Spectrum Math Grade 7 Chapter 2 Posttest Answer Key

This handy Spectrum Math Grade 7 Answer Key Chapter 2 Posttest provides detailed answers for the workbook questions

Spectrum Math Grade 7 Chapter 2 Posttest Answers Key

Check What You Learned

Multiplying and Dividing Rational Numbers

Rewrite each expression using the distributive property.

Question 1.
a. 7 × (10 + a) =
_____________
Answer: (7 × 10) + (7 × a)
7 × (10 + a) = (7 × 10) + (7 × a)
The distributive property states that multiplying the sum of two or more addends by a number yields the same outcome as multiplying each addend separately by the number and combining the resulting products.

b. (2 × c) + (2 × d) =
_____________
Answer: 2 × (c + d)
(2 × c) + (2 × d) = 2 × (c + d)
The distributive property states that multiplying the sum of two or more addends by a number yields the same outcome as multiplying each addend separately by the number and combining the resulting products.

Question 2.
a. (y × 2) + (y × 6) =
_____________
Answer: y × (2 + 6)
(y × 2) + (y × 6) = y × (2 + 6)
The distributive property states that multiplying the sum of two or more addends by a number yields the same outcome as multiplying each addend separately by the number and combining the resulting products.

b. 5 × (k + 4) =
_____________
Answer: (5 × k) + (5 × 4)
5 × (k + 4) = (5 × k) + (5 × 4)
The distributive property states that multiplying the sum of two or more addends by a number yields the same outcome as multiplying each addend separately by the number and combining the resulting products.

Identify the property described as commutative, associative, identity, or zero.

Question 3.
When three or more numbers are multiplied together, the product is the same regardless of how the factors are grouped. _________
Answer: associative property
When three or more numbers are multiplied together, the product is the same regardless of how the factors are grouped is called associative property.
According to the associative principle of multiplication, when multiplying three integers, the outcome will always be the same regardless of how the numbers are grouped. If there are three numbers, x, y and z, the associative property of multiplication implies that x × (y × z) = (x × y) × z.

Question 4.
When zero is divided by any number, the quotient is always 0. _________
Answer: zero property
When zero is divided by any number, the quotient is always 0 is called zero property.
According to the zero property of division, if 0(zero) is divided by any other number, the result will be zero. If there is a number, x then the zero property of division implies that 0 ÷ x = 0.

Question 5.
The product of any number and 1 is that number. ___________
Answer: Identity Property
The product of any number and 1 is that number is called Identity Property.
According to the identity property of multiplication, if a number is multiplied by 1 (one), the result will be the original number. This property is applied when numbers are multiplied by 1. If there is a number, x then the identity property implies that x × 1 = x.

Question 6.
When two numbers are multiplied together, the product is the same regardless of the order of the factors. ________
Answer: Commutative Property
When two numbers are multiplied together, the product is the same regardless of the order of the factors is called Commutative Property.
According to the commutative property of multiplication, changing the order of the numbers we are multiplying does not change the product. If there are two numbers, x and y, the commutative property of multiplication implies that x × y = y × x.

Question 7.
a. y × x = x × y
____________
Answer: Commutative Property
y × x = x × y
The above expression is the example for Commutative Property.
When two numbers are multiplied together, the product is the same regardless of the order of the factors is called Commutative Property.
According to the commutative property of multiplication, changing the order of the numbers we are multiplying does not change the product. If there are two numbers, x and y, the commutative property of multiplication implies that x × y = y × x.

b. (a × b) × c = a × (b × c)
____________
Answer: associative property
(a × b) × c = a × (b × c)
The above expression is the example for associative property.
According to the associative principle of multiplication, when multiplying three integers, the outcome will always be the same regardless of how the numbers are grouped. If there are three numbers, x, y and z, the associative property of multiplication implies that x × (y × z) = (x × y) × z.

Question 8.
a. 5 × 1 = 5
____________
Answer: Identity Property
5 × 1 = 5
The above expression is the example for Identity Property.
According to the identity property of multiplication, if a number is multiplied by 1 (one), the result will be the original number. This property is applied when numbers are multiplied by 1. If there is a number, x then the identity property implies that x × 1 = x.

b. 0 ÷ 6 = 0
____________
Answer: Zero Property
0 ÷ 6 = 0
The above expression is the example for zero property.
When zero is divided by any number, the quotient is always 0 is called zero property.
According to the zero property of division, if 0(zero) is divided by any other number, the result will be zero. If there is a number, x then the zero property of division implies that 0 ÷ x = 0.

Change each rational number into a decimal using long division. Place a line over digits which repeat.

Question 9.
a. \(\frac{2}{9}\)
_________________________
Answer: 0.222
Spectrum Math Grade 7 Chapter 2 Posttest Answers Key
Rational numbers can be converted into decimals using long division. All fractions will be turned into decimals that either terminate or repeat. Repeating decimals can be given as a same pattern of numbers will get when we perform division. A line will be placed above the digits which are repeating.
Here, If we divide 2 by 9, we will get repeating decimal 0.222, so a line was indicated above 2.
Therefore, \(\frac{2}{9}\) = 0.222

b. \(\frac{4}{9}\) =
_______________________________
Answer: 0.444
Spectrum Math Grade 7 Chapter 2 Posttest Answers Key
Rational numbers can be converted into decimals using long division. All fractions will be turned into decimals that either terminate or repeat. Repeating decimals can be given as a same pattern of numbers will get when we perform division. A line will be placed above the digits which are repeating.
Here, If we divide 4 by 9, we will get repeating decimal 0.444, so a line was indicated above 4.
Therefore, \(\frac{4}{9}\) = 0.444

Question 10.
a. \(\frac{1}{11}\)
_____________________________
Answer: 0.0909
Spectrum Math Grade 7 Chapter 2 Posttest Answers Key
Rational numbers can be converted into decimals using long division. All fractions will be turned into decimals that either terminate or repeat. Repeating decimals can be given as a same pattern of numbers will get when we perform division. A line will be placed above the digits which are repeating.
Here, If we divide 1 by 11, we will get repeating decimal 0.0909, so a line was indicated above 09.
Therefore, \(\frac{1}{11}\) = 0.0909

b. \(\frac{2}{5}\) =
________________
Answer:
Spectrum Math Grade 7 Chapter 2 Posttest Answers Key
Rational numbers can be converted into decimals using long division. All fractions will be turned into decimals that either terminate or repeat. Repeating decimals can be given as a same pattern of numbers will get when we perform division. A line will be placed above the digits which are repeating.
Here, If we divide 2 by 5, we will get terminating decimal 0.4
Therefore, \(\frac{2}{5}\) = 0.4

Multiply or divide. Write answers in simplest form.

Question 11.
a. \(\frac{3}{4}\) × \(\frac{1}{6}\) = ____
Answer: \(\frac{1}{8}\)
\(\frac{3}{4}\) × \(\frac{1}{6}\)
Reduce the above fractions into simplest form if possible.
Then, multiply the numerators and denominators separately.
= \(\frac{3 × 1}{4 × 6}\)
Divide 3 in numerator and 6 in denominator by 3, which is a common factor
= \(\frac{1 × 1}{4 × 2}\)
= \(\frac{1}{8}\)
Therefore ,\(\frac{3}{4}\) × \(\frac{1}{6}\) = \(\frac{1}{8}\)

b. \(\frac{5}{7}\) × \(\frac{2}{3}\) = ____
Answer: \(\frac{10}{21}\)
\(\frac{5}{7}\) × \(\frac{2}{3}\)
Reduce the above fractions into simplest form if possible.
Then, multiply the numerators and denominators separately.
= \(\frac{5 × 2}{7 × 3}\)
= \(\frac{10}{21}\)
Therefore, \(\frac{5}{7}\) × \(\frac{2}{3}\) = \(\frac{10}{21}\)

c. 5\(\frac{1}{2}\) × 1\(\frac{1}{4}\) = ____
Answer:6\(\frac{7}{8}\)
5\(\frac{1}{2}\) × 1\(\frac{1}{4}\)
Convert the above numbers into improper fractions to make calculations easy
=\(\frac{11}{2}\) × \(\frac{5}{4}\)
Reduce the above fractions into simplest form if possible.
Then, multiply the numerators and denominators separately.
= \(\frac{11 × 5}{2 × 4}\)
= \(\frac{55}{8}\)
=  6\(\frac{7}{8}\)
Therefore, 5\(\frac{1}{2}\) × 1\(\frac{1}{4}\) =6\(\frac{7}{8}\)

Question 12.
a. 5\(\frac{1}{4}\) ÷ \(\frac{1}{6}\) = ____
Answer: 31\(\frac{1}{2}\)
5\(\frac{1}{4}\) ÷ \(\frac{1}{6}\)
Convert the above numbers into improper fractions to make calculations easy
= \(\frac{21}{4}\) ÷ \(\frac{1}{6}\)
To divide by a fraction, multiply by its reciprocal.
Here, the reciprocal of \(\frac{1}{6}\) = \(\frac{6}{1}\)
\(\frac{21}{4}\) ÷ \(\frac{1}{6}\) = \(\frac{21}{4}\) × \(\frac{6}{1}\)
Reduce the above fractions into simplest form if possible.
Then, multiply the numerators and denominators separately.
= \(\frac{21 × 6}{4 × 1}\)
Divide 6 in numerator and 4 in denominator by 2, which is a common factor
= \(\frac{21 × 3}{2 × 1}\)
= \(\frac{63}{2}\)
= 31\(\frac{1}{2}\)
Therefore, 5\(\frac{1}{4}\) ÷ \(\frac{1}{6}\) = 31\(\frac{1}{2}\)

b. 6\(\frac{4}{7}\) ÷ 12 = ____
Answer: \(\frac{23}{42}\)
6\(\frac{4}{7}\) ÷ 12
Convert the above numbers into improper fractions to make calculations easy
= \(\frac{46}{7}\) ÷ \(\frac{12}{1}\)
To divide by a fraction, multiply by its reciprocal.
Here, the reciprocal of \(\frac{12}{1}\) = \(\frac{1}{12}\)
So, \(\frac{46}{7}\) ÷ \(\frac{12}{1}\) = \(\frac{46}{7}\) × \(\frac{1}{12}\)
Reduce the above fractions into simplest form if possible.
Then, multiply the numerators and denominators separately.
= \(\frac{46 × 1}{7 × 12}\)
Divide 46 in numerator and 12 in denominator by 2, which is a common factor
= \(\frac{23 × 1}{7 × 6}\)
= \(\frac{23}{42}\)
Therefore, 6\(\frac{4}{7}\) ÷ 12 = \(\frac{23}{42}\)

c. 1\(\frac{1}{2}\) ÷ \(\frac{3}{5}\) = ____
Answer: 2\(\frac{1}{2}\)
1\(\frac{1}{2}\) ÷ \(\frac{3}{5}\)
Convert the above numbers into improper fractions to make calculations easy
= \(\frac{3}{2}\) ÷ \(\frac{3}{5}\)
To divide by a fraction, multiply by its reciprocal.
Here, the reciprocal of \(\frac{3}{5}\) = \(\frac{5}{3}\)
So, \(\frac{3}{2}\) ÷ \(\frac{3}{5}\) = \(\frac{3}{2}\)  ×  \(\frac{5}{3}\)
Reduce the above fractions into simplest form if possible.
Then, multiply the numerators and denominators separately.
= \(\frac{3 × 5}{2 × 3}\)
Divide 3 in numerator and 3 in denominator by 3, which is a common factor
= \(\frac{1 × 5}{2 × 1}\)
= \(\frac{5}{2}\)
= 2\(\frac{1}{2}\)
Therefore, 1\(\frac{1}{2}\) ÷ \(\frac{3}{5}\)= 2\(\frac{1}{2}\)

Question 13.
a. 7 × (-6) = ____
Answer: -42
7 × (-6) = -42
The product of two positive integers is always positive.
The product of two negative integers is always positive.
The product of one positive and one negative integer is always negative.
Here -6 is a negative integer and 7 is a positive integer therefore the result of their product is a negative integer.

b. 3 × (-4) = ____
Answer: -12
3 × (-4) = -12
The product of two positive integers is always positive.
The product of two negative integers is always positive.
The product of one positive and one negative integer is always negative.
Here -4 is a negative integer and 3 is a positive integer therefore the result of their product is a negative integer.

c. -5 × (-2) = ____
Answer: 10
-5 × (-2) = 10
The product of two positive integers is always positive.
The product of two negative integers is always positive.
The product of one positive and one negative integer is always negative.
Here -2 and -5 both are negative integers hence their product is also a negative integer.

Question 14.
a. 12 ÷ (-4) = ____
Answer: -3
let 12 ÷ (-4) = x
As division and multiplication are inverse operations, the above equation can be written as follows
12 = x × (-4)
x = -3
12 ÷ (-4) = -3
Therefore Inverse operation: -3
The quotient of two integers with the same sign is positive and the quotient of two integers with different signs is negative. Here -18 and 9 both are having different sign hence the result would be negative.

b. -15 ÷ (-5) = ____
Answer: 2
let -15 ÷ (-5) = x
As division and multiplication are inverse operations, the above equation can be written as follows
-15 = x × (-5)
x = 2
-15 ÷ (-5) = 2
Therefore Inverse operation: 2
The quotient of two integers with the same sign is positive and the quotient of two integers with different signs is negative. Here -40 and -4 both are having same sign hence the result would be positive.

c. -21 ÷ 7 = ____
Answer: -3
let -21 ÷ 7 = x
As division and multiplication are inverse operations, the above equation can be written as follows
-21 = x × 7
x = -3
-21 ÷ 7 = -3
Therefore Inverse operation: -3
The quotient of two integers with the same sign is positive and the quotient of two integers with different signs is negative. Here -18 and 9 both are having different sign hence the result would be negative.

Solve each problem.

Question 15.
A bucket that holds 5\(\frac{1}{4}\) if gallons of water is being used to fill a tub that can hold 34\(\frac{1}{8}\) gallons. How many buckets will be needed to fill the tub?
buckets are needed to fill the tub.
Answer: 6\(\frac{1}{2}\)
A bucket that holds 5\(\frac{1}{4}\) if gallons of water is being used to fill a tub that can hold 34\(\frac{1}{8}\) gallons.
number of buckets needed to fill the tub = 34\(\frac{1}{8}\) ÷ 5\(\frac{1}{4}\)
Convert the above numbers into improper fractions to make the calculations easy.
34\(\frac{1}{8}\) = \(\frac{273}{8}\)
5\(\frac{1}{4}\) = \(\frac{21}{4}\)
So, 34\(\frac{1}{8}\) ÷ 5\(\frac{1}{4}\) = \(\frac{273}{8}\) ÷ \(\frac{21}{4}\)
To divide by a fraction, multiply by its reciprocal.
Here, the reciprocal of \(\frac{21}{4}\) = \(\frac{4}{21}\)
Hence, \(\frac{273}{8}\) × \(\frac{4}{21}\)
Reduce the above fractions into simplest form if possible.
Then, multiply the numerators and denominators separately.
= \(\frac{273 × 4}{8 × 21}\)
Divide 4 in numerator and 8 in denominator with 4, which is a common factor
= \(\frac{273 × 1}{2 × 21}\)
Divide 273 in numerator and 21 in denominator with 21, which is a common factor
= \(\frac{13 × 1}{2 × 1}\)
= \(\frac{13}{2}\)
= 6\(\frac{1}{2}\)
Therefore, 6\(\frac{1}{2}\) buckets are needed to fill the tub.

Question 16.
A black piece of pipe is 8\(\frac{1}{3}\) centimeters long. A silver piece of pipe is 2\(\frac{3}{5}\) times longer. How long is the silver piece of pipe?
The silver piece is ______ centimeters long.
Answer: 21\(\frac{2}{3}\)
A black piece of pipe is 8\(\frac{1}{3}\) centimeters long. A silver piece of pipe is 2\(\frac{3}{5}\) times longer.
The number of centimeters the silver piece is 8\(\frac{1}{3}\) × 2\(\frac{3}{5}\)
Convert the above numbers into improper fractions to make the calculations easy.
8\(\frac{1}{3}\)= \(\frac{25}{3}\)
2\(\frac{3}{5}\) = \(\frac{13}{5}\)
So, 8\(\frac{1}{3}\) × 2\(\frac{3}{5}\) = \(\frac{25}{3}\) × \(\frac{13}{5}\)
Reduce the above fractions into simplest form if possible.
Then, multiply the numerators and denominators separately.
= \(\frac{25 × 13}{3 × 5}\)
Divide 25 in numerator and 5 in denominator with 5, which is a common factor
= \(\frac{5 × 13}{3 × 1}\)
= \(\frac{65}{3}\)
= 21\(\frac{2}{3}\)
Therefore, The silver piece is21\(\frac{2}{3}\) centimeters long.

Question 17.
One section of wood is 3\(\frac{5}{8}\) meters long. Another section is twice that long. When the two pieces are put together, how long is the piece of wood that is created?
The piece of wood is _____ meters long.
Answer: 10\(\frac{7}{8}\)
One section of wood is 3\(\frac{5}{8}\) meters long.
Convert the above number into improper fraction. Then, 3\(\frac{5}{8}\)  = \(\frac{29}{8}\)
Another section is twice that long i.e. 2 × 3\(\frac{5}{8}\) = 2 × \(\frac{29}{8}\)
= \(\frac{2}{1}\) × \(\frac{29}{8}\)
Reduce the above fractions into simplest form if possible.
Then, multiply the numerators and denominators separately.
= \(\frac{2 × 29}{1 × 8}\)
= \(\frac{58}{8}\)
When the two pieces are put together, the piece of wood that is created = \(\frac{29}{8}\) + \(\frac{58}{8}\)
= \(\frac{29+58}{8}\)
= \(\frac{87}{8}\)
= 10\(\frac{7}{8}\)
Therefore, The piece of wood is 10\(\frac{7}{8}\) meters long.

Question 18.
Danielle wants to fill a box with dirt to start a garden. If the box is 2\(\frac{1}{5}\) feet long, by 1\(\frac{1}{3}\) feet wide, and 1\(\frac{1}{2}\) feet deep, how much dirt does Danielle need to fill up the box for her garden?
Danielle needs ____ cubic feet of dirt.
Answer: 4\(\frac{4}{10}\)
If the box is 2\(\frac{1}{5}\) feet long, by 1\(\frac{1}{3}\) feet wide, and 1\(\frac{1}{2}\) feet deep
The amount of dirt needed to fill the box = 2\(\frac{1}{5}\) × 1\(\frac{1}{3}\) × 1\(\frac{1}{2}\)
Convert the above number into improper fraction to make calculations easy.
= \(\frac{11}{5}\) × \(\frac{4}{3}\) × \(\frac{3}{2}\)
Reduce the above fractions into simplest form if possible.
Then, multiply the numerators and denominators separately.
= \(\frac{11 × 4 × 3}{5 × 3 × 2}\)
Divide 3 in numerator and 3 in denominator with 3, which is a common factor
= \(\frac{11 × 4 × 1}{5 × 1 × 2}\)
= \(\frac{44}{10}\)
= 4\(\frac{4}{10}\)
Therefore, Danielle needs 4\(\frac{4}{10}\) cubic feet of dirt.

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