Go through the Math in Focus Grade 7 Workbook Answer Key Chapter 3 Review Test to finish your assignments.

## Math in Focus Grade 7 Course 2 A Chapter 3 Review Test Answer Key

**Concepts and Skills
**

**Simplify each expression.**

Question 1.

1.4w – 0.6w

Answer:

1.4w – 0.6w = 0.8w.

Explanation:

1.4w – 0.6w

= 0.8w.

Question 2.

\(\frac{3}{4}\)m + \(\frac{4}{5}\)m

Answer:

\(\frac{3}{4}\)m + \(\frac{4}{5}\)m = \(\frac{31}{20}\)m.

Explanation:

\(\frac{3}{4}\)m + \(\frac{4}{5}\)m

LCD of 4 n 5 = 20.

= [(3 × 5) + (4 × 4)]m ÷ 20

= (15 + 16)m ÷ 20

= 31m ÷ 20 or \(\frac{31}{20}\)m.

Question 3.

\(\frac{1}{6}\)y + \(\frac{1}{2}\)y + \(\frac{1}{3}\)y

Answer:

\(\frac{1}{6}\)y + \(\frac{1}{2}\)y + \(\frac{1}{3}\)y = y.

Explanation:

\(\frac{1}{6}\)y + \(\frac{1}{2}\)y + \(\frac{1}{3}\)y = y.

LCD of 6 n 2 = 6.

= [(1 × 1) + (1 × 3)]y ÷ 6] + \(\frac{1}{3}\)y

= [(1 + 3)y ÷ 6] + \(\frac{1}{3}\)y

= (4y ÷ 6) + \(\frac{1}{3}\)y

= \(\frac{2}{3}\)y+ \(\frac{1}{3}\)y

= (2 + 1)y ÷ 3

= 3y ÷ 3

= y.

Question 4.

1.8m – 0.2m – 7m

Answer:

1.8m – 0.2m – 7m = -5.4m.

Explanation:

1.8m – 0.2m – 7m

= 1.6m – 7m

= -5.4m.

Question 5.

1.3a – 0.8b + 2.2b – a

Answer:

1.3a – 0.8b + 2.2b – a = 0.3a + 1.4b.

Explanation:

1.3a – 0.8b + 2.2b – a

= 1.3a – a – 0.8b + 2.2b

= 0.3a – 0.8b + 2.2b

= 0.3a + 1.4b.

Question 6.

1 + \(\frac{1}{5}\)a + \(\frac{3}{5}\)b + \(\frac{4}{5}\)a

Answer:

1 + \(\frac{1}{5}\)a + \(\frac{3}{5}\)b + \(\frac{4}{5}\)a = 1 + a + \(\frac{3}{5}\)b.

Explanation:

1 + \(\frac{1}{5}\)a + \(\frac{3}{5}\)b + \(\frac{4}{5}\)a

= 1 + \(\frac{1}{5}\)a + \(\frac{4}{5}\)a + \(\frac{3}{5}\)b

= 1 + [(1 + 4)a ÷ 5] + \(\frac{3}{5}\)b

= 1 + (5a ÷ 5) + \(\frac{3}{5}\)b

= 1 + a + \(\frac{3}{5}\)b.

**Expand each expression. Then simplify when you can.
**Question 7.

1.2(2p – 3)

Answer:

1.2(2p – 3) = 2.4p – 3.6.

Explanation:

1.2(2p – 3)

= (1.2 × 2p) – (3 × 1.2)

= 2.4p – 3.6.

Question 8.

\(\frac{1}{3}\)(12p + 9q)

Answer:

\(\frac{1}{3}\)(12p + 9q) = 4p + 3q.

Explanation:

\(\frac{1}{3}\)(12p + 9q)

= [12p × \(\frac{1}{3}\)] + [ 9q × \(\frac{1}{3}\)]

= 4p + 3q.

Question 9.

\(\frac{1}{5}\)\(\left(\frac{t}{3}+\frac{1}{2}\right)\)

Answer:

\(\frac{1}{5}\)\(\left(\frac{t}{3}+\frac{1}{2}\right)\) = \(\frac{t}{15}\) + \(\frac{1}{10}\).

Explanation:

\(\frac{1}{5}\)\(\left(\frac{t}{3}+\frac{1}{2}\right)\)

= [\(\frac{t}{3}\) × \(\frac{1}{5}\)] + [\(\frac{1}{2}\) × \(\frac{1}{5}\)]

= \(\frac{t}{15}\) + \(\frac{1}{10}\).

Question 10.

-4(-2q + 2.5)

Answer:

-4(-2q + 2.5) = 8q – 10.

Explanation:

-4(-2q + 2.5)

= (-4 × -2q) + (2.5 × -4)

= 8q – 10.

Question 11.

–\(\frac{2}{3}\)(6x + 3)

Answer:

–\(\frac{2}{3}\)(6x + 3) = 4x – 2.

Explanation:

–\(\frac{2}{3}\)(6x + 3)

= (6x × –\(\frac{2}{3}\)) + (3 × –\(\frac{2}{3}\))

= (2x × 2) + (1 × -2)

= 4x – 2.

Question 12.

-0.5(2m – 4n)

Answer:

-0.5(2m – 4n) = -m + 0.5n.

Explanation:

-0.5(2m – 4n)

= (2m × -0.5) – (1n × -0.5)

= -1m – (-0.5n)

= -m + 0.5n.

Question 13.

3(a + 3) + 2a

Answer:

3(a + 3) + 2a = 5a + 9.

Explanation:

3(a + 3) + 2a

= [(3 × a) + (3 × 3)] + 2a

= 3a + 9 + 2a

= 3a + 2a + 9

= 5a + 9.

Question 14.

4(2p – 3) – 3(p + 2)

Answer:

4(2p – 3) – 3(p + 2) = 5p – 6.

Explanation:

4(2p – 3) – 3(p + 2)

= [(2p × 4) – (3 × 4)] – [(3 × p) + (2 × 3)]

= (8p – 12) – (3p + 6)

= 8p – 12 – 3p – 6

= 8p – 3p – 12 – 6

= 5p – 6.

Question 15.

2.5(m – 2) + 5.6m

Answer:

2.5(m – 2) + 5.6m = 8.1m – 5.

Explanation:

2.5(m – 2) + 5.6m

= [(m × 2.5) – (2 × 2.5)] + 5.6m

= (2.5m – 5) + 5.6m

= 2.5m – 5 + 5.6m

= 2.5m + 5.6m – 5

= 8.1m – 5.

Question 16.

4(0.6n – 3) – 0.2(2n – 3)

Answer:

4(0.6n – 3) – 0.2(2n – 3) = 2n – 12.6.

Explanation:

4(0.6n – 3) – 0.2(2n – 3)

= 2.4n – 12 – 0.4n + 0.6

= 2.4n – 0.4n – 12 – 0.6

= 2n – 12.6.

**Factor each expression.
**Question 17.

4t – 20s

Answer:

4t – 20s = 4(t – 5s).

Explanation:

4t – 20s

= 4(t – 5s).

Question 18.

-6p – 21q

Answer:

-6p – 21q = -3(2p + 7q).

Explanation:

-6p – 21q

= -3(2p + 7q).

Question 19.

8i + 12 + 4j

Answer:

8i + 12 + 4j = 4(2i + 3 + j).

Explanation:

8i + 12 + 4j

= 4(2i + 3 + j)

Question 20.

6a + 10b – 20

Answer:

6a + 10b – 20 = 2(3a + 5b – 10).

Explanation:

6a + 10b – 20

= 2(3a + 5b – 10).

Question 21.

-9m – 3n – 6

Answer:

-9m – 3n – 6 = -3(3m + n + 2).

Explanation:

-9m – 3n – 6

= -3(3m + n + 2).

Question 22.

-15x – 6 – 12y

Answer:

-15x – 6 – 12y = -3(5x + 2 + 4y).

Explanation:

-15x – 6 – 12y

= -3(5x + 2 + 4y).

**Translate each verbal description into an algebraic expression. Then simplify when you can.
**Question 23.

One-fourth x less than the sum of 7 and 2x.

Answer:

\(\frac{x}{4}\) – (7 + 2x) = – \(\frac{7}{4}\)x – 7.

Explanation:

One-fourth x less than the sum of 7 and 2x.

=> \(\frac{x}{4}\) – (7 + 2x)

=> \(\frac{x}{4}\) – 7 – 2x

=> \(\frac{x}{4}\) – 2x – 7.

=> [(x – 8x) ÷ 4 ] – 7.

=> (-7x ÷ 4) – 7

=> – \(\frac{7}{4}\)x – 7.

Question 24.

4 times 5y divided by 18.

Answer:

(4 × 5y ) ÷ 18 = \(\frac{10}{9}\)y.

Explanation:

4 times 5y divided by 18

=> (4 × 5y ) ÷ 18

=> 20y ÷ 18

=> 10y ÷ 9 or \(\frac{10}{9}\)y.

Question 25.

Five-ninths of (3p + 1) subtracted from one-third of (q + p).

Answer:

[\(\frac{5}{9}\) × (3p + 1) ] – [\(\frac{1}{3}\) × (q + p)] = 2p – \(\frac{1}{3}\)q + \(\frac{5}{9}\)

Explanation:

Five-ninths of (3p + 1) subtracted from one-third of (q + p)

=> [\(\frac{5}{9}\) × (3p + 1) ] – [\(\frac{1}{3}\) × (q + p)]

=> \(\frac{15}{9}\)p + \(\frac{5}{9}\) – \(\frac{1}{3}\)q + \(\frac{1}{3}\)p.

=> \(\frac{15}{9}\)p + \(\frac{1}{3}\)p – \(\frac{1}{3}\)q + \(\frac{5}{9}\)

= LCD of 9 n 3 = 9.

= [(15 + 3)p ÷ 9 ] – \(\frac{1}{3}\)q + \(\frac{5}{9}\)

= (18p ÷ 9) – \(\frac{1}{3}\)q + \(\frac{5}{9}\)

= (2p ÷ 1) – \(\frac{1}{3}\)q + \(\frac{5}{9}\)

= 2p – \(\frac{1}{3}\)q + \(\frac{5}{9}\)

**Problem Solving
**

**Solve. Show your work.**

Question 26.

After 14 boys leave a concert, the ratio of boys to girls is 3 : 10. If there are p girls at the concert, write an algebraic expression for the number of boys at the beginning of the concert in terms of p.

Answer:

An algebraic expression for the number of boys at the beginning of the concert in terms of p =

0.3 p + 14.

Explanation:

Number of girls in the concert = p.

After 14 boys leave a concert, the ratio of boys to girls is 3 : 10.

Let total number of boys be b.

Ratio of boys after 14 boys leave:

=> \(\frac{b – 14}{p}\) = \(\frac{3}{10}\)

=> 3 × p = 10( b – 14)

=> 3 p = 10 b – 140

=> 10 b = 3 p + 140

= > b = 0.3 p + 14