Math in Focus Grade 7 Chapter 3 Review Test Answer Key

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

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