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Go through the Spectrum Math Grade 6 Answer Key Chapter 3 Lesson 3.2 Solving Ratios and get the proper assistance needed during your homework.

Spectrum Math Grade 6 Chapter 3 Lesson 3.2 Solving Ratios Answers Key

A proportion can be used in problem solving.
The ratio of apples to oranges is 4 to 5. There are 20 oranges in the basket.
How many apples are there?
\(\frac{4}{5}\) = \(\frac{n}{20}\) Set up a proportion, using n for the missing number.
4 × 20 = 5 × n Cross-multiply.
\(\frac{80}{5}\) = n Solve for n.
16 = n There are 16 apples.

Solve.
Question 1.
a. \(\frac{1}{3}\) = \(\frac{n}{24}\) ___________
Answer:
The value of n in \(\frac{1}{3}\) = \(\frac{n}{24}\) is 8.

Explanation:
\(\frac{1}{3}\) = \(\frac{n}{24}\)
=> 1 × 24 = n × 3
=> 24 = 3n
=> 24 ÷ 3 = n
=> 8 = n.

b. \(\frac{4}{9}\) = \(\frac{n}{36}\) ___________
Answer:
The value of n in \(\frac{4}{9}\) = \(\frac{n}{36}\) is 16.

Explanation:
\(\frac{4}{9}\) = \(\frac{n}{36}\)
=> 4 × 36 = n × 9
=>144 = 9n
=> 144 ÷ 9 = n
=> 16 = n.

c. \(\frac{5}{45}\) = \(\frac{n}{9}\) ___________
Answer:
The value of n in \(\frac{5}{45}\) = \(\frac{n}{9}\) is 1.

Explanation:
\(\frac{5}{45}\) = \(\frac{n}{9}\)
=> 5 × 9 = n × 45
=> 45 = 45n
=> 45 ÷ 45 = n
=> 1 = n.

Question 2.
a. \(\frac{3}{5}\) = \(\frac{n}{15}\) ___________
Answer:
The value of n in \(\frac{3}{5}\) = \(\frac{n}{15}\) is 9.

Explanation:
\(\frac{3}{5}\) = \(\frac{n}{15}\)
=> 3 × 15 = n × 5
=> 45 = 5n
=> 45 ÷ 5 = n
=> 9 = n.

b. \(\frac{10}{70}\) = \(\frac{n}{7}\) ___________
Answer:
The value of n in \(\frac{10}{70}\) = \(\frac{n}{7}\) is 1.

Explanation:
\(\frac{10}{70}\) = \(\frac{n}{7}\)
=> 10 × 7 = n × 70
=> 70 = 70n
=> 70 ÷ 70 = n
=> 1 = n.

c. \(\frac{25}{40}\) = \(\frac{n}{16}\) ___________
Answer:
The value of n in \(\frac{25}{40}\) = \(\frac{n}{16}\) is 10.

Explanation:
\(\frac{25}{40}\) = \(\frac{n}{16}\)
=> 25 × 16 = n × 40
=> 400 = 40n
=> 400 ÷ 40 = n
=> 10 = n.

Question 3.
a. \(\frac{7}{12}\) = \(\frac{n}{36}\) ___________
Answer:
The value of n in \(\frac{7}{12}\) = \(\frac{n}{36}\) is 21.

Explanation:
\(\frac{7}{12}\) = \(\frac{n}{36}\)
=> 7 × 36 = n × 12
=> 252 = 12n
=> 252 ÷ 12 = n
=> 21 =n.

b. \(\frac{13}{26}\) = \(\frac{n}{4}\) ___________
Answer:
The value of n in \(\frac{13}{26}\) = \(\frac{n}{4}\) is 2.

Explanation:
\(\frac{13}{26}\) = \(\frac{n}{4}\)
=> 13 × 4 = n × 26
=> 52 = 26n
=> 52 ÷ 26 = n
=> 2 = n.

c. \(\frac{7}{1}\) = \(\frac{n}{3}\) ___________
Answer:
The value of n in \(\frac{7}{1}\) = \(\frac{n}{3}\) is 21.

Explanation:
\(\frac{7}{1}\) = \(\frac{n}{3}\)
=> 7 × 3 = n × 1
=> 21 = 1n
=> 21 ÷ 1 = n
=> 21 = n.

Question 4.
a. \(\frac{8}{5}\) = \(\frac{n}{40}\) ___________
Answer:
The value of n in \(\frac{8}{5}\) = \(\frac{n}{40}\) is 64.

Explanation:
\(\frac{8}{5}\) = \(\frac{n}{40}\)
=> 8 × 40 = n × 5
=> 320 = 5n
=> 320 ÷ 5 = n
=> 64 = n.

b. \(\frac{2}{6}\) = \(\frac{n}{33}\) ___________
Answer:
The value of n in \(\frac{2}{6}\) = \(\frac{n}{33}\) is 11.

Explanation:
\(\frac{2}{6}\) = \(\frac{n}{33}\)
=> 2 × 33 = n × 6
=> 66 = 6n
=> 66 ÷ 6 = n
=> 11 = n.

c. \(\frac{5}{13}\) = \(\frac{n}{39}\) ___________
Answer:
The value of n in latex]\frac{5}{13}[/latex] = \(\frac{n}{39}\) is 15.

Explanation:
latex]\frac{5}{13}[/latex] = \(\frac{n}{39}\)
=> 5 × 39 = n × 13
=> 195 = 13n
=> 195 ÷ 13 = n
=> 15 = n.

Question 5.
a. \(\frac{5}{6}\) = \(\frac{n}{18}\) ___________
Answer:
The value of n in \(\frac{5}{6}\) = \(\frac{n}{18}\) is 15.

Explanation:
\(\frac{5}{6}\) = \(\frac{n}{18}\)
=> 5 × 18 = n × 6
=> 90 = 6n
=> 90 ÷ 6 = n
=> 15 = n.

b. \(\frac{9}{8}\) = \(\frac{n}{32}\) ___________
Answer:
The value of n in \(\frac{9}{8}\) = \(\frac{n}{32}\) is 36.

Explanation:
\(\frac{9}{8}\) = \(\frac{n}{32}\)
=> 9 × 32 = n × 8
=> 288 = 8n
=> 288 ÷ 8 = n
=> 36 = n.

c. \(\frac{2}{3}\) = \(\frac{n}{15}\) ___________
Answer:
The value of n in \(\frac{2}{3}\) = \(\frac{n}{15}\) is 10.

Explanation:
\(\frac{2}{3}\) = \(\frac{n}{15}\)
=> 2 × 15 = n × 3
=> 30 = 3n
=> 30 ÷ 3 = n
=> 10 = n.

A proportion can be used in problem-solving.
The ratio of apples to oranges is 4 to 5. There are 20 oranges in the basket. How many apples are there?
\(\frac{4}{5}\) = \(\frac{4}{20}\) ___________ Set up a proportion, using n for the missing number.
4 × 20 = 5 × n Cross-multiply
\(\frac{80}{5}\) =n Solve for n
16 = n There are 16 apples.

Solve.
Question 1.
a. \(\frac{1}{3}\) = \(\frac{n}{24}\) _________
Answer:
The value of n in \(\frac{1}{3}\) = \(\frac{n}{24}\) is 8.

Explanation:
\(\frac{1}{3}\) = \(\frac{n}{24}\)
=> 1 × 24 = n × 3
=> 24 = 3n
=> 24 ÷ 3 = n
=> 8 = n.

b. \(\frac{4}{9}\) = \(\frac{n}{36}\) _________
Answer:
The value of n in \(\frac{4}{9}\) = \(\frac{n}{36}\) is 16.

Explanation:
\(\frac{4}{9}\) = \(\frac{n}{36}\)
=> 4 × 36 = n × 9
=> 144 = 9n
=> 144 ÷ 9 = n
=> 16 = n.

c. \(\frac{5}{45}\) = \(\frac{n}{9}\) _________
Answer:
The value of n in \(\frac{5}{45}\) = \(\frac{n}{9}\) is 1.

Explanation:
\(\frac{5}{45}\) = \(\frac{n}{9}\)
= 4 × 9 = n × 45
=> 45 = 45n
=> 45 ÷ 45 = n
=> 1 = n.

Question 2.
a. \(\frac{3}{5}\) = \(\frac{n}{15}\) _________
Answer:
The value of n in \(\frac{3}{5}\) = \(\frac{n}{15}\) is 9.

Explanation:
\(\frac{3}{5}\) = \(\frac{n}{15}\)
=> 3 × 15 = n × 5
=> 45 = 5n
=> 45 ÷ 5 = n
=> 9 = n.

b. \(\frac{10}{70}\) = \(\frac{n}{7}\) _________
Answer:
The value of n in \(\frac{10}{70}\) = \(\frac{n}{7}\) is 1.

Explanation:
\(\frac{10}{70}\) = \(\frac{n}{7}\)
=> 10 × 7 = n × 70
=> 70 = 70n
=> 70 ÷ 70 = n
=> 1 = n.

c. \(\frac{25}{40}\) = \(\frac{n}{16}\) _________
Answer:
The value of n in \(\frac{25}{40}\) = \(\frac{n}{16}\) is 10.

Explanation:
\(\frac{25}{40}\) = \(\frac{n}{16}\)
=> 25 × 16 = n × 40
=> 400 = 40n
=> 400 ÷ 40 = n
=> 10 = n.

Question 3.
a. \(\frac{7}{12}\) = \(\frac{n}{36}\) _________
Answer:
The value of n in \(\frac{7}{12}\) = \(\frac{n}{36}\) is 21.

Explanation:
\(\frac{7}{12}\) = \(\frac{n}{36}\)
=> 7 × 36 = n × 12
=> 252 = 12n
=> 252 ÷ 12 = n
=> 21 = n.

b. \(\frac{13}{26}\) = \(\frac{n}{4}\) _________
Answer:
The value of n in \(\frac{13}{26}\) = \(\frac{n}{4}\) is 2.

Explanation:
\(\frac{13}{26}\) = \(\frac{n}{4}\)
=> 13 × 4 = n × 26
=> 52 = 26n
=> 52 ÷ 26 = n
=> 2 = n.

c. \(\frac{7}{1}\) = \(\frac{n}{3}\) _________
Answer:
The value of n in \(\frac{7}{1}\) = \(\frac{n}{3}\) is 21.

Explanation:
\(\frac{7}{1}\) = \(\frac{n}{3}\)
= 7 × 3 = n × 1
=> 21 = 1n
=> 21 ÷ 1 = n
=> 21 = n.

Question 4.
a. \(\frac{8}{5}\) = \(\frac{n}{40}\) _________
Answer:
The value of n in \(\frac{8}{5}\) = \(\frac{n}{40}\) is 64.

Explanation:
\(\frac{8}{5}\) = \(\frac{n}{40}\)
=> 8 × 40 = n × 5
=> 320 = 5n
=> 320 ÷ 5 = n
=> 64 = n.

b. \(\frac{2}{6}\) = \(\frac{n}{33}\) _________
Answer:
The value of n in \(\frac{2}{6}\) = \(\frac{n}{33}\) is 11.

Explanation:
\(\frac{2}{6}\) = \(\frac{n}{33}\)
=> 2 × 33 = n × 6
=> 66 = 6n
=> 66 ÷ 6 = n
=> 11 = n.

c. \(\frac{5}{13}\) = \(\frac{n}{39}\) _________
Answer:
The value of n in \(\frac{5}{13}\) = \(\frac{n}{39}\) is 15.

Explanation:
\(\frac{5}{13}\) = \(\frac{n}{39}\)
=> 5 × 39 = n × 13
=> 195 = 13n
=> 195 ÷ 13 = n
=> 15 = n.

Question 5.
a. \(\frac{5}{6}\) = \(\frac{n}{18}\) _________
Answer:
The value of n in \(\frac{5}{6}\) = \(\frac{n}{18}\) is 15.

Explanation:
\(\frac{5}{6}\) = \(\frac{n}{18}\)
=> 5 × 18 = n × 6
=> 90 = 6n
=> 90 ÷ 6 = n
=> 15 = n.

b. \(\frac{9}{8}\) = \(\frac{n}{32}\) _________
Answer:
The value of n in \(\frac{9}{8}\) = \(\frac{n}{32}\) is 36.

Explanation:
\(\frac{9}{8}\) = \(\frac{n}{32}\)
=> 9 × 32 = n × 8
=> 288 = 8n
=> 288 ÷ 8 = n
=> 36 = n.

c. \(\frac{2}{3}\) = \(\frac{n}{15}\) _________
Answer:
The value of n in \(\frac{2}{3}\) = \(\frac{n}{15}\) is 10.

Explanation:
\(\frac{2}{3}\) = \(\frac{n}{15}\)
=> 2 × 15 = n × 3
=> 30 = 3n
=> 30 ÷ 3 = n
=> 10 = n.

The missing number can appear any place in a proportion. Solve the same way.
\(\frac{2}{3}\) = \(\frac{6}{n}\)
3 × 6 = 2 × n
\(\frac{18}{2}\) = n
9 = n

\(\frac{3}{5}\) = \(\frac{n}{10}\)
3 × 10 = 5 × n
\(\frac{30}{5}\) = n
6 = n

\(\frac{3}{n}\) = \(\frac{6}{8}\)
3 × 8 = 6 × n
\(\frac{24}{6}\) = n
4 = n

\(\frac{n}{4}\) = \(\frac{3}{6}\)
4 × 3 = 6 × n
\(\frac{12}{6}\) = n
2 = n

Solve.
Question 1.
a. \(\frac{n}{3}\) = \(\frac{3}{9}\) _________
Answer:
The value of n in \(\frac{n}{3}\) = \(\frac{3}{9}\) is 1.

Explanation:
\(\frac{n}{3}\) = \(\frac{3}{9}\)
=> n × 9 = 3 × 3
=> 9n = 9
=> n = 9 ÷ 9
=> n = 1.

b. \(\frac{5}{3}\) = \(\frac{15}{n}\) _________
Answer:
The value of n in \(\frac{5}{3}\) = \(\frac{15}{n}\) is 9.

Explanation:
\(\frac{5}{3}\) = \(\frac{15}{n}\)
=> 5 × n = 15 × 3
=> 5n = 45
=> n = 45 ÷ 5
=> n = 9.

c. \(\frac{2}{n}\) = \(\frac{1}{4}\) _________
Answer:
The value of n in \(\frac{2}{n}\) = \(\frac{1}{4}\) is 8.

Explanation:
\(\frac{2}{n}\) = \(\frac{1}{4}\)
= 2 × 4 = 1 × n
=> 8 = 1n
=> 8 ÷ 1 = n
=> 8 = n.

Question 2.
a. \(\frac{15}{30}\) = \(\frac{2}{n}\) _________
Answer:
The value of n in \(\frac{15}{30}\) = \(\frac{2}{n}\) is 4.

Explanation:
\(\frac{15}{30}\) = \(\frac{2}{n}\)
=> 15 × n = 2 × 30
=> 15n = 60
=> n = 60 ÷ 15
=> n = 4.

b. \(\frac{4}{6}\) = \(\frac{n}{24}\) _________
Answer:
The value of n in \(\frac{4}{6}\) = \(\frac{n}{24}\) is 16.

Explanation:
\(\frac{4}{6}\) = \(\frac{n}{24}\)
=> 4 × 24 = n × 6
=> 96 = 6n
=> 96 ÷ 6 = n
=> 16 = n.

c. \(\frac{n}{7}\) = \(\frac{15}{21}\) _________
Answer:
The value of n in \(\frac{n}{7}\) = \(\frac{15}{21}\) is 5.

Explanation:
\(\frac{n}{7}\) = \(\frac{15}{21}\)
=> n × 21 = 15 × 7
=> 21n = 105
=> n = 105 ÷ 21
=> n = 5.

Question 3.
a. \(\frac{6}{n}\) = \(\frac{15}{20}\) _________
Answer:
The value of n in \(\frac{6}{n}\) = \(\frac{15}{20}\) is 8.

Explanation:
\(\frac{6}{n}\) = \(\frac{15}{20}\)
=> 6 × 20 = 15 × n
=> 120 = 15n
=> 120 ÷ 15 = n
=> 8 = n.

b. \(\frac{n}{12}\) = \(\frac{9}{18}\) _________
Answer:
The value of n in \(\frac{n}{12}\) = \(\frac{9}{18}\) is 6.

Explanation:
\(\frac{n}{12}\) = \(\frac{9}{18}\)
=> n × 18 = 9 × 12
=> 18n = 108
=> n = 108 ÷ 18
=> n = 6.

c. \(\frac{9}{2}\) = \(\frac{27}{n}\) _________
Answer:
The value of n in \(\frac{9}{2}\) = \(\frac{27}{n}\) is 6.

Explanation:
\(\frac{9}{2}\) = \(\frac{27}{n}\)
=> 9 × n = 27 × 2
=> 9n = 54
=> n = 54 ÷ 9
=> n = 6.

Question 4.
a. \(\frac{7}{9}\) = \(\frac{n}{63}\) _________
Answer:
The value of n in \(\frac{7}{9}\) = \(\frac{n}{63}\) is 49.

Explanation:
\(\frac{7}{9}\) = \(\frac{n}{63}\)
=> 7 × 63 = n × 9
=> 441 = 9n
=> 441 ÷ 9 = n
=> 49 = n.

b. \(\frac{15}{n}\) = \(\frac{12}{4}\) _________
Answer:
The value of n in \(\frac{15}{n}\) = \(\frac{12}{4}\) is 5.

Explanation:
\(\frac{15}{n}\) = \(\frac{12}{4}\)
=> 15 × 4 = 12 × n
=> 60 = 12n
=> 60 ÷ 12 = n
=> 5 = n.

c. \(\frac{40}{100}\) = \(\frac{n}{25}\) _________
Answer:
The value of n in \(\frac{40}{100}\) = \(\frac{n}{25}\) is 10.

Explanation:
\(\frac{40}{100}\) = \(\frac{n}{25}\)
=> 40 × 25 = n × 100
=> 1000 = 100n
=> 1000 ÷ 100 = n
=> 10 = n.

Question 5.
a. \(\frac{35}{n}\) = \(\frac{4}{8}\) _________
Answer:
The value of n in \(\frac{35}{n}\) = \(\frac{4}{8}\) is 70.

Explanation:
\(\frac{35}{n}\) = \(\frac{4}{8}\)
=> 35 × 8 = 4 × n
=> 280 = 4n
=> 280 ÷ 4 = n
=> 70 = n.

b. \(\frac{16}{4}\) = \(\frac{36}{n}\) _________
Answer:
The value of n in \(\frac{16}{4}\) = \(\frac{36}{n}\) is 9.

Explanation:
\(\frac{16}{4}\) = \(\frac{36}{n}\)
=> 16 × n = 36 × 4
=>16n = 144
=> n = 144 ÷ 16
=> n = 9.

c. \(\frac{n}{12}\) = \(\frac{25}{30}\) _________
Answer:
The value of n in \(\frac{n}{12}\) = \(\frac{25}{30}\) is 10.

Explanation:
\(\frac{n}{12}\) = \(\frac{25}{30}\)
=> n × 30 = 25 × 12
=> 30n = 300
=> n = 300 ÷ 30
=> n = 10.

This handy Spectrum Math Grade 7 Answer Key Chapter 3 Pretest provides detailed answers for the workbook questions

Spectrum Math Grade 7 Chapter 3 Pretest Answers Key

Expressions, Equations, and Inequalities

Rewrite each expression using the property indicated.

Question 1.
a. associative: (5 + 6) + 7
_____________
Answer: 5 + (6 + 7)
According to the associative principle of addition, when adding 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 addition implies that x + (y + z) = (x + y) + z. The grouping of addends does not change the sum.
(5 + 6) + 7 = 5 + (6 + 7)
18 = 18

b. identity: 56 × 1
_____________
Answer: 56
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.
56 × 1 = 56

Question 2.
a. zero: 0 ÷ 4
Answer: 0
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.
0 ÷ 4 = 0

b. commutative: 8 × 9
Answer: 9 × 8
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.
8 × 9 = 9 × 8
72 = 72

Question 3.
a. distributive: 3 × (5 – 2)
_____________
Answer: (3 × 5) – (3 × 2)
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. a × (b – c) = (a × b) – (a × c) is the rule for distributive property.
3 × (5 – 2) = (3 × 5) – (3 × 2)
9 = 9

b. associative: (7 × 2) × 3
_____________
Answer: 7 × (2 × 3)
According to the associative principle of addition, when adding 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 addition implies that x × (y × z) = (x × y) × z. The grouping of addends does not change the sum.
(7 × 2) × 3 = 7 × (2 × 3)
42 = 42

Write each phrase as an expression or equation.

Question 4.
a. five less than a number
_____________
Answer: x -5
Let x be the number.
The expression for ‘five less than a number’ can be given as x -5

b. eight more than a number
___________________
Answer: x + 8
Let x be the number.
The expression for ‘eight more than a number’ can be given as x + 8

Question 5.
a. number divided by six
_____________
Answer: x ÷ 6
Let x be the number.
The expression for ‘number divided by six’ can be given as x ÷ 6

b. the product of two and a number
_____________
Answer: 2 × x
Let x be the number.
The expression for ‘the product of two and a number’ can be given as 2 × x

Question 6.
a. the sum of 3 and a number is 12
_____________
Answer: 3 + x = 12
Let x be the number.
The equation for ‘the sum of 3 and a number is 12’ can be given as 3 + x = 12

b. six less than a number is nineteen
_____________
Answer: x – 6 = 19
Let x be the number.
The equation for ‘six less than a number is nineteen’ can be given as x – 6 = 19

Question 7.
a. thirty divided by a number is three
_____________
Answer: 30 ÷ x = 3
Let x be the number.
The equation for ‘thirty divided by a number is three’ can be given as 30 ÷ x = 3

b. the product of 5 and a number is fifteen
_____________
Answer: 5 ×  x = 15
Let x be the number.
The equation for ‘the product of 5 and a number is fifteen’ can be given as 5 ×  x = 15

Question 8.
a. the product of 5 and a number
_____________
Answer: 5 × x
Let x be the number.
The expression for ‘the product of 5 and a number’ can be given as 5 × x

b. the sum of 6 and a number is 16
_____________
Answer: 6 + x= 16
Let x be the number.
The equation for ‘the sum of 6 and a number is 16′ can be given as 6 + x= 16

Question 9.
a. 19 less than a number
_____________
Answer: 19 < x
Let x be the number.
The expression for ’19 less than a number’ can be given as 19 < x

b. 27 divided by a number is 9
_____________
Answer: 27 ÷ x = 9
Let x be the number.
The equation for ’27 divided by a number is 9′ can be given as 27 ÷ x = 9

Question 10.
a. 12 less than a number is 5
_____________
Answer: 12 < x = 5
Let x be the number.
The equation for ’12 less than a number is 5′ can be given as 12 < x = 5

b. the product of 6 and a number is 72
_____________
Answer: 6 ×  x = 72
Let x be the number.
The equation for ‘the product of 6 and a number is 72’ can be given as 6 ×  x = 72

Solve each problem.

Question 11.
Alicia had $22 to spend on pencils. If each pencil costs $ 1.50, how many pencils can she buy?

Let p represent the number of pencils.
Equation or Inequality: _____________
Alicia can buy ______________________ pencils.
Answer: Inequality: p × $ 1.50 = $22
Alicia can buy 14 pencils.
Alicia had $22 to spend on pencils.
each pencil costs $ 1.50
Let p represent the number of pencils.
Inequality: p × $ 1.50 = $22
p = \(\frac{22}{1.5}\)
p = 14.666
Therefore, Alicia can buy 14 pencils.

Question 12.
The sum of three consecutive numbers is 51. What is the smallest of these numbers?
Let n represent the smallest number of the set.
Equation or Inequality: ___________
The smallest of these numbers is _________.
Answer: Equation: x + x+1+ x+2 = 51
The smallest of these numbers is 16
The sum of three consecutive numbers is 51.
Let the numbers be x, x+1 and x+2.
Then the sum is : Equation: x + x+1+ x+2 = 51
3x + 3 = 51
3x = 51 – 3 = 48
x = 48 ÷ 3
x = 16
Therefore, The smallest of these numbers is 16

Question 13.
Mark bought 8 boxes. A week later, half of all his boxes were destroyed in a fire. There are now only 20 boxes left. With how many did he start?
Let b represent how many boxes he started with.
Equation or inequality: ___________
Mark began with ____________________ boxes.
Answer: Equation: \(\frac{b + 8}{2}\) = 20
Mark began with 32 boxes.
Let b represent how many boxes he started with.
Mark bought 8 boxes.
So he have b + 8 books at present.
A week later, half of all his boxes were destroyed in a fire.
There are now only 20 boxes left.
The number boxed Mark began with = \(\frac{b + 8}{2}\) = 20
b + 8 = 20 × 2
b + 8 = 40
b = 40 – 8
b = 32
Therefore, Mark began with 32 boxes.

Question 14.
Jillian sold half of her comic books and then bought 15 more. She now has 30. With how many did she begin?

Let c represent the number of comic books with which she began.
Equation or inequality: ___________
Jillian began with ____________________ CDs.
Answer: Equation: \(\frac{c}{2}\) + 15 = 30
Jillian began with 30 CDs.
Jillian sold half of her comic books and then bought 15 more.
She now has 30.
Let c represent the number of comic books with which she began.
The number of books she begin with = \(\frac{c}{2}\) + 15 = 30
\(\frac{c}{2}\)  = 30 -15
\(\frac{c}{2}\) = 15
c = 15 × 2
c =30
Therefore, Jillian began with 30 CDs.

Question 15.
On Tuesday, Shan ice bought 5 new pens. On Wednesday, half of all the pens that she had were accidentally thrown away. On Thursday, there were only 16 left. How many did she have on Monday?
Let p represent the number of pens she had on Monday.
Equation or inequality: ___________
Shanice had ___________________ pens on Monday.
Answer: Equation: \(\frac{p+5}{2}\) = 16
Shanice had 27 pens on Monday.
Let p represent the number of pens she had on Monday.
On Tuesday, Shan ice bought 5 new pens. So she had p + 5 pens on Tuesday
On Wednesday, half of all the pens that she had were accidentally thrown away. So, she had \(\frac{p+5}{2}\)
On Thursday, there were only 16 left.
\(\frac{p+5}{2}\) = 16
p + 5 = 16 × 2
p + 5 = 32
p = 32- 5
p = 27
Therefore, Shanice had 27 pens on Monday.

This handy Spectrum Math Grade 7 Answer Key Chapter 3 Lesson 3.4 Using Variables to Solve Problems provides detailed answers for the workbook questions.

Spectrum Math Grade 7 Chapter 3 Lesson 3.4 Using Variables to Solve Problems Answers Key

Write an equation to represent the problem, using the variable n for the unknown number. Then, solve for the value of the variable. Look at the following problem as an example.
George and Cindy are saving for bicycles. Cindy has saved $ 15 less than twice as much as George has saved. Together, they have saved $ I 20, How much did each of them save?
Let n stand for the amount George has saved. What stands for the amount Cindy has saved? 2n – 15 What equals the total amount? n + (2n – 15) = 120 Simplify: 3n – 15 = 120 Solve.
How much has George saved? $45
How much has Cindy saved? $75

Solve each problem.

Question 1.
Nate and Laura picked apples. Laura picked 5 as many as Nate picked. Together they picked 90 apples. How many did each of them pick?

Let n stand for the number Nate picked.
Equation: _____________
How many apples did Nate pick? ______________
How many apples did Laura pick? ____________
Answer: Equation: 5n + n = 90
How many apples did Nate pick? 15
How many apples did Laura pick? 75
Let n stand for the number Nate picked.
According to the question, number of apples picked by Laura = 5n
Together they picked 90 apples. So, 5n + n = 90
6n = 90
n = \(\frac{90}{6}\)
n = 15
Therefore, number of apples picked by Nate = 15
and, number of apples picked by Laura = 5 × 15 = 75

Question 2.
Jordan travels \(\frac{3}{4}\) of a mile longer to school each day than Harrison does. Combined, they travel 5\(\frac{1}{4}\) miles to school. How far does each travel?
Let n stand for the distance Jordan travels.
Equation: n – \(\frac{3}{4}\) + n = 5\(\frac{1}{4}\)
How far does Jordan travel?3 miles
How far does Harrison travel?2\(\frac{1}{4}\) miles
Answer: Equation: _____________
How far does Jordan travel? _____________
How far does Harrison travel?
Jordan travels \(\frac{3}{4}\) of a mile longer to school each day than Harrison does.
Let n stand for the distance Jordan travels.
and distance Harrison travels = n – \(\frac{3}{4}\)
Together they travelled 5\(\frac{1}{4}\) miles to school.
n – \(\frac{3}{4}\) + n = 5\(\frac{1}{4}\)
5\(\frac{1}{4}\) converting this into improper fraction. Then, it becomes \(\frac{21}{4}\)
2n – \(\frac{3}{4}\) = \(\frac{21}{4}\)
2n = \(\frac{21}{4}\) + \(\frac{3}{4}\)
2n = \(\frac{21 + 3}{4}\)
2n = \(\frac{24}{4}\)
2n = \(\frac{6}{1}\)
So, n = 3 miles
Therefore, Jorden travels = 3 miles
Harrison travels = 3 – \(\frac{3}{4}\)
= \(\frac{12 – 3}{4}\)
= \(\frac{9}{4}\)
= 2\(\frac{1}{4}\) miles

Question 3.
Two jackets have a combined cost of $98. Jacket A costs $ 12 less than Jacket B. How much does each jacket cost?
Let n stand for the cost of Jacket A.
Equation: __________
Jacket A costs ____________
Jacket B costs __________
Answer: Equation: n + n + $ 12 = $98
Jacket A costs = $43
Jacket B costs = $55
Let n stand for the cost of Jacket A.
Jacket A costs $ 12 less than Jacket B
So, the cost of jacket B = n + $ 12
Two jackets have a combined cost of $98
Therefore, n + n + $ 12 = $98
2n + $12 = $98
2n =$98 – $12
2n = $86
n = \(\frac{$86}{2}\)
n = $43
Therefore, the cost of jacket A = $43
and , the cost of jacket B = $43 + $12 = $55

Solve each problem.

Question 1.
William purchased a new car. The total price he will pay for the car, including interest, is $ 17,880. If he splits his car payments over 60 months, how much will he pay each month?

Let p represent each payment.
Equation: __________
William will pay _____ each month.
Answer: Equation: p =\(\frac{$ 17,880}{60}\)
William will pay $ 298 each month.
The total price he will pay for the car, including interest, is $ 17,880.
Duration of car payments = 60 months
Let p represent each payment.
So, payment done by William each month can be given as total price divided by number of months, which is p =\(\frac{$ 17,880}{60}\)
After simplification, by dividing 17880 with 60, it becomes, p = $ 298.
So, William pays $ 298 every month.

Question 2.
Tracy has $ 1.55 in quarters and dimes. If she has 3 quarters, how many dimes does she have?
Let d represent the number of dimes.
Equation: __________
Tracy has _____ dimes.
Answer: Equation:3 × 25 + d × 10 = 155
Tracy has 8 dimes.
Tracy has $ 1.55 in quarters and dimes.
1 quarter = 25 cents
1 dime = 10 cents
Tracy has 3 quarters
Let d represent the number of dimes.
Therefore, 3 × 25 + d × 10 = 155 (as 1.55 dollars = 155 cents)
75 + 10d = 155
10d = 155 – 75
10d = 80
d = \(\frac{80}{10}\)
d = 8
Therefore, number of dimes Tracy as 8

Question 3.
Kavon is saving money to buy a bicycle that costs $ 150. He has been saving his $5 weekly allowance for the last 8 weeks and he saved $50 from his birthday money. How much more money does Kavon need to buy his bicycle?

Let m represent the money Kavon needs.
Equation: ___________
Kavon needs __________
Answer: Equation: m + $90 = $150
Kavon needs $60
Kavon is saving money to buy a bicycle that costs $ 150.
He has been saving his $5 weekly allowance for the last 8 weeks and he saved $50 from his birthday money.
So, total money he has right now = $5 × 8 + $50 = $40 + $50 = $90
Let m represent the money Kavon needs.
The more money Kavon needs to buy his bicycle = m + $90 = $150
m = $150 – $90
m = $60
Therefore, Kavon needs $60 more to buy his bicycle

Question 4.
Lincoln Middle School won their football game last week by scoring 23 points. If they scored two 7-point touchdowns, how many 3-point field goals did they score?
Let f represent the number of field goals.
Equation: ___________
They scored ________ field goals.
Answer: Equation: 2 × 7 + 3 × f = 23
They scored 3 field goals.
Lincoln Middle School won their football game last week by scoring 23 points.
They scored two 7-point touchdowns an we need to find the number of 3-point field goals
Let f represent the number of field goals.
So, number of field goals = 2 × 7 + 3 × f = 23
14 + 3f = 23
3f = 23 – 14
3f = 9
Hence, f = 3
Therefore, they have scored 3 3-point field goals

Question 5.
Walker is reading a book that is 792 pages. He reads 15 pages a day during the week, and 25 pages a day during the weekend. After 5 weeks of reading, how many pages does Walker still have left to read before he finishes the book?
Let r represent the pages left to read.
Equation: __________
Walker has _____ pages left to read.
Answer: Equation: r + 625 = 792
Walker has 167 pages left to read.
Walker is reading a book that is 792 pages.
He reads 15 pages a day during the week, and 25 pages a day during the weekend. So, he reads 15 × 5 + 25 × 2 pages in a week, which is 125 pages
After 5 weeks of reading, number of pages he completed  = 5 × 125 = 625 pages
Let r represent the pages left to read.
Therefore, number of pages left by Walker to read  = r + 625 = 792
r = 792 – 625
r = 167 pages

Solve each problem.

Question 1.
Peaches are on sale at the farmer’s market for $ 1.75 per pound. If Ida buys $8.75 worth of peaches, how many pounds of peaches did she buy?

Let p represent pounds of peaches.
Equation: ___________
Ida bought ____ pounds of peaches.
Answer: Equation: p × $ 1.75 = $8.75
Ida bought 5 pounds of peaches.
Peaches are on sale at the farmer’s market for $ 1.75 per pound.
Ida bought $8.75 worth of peaches
Let p represent pounds of peaches.
Number of pounds of peaches she bought = p × $ 1.75 = $8.75
p = \(\frac{$8.75}{$ 1.75}\)
After simplifying the above statement,
p = 5
Therefore, She bought 5 pounds of peaches

Question 2.
Kylie makes $8.50 an hour working at a restaurant. If she brings home $ 170 in her paycheck, how many hours did she work?
Let h represent the number of hours Kylie works.
Equation: h × $8.50 = $ 170
Kylie worked 20 hours.
Answer: Equation: ____________
Kylie worked _____ hours.
Kylie makes $8.50 an hour working at a restaurant.
she brings home $ 170 in her paycheck
Let h represent the number of hours Kylie works
Number of hours she worked  = h × $8.50 = $ 170
h = \(\frac{ $ 170}{$8.50}\)
After simplifying the above statement,
h = 20 hours
Therefore, she worked for 20 hours

Question 3.
Larry and 3 friends went to a basketball game. They Daid $5.00 for each of their tickets and each bought a Dag of candy. If they spent a total of $28, how much was each bag of candy?
Let c represent bags of candy.
Equation: __________
Each bag of candy cost _____.
Answer: Equation: $20.00 + 4 × c = $28.00
Each bag of candy cost $2.00
Larry and 3 friends went to a basketball game.
They Paid $5.00 for each of their tickets
So, amount they spent on tickets = $5.00 × 4 = $20.00
They spent total of $28.00
Let c represent bags of candy.
Cost of each candy bag = $20.00 + 4 × c = $28.00
4c = $28.00 – $20.00
4c = $8.00
c = \(\frac{ $8.00}{4}\)
c = $2

Question 4.
Three hoses are connected end to end. The first hose is 6.25 feet. The second hose is 6.5 feet. If the length of all 3 hoses when connected is 20 feet, how long is the third hose?
Let h represent the length of the third hose.
Equation: _________
The third hose is ____ feet long.
Answer: Equation: 6.25 + 6.5 +h = 20
The third hose is 7.25 feet long.
Three hoses are connected end to end.
The first hose is 6.25 feet. The second hose is 6.5 feet.
the length of all 3 hoses when connected is 20 feet
Let h represent the length of the third hose.
Therefore, the length of third hose = 6.25 + 6.5 +h = 20
12.75 + h = 20
h = 20 – 12.75
h = 7.25
So, the length of third hose = 7.25 feet

Question 5.
Quinn and her mom went to the movies. They paid $ 10.50 for each of their tickets and each bought
a tub of popcorn. If they spent a total of $38.50, how much was each tub of popcorn?
Let p represent tubs of popcorn.
Equation: _____________
Each tub of popcorn cost ______.
Answer: Equation: $ 21 + 2 × p =  $38.50
Each tub of popcorn cost $8.75
Quinn and her mom went to the movies.
They paid $ 10.50 for each of their tickets and each bought a tub of popcorn.
The amount they spent on their tickets = 2 × $ 10.50 = $ 21
They spent total amount of $38.50
Let p represent tubs of popcorn.
Cost of each tub of popcorn = $ 21 + 2 × p =  $38.50
2p = $38.50 – $21
2p = $17.50
p = \(\frac{ $17.50}{2}\)
p =  $8.75
Therefore, each tub of popcorn costs $8.75

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

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

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

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:

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.

This handy Spectrum Math Grade 7 Answer Key Chapter 3 Lesson 3.5 Using Variables to Express Inequalities provides detailed answers for the workbook questions.

Spectrum Math Grade 7 Chapter 3 Lesson 3.5 Using Variables to Express Inequalities Answers Key

An inequality is a mathematical sentence that states that Iwo expressions are not equal.
2 × 5 > 6
Inequalities can be solved the same way as you solve equations.
-4 × x ≥ -4
-4 × x ÷ (-4) ≥ -4 ÷ (-4)
x ≤ 1

When you multiply or divide by a negative number, you must flip the inequality sign.

Solve each inequality and graph its solution.

Question 1.
a. -4 × m > 20

Answer:
An inequality is a mathematical sentence that states that Iwo expressions are not equal.
-4 × m > 20
-4 × m ÷ (-4) > 20 ÷ (-4)
m < -5
When you multiply or divide by a negative number, you must flip the inequality sign.

b. \(\frac{v}{5}\) ≤ –\(\frac{3}{5}\)

Answer:
An inequality is a mathematical sentence that states that Iwo expressions are not equal.
\(\frac{v}{5}\) ≤ –\(\frac{3}{5}\)
\(\frac{v}{5}\) × 5 ≤ –\(\frac{3}{5}\) × 5
v ≤ -3

Question 2.
a. 15 × x ≤ 15

Answer:
An inequality is a mathematical sentence that states that Iwo expressions are not equal.
15 × x ≤ 15
15 × x ÷15 ≤ 15÷15
x ≤ 1

b. h ÷ 6 < -12

Answer:
An inequality is a mathematical sentence that states that Iwo expressions are not equal.
h ÷ 6 < -12
h ÷ 6 × 6< -12 × 6
h < -72

Question 3.
a. -10a < -70

Answer:
An inequality is a mathematical sentence that states that Iwo expressions are not equal.
-10a < -70
-10a ÷ (-10) < -70 ÷ (-10)
a > 7
When you multiply or divide by a negative number, you must flip the inequality sign.

b. n ÷ 2 ≥ 2

Answer:
An inequality is a mathematical sentence that states that Iwo expressions are not equal.
n ÷ 2 ≥ 2
n ÷ 2 × 2 ≥ 2 ×2
n ≥ 4

Word problems can be solved by creating inequality statements.
Aria has $55 to spend on flowers. She wants to buy two rose bushes, which will cost $20, and spend the rest of her money on lilies. Each lily costs $ 10. Write an inequality to show how many lilies Aria can buy.
Let l represent the number of lilies she can buy.
Inequality: $ 10 × l + $20 ≤ $55
$10 × l + $20 – $20 ≤ $55 – $20
$10 × l ÷ $10 ≤ $35 ÷ $10
1 ≤ 3.5
Aria can buy 3 lilies.

Solve each problem by creating an inequality.

Question 1.
Andrew had $20 to spend at the fair. If he paid $5 to get into the fair, and rides cost $2 each, what is the maximum number of rides he could go on?

Let r represent the number of rides.
Inequality: $5 + r×$2 ≤ $20
Andrew could go on 7 rides.
Inequality: __________
Andrew could go on _____ rides.
Answer:
Andrew had $20 to spend at the fair.
he paid $5 to get into the fair, and rides cost $2 each.
Let r represent the number of rides.
Inequality: $5 + r×$2 ≤ $20
Add -$5 on both sides
-$5 + $5 + r×$2 ≤ $20 -$5
r×$2 ≤ $15
Divide $2 on both sides
r×$2 ÷$2≤ $15÷$2
r ≤ $7.5
Therefore, Andrew could go on 7 rides.

Question 2.
Sandra has $75 to spend on a new outfit. She finds a sweater that costs twice as much the skirt. What is the most the skirt can cost?
Let s represent the cost of the skirt.
Inequality: _________
The most the skirt can cost is __________
Answer:
Inequality: s + 2s ≤ $75
The most the skirt can cost is $25
Sandra has $75 to spend on a new outfit.
She finds a sweater that costs twice as much the skirt.
Let s represent the cost of the skirt.
Inequality: s + 2s ≤ $75
3s ≤ $75
divide 3 on both sides
3s ÷3≤ $75÷3
s ≤ $25
Therefore, The most the skirt can cost is $25

Question 3.
Alan earns $7.50 per hour at his after-school jobs. He is saving money to buy a skateboard that costs $ I 20. How many hours will he have to work to earn enough money for the skateboard?

Let h represent the number of hours Alan will have to work.
Inequality: _________
Alan will have to work _____ hours.
Answer:
Inequality: $7.50 × h ≤ $ 120
Alan will have to work 16 hours.
Alan earns $7.50 per hour at his after-school jobs.
He is saving money to buy a skateboard that costs $ 120.
Let h represent the number of hours Alan will have to work.
Inequality: $7.50 × h ≤ $ 120
Divide $7.50  by both sides
$7.50 × h ÷ $7.50 ≤ $ 120 ÷$7.50
h ≤ $16
Alan will have to work 16 hours.

Solve each problem by creating an inequality.

Question 1.
Blue Bird Taxi charges a $2.00 flat rate in addition to $0.55 per mile. Marcy only has $ 10 to spend on a taxi ride. What is the farthest she can ride without going over her limit?
Let d equal the distance Marcy can travel.
Inequality: ____________
Marcy can travel ____ miles without going over her limit.
Answer:
Inequality: 2.00 + 0.55 × d ≤ 10
Marcy can travel 14.54 miles without going over her limit.
Blue Bird Taxi charges a $2.00 flat rate in addition to $0.55 per mile.
Marcy only has $ 10 to spend on a taxi ride.
Let d equal the distance Marcy can travel.
Inequality: 2.00 + 0.55 × d ≤ 10
Add -2.00 on both sides
2.00 + 0.55 × d -2.00 ≤ 10 -2.00
0.55 × d≤  8
divide by 0.55 on both sides
0.55 × d ÷0.55≤  8÷0.55
d ≤ 14.54
Marcy can travel 14.54 miles without going over her limit.

Question 2.
The school store is selling notebooks for $ 1.50 and T-shirts for $ 10.00 to raise money for the school. They have a goal of raising $250 to buy supplies for the science lab. If they have sold 60 notebooks, how many T-shirts will they need to sell to reach their goal?
Let t equal the number of T-shirts.
Inequality: _______
They need to sell _____ T-shirts.
Answer:
Inequality: 60 × $ 1.50 + t × $ 10.00 ≤ $250
They need to sell 16 T-shirts.
The school store is selling notebooks for $ 1.50 and T-shirts for $ 10.00 to raise money for the school.
They have a goal of raising $250 to buy supplies for the science lab.
They have sold 60 notebooks
Let t equal the number of T-shirts.
Inequality: 60 × $ 1.50 + t × $ 10.00 ≤ $250
$90 + $10t ≤ $250
Add -$90 on both sides
$90 -$90 + $10t ≤ $250 – $90
$10t ≤ $160
Divide $10 on both sides
$10t ÷$10 ≤ $160÷$10
t ≤ $16
Therefore, they need to sell 16 T-shirts.

Question 3.
There are 178 7th grade students and 20 chaperones going on the field trip to the aquarium. Each bus holds 42 people. How many buses will the group have to take?

Let b represent the number of buses.
Inequality: ___________
They will need to take ____ buses.
Answer:
Inequality: b × 42 ≥ 198
They will need to take 5 buses.
There are 178 7th grade students and 20 chaperones going on the field trip to the aquarium.
Total number of people that can go to the trip = 178 + 20 = 198
Each bus holds 42 people.
Let b represent the number of buses.
Inequality: b × 42 ≥ 198
divide by 42 on both sides
42b ÷ 42≥198 ÷42
b ≥  4.71
They will need to take 5 buses.

Question 4.
Sofia’s parents gave her an allowance for summer camp of $ 125. If she is going to be at camp for 6 weeks, what is the most she can spend each week while she is at camp?
Let m represent the amount Sofia can spend each week.
Inequality: __________
The most Sofia can spend each week is ______.
Answer:
Inequality: 6×m ≥$ 125
The most Sofia can spend each week is $20.833
Sofia’s parents gave her an allowance for summer camp of $ 125.
she is going to be at camp for 6 weeks
Let m represent the amount Sofia can spend each week.
Inequality: 6×m ≥$ 125
Divide 6 on both sides
6×m ÷6 ≥ $ 125 ÷6
m ≥ $20.833
The most Sofia can spend each week is $20.833

Question 5.
The cell phone company allows all users 450 text messages a month. Any text messages over the allowed amount are charged $0.25 per message. Craig only has $26 extra to spend on his cell phone bill. How many messages can he go over the allowed amount for the month without breaking his budget of $26?

Let p represent the amount of text messages Craig can go over.
Inequality: p × $0.25 ≤$26
Craig can send and receive 104 extra text messages without breaking his budget of $26.
Inequality: __________
Craig can send and receive ______ extra text messages without breaking his budget of $26.
Answer:
The cell phone company allows all users 450 text messages a month.
Any text messages over the allowed amount are charged $0.25 per message.
Craig only has $26 extra to spend on his cell phone bill.
Let p represent the amount of text messages Craig can go over.
Inequality: p × $0.25 ≤$26
divide by $0.25 on both sides
p × $0.25 ÷ $0.25≤ $26 ÷ $0.25
p ≤ 104
Therefore, Craig can use extra of 104 messages
In total, he can use 450 + 104 = 554 messages