This handy Spectrum Math Grade 7 Answer Key Chapter 4 Posttest provides detailed answers for the workbook questions
Spectrum Math Grade 7 Chapter 4 Posttest Answers Key
Check What You Learned
Ratios and Proportional Relationships
Solve each proportion.
Question 1.
a. \(\frac{3}{2}\) = \(\frac{n}{6}\) _________
Answer: n = 9
\(\frac{3}{2}\) = \(\frac{n}{6}\)
In order to solve the above proportion and get the value of n, perform cross multiplication
As product of means = product of extremes
2 × n = 3 × 6
2n = 18
n = \(\frac{18}{2}\)
Therefore, n = 9
b. \(\frac{17}{34}\) = \(\frac{1}{n}\) _________
Answer: n = 2
\(\frac{17}{34}\) = \(\frac{1}{n}\)
In order to solve the above proportion and get the value of n, perform cross multiplication
As product of means = product of extremes
17 × n = 34 × 1
17n = 34
n = \(\frac{34}{17}\)
Therefore, n = 2
c. \(\frac{n}{16}\) = \(\frac{6}{4}\) _________
Answer: n = 24
\(\frac{n}{16}\) = \(\frac{6}{4}\)
In order to solve the above proportion and get the value of n, perform cross multiplication
As product of means = product of extremes
4 × n = 16 × 6
4n = 96
n = \(\frac{96}{4}\)
Therefore, n = 24
Question 2.
a. \(\frac{7}{n}\) = \(\frac{21}{12}\) _________
Answer: n = 4
\(\frac{7}{n}\) = \(\frac{21}{12}\)
In order to solve the above proportion and get the value of n, perform cross multiplication
As product of means = product of extremes
21 × n = 7 × 12
21n = 84
n = \(\frac{84}{21}\)
Therefore, n = 4
b. \(\frac{5}{8}\) = \(\frac{n}{40}\) _________
Answer: n = 25
\(\frac{5}{8}\) = \(\frac{n}{40}\)
In order to solve the above proportion and get the value of n, perform cross multiplication
As product of means = product of extremes
8 × n = 5 × 40
8n = 200
n = \(\frac{200}{8}\)
Therefore, n = 25
c. \(\frac{1}{2}\) = \(\frac{56}{n}\) _________
Answer: n = 112
\(\frac{1}{2}\) = \(\frac{56}{n}\)
In order to solve the above proportion and get the value of n, perform cross multiplication
As product of means = product of extremes
1 × n = 56 × 2
Therefore, n = 112
Circle the ratios that are equal. Show your work.
Question 3.
a. \(\frac{15}{20}\), \(\frac{3}{4}\)
Answer:
\(\frac{15}{20}\), \(\frac{3}{4}\)
15 × 4 = 60
20 × 3 = 60
A ratio is a comparison of two numbers. A proportion expresses the equality of two ratios.
In order to determine whether the two ratios are equal or not, cross multiply both the ratios. If the results after cross multiplication is same on both sides, then they are said to be equal. Otherwise they are said to be unequal ratios.
In the above given ratios the result is same after cross multiplication.
Therefore, \(\frac{15}{20}\), \(\frac{3}{4}\) are equal ratios.
b. \(\frac{8}{12}\), \(\frac{10}{14}\)
Answer:
\(\frac{8}{12}\), \(\frac{10}{14}\)
8 × 14 = 112
12 × 10 = 120
A ratio is a comparison of two numbers. A proportion expresses the equality of two ratios.
In order to determine whether the two ratios are equal or not, cross multiply both the ratios. If the results after cross multiplication is same on both sides, then they are said to be equal. Otherwise they are said to be unequal ratios.
In the above given ratios the result is same after cross multiplication.
Therefore, \(\frac{8}{12}\), \(\frac{10}{14}\) are not equal ratios.
c. \(\frac{4}{3}\), \(\frac{16}{12}\)
Answer:
\(\frac{4}{3}\), \(\frac{16}{12}\)
4 × 12 = 48
3 × 16 = 48
A ratio is a comparison of two numbers. A proportion expresses the equality of two ratios.
In order to determine whether the two ratios are equal or not, cross multiply both the ratios. If the results after cross multiplication is same on both sides, then they are said to be equal. Otherwise they are said to be unequal ratios.
In the above given ratios the result is same after cross multiplication.
Therefore, \(\frac{4}{3}\), \(\frac{16}{12}\) are equal ratios.
Find the constant of proportionality for each set of values.
Question 4.
a.
k = _____
Answer: k = 5
Step 1: Set up an equation in which the constant (k) is equal to y ÷ x.
Step 2: Check the equation across multiple points to verify the constant.
Step 3: 5 ÷ 1 = 5 ,
10 ÷ 2 = 5 ,
15 ÷ 3 = 5,
20 ÷ 4 = 5
Therefore, k = 5
b.
k = _____
Answer: k = 5
Step 1: Set up an equation in which the constant (k) is equal to y ÷ x.
Step 2: Check the equation across multiple points to verify the constant.
Step 3: 10 ÷ 2 = 5 ,
20÷ 4 = 5 ,
30 ÷ 6 = 5,
40 ÷ 8 = 5
Therefore, k = 5
Find the constant of proportionality.
Question 5.
k = _____
Answer: k = -0.75
When proportional relationships are graphed, the points the line runs through can be used to find the constant of proportionality.
This line runs through points (3,4) and (7,1).
First, find the proportion of this relationship by choosing one point and inserting its coordinates into the proportion equation.
k = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\) k = \(\frac{1-4}{7-3}\) = \(\frac{-3}{4}\) = -0.75
The constant of proportionality for this line is -0.75.
Therefore, k = -0.75
Solve each problem.
Question 6.
Lisa ran 3\(\frac{1}{2}\) miles in 21 minutes. At that rate, how long would it take her to run 5 miles?
It would take Lisa ____ minutes to run 5 miles.
Answer: It would take Lisa 30 minutes to run 5 miles.
Lisa ran 3\(\frac{1}{2}\) miles in 21 minutes. At that rate, how long would it take her to run 5 miles?
Let x be the time taken by her to cover 5 miles
First simplify 3\(\frac{1}{2}\) which is \(\frac{7}{2}\)
Therefore, 3\(\frac{1}{2}\) ÷ 5 = \(\frac{21}{x}\)
\(\frac{7}{2}\) ÷ 5 = \(\frac{21}{x}\)
\(\frac{7}{10}\) = \(\frac{21}{x}\)
By cross multiplication,
7 × x = 10 × 21
x = \(\frac{210}{7}\)
Therefore, x = 30 minutes
It would take Lisa 30 minutes to run 5 miles.
Question 7.
Manuel biked 12\(\frac{1}{4}\) if miles in 45 minutes. At that rate, how far could he go in I hour?
Manuel could bike ____ miles in 1 hour.
Answer: Manuel could bike 16\(\frac{1}{3}\) miles in 1 hour.
Manuel biked 12\(\frac{1}{4}\) if miles in 45 minutes. At that rate, how far could he go in I hour?
Let x be the miles covered by him in 1 hour, which is 60 minutes.
First simplify 12\(\frac{1}{4}\) which is \(\frac{49}{4}\)
Therefore, 12\(\frac{1}{4}\) ÷ x = \(\frac{45}{60}\)
\(\frac{49}{4}\) ÷ x = \(\frac{45}{60}\)
\(\frac{49}{4x}\) = \(\frac{45}{60}\)
\(\frac{49}{4x}\) = \(\frac{3}{4}\)
By cross multiplication,
3 × 4x = 49 × 4
x = \(\frac{49}{3}\) = 16\(\frac{1}{3}\)
Therefore, Manuel could bike 16\(\frac{1}{3}\) miles in 1 hour.
Question 8.
A recipe to make 5 cupcakes calls for 10 tablespoons of sugar. Alicia wants to make 10 cupcakes using this recipe. What equation will she need to use to find out how many tablespoons of sugar to use?
Equation: _________
Answer: Equation: \(\frac{5}{10}\) = \(\frac{10}{x}\)
A recipe to make 5 cupcakes calls for 10 tablespoons of sugar. Alicia wants to make 10 cupcakes using this recipe.
Let x be the number tablespoons of sugar for 10 cupcakes
Equation: \(\frac{5}{10}\) = \(\frac{10}{x}\)
By cross multiplication,
5 × x = 10 × 10
5x = 100
x = \(\frac{100}{5}\) = 20
Therefore, Alicia wants 20 tablespoons of sugar to make 10 cupcakes
Question 9.
Luis has $660 in his savings account earning 4\(\frac{1}{2}\) % interest. How much interest will he earn in 2 years? How much money will be in the account?
Luis will earn ____ in interest.
He will have a total of ____ in his account.
Answer: Luis will earn $59.4 in interest.
He will have a total of $719.4 in his account.
Luis has $660 in his savings account earning 4\(\frac{1}{2}\) % interest. How much interest will he earn in 2 years
Proportional relationships can be used to solve ratio and percent problems.
Now, write a proportion using the above information
x = 4\(\frac{1}{2}\) %× 660 × 2
x = \(\frac{9}{2}\) %× 660 × 2
x = \(\frac{9}{2}\) × \(\frac{1}{100}\) × 660 × 2
x = \(\frac{9}{200}\) × 660 × 2
After simplification,
x = $59.4
Total amount = $660 + $59.4 = $719.4
Luis will earn $59.4 in interest.
He will have a total of $719.4 in his account.
Question 10.
Mrs. Cole borrowed $ 1,200 for 6 months year) at 3\(\frac{1}{4}\) % interest. How much interest will she pay? What is the total amount she will pay?
Mrs. Cole will pay ____ in interest.
She will pay a total of ____.
Answer: Mrs. Cole will pay $ 19.5 in interest.
She will pay a total of $ 1219.5
Mrs. Cole borrowed $ 1,200 for 6 months year at 3\(\frac{1}{4}\) % interest.
6 months in year can be indicated as \(\frac{1}{2}\) fraction
Proportional relationships can be used to solve ratio and percent problems.
Now, write a proportion using the above information
x = 3\(\frac{1}{4}\) %× 1200 × \(\frac{1}{2}\)
x = \(\frac{13}{4}\) %× 1200 × \(\frac{1}{2}\)
x = \(\frac{13}{4}\) × \(\frac{1}{100}\) × 1200 × \(\frac{1}{2}\)
x = \(\frac{13}{400}\) × 1200 × \(\frac{1}{2}\)
x = \(\frac{13 × 1200 × 1}{400 × 2}\)
x = \(\frac{15600}{800}\)
After simplification,
x = $ 19.5
Total amount = 1200 + 19.5 = 1219.5
Therefore, Mrs. Cole will pay $ 19.5 in interest.
She will pay a total of $ 1219.5
Question 11.
Flo worked for 9 hours and has earned $108.00. She is planning to work 40 hours this week. What equation will she need to use to find out how much she will be paid?
Equation: _________
Answer: Equation: \(\frac{9}{40}\) = \(\frac{108}{x}\)
Flo worked for 9 hours and has earned $108.00. She is planning to work 40 hours this week.
Let x be the amount she earned to work for 40 hours
Equation: \(\frac{9}{40}\) = \(\frac{108}{x}\)
By cross multiplication,
9 × x = 40 × 108
9x = 4320
Therefore, x = $480
Question 12.
Ansley went for a long hike and burned 452 calories in 2\(\frac{1}{4}\) hours. Bobbi decided to go for a jog and burned 1,045 calories in 3\(\frac{1}{2}\); hours. Who burned the most calories per hour?
Let a represent Ansley’s and b represent Bobbi’s calories burned.
Equivalent Ratio 1: __________
Equivalent Ratio 2: __________
______ burned the most calories per hour.
Answer: Equivalent Ratio 1: \(\frac{452}{2 \frac{1}{4}}\) = \(\frac{c}{t}\) a
Equivalent Ratio 2: \(\frac{1045}{3 \frac{1}{2}}\) = \(\frac{c}{t}\) b
Bobbi burned the most calories per hour.
Ansley went for a long hike and burned 452 calories in 2\(\frac{1}{4}\) hours. Bobbi decided to go for a jog and burned 1,045 calories in 3\(\frac{1}{2}\); hours
Let a represent Ansley’s calories burned and b represent Bobbi’s calories burned.
Equivalent Ratio 1: \(\frac{452}{2 \frac{1}{4}}\) = \(\frac{c}{t}\) a
\(\frac{452}{ \frac{9}{4}}\) = \(\frac{c}{t}\) a
= \(\frac{1808}{9}\) = 200.888 calories per hour.
Equivalent Ratio 2: \(\frac{1045}{3 \frac{1}{2}}\) = \(\frac{c}{t}\) b
\(\frac{1045}{ \frac{7}{2}}\) = \(\frac{c}{t}\) b
= \(\frac{2090}{7}\) = 298.571 calories per hour.
Therefore, by comparing the above equivalent ratios, Bobbi burned the most calories per hour.