Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 7 Review Test to score better marks in the exam.

## Math in Focus Grade 6 Course 1 A Chapter 7 Review Test Answer Key

**Concepts and Skills**

**Write an algebraic expression for each of the following.**

Question 1.

A number that is 5 more than twice x

Answer:

2x + 5

Explanation:

Let’s take a number as variable x.

A number that is 5 more than twice x

2x+ 5.

A number that is 5 more than twice x

is written as 2x+5

Question 2.

The total cost, in dollars, of 4 pencils and 5 pens if each pencil costs w cents and each pen costs 2w cents

Answer:

4w + 10w = 14w

Explanation:

The total cost, in dollars, of 4 pencils and 5 pens

given, each pencil costs w cents and each pen costs 2w cents

pencils = w cents

4 pencils = 4w cents

pen = 2w cents

5 pens = 10w cents

Question 3.

The length of a side of a square whose perimeter is r units

Answer:

\(\frac{r}{4}\)

Explanation:

perimeter of a square = side x 4

side = perimeter / 4

perimeter is given as r

side = r/4

Question 4.

The perimeter of a rectangle whose sides are of lengths (3z + 2) units and (2z + 3) units

Answer:

(10z + 10) units

Explanation:

The perimeter of a rectangle whose sides are of lengths (3z + 2) units and (2z + 3) units

perimeter of a rectangle = 2 [(3z + 2) + (2z + 3)] units

=2 [ 5z + 5]

= 10z + 10

**Evaluate each expression for the given value of the variable.**

Question 5.

3(x + 4) – \(\frac{x}{2}\) when x = 2

Answer:

17

Explanation:

3(x + 4) – \(\frac{x}{2}\) when x = 2

= 3(2 + 4) – \(\frac{2}{2}\)

= [(3 x 6) – 1]

= 18 – 1 = 17

Question 6.

\(\frac{5 p+9}{2}\) + \(\frac{2 p+5}{3}\) when p = 5

Answer:

22

Explanation:

\(\frac{5 p+9}{2}\) + \(\frac{2 p+5}{3}\) when p = 5

= \(\frac{5 x 5 + 9}{2}\) + \(\frac{2 x 5 + 5}{3}\)

= \(\frac{25+9}{2}\) + \(\frac{10+5}{3}\)

= \(\frac{34}{2}\) + \(\frac{15}{3}\)

= \(\frac{(3×34)+(2×15)}{6}\)

= \(\frac{102+30}{6}\)

= \(\frac{132}{6}\)

= 22

**Simplify each expression.**

Question 7.

24k + 11 – 5k – 4

Answer:

19k + 7

Explanation:

24k + 11 – 5k – 4

= 19k + 7

Question 8.

10 + 13h – 6 – 4h + 9 + 12h

Answer:

13 + 21h

Explanation:

10 + 13h – 6 – 4h + 9 + 12h

= 10 – 6 + 9 + 13h – 4h + 12h

= 4 + 9 + 9h + 12h

= 13 + 21h

**Expand each expression.**

Question 9.

5(m + 3) + 2(m + 8)

Answer:

7m + 31

Explanation:

5(m + 3) + 2(m + 8)

= (5m + 15) + (2m + 16)

= 7m + 31

Question 10.

9(x + 2) + 4(5 + x)

Answer:

13x + 38

Explanation:

9(x + 2) + 4(5 + x)

= 9x + 18 + 20 + 4x

= 13x + 38

**Factor each expression.**

Question 11.

5a – 25

Answer:

5(a – 5)

Explanation:

5a – 25

Take 5 as a common factor

5(a – 5)

Question 12.

28 – 7x

Answer:

7(4 – x)

Explanation:

28 – 7x

Take 7 as a common factor

7(4 – x)

Question 13.

12z + 28 – 7z – 3

Answer:

5(z + 5)

Explanation:

12z + 28 – 7z – 3

= 5z – 25

= 5(z + 5)

**State whether each pair of expressions are equivalent.**

Question 14.

3(x + 5) and 5(x + 3)

Answer:

No, pair of expression is not equivalent.

Explanation:

3(x + 5) and 5(x + 3)

3x + 15 = 5x + 15

Question 15.

6y – 26 and 2(3y – 13)

Answer:

Yes, pair of expression equivalent.

Explanation:

6y – 26 and 2(3y – 13)

6y – 26 = 6y – 26

Question 16.

18 – 12p and 3(5 + 6p) + 3(2p + 1)

Answer:

No, pair of expression is not equivalent.

Explanation:

18 – 12p and 3(5 + 6p) + 3(2p + 1)

18 – 12p = 15 + 18p + 6p + 3

18 – 12p = 18 + 24p

Question 17.

15 – 5q and 5(q – 3)

Answer:

No, pair of expression is not equivalent.

Explanation:

15 – 5q and 5(q – 3)

15 – 5q = 5q – 15

**Problem Solving**

**Solve. Show your work.**

Question 18.

Juan is g years old and Eva is 2 years younger than Juan.

a) Find the sum of their ages in terms of g.

Answer:

(2g – 2) years

Explanation:

Juan is g years old

Eva is 2 years younger than Juan = (g – 2)

the sum of their ages in terms of g,

g + (g – 2)

So, (2g – 2) years

b) Find the sum of their ages in g years’ time, in terms of g.

Answer:

(4g – 2) years

Explanation:

Juan is g years old

Eva is 2 years younger than Juan = (g – 2)

the sum of their ages in g years’ time, in terms of g.

Question 19.

Parker bought 4 times as many marbles as Molly. Cole bought 5 fewer marbles than Parker. If Molly bought p marbles, how many marbles did Cole buy?

Answer:

(4p – 5) marbles.

Explanation:

Number of marbles Molly bought = p

Parker bought 4 times as many marbles as Molly = 4p

Cole bought 5 fewer marbles than Parker = (4p – 5)

Number of marbles bought by Cole = (4p – 50

Question 20.

Andrea baked h muffins. Violet baked 8 fewer muffins than Andrea. Find the average number of muffins baked by both girls, in terms of h.

Answer:

(h – 4) muffins

Explanation:

Andrea baked h muffins.

Violet baked 8 fewer muffins than Andrea = (h – 8)

Average number of muffins baked by both girls, in terms of h

h + (h – 8) ÷ 2

= (2h – 8) ÷ 2

= (h – 4)

Question 21.

A square garden has a side length that is 3 meters shorter than the length of a rectangular garden. Find the perimeter of the rectangular garden in terms of y.

Answer:

(8y + 8) meters

Explanation:

A square garden has a side length that is 3 meters shorter than the length of a rectangular garden.

side of square = perimeter/4

= 8y/4 = 2y

3 meters shorter than the length of a rectangular garden

= 2y + 3

Perimeter of a rectangle garden = 2 [ length + width]

Perimeter of a rectangle garden = 2 [ 2y + 3 + 2y + 1]

=2[4y+4]

=8y + 8

Question 22.

Mrs. Roberts sewed m shirts using 2 yards of cloth for each shirt. She also sewed (m + 2) dresses, using 5 yards of cloth for each dress.

a) How much cloth did she use altogether? Give your answer in terms of m.

Answer:

(7m + 10) yards

Explanation:

Mrs. Roberts sewed m shirts using 2 yards of cloth for each shirt.

She also sewed (m + 2) dresses, using 5 yards of cloth for each dress.

2m + 5 x (m+2) =

= 2m + 5m + 10

= 7m + 10 yards cloth used

b) If m = 7, find how much more cloth she used to sew the dresses than the shirts.

Answer:

31 yards

Explanation:

m = 7

5(m+2) – 2m

=5(7 + 2 ) – 2 x 7

= 5 x 9 – 14

= 45 – 14

= 31

Question on 23.

A glass jug can hold (6p + 8) quarts more water than a plastic container. 2 glass jugs and 2 plastic containers contain 56p quarts of water in all.

a) How much water can the plastic container hold? Give your answer in terms of p.

Answer:

(11p – 4) quarts

Explanation:

Let X = quarts the plastic container can hold

Then the jug can hold (6p+8) + X

2 jugs and 2 plastic containers = 56p quarts ==>

2(6p + 8 + X) + 2X = 56p

12p + 16 + 2X + 2X = 56p

4X = 44p – 16

X = 11p – 4

The plastic container can hold 11p – 4 quarts of water

b) If p = 3, find how much water can 1 glass jug and 1 plastic container hold in all.

Answer:

84 quarts

Explanation:

A glass jug can hold (6p + 8) quarts

Question 24.

Mr. Lee can paint 20 chairs in t hours. He uses 3 liters of paint for every 12 chairs that he painted.

a) Find, in terms of t, the number of chairs that he can paint in 3 hours.

Answer:

\(\frac{60}{t}\) chairs

Explanation:

Mr. Lee can paint 20 chairs in t hours

let X number of chairs that he can paint in 3 hours.

20 => t

X => 3

3 x 20 = X x t

Xt = 60

X = \(\frac{60}{t}\) chairs

b) Find, in terms of t, the time taken by Mr. Lee to paint 7 chairs.

Answer:

\(\frac{7}{20}\) t hours

Explanation:

20 => t

7 => X

X x 20 = 7 x t

X = \(\frac{7}{20}\) t

c) If t = 4, find the amount of paint Mr. Lee has used after painting for 4 hours.

Answer:

5 liters

Explanation:

He uses 3 liters of paint for every 12 chairs that he painted.

Mr. Lee can paint 20 chairs in t hours

20 => t

if t = 4

Mr. Lee can paint 20 chairs in 4 hours

He uses 3 liters of paint for every 12 chairs that he painted.

x ltrs for 20 chairs

3 => 12

x => 20

3 x 20 = x x 12

x = 60 / 12

x = 5 ltrs