Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 4 Review Test to score better marks in the exam.

## Math in Focus Grade 7 Course 2 A Chapter 4 Review Test Answer Key

**Concepts and Skills**

**Solve each equation.**

Question 1.

8x – 7 = 17

Answer:

**x = 3**

Explanation:

Given, 8x – 7 = 17

Add 7 on both sides.

8x – 7 + 7 = 17 + 7

8x = 24

Divide 8 on both sides.

8x ÷ 8 = 24 ÷ 8

x = 3

Question 2.

4 – 6x = 8

Answer:

**x = -(2/3)**

Explanation:

Given, 4 – 6x = 8

Subtract 4 on both sides.

4 – 6x – 4 = 8 – 4

-6x = 4

Divide -6 on both sides.

-6x ÷ -6 = 4 ÷ -6

x = -(2/3)

Question 3.

6 – \(\frac{y}{3}\) = 0

Answer:

**y = 18**

Explanation:

Give, 6 – (y/3) = 0

Subtract 6 on both sides.

6 – (y/3) – 6= 0 – 6

-(y/3) = -6

(y/3) = 6

Multiply 3 on both sides.

(y/3) × 3 = 6 × 3

y = 18

Question 4.

3 – 3.6x = 4.2

Answer:

**x = -0.333**

Explanation:

Given, 3 – 3.6x = 4.2

Subtract 3 on both sides.

3 – 3.6x – 3 = 4.2 – 3

-3.6x = 1.2

Divide -3.6 on both sides.

-3.6x ÷ -3.6 = 1.2 ÷ -3.6

x = -0.333

Question 5.

7x – 5 = 3x + 4

Answer:

**x = (3/2)**

Explanation:

Given, 7x – 5 = 3x + 4

Add 5 on both sides.

7x – 5 + 5 = 3x + 4 + 5

7x = 3x + 9

Subtract 3x on both sides.

7x – 3x = 3x + 9 – 3x

4x = 9

Divide 4 on both sides.

4x ÷ 4 = 9 ÷ 4

x = (9/4)

x = (3/2)

Question 6.

\(\frac{7}{10}\)y – \(\frac{1}{5}\) = \(\frac{3}{5}\)y + \(\frac{6}{5}\)

Answer:

**y = 14**

Explanation:

(7/10)y – (1/5) = (3/5)y + (6/5)

Add (1/5) on the both sides.

(7/10)y – (1/5) + (1/5) = (3/5)y + (6/5) + (1/5)

(7/10)y = (3/5)y + (7/5)

Subtract (3/5)y on both sides.

(7/10)y – (3/5)y = (3/5)y +(7/5) – (3/5)y

(7/10)y – ((3×2)/10)y =(7/5)

(7/10)y – (6/10)y =(7/5)

(y/10) = (7/5)

Multiply 1y = 140 on both sides.

(y/10) × 10 = (7/5) × 10

y = 14

Question 7.

3.4y – 5.2 – 3y = 2

Answer:

**y = 18**

Explanation:

Given, 3.4y – 5.2 – 3y = 2

0.4y – 5.2 = 2

Add 5.2 on both sides.

0.4y – 5.2 + 5.2 = 2 + 5.2

0.4y = 7.2

Divide 0.4 on both sides.

0.4y ÷ 0.4 = 7.2 ÷ 0.4

y = 18

Question 8.

15y – 4(2y – 3) = -2

Answer:

**y = -2**

Explanation:

Given, 15y – 4(2y – 3) = -2

15y – 8y + 12 = -2

7y + 12 = -2

Subtract 12 on both sides.

7y + 12 – 12= -2 – 12

7y = -14

Divide 7 on both sides.

7y ÷ 7= -14 ÷ 7

y = -2

Question 9.

\(\frac{1}{4}\)(x + 3) + \(\frac{3}{8}\)x = \(\frac{13}{4}\)

Answer:

**x = 4**

Explanation:

Given, (1/4)(x + 3) + (3/8)x = (13/4)

(1/4)x + (3/4) + (3/8)x = (13/4)

Subtract (3/4) on both sides.

(1/4)x + (3/4) + (3/8)x – (3/4) = (13/4) – (3/4)

(1/4)x + (3/8)x =(13-3)/4

(2 + 3)x/8 = 10/4

(5/8)x = (10/4)

Multiply 8 on both sides.

(5/8)x × 8 = (10/4) × 8

5x = 20

Divide 5 on both sides.

5x ÷ 5 = 20 ÷ 5

**x = 4**

Question 10.

0.4(x + 0.7) = 0.6x – 4.2

Answer:

**x = 22.4**

Explanation:

Given, 0.4(x + 0.7) = 0.6x – 4.2

0.4x + 0.28 =0.6x – 4.2

Add 4.2 on the both sides.

0.4x +0.28 + 4.2 =0.6x – 4.2 + 4.2

0.4x + 4.48 = 0.6x

Subtract 0.4x on the both sides.

0.4x + 4.48 – 0.4x = 0.6x – 0.4x

4.48 = 0.2x

0.2x = 4.48

Divide o.2 on both sides.

0.2x ÷ 0.2 = 4.48 ÷ 0.2

x = 22.4

**Solve each inequality. Graph each solution set.**

Question 11.

4x – 3 > 1

Answer:

**x > 1**

Explanation:

Given, 4x – 3 > 1

Add 3 on both sides.

4x – 3 + 3 > 1 + 3

4x > 4

x > 1

Question 12.

6 ≤ 1 – 5x

Answer:

**-1 ≤ x**

Explanation:

Given, 6 ≤ 1 – 5x

Subtract 6 on both sides.

6 – 6 ≤ 1 – 5x – 6

0 ≤ -5x – 5

0 ≤ -5 (x + 1)

Divide -5 on both sides.

0 ÷ -5 ≤ -5 (x + 1) ÷ -5

0 ≤ (x + 1)

-1 ≤ x

Question 13.

\(\frac{2}{3}\) – \(\frac{x}{6}\) ≥ –\(\frac{1}{2}\)

Answer:

**x ≤ 1**

Explanation:

(2/3) – (x/6) ≥ (1/2)

Subtract (2/3) on both sides.

(2/3) – (x/6) – (2/3) ≥ (1/2) – (2/3)

-(x/6) ≥ (3 – 4/6)

-(x/6) ≥ -(1/6)

Multiply -6 on both sides.

-(x/6) × (-6) ≥ -(1/6) × (-6)

x ≤ 1

Question 14.

-6.9 < 8.1 – 1.5x

Answer:

**x > 10**

Explanation:

Given, -6.9 < 8.1 – 1.5x

Subtract 8.1 on both sides.

-6.9 – 8.1 < 8.1 – 1.5x – 8.1

-15 < -1.5x

1.5x > 15

Divide 1.5 on both sides.

1.5x ÷ 1.5 > 15 ÷ 1.5

x > 10

Question 15.

9y – 5 ≤ 4y + 15

Answer:

**y ≤ 4**

Explanation:

Given, 9y – 5 ≤ 4y + 15

Subtract 4y on both sides.

9y – 5 – 4y ≤ 4y + 15 – 4y

5y – 5 ≤ 15

Add 5 on both sides.

5y – 5 + 5 ≤ 15 + 5

5y ≤ 20

Divide 5 on both sides.

5y ÷ 5 ≤ 20 ÷ 5

y ≤ 4

Question 16.

\(\frac{7}{9}\)x – \(\frac{2}{3}\) > \(\frac{1}{6}\)x + 3

Answer:

**x > 6**

Explanation:

Given, (7/9)x – (2/3) > (1/6)x + 3

Subtract (1/6)x on both sides.

(7/9)x – (2/3) – (1/6)x > (1/6)x + 3 – (1/6)x

(14 – 3)/18x – (2/3) > 3

11/18x – (2/3) > 3

Add (2/3) on both sides.

11/18x- (2/3) + (2/3) > 3 + (2/3)

11/18x > (11/3)

Divide 11 on both sides.

11/18x ÷ 11 > (11/3) ÷ 11

(1/18)x > (1/3)

Multiply 3 on both sides.

(1/18)x × 3 > (1/3) × 3

(1/6)x > 1

Multiply 6 on both sides.

(1/6)x × 6 > 1 × 6

x > 6

Question 17.

12.9 < 0.3(5.3 – x)

Answer:

Question 18.

3(x + 1) > 5x + 7

Answer:

Question 19.

\(\frac{1}{5}\)(4x – 1) ≥ \(\frac{2}{3}\)x + \(\frac{3}{5}\)

Answer:

**x ≥ 6**

Explanation:

Given, (1/5)(4x – 1) ≥ (2/3)x + (3/5)

Multiply 5 on both sides

(1/5)(4x – 1) × 5 ≥ ((2/3)x + (3/5)) × 5

(4x – 1) ≥ (10/3)x + 3

Add 1 on both sides.

(4x – 1) + 1 ≥ (10/3)x + 3 + 1

4x ≥ (10/3)x + 4

Subtract (10/3)x on both sides.

4x – (10/3)x ≥ (10/3)x + 4 – (10/3)x

(12 – 10)x/3 ≥ 4

(2/3)x ≥ 4

Multiply 3 on both sides.

(2/3)x × 3 ≥ 4 × 3

2x ≥ 12

Divide 2 on both sides.

2x ÷ 2 ≥ 12 ÷ 2

x ≥ 6

Question 20.

4(3 – 0.1x) ≤ 15 – 0.6x

Answer:

**x ≤ 15**

Explanation:

Given, 4(3 – 0.1x) ≤ 15 – 0.6x

12 -0.4x ≤ 15 – 0.6x

Subtract 12 on both sides.

12 -0.4x – 12 ≤ 15 – 0.6x – 12

-0.4x ≤ 3 -0.6x

Add 0.6x on both sides.

-0.4x + 0.6x ≤ 3 -0.6x + 0.6x

0.2x ≤ 3

Divide 0.2 on both sides.

0.2x ÷ 0.2 ≤ 3 ÷ 0.2

x ≤ 15

**Problem Solving**

**Write an equation for questions 21 to 25. Solve and show your work.**

Question 21.

Aiden wrote a riddle: Five less than \(\frac{1}{5}\) times a number is the same as the sum of the number and \(\frac{1}{3}\). Find the number.

Answer:

**x = -(20/3)**

Explanation:

Let us consider the number be ‘x’.

(1/5)x – 5 = x + (1/3)

Add 5 on both sides.

(1/5)x – 5 + 5 = x + (1/3) + 5

(1/5)x = x + (16/3)

Subtract (1/5)x on both sides.

(1/5)x – (1/5)x = x + (16/3) – (1/5)x

0 = (4/5)x + (16/3)

(4/5)x = (- 16/3)

Multiply 5 on both sides.

(4/5)x × 5 = (- 16/3) × 5

4x = – 80/3

Divide 4 on both sides.

4x ÷ 4 = (- 80/3) ÷ 4

**x = -(20/3)**

Question 22.

Mary is 6 years older than her sister Kelly. The sum of their ages is 48. How old is Kelly?

Answer:

**Kelly’s age, x = 21**

Explanation:

Let us consider the age of the Kelly be ‘x’.

Age of Mary is x +6

Sum of their ages = 48

x +x + 6 = 48

2x + 6 = 48

Subtract 6 on both sides.

2x + 6 – 6 = 48 – 6

2x = 42

Divide 2 on both sides.

2x ÷ 2 = 42 ÷ 2

x = 21

Question 23.

The sum of the page numbers of two facing pages in a book is 145. What are the page numbers?

Answer:

x = 72; x + 1 = 73.

**72, 73 are the page numbers.**

Explanation:

Let us consider the numbers be x and x+1.

x + x +1 = 145

2x + 1 = 145

Subtract 1 on both sides.

2x + 1 – 1 = 145 – 1

2x = 144

Divide 2 on both sides.

2x ÷ 2 = 144 ÷ 2

x = 72

x + 1 = 73

The sum of numbers of two facing pages in a book = 145

Question 24.

The perimeter of an equilateral triangle is 6\(\frac{3}{4}\) inches. Find the length of each side of the equilateral triangle.

Answer:

**Length of each side, a = (3/2)**

Explanation:

Given, The perimeter of an equilateral triangle is 6(3/4).

3a = 6(3/4)

Divide 3 on both sides.

3a ÷ 3 = 6(3/4) ÷ 3

a = (6/4)

a = (3/2)

Question 25.

The sum of the interior angle measures of a quadrilateral is 360°. The measure of angle A is three times the measure of angle D. The measure of angle 6 is four times that of angle D. The measure of angle C is 24° more than angle B. Find the measure of each angle of the quadrilateral.

Answer:

**Write an inequality for each question. Solve and show your work.**

Question 26.

Laura wants the average amount of money she spends each day on her four-day vacation to be no more than $64. On the first three days, she spends $71, $62, and $59. What is the greatest amount of money she can spend on the fourth day?

Answer:

**64**

Explanation:

Given, Mean = ax=n(Mean)−(a1+...+an)

= 4(64) – (71 + 62 + 59)

= 256 – 192

= 64

Question 27.

Kevin plans to sign up for p hours of training at a culinary school. The school offers two payment options as shown below.

For how many hours of training is Option B less expensive than Option A?

Answer:

**x < 20, 20 hours.**

Explanation:

Let us consider number of hours be ‘x’.

18x + 0 < 8x +200

18x < 8x + 200

Subtract 8x on both sides.

18x – 8x < 8x + 200 – 8x

10x < 200

Divide 10 on both sides.

10x ÷ 10 < 200 ÷ 10

x < 20

Question 28.

Peter has found a job in a computer store. As shown below, he has two options for how he will be paid. The commission he makes for Option B is based on his weekly sales. For example, if his sales total $1,000 a week, he receives his base salary of $250 plus 8% of $1,000.

Peter is thinking about Option B. What would his weekly sales need to be for him to make at least as much as he would for Option A?

Answer:

**The sales should be nearly above 4000$ per week to get more than Option A. Option A is better than Option B.**

Explanation:

Let us consider ‘x’ be the number of sales.

600 ≥ 250 + 0.08x

Subtract 250 on both sides.

600 – 250 ≥ 250 + (0.08×1000)x – 250

350 ≥ 80x

Divide 80 on both sides.

350 ÷ 80 ≥ 80x ÷ 80

4 ≥ x

The sales should be nearly above 4000$ per week to get more than Option A.

so Option A is better than Option B.

Question 29.

The school events committee is planning to buy a banner and some helium balloons for graduation night. A store charges them $35 for the banner and $3.50 for each helium balloon. If the committee has at most $125 to spend, how many helium balloons can they buy?

Answer:

**x ≤ 25**

Explanation:

Let us consider number of helium balloons be ‘x’.

35 + 3.5x ≤ 125

Subtract 35 on both sides.

35 + 3.5x – 35 ≤ 125 – 35

3.5x ÷ 3.5 ≤ 90 ÷ 3.5

Divide 3.5 on both sides.

3.5x ≤ 90

x ≤ 25

Question 30.

The coach of the field hockey team can spend at most $475 on new team uniforms. The coach will order the uniforms online and pay a mailing cost of $6.50. If each uniform costs $29, how many uniforms can the coach order?

Answer:

**x ≤ 16, 16 uniforms.**

Explanation:

Let us consider the number of uniforms be ‘x’.

6.5 + 29x ≤ 475

Subtract 6.5 on both sides.

6.5 + 29x – 6.5 ≤ 475 – 6.5

29x ≤ 468.5

Divide 29 on both sides.

29x ÷ 29 ≤ 468.5 ÷ 29

x ≤ 16