Math in Focus Grade 6 Chapter 7 Review Test Answer Key

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.
Math in Focus Grade 6 Chapter 7 Review Test Answer Key 1
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

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