gives the cost in dollars of c comic books and g graphic novels. Next month, the store’s owner plans to increase the price of each graphic novel by $3. Write an expression that will give the cost of c comic books and g graphic novels next month.
Type below:
_____________

Answer: 2c + 11g

Explanation:
Look for Structure At a bookstore, the expression 2c + 8g gives the cost in dollars of c comic books and g graphic novels.
Next month, the store’s owner plans to increase the price of each graphic novel by $3.
Here we have to add $3 to 8 g = 3g + 8g = 11g
Sum of cost of c comic books and g graphic novels
Thus the expression is 2c + 11g

Unlock the Problem – Page No. 372

Question 9.
Martina signs up for the cell phone plan described at the right. Write an expression that gives the total cost of the plan in dollars if Martina uses it for m months.
Go Math Grade 6 Answer Key Chapter 7 Exponents img 4
a. What information do you know about the cell phone plan?
Type below:
_____________

Answer: Pay a low monthly fee of $50. Receive $10 off your first month’s fee.

Question 9.
b. Write an expression for the monthly fee in dollars for m months.
Type below:
_____________

Answer:
M is the number of months.
50 × m
Given that $10 off on first-month fee.
50m + (50-10)
50m + $40

Question 9.
c. What operation can you use to show the discount of $10 for the first month?
Type below:
_____________

Answer: We have to use subtraction operations to show a discount of $10 for the first month.

Question 9.
d. Write an expression for the total cost of the plan in dollars for m months
Type below:
_____________

Answer: 50m + 40

Question 10.
A group of n friends evenly share the cost of dinner. The dinner costs $74. After dinner, each friend pays $11 for a movie. Write an expression to represent what each friend paid for dinner and the movie.
Type below:
_____________

Answer: 74 ÷ n + 11n

Explanation:
Given,
A group of n friends evenly share the cost of dinner.
The dinner costs $74. After dinner, each friend pays $11 for a movie.
The word share represents the division operation.
That means we have to divide 74 by n.
74 ÷ n
After that n friends paid $11 for the movie
Multiply 11 with n.
Thus the expression to represent what each friend paid for dinner and the movie is 74 ÷ n + 11n

Question 11.
A cell phone company charges $40 per month plus $0.05 for each text message sent. Select the expressions that represent the cost in dollars for one month of cell phone usage and sending m text messages. Mark all that apply.
Options:
a. 40m + 0.05
b. 40 + 0.05m
c. 40 more than the product of 0.05 and m
d. the product of 40 and m plus 0.05

Answer: 40 + 0.05m

Explanation:
A cell phone company charges $40 per month plus $0.05 for each text message sent.
Let m represent the messages sent.
40m + 0.05m
Thus the answer is option B.

Write Algebraic Expressions – Page No. 373

Write an algebraic expression for the word expression.

Question 1.
13 less than p
Type below:
_____________

Answer: 13 – p

Explanation:
Less than is nothing but subtraction.
So the expression for 13 less than p is 13 – p

Go Math Grade 6 Answer Key Chapter 7 Exponents Page 373 Q2

Question 3.
6 more than the difference between b and 5
Type below:
_____________

Answer: 6 + (b – 5)

Explanation:
More than is nothing but addition and difference mean subtraction.
The expression for 6 more than the difference of b and 5 is 6 + (b – 5)

Question 4.
the sum of 15 and the product of 5 and v
Type below:
_____________

Answer: 15 + 5v

Explanation:
Product is nothing but multiplication and sum is nothing but an addition.
So, the expression for the sum of 15 and the product of 5 and v is 15 + 5 × v

Question 5.
the difference of 2 and the product of 3 and k
Type below:
_____________

Answer: 2 – 3k

Explanation:
The difference means subtraction and Product are nothing but the multiplication
So, the difference between 2 and the product of 3 and k is 2 – 3 × k
2 – 3k

Go Math Grade 6 Answer Key Chapter 7 Exponents Page 373 Q6

Question 7.
the quotient of m and 7
Type below:
_____________

Answer: m ÷ 7

Explanation:
Given the quotient of m and 7
That means we have to divide m by 7.
Thus the answer is m ÷ 7

Question 8.
9 more than 2 multiplied by f
Type below:
_____________

Answer: 9 + 2f

Explanation:
9 more than 2 multiplied by f
We have to add 9 to 2 × f
So, the expression is 9 + 2f

Question 9.
6 minus the difference between x and 3
Type below:
_____________

Answer: 6 – (x – 3)

Explanation:
First, subtract 3 from x
The expression for 6 minus the difference of x and 3 is 6 – (x – 3)

Question 10.
10 less than the quotient of g and 3
Type below:
_____________

Answer: 10 – (g ÷ 3)

Explanation:
The quotient of g and 3 is nothing but dividing g by 3
g ÷ 3
Now subtract g ÷ 3 from 10.
So, the expression for 10 less than the quotient of g and 3 is 10 – (g ÷ 3)

Question 11.
the sum of 4 multiplied by a and 5 multiplied by b
Type below:
_____________

Answer: 4a + 5b

Explanation:
First, multiply 4 with a and then multiply 5 with b
After that add both expressions.
4a + 5b
So, the sum of 4 multiplied by a and 5 multiplied by b is 4a + 5b

Question 12.
14 more than the difference between r and s
Type below:
_____________

Answer: 14 + (r – s)

Explanation:
Subtract r and s
And then add 14 to that r -s
14 + (r – s)

Problem Solving

Question 13.
Let h represent Mark’s height in inches. Suzanne is 7 inches shorter than Mark. Write an algebraic expression that represents Suzanne’s height in inches.
Type below:
_____________

Answer: h – 7

Explanation:
Let h represent Mark’s height in inches. Suzanne is 7 inches shorter than Mark.
That means we have to subtract 7 from h.
i.e., h – 7
Thus Suzanne’s height is h – 7 inches.

Question 14.
A company rents bicycles for a fee of $10 plus $4 per hour of use. Write an algebraic expression for the total cost in dollars for renting a bicycle for h hours.
Type below:
_____________

Answer: 10 + 4h

Explanation:
A company rents bicycles for a fee of $10 plus $4 per hour of use.
Multiply 4 with hours
And then 10 to 4h
10 + 4h
Thus the total cost in dollars for renting a bicycle for h hours is 10 + 4h

Question 15.
Give an example of a real-world situation involving two unknown quantities. Then write an algebraic expression to represent the situation.
Type below:
_____________

Answer:
Cooper bikes so many miles per day and does it for 7 months.
The expression for the question is 6m × 7

Lesson Check – Page No. 374

Question 1.
The female lion at a zoo weighs 190 pounds more than the female cheetah. Let c represent the weight in pounds of the cheetah. Write an expression that gives the weight in pounds of the lion.
Type below:
_____________

Answer: c + 190

Explanation:
Given that, The female lion at a zoo weighs 190 pounds more than the female cheetah.
Let c represent the weight in pounds of the cheetah.
We have to add 190 to the weight in pounds of the cheetah.
That means c + 190
Thus the expression that gives the weight in pounds of the lion is c + 190.

Question 2.
Tickets to a play cost $8 each. Write an expression that gives the ticket cost in dollars for a group of g girls and b-boys.
Type below:
_____________

Answer: 8 × (g + b)

Explanation:
First add girls group and boys group.
g + b
And then multiply 8 with the group of girls and boys.
8 × (g + b)
So, the expression that gives the ticket cost in dollars for a group of g girls and b-boys is 8 × (g + b).

Spiral Review

Go Math Grade 6 Answer Key Chapter 7 Exponents Page 374 Q3

Question 4.
There are 32 peanuts in a bag. Elliott takes 25% of the peanuts from the bag. Then Zaire takes 50% of the remaining peanuts. How many peanuts are left in the bag?
_______ peanuts

Answer: 12

Explanation:
First, we have to find 25% of 32.
25% of 32 its 0.25 × 32=8
Now we have to subtract 32 and 8
32 – 8=24
Now we have to find 50% of 24
50% of 24 = 12
24-12=12.
Thus 12 peanuts are left in the bag.

Go Math Grade 6 Answer Key Chapter 7 Exponents Page 374 Q5

Question 6.
Write an expression using exponents that represent the area of the figure in square centimeters
Go Math Grade 6 Answer Key Chapter 7 Exponents img 5
Type below:
_____________

Answer: 72 – 22

Explanation:
The area of the square is 7 cm × 7 cm = 72
The area of the square is 2 cm × 2 cm = 22
Now subtract a small square from the large square.
The expression that represents the area of the figure is 72 – 22

Share and Show – Page No. 377

Identify the parts of the expression. Then, write a word expression for the numerical or algebraic expression.

Question 1.
7 × (9 ÷ 3)
Type below:
_____________

Answer:
The quotient of 9 and 3 and then multiply by 7.
Word expression: Product of 7 with the quotient of 9 and 3.

Go Math Grade 6 Answer Key Chapter 7 Exponents Page 377 Q2

On Your Own

Practice: Copy and Solve Identify the parts of the expression. Then write a word expression for the numerical or algebraic expression.

Question 3.
8 + (10 − 7)
Type below:
_____________

Answer:
Subtraction is the difference between 10 and 7. Addition to the subtraction of 10 and 7.
Word expression: Add 8 to the difference between 10 and 7.

Question 4.
1.5 × 6 + 8.3
Type below:
_____________

Answer:
The addition is the sum of 6 and 8.3 and then multiply the sum to 1.5.
Word expression: 1.5 times the sum of 6 and 8.3

Question 5.
b + 12x
Type below:
_____________

Answer:
Product of 12 and x. Add b to the product of 12 and x.
Word expression: Sum of b to the product of 12 and x.

Question 6.
4a ÷ 6
Type below:
_____________

Answer:
The division is the quotient of 4a and 6. Multiply 4 and a. The expression is the product of 4 and divided by 6.
Word expression: The quotient of the products 4 and a and 6.

Identify the terms of the expression. Then, give the coefficient of each term.

Question 7.
k − \(\frac{1}{3}\)d
Type below:
_____________

Answer:
The terms of the expression are k and \(\frac{1}{3}\)d
Coefficients – 1 and \(\frac{1}{3}\)

Question 8.
0.5x + 2.5y
Type below:
_____________

Answer:
The terms of the expression are 0.5x and 2.5y
Coefficients – 0.5 and 2.5

Question 9.
Connect Symbols and Words Ava said she wrote an expression with three terms. She said the first term has a coefficient 7, the second term has a coefficient 1, and the third term has a coefficient 0.1. Each term involves a different variable. Write an expression that could be the expression Ava wrote
Type below:
_____________

Answer:
Connect Symbols and Words Ava said she wrote an expression with three terms.
She said the first term has the coefficient 7, the second term has a coefficient 1, and the third term has a coefficient 0.1.
The expression for the first term is 7x
The expression for the second term is 1y
The expression for the third term is 0.1z
7x + y + 0.1z

Problem Solving + Applications – Page No. 378

Use the table for 10–12.
Go Math Grade 6 Answer Key Chapter 7 Exponents img 6

Question 10.
A football team scored 2 touchdowns and 2 extra points. Their opponent scored 1 touchdown and 2 field goals. Write a numerical expression for the points scored in the game.
Type below:
_____________

Answer:
A football team scored 2 touchdowns and 2 extra points.
2 touchdowns = 2 × 6
2 extra points = 2 × 1
Their opponent scored 1 touchdown and 2 field goals.
1 touchdown = 1 × 6
2 field goals = 2 × 3
Thue the numerical expression is 12 + 2 + 6 + 6
14 + 12
The numerical expression for the points scored in the game is 14 + 12.

Question 11.
Write an algebraic expression for the number of points scored by a football team that makes t touchdowns, f field goals, and e extra points
Type below:
_____________

Answer: 6t + 3f + e

Explanation:
The number of points scored by a football team that makes t touchdowns, f field goals, and e extra points.
The table shows that touchdown has 6 points, field goal has 3 points and extra point has 1 point.
So we need to add all the points to make the expressions
That means 6t + 3f + e

Question 12.
Identify the parts of the expression you wrote in Exercise 11.
Type below:
_____________

Question 13.
Give an example of an expression involving multiplication in which one of the factors is a sum. Explain why you do or do not need parentheses in your expression
Type below:
_____________

Answer: 6 × 2 + 3
In this expression, there is no need for parentheses because there are no exponents or multiple operations.

Question 14.
Kennedy bought a pounds of almonds at $5 per pound and p pounds of peanuts at $2 per pound. Write an algebraic expression for the cost of Kennedy’s purchase.
Type below:
_____________

Answer: 5 + 2p = x

Explanation:
Kennedy bought a pounds of almonds at $5 per pound and p pounds of peanuts at $2 per pound.
We have to multiply p with $2 per pound.
The algebraic expression for the cost of Kennedy’s purchase is the sum of 5 and the product of p and 2
Thus the expression is 5 + 2p = x

Identify Parts of Expressions – Page No. 379

Identify the parts of the expression. Then write a word expression for the numerical or algebraic expression.

Question 1.
(16 − 7) ÷ 3
Type below:
_____________

Answer:
Subtraction is the difference between 16 and 7. The division is the quotient of the difference and 3
Word expression: the quotient of the difference 16 and 7 and 3.

Go Math Grade 6 Answer Key Chapter 7 Exponents Page 379 Q2

Identify the terms of the expression. Then give the coefficient of each term.

Question 3.
11r + 7s
Type below:
_____________

Answer:
The terms of the expression are 11r and 7s
The coefficient of each term is 11 and 7.

Question 4.
6g − h
Type below:
_____________

Answer:
The terms of the expression are 6g and h
The coefficient of each term is 6 and 1.

Problem Solving

Question 5.
Adam bought granola bars at the store. The expression 6p + 5n gives the number of bars in p boxes of plain granola bars and n boxes of granola bars with nuts. What are the terms of the expression?
Type below:
_____________

Answer:
Adam bought granola bars at the store.
The expression 6p + 5n gives the number of bars in p boxes of plain granola bars and n boxes of granola bars with nuts.
The terms of the expression are 6p and 5n.

Question 6.
In the sixth grade, each student will get 4 new books. There is one class of 15 students and one class of 20 students. The expression 4 × (15 + 20) gives the total number of new books. Write a word expression for the numerical expression.
Type below:
_____________

Answer:
In the sixth grade, each student will get 4 new books.
There is one class of 15 students and one class of 20 students.
The expression 4 × (15 + 20) gives the total number of new books.
The product of 4 the sum of 15 and 20.

Question 7.
Explain how knowing the order of operations helps you write a word expression for a numerical or algebraic expression.
Type below:
_____________

Answer: Because if you don’t know and use the order of