Math in Focus

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Cumulative Review Chapters 4-7 to score better marks in the exam.

Math in Focus Grade 6 Course 1 A Cumulative Review Chapters 4-7 Answer Key

Concepts and Skills

Write each ratio in simplest form. (Lesson 4.2)

Question 1.
12 : 28
Answer:
4 : 7
Explanation:
To find the simplest form of ratios we can use,
prime factor method or greatest common factor method.
12 : 28
common factor for both the numbers is 3
so, divided both sides with 3
12 : 28 = 4 : 7

Question 2.
32 : 18
Answer:
16 : 9
Explanation:
To find the simplest form of ratios we can use,
prime factor method or greatest common factor method.
32 : 18
common factor for both the numbers is 2
so, divided both sides with 2
32 : 18 = 16 : 9

Question 3.
36 : 81
Answer:
4 : 9
Explanation:
To find the simplest form of ratios we can use,
prime factor method or greatest common factor method.
36 : 81
common factor for both the numbers is 9
so, divided both sides with 9
36 : 81 = 4 : 9

Question 4.
64 : 40
Answer:
8 : 5
Explanation:
To find the simplest form of ratios we can use,
prime factor method or greatest common factor method.
64 : 40
common factor for both the numbers is 8
so, divided both sides with 8
64 : 40 = 8 : 5

Find the missing term in each pair of equivalent ratios. (Lesson 4.2)

Question 5.
35 : 25 = : 5
Answer:
Missing term is 7
Explanation:
product of extremes = product of means
a:b = c:d
ab = cd
ad = bc
a = 35, b = 25, c = x, d = 5
35 x 5 = 25x
25x = 175
x = 175 ÷ 25
x = 7

Question 6.
48 : 33 = 16 :
Answer:
Missing term is 11
Explanation:
product of extremes = product of means
a:b = c:d
ab = cd
ad = bc
a = 48, b = 33, c = 16, d = x
48x  = 33 x 16
48x = 528
x = 528 ÷ 48
x = 11

Question 7.
18 : = 9 : 12
Answer:
Missing term is 24
Explanation:
product of extremes = product of means
a:b = c:d
ab = cd
ad = bc
a = 18, b = x, c = 9, d = 12
18 x 12 = 9x
9x = 216
x = 216 ÷ 9
x = 24

Question 8.
: 24 = 21 : 14
Answer:
Missing term is 36
Explanation:
product of extremes = product of means
a:b = c:d
ab = cd
ad = bc
a = x, b = 24, c = 21, d = 14
14x = 24 x 21
14x = 504
x = 504 ÷ 14
x = 36

Express each percent as a fraction in simplest form. (Lesson 6.1)

Question 9.
58%
Answer:
= \(\frac{29}{50}\)
Explanation:
= \(\frac{58}{100}\)
= \(\frac{29}{50}\)

Question 10.
24\(\frac{1}{3}\)%
Answer:
2433.33
Explanation:
24 \(\frac{1}{3}\) x 100
= \(\frac{73}{3}\) x 100
= \(\frac{73 x 100}{3}\)
= \(\frac{7300}{3}\)
= 2433.33
Question 11.
9.4%
Answer:
\(\frac{29}{50}\)
Explanation:
9.4% = \(\frac{9.4}{100}\)
multiply the numerator and denominator with 10,
as numerator has decimal
= \(\frac{94}{1000}\)
Divided numerator and denominator with 2
= \(\frac{47}{500}\)

Express each percent as a decimal. (Lesson 6.1)

Question 12.
37%
Answer:
0.37
Explanation:
37% = \(\frac{37}{100}\)
= 0.37

Question 13.
67%
Answer:
0.67
Explanation:
67% = \(\frac{67}{100}\)
= 0.67

Question 14.
8.9%
Answer:
0.089
Explanation:
8.9% = \(\frac{8.9}{100}\)
= 0.089

Express each fraction as a percent. (Lesson 6.2)

Question 15.
\(\frac{19}{20}\)
Answer:
95%
Explanation:
\(\frac{19}{20}\)
= \(\frac{19}{20}\) × 100%
= \(\frac{1900}{20}\) %
= 95%

Question 16.
\(\frac{23}{25}\)
Answer:
92%
Explanation:
\(\frac{23}{25}\)
= \(\frac{23}{25}\) × 100%
= \(\frac{2300}{25}\) %
= 92%

Question 17.
\(\frac{360}{400}\)
Answer:
90%
Explanation:
\(\frac{360}{400}\)
= \(\frac{360}{400}\) × 100%
= \(\frac{36000}{400}\) %
= 90%

Express each decimal as a percent. (Lesson 6.2)

Question 18.
0.07
Answer:
7%
Explanation:
Express 0.04 as a percent.
Converting from a decimal to a percentage is done by multiplying the decimal value by 100 and adding %.
0.07 = \(\frac{7}{100}\)
= 0.07 x 100 = 7

Question 19.
0.62
Answer:
62%
Explanation:
Express 0.62 as a percent.
Converting from a decimal to a percentage is done by multiplying the decimal value by 100 and adding %.
0.62= \(\frac{62}{100}\)
= 0.62 x 100
= 62

Question 20.
0.8
Answer:
80%
Explanation:
Express 0.8 as a percent.
Converting from a decimal to a percentage is done by multiplying the decimal value by 100 and adding %.
0.8 = \(\frac{8}{100}\)
= 0.8 x 100
= 80

Find the quantity represented by each percent. (Lesson 6.3)

Question 21.
55% of 600 liters
Answer:
330 liters
Explanation:
55% of 600 liters
= \(\frac{55}{100}\) × 600
= 330
66% of 740 is 330.

Question 22.
73% of $3,900
Answer:
$2,847
Explanation:
73% of $3,900
= \(\frac{73}{100}\) × 3,900
= 2,847
73% of 740 is $2,847.

Write an algebraic expression for each of the following. (Lesson 7.1)

Question 23.
A number that is 7 more than 3 times x
Answer:
3x + 7
Explanation:
x + 7 is an algebraic expression in terms of x.
x and 7 are the terms of expression.
A number that is 7 more than 3 times x
3x + 7

Question 24.
The total cost, in cents, of 8 apples and 9 oranges if each apple costs 2y cents and each orange costs 3y cents
Answer:
43y cents
Explanation:
each apple costs 2y cents
each orange costs 3y cents
8 x 2y + 9 x 3y
= 16y + 27y
= 43y

Question 25.
The perimeter of a rectangle whose sides are of lengths (4z + 3) units and (3z + 5) units
Answer:
14z + 16 units
Explanation:
The perimeter of a rectangle = 2(Length + Breadth)
= 2 ( (4z + 3) + (3z + 5))
= 2 (4z + 3z + 3 + 5)
= 2 (7z + 8)
= 14z + 16

Evaluate each expression for the given value of the variable. (Lesson 7.2)

Question 26.
4(x + 5) – \(\frac{2x}{3}\) when x = 6
Answer:
40
Explanation:
by substituting  x value = 6 in the given algebraic equation
4(x + 5) – \(\frac{2x}{3}\)
= 4(6 + 5) – \(\frac{2 x 6}{3}\)
= (24 + 20) – \(\frac{12}{3}\)
= 44 – \(\frac{12}{3}\)
= (132 – 12) ÷ 3
= 120 ÷ 3
= 40

Question 27.
\(\frac{6 y+7}{3}\) + \(\frac{3 y+4}{4}\) when y = 8
Answer:
\(\frac{152}{6}\)  or 25.33
Explanation:
\(\frac{6 y+7}{3}\) + \(\frac{3 y+4}{4}\) when y = 8
= \(\frac{6 x 8+7}{3}\) + \(\frac{3 x 8+4}{4}\)
= \(\frac{48+7}{3}\) + \(\frac{24+4}{4}\)
= \(\frac{55}{3}\) + \(\frac{28}{4}\)
= \(\frac{304}{12}\)
= \(\frac{152}{6}\)  or 25.33

Simplify each expression. (Lesson 7.3)

Question 28.
36a + 12 – 6a – 7
Answer:
30a + 5
Explanation:
36a + 12 – 6a – 7
= 36a – 6a + 12 – 7
= 30a + 5

Question 29.
21 + 34b – 8 – 5b + 8 + 23b
Answer:
Explanation:
21 + 34b – 8 – 5b + 8 + 23b
= 21 – 8 + 8 + 34b – 5b + 23b
= 21 + 52b

Expand each expression. (Lesson 7.4)

Question 30.
6(m + 4) + 3(m + 9)
Answer:
9m + 51
Explanation:
6(m + 4) + 3(m + 9)
= 6m + 24 + 3m + 27
= 9m + 51

Question 31.
8(n + 4) + 5(6 + n)
Answer:
13n + 62
Explanation:
8(n + 4) + 5(6 + n)
= 8n + 32 + 30 + 5n
= 13n + 62

Factor each expression. (Lesson 7.4)

Question 32.
6a – 42
Answer:
6(a – 7)
Explanation:
6a – 42
In the above expression 6 is a common factor
= 6(a – 7)

Question 33.
56 – 8b
Answer:
8(7 – b)
Explanation:
56 – 8b
In the above expression 8 is a common factor
= 8(7 – b)

Question 34.
23c + 37 – 5c + 8
Answer:
9(2c + 5)
Explanation:
23c + 37 – 5c + 8
= 23c – 5c + 37 + 8
= 18c + 45
= 9(2c + 5)

Question 35.
58 + 40d – 9 – 19d
Answer:
7(7 – 4d)
Explanation:
58 + 40d – 9 – 19d
= 58 – 9 + 40d – 19d
= 49 – 28d
= 7(7 – 4d)

Problem Solving

Solve. Show your work.

Question 36.
A factory produces 550 bottles of water in 25 minutes. How many bottles can it produce in 3 minutes? (Chapter 5)
Answer:
66 water bottles
Explanation:
A factory produces 550 bottles of water in 25 minutes.
Number of bottles can it produce in 3 minutes
= (550 x 3) ÷ 25
= 1650 ÷ 25 = 66

Question 37.
A machine can stamp 75 caps per minute. At this rate, how long will it take to stamp 3,000 caps? (Chapter 5)
Answer:
40 minutes
Explanation:
A machine can stamp 75 caps per minute
At this rate, how long will it take to stamp 3,000 caps
= 3000 ÷ 75
= 40 minutes

Question 38.
There were 65 students in the hall. 80% of them were girls. How many girls were in the hall? (Chapter 6)
Answer:
52 girls
Explanation:
There were 65 students in the hall.
80% of them were girls.
Number of  girls in the hall
= 80% 0f 65
= \(\frac{80}{100}\) X 65
= \(\frac{5200}{100}\)
= 52

Question 39.
Kenneth had $1,800. He spent 34% of it on a watch. How much did he pay for the watch? (Chapter 6)
Answer:
$612
Explanation:
Kenneth had $1,800.
He spent 34% of it on a watch.
Amount he paid for the watch
= 34% 0f 1800
= \(\frac{34}{100}\) X 1800
= 34 X 18 = 612

Question 40.
28% of a number is 168. Find the number. (Chapter 6)
Answer:
number is 200
Explanation:
28% of a number is 168
28% of x = 168
\(\frac{28}{100}\)x = 168
\(\frac{28x}{100}\) = 168
28x = 168 x 100
28x = 16800
x = 16800 ÷ 28
x = 600

Question 41.
140% of a number is 364. Find the number. (Chapter 6)
Answer:
number is 260
Explanation:
140% of a number is 364
140% of x = 364
\(\frac{140}{100}\)x = 364
\(\frac{140x}{100}\) = 364
140x = 364 x 100
140x = 36400
x = 36400 ÷ 140
x = 260

Question 42.
Of the 90 students who sat for a test, 24 students passed. Find the ratio of the number of students who passed the test to the number of students who did not. Give your answer in simplest form. (Chapter 4)
Answer:
4 : 11
Explanation:
Of the 90 students who sat for a test, 24 students passed.
passed = 90 – 24 = 66
So, the ratio of the number of students who passed the test to the number of students who did not.
24 : 66 = 4 : 11
Question 43.
On Saturday, Aaron spent $108. On Sunday, he spent $54 more than what he spent on Saturday. Find the ratio of Aaron’s spending on Saturday to his spending on both Saturday and Sunday. Give your answer in simplest form. (Chapter 4)
Answer:
2 : 3
Explanation:
On Saturday, Aaron spent $108.
On Sunday, he spent $54 more than what he spent on Saturday.
$108 + $54 = $162
So, the ratio of Aaron’s spending on Saturday to his spending on both Saturday and Sunday.
108 : 162 = 2 : 3

Question 44.
A dalmatian weighs 72 pounds. A bullmastiff is 24 pounds heavier than the dalmatian. A bulldog is 12 pounds lighter than the bullmastiff. Find the ratio of the bulldog’s weight to the dalmatian’s weight. Give your answer in simplest form. (Chapter 4)
Answer:
7 : 6
Explanation:
A dalmatian weighs 72 pounds.
A bullmastiff is 24 pounds heavier than the dalmatian.
72 pounds + 24 pounds = 96 pounds
A bulldog is 12 pounds lighter than the bullmastiff.
96 pounds – 12 pounds = 84
The ratio of the bulldog’s weight to the dalmatian’s weight.
84 : 72 = 7 : 6

Question 45.
The ratio of the number of left-handers to the number of right-handers in a middle school is 6 : 15. If there are 120 left-handers, how many right-handers are there? (Chapter 4)
Answer:
300 right-handers
Explanation:
The ratio of the number of left-handers to the number of right-handers in a middle school is 6 : 15.
If there are 120 left-handers,
then how many right-handers are there
6 : 15 :: 120 : x
product of extremes = product of means
a:b = c:d
\(\frac{a}{b}\) = \(\frac{c}{d}\)
ad = bc
a = 6, b = 15, c = 120, d = x
6x = 15 x 120
4x = 1800
x = \(\frac{1800}{6}\)
x = 300

Question 46.
Mrs. Jackson gave a sum of money to her son and daughter in the ratio 8 : 9. Her daughter received $3,060. How much did Mrs. Jackson give to her two children in all? (Chapter 4)
Answer:
$5,780
Explanation:
Mrs. Jackson gave a sum of money to her son and daughter in the ratio 8 : 9.
Her daughter received $3,060.
Her son received
8 : 9 = 3060 : x
\(\frac{8}{9}\) = 3060
9x = 3060 x 8
x = 24480 ÷  9
x = 2720
Total amount Mrs. Jackson give to her two children in all
= 3060 + 2720
= 5780

Question 47.
The ratio of the number of boys to the number of girls in a school is 7 : 9. If there are 1,248 students in the school, how many girls are there? (Chapter 4)
Answer:
702 girls
The ratio of the number of boys to the number of girls in a school is 7 : 9.
If there are 1,248 students in the school,
Let the number of boys be 7x and girls be 9x
Total number of students = 7x + 9x = 1248
=> 16x = 1248
x = 1248 ÷ 16
x = 78
Number of girls = 9x
= 9 x 78 = 702

Question 48.
A sum of money was shared among Daniel, Elliot, and Frank in the ratio 5 : 7 : 8. If Frank’s share was $2,781 more than Daniel’s share, what was the original sum of money shared among the three men? (Chapter 4)
Answer:
$18540
Explanation:
Let 5x = Daniel’s share
7x = Elliot’s share
8x = Frank’s share
8x = 5x + 2781
3x = 2781
x = 927
Daniel’s share = 5x
= 5 × 927 = $4635
Elliot’s share = 7x
= 7 × 927 = $6489
Frank’s share = 8x
= 8 × 927 = $7416
Amount shared among all = 4635 + 6489 + 7416 = $18540

Question 49.
A tank is filled with water at a rate of 2.4 liters per minute. At this rate, how long will it take to fill the tank with 84 liters of water? (Chapter 5)
Answer:
35 minutes
Explanation:
A tank is filled with water at a rate of 2.4 liters per minute.
At this rate, how long will it take to fill the tank with 84 liters of water
1 min = 2.4 L
84 L = x
x = 84 ÷ 2.4
x = 35min

Question 50.
A farmer uses 920 grams of grains to feed 8 chickens. (Chapter 5)
a) At this rate, how many grams of grains does he use to feed 100 chickens?
Answer:
11500 grams
Explanation:
number of grams used to feed 8 chickens = 920g
number of grams used to feed 1 chicken = 920 ÷ 8 = 115g
number of grams used to feed 100 chickens = 100 × grams used for 1 chicken
= 100 × 115 = 11500g

b) At this rate, how many chickens can he feed using 48.3 kilograms of grains?
Answer:
420 chickens
Explanation:
Grams used = 48.3 × 1000 = 48300g
1kg = 1000g
number of grams used to feed 1 chicken = 115g
number of chickens = grams used ÷ number of grams used to feed 1 chicken
= 48300 ÷ 115
= 420 chickens

Question 51.
The distance between School A and School B is 360 kilometers. (Chapter 5)
a) If a bus leaves School A at 9:30 A.M. and reaches School B at 2:30 P.M., what is its speed in kilometers per hour?
Answer:
A to B = 360Km
Explanation:
start time at  A 9:30 AM
reaching time at B 2:30 PM
the difference = 5Hours
Speed = \(\frac{Distance}{Time}\)
= \(\frac{360}{5}\)
= 72 kilometers per hour

b) If a car travels at a speed of 80 kilometers per hour, how long will it take to travel from School A to School B?
Answer:
400 kilometer
Explanation:
start time at  A 9:30 AM
reaching time at B 2:30 PM
speed of car is 80kilometer per hour
Speed = \(\frac{Distance}{Time}\)
80 = \(\frac{Distance}{5}\)
80 × 5 = Distance
Distance travels by a car = 400 kilometers

Question 52.
A truck traveled from Town A to Town B, and then to Town C. The truck took 3 hours to travel from Town A to Town B at an average speed of 42 kilometers per hour. It then traveled from Town B to Town C at an average speed of 68 kilometers per hour. The truck took a total of 5 hours to travel from Town A to Town C. (Chapter 5)
a) What is the distance between Town A and Town B?
Answer:
126km
Explanation:
Town A to Town B
time = 3hours and (speed = 42 kmph)
the distance between Town A and Town B
Distance = Speed x time
= 42 × 3
= 126 kilometers
the distance between Town A and Town B is 126 km

b) What was average speed of the truck for the whole journey?
Answer:
55 kilometers per hour
Explanation:
(42 + 68)/2
= 110/2
= 55 kilometers per hour

Question 53.
Russell bought a CD player that cost $120 before it went on sale for a 15% discount. If he paid 5% sales tax on the sale price, how much did Russell pay for the CD player? (Chapter 6)

Answer:
$107.10
Explanation:
Russell bought a CD player that cost $120,
before it went on sale for a 15% discount.
If he paid 5% sales tax on the sale price, how much did Russell pay for the CD player
(120 × 15)/100
= 18
= 120 – 18
= 102
= (102 × 5)/100
= 510/100
= 5.1
Russell pay for the CD player
= $102 + $5.1
= $107.10

Question 54.
Last year, Jane had $2,800 in her bank account. This year, her savings increased by 4.5%. $he plans to increase her savings by 5% next year. How much will Jane have in her account after the two-year period? (Chapter 6)
Answer:
$3,072.30
Explanation:
Jane had $2,800 in her bank account.
(2,800 × 4.5)/100
=12,600/100
=126
2,800 + 126 = 2,926
= (2,926 × 5 )/100
= 14,630/100
= 146.3
2926 + 146.3
= 3,072.3

Question 55.
Two jackets were on sale. The first jacket cost $239 and was marked down 40%. The second jacket cost $159 and was given a 20% discount. Which of the two jackets cost less during the sale? Justify your reasoning. (Chapter 6)
Answer:
Second jacket
Explanation:
The first jacket cost $239 and was marked down 40%.
100 – 40 = 60%
60% of 239
(60 ÷ 100) × 239 = 143.40
The second jacket cost $159 and was given a 20% discount
100 – 20 = 80%
80% of 159
(80 ÷ 100) × 159 = 127.20
By compare the price of both the jackets,
second jacket price is less than the first one.

Question 56.
One year, the price of a motorcycle was $3,000. It increased to $3,550 another year. What is the percent increase in its price? Round your answer to 2 decimal places. (Chapter 6)
Answer:
18.33 or 18%
Explanation:
the price of a motorcycle was $3,000.
It increased to $3,550 another year.
the price in second year increased = 3550 – 3000 = 550
the percent increase in its price
1 – (3550 ÷ 3000)
= 1 – 1.1833
= 1833 ÷ 100 = 18.33

Question 57.
Leonard has 5 times as many stamps as Alison. Melissa has 7 more stamps than Leonard. If Alison has q stamps, how many stamps does Melissa have? (Chapter 7)
Answer:
5q + 7 stamps
Explanation:
A ⇒ q
L ⇒ 5q
M ⇒5q + 7
Melissa have 5q + 7 stamps

Question 58.
Tim is k years old and Jennifer is 4 years younger than Tim. Find their average age in 3k years’ time, in terms of k. (Chapter 7)
Answer:
(4k – 2) years
Explanation:
Tim k years
Jennifer is k – 4
average age in 3k years
= (k + 3k +k – 4 + 3k)/2
= 8k – 4/2
= 2(4k – 2)/2
= 4k – 2

Question 59.
The length of a rectangular garden is 7 meters longer than its width. If its perimeter is (12x + 14) meters, find its length in terms of x. (Chapter 7)
Answer:
(3x + 7)m
Explanation:
Perimeter of rectangle = 2 (length +width)
12x + 14 = 2 (x + 7 + x)
Perimeter = 2 width + 2 length
12x + 14  = 2 length + 2x
12x – 2x + 14 = 2 length
10x + 14 = 2 length
length = 2(5x+ 7)/2
= 5x+7

Question 60.
A cupboard weighs (5y + 6) pounds more than a table. 3 cupboards and 4 tables weigh (29y + 18) pounds in all. (Chapter 7)
a) Find the weight of the table in terms of y.
Answer:
2y lb
Explanation:
29y + 18 = 3 cupboards + 4 tables
29y + 18 = 3[( 5y + 6 ) + table] + 4 tables
29y + 18 =15y + 18 + 3 tables + 4 tables
29y – 15y +18 – 18 = 7 tables
14y = 7 tables
table = 2y

b) If y = 15, find the total weight of 2 cupboards and 3 tables.
Answer:
312 lb
Explanation:
y = 15
2[( 5y + 6 ) + table] + 3 tables
2[( 5y + 6 ) + 2y] + 3 × 2y
for y = 15
10y + 12 + 4y + 6y
20y +12
= 20(15) + 12
= 300 + 12
= 312 lbs

Question 61.
A factory can produce 45 containers of yogurt in t minutes. 950 grams of fruit are used for every 25 containers of yogurt. (Chapters 5, 7)
a) Find, in terms of t, the number of containers of yogurt that the factory can produce in 5 minutes.
Answer:
\(\frac{225}{t}\) containers
Explanation:
45 containers in t minutes,
950 grams of fruit are used for every 25 containers of yogurt.
(45 × 5) ÷  t
= \(\frac{225}{t}\)min

b) Find, in terms of t, the time taken by the factory to produce 100 containers of yogurt.
Answer:
\(\frac{20}{9}\)min or 2.22min
Explanation:
45 containers in t minutes,
time taken for 100 containers
Time = (100 containers ÷ 45 containers) × t
= \(\frac{20}{9}\)min or 2.22min

c) If t = 6, find the amount of fruit used in producing containers of yogurt in 8 minutes.
Answer:
2,280 grams
Explanation:
If t = 6min
45 containers in 6min
In 8 min number of containers produced
(45/6min) × 8min = 60 Containers.
the amount of fruit used in producing containers of yogurt in 8 minutes
=(950grams/25cont) × 60 = 2280 Grams

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 7 Lesson 7.4 Expanding and Factoring Algebraic Expressions to score better marks in the exam.

Math in Focus Grade 6 Course 1 A Chapter 7 Lesson 7.4 Answer Key Expanding and Factoring Algebraic Expressions

Math in Focus Grade 6 Chapter 7 Lesson 7.4 Guided Practice Answer Key

Expand each expression.

Question 1.
3(x + 4)
Answer:
3x+12
Explanation:
3(x + 4)
= 3x + 3 x 4
= 3x + 12

Question 2.
6(2x + 3)
Answer:
12x + 18
Explanation:
6(2x + 3)
= 6 X 2x  +  6 x 3
= 12x + 18

Question 3.
2(7 + 6x)
Answer: 14+ 12x
Explanation:
2(7 + 6x)
= 2 x 7 + 6x X 2
= 14 + 12x

Question 4.
5(y – 3)
Answer:
5y – 15
Explanation:
5(y – 3)
= 5y-15

Question 5.
4(4y – 1)
Answer:
16y – 4
Explanation:
4(4y – 1)
= 16y – 4

Question 6.
9(5x – 2)
Answer:
45x – 18
Explanation:
9(5x – 2)
= 9 x 5x – 2 x 9
= 45x – 18

State whether each pair of expressions are equivalent.

Question 7.
6(x + 5) and 6x + 30
Answer:
Yes, pair of expressions are equivalent
Explanation:
6(x + 5) and 6x + 30
6(x + 5) = 6x + 30
6(x + 5) and 6x + 30 are equal.

Question 8.
7(x + 3) and 21 + 7x
Answer:
Yes, pair of expressions are equivalent
Explanation:
7(x + 3) and 21 + 7x
7(x + 3) = 7x + 21
7x + 21 and 21 + 7x are equal.

Question 9.
4(y – 4) and 4y – 4
Answer:
NO, pair of expressions are not equivalent
Explanation:
4(y – 4) and 4y – 4
4(y – 4) = 4y – 16
4y – 16 is not equal to 4y – 4

Question 10.
3(y – 6) and 18 – 3y
Answer:
NO, pair of expressions are not equivalent
Explanation:
3(y – 6) and 18 – 3y
3(y – 6) = 3y – 18
3y – 18 is not equal to 18 – 3y

Hands-On Activity

Materials

• paper
• ruler
• scissors

RECOGNIZE THAT EXPANDED EXPRESSIONS ARE EQUIVALENT

STEP 1: Draw a rectangle that is 8 centimeters wide and more than 3 centimeters long on a piece of paper. Then cut out the rectangle.
Answer:

STEP 2: Find the area of the rectangle in terms of p.
Answer:
8p + 24 cm
Explanation:
Area of rectangle = Length x Breadth
= (p+3) x 8
= 8p + 24 cm
STEP 3: Then cut the rectangle into two rectangles A and B as shown.

STEP 4: Find the areas of rectangle A and rectangle B.
Answer:
24cm
Explanation:
Area of rectangle A = Length x Breadth
= p x 8
= 8p cm
Area of rectangle B = Length x Breadth
= 3 x 8
= 24 cm

STEP 5: Using your answers found in STEP 2 and STEP 4, state how the three areas are related.
Answer:
8p + 24 cm
Explanation:
Area of rectangle = Area of rectangle A + Area of rectangle B
8p + 24 cm = 8p + 24 cm

STEP 6: You may repeat the activity using rectangles of other sizes.
can be done with different size of rectangles

Factor each expression.

Question 11.
3x + 3
Answer:
3,(x + 1)
Explanation:
In the above expression 3 is common factor.
3 and (x + 1) are the factors of 3x + 3

Question 12.
4x + 6
Answer:
2 and (2x + 3)
Explanation:
In the above expression 2 is common factor.
2,(2x + 3) are the factors of 4x + 6

Question 13.
8 + 6y
Answer:
2 and (4 + 3y)
Explanation:
8+6y
In the above expression 2 is common factor.
2 and (4 + 3y) are the factors of 8 + 6y

Question 14.
5y – 10
Answer:
5, (y-2)
Explanation:
In the above expression 5 is common factor.
5y – 10 = 5(y – 2)

Question 15.
4 – 10z
Answer:
2(2 – 5z)
Explanation:
In the above expression 2 is common factor.
4 – 10z = 2(2 – 5z)

Question 16.
12 – 8x
Answer:
4, (3-2x)
Explanation:
In the above expression 4 is common factor.
12 – 8x = 4(3 – 2x)

Question 17.
8f + 6
Answer:
2 ,(4f+3)
Explanation:
In the above expression 2 is common factor.
8f + 6 = 2(4f + 3)

Question 18.
12t – 8
Answer:
4,(3f – 2)
Explanation:
In the above expression 4 is common factor.
12t – 8 = 4(3f – 2)

Question 19.
15 + 5q
Answer:
5, (3+q)
Explanation:
In the above expression 5 is common factor.
15 + 5q = 5(3 + q)

Question 20.
32m – 40
Answer:
8 and (4m-5)
Explanation:
In the above expression 8 is common.
8 and (4m – 5)are the factors of 32m – 40

State whether each pair of expressions are equivalent.

Question 21.
8x + 6 and 2(4x + 3)
Answer:
Yes, pair of expressions are equivalent.
Explanation:
8x + 6 and 2(4x + 3)
8x + 6  = 2(4x + 3)
2(4x + 3) and 2(4x + 3) are equal.

Question 22.
3(y + 6) and 18 + 3y
Answer:
Yes, pair of expressions are equivalent.
Explanation:
3(y + 6) and 18 + 3y
3(y + 6) = 3y + 18
3y + 18 = 18 + 3y are equal.

Question 23.
5x – 10 and 5(x – 5)
Answer:
No, pair of expressions are not equivalent.
Explanation:
5x – 10 and 5(x – 5)
5x – 10 = 5(x – 2)
5(x – 2) and 5(x – 5) are not equal.

Question 24.
4(y – 4) and 16 – 4y
Answer:
No, pair of expressions are not equivalent.
Explanation:
4(y – 4) and 16 – 4y
4(y – 4) = 4y – 16
4y – 16 is not equivalent to 16 – 4y
4(y – 4) and 16 – 4y are not equivalent.

Question 25.
3(x + 5) and 15 + 3x
Answer:
Yes, pair of expressions are not equivalent.
Explanation:
3(x + 5) and 15 + 3x
3(x + 5) = 3x + 15
3x + 15 is equal to 15 + 3x
3(x + 5) and 15 + 3x are equivalent expressions.

Question 26.
12 – 8y and 4(2y – 3)
Answer:
No, pair of expressions are not equivalent.
Explanation:
12 – 8y and 4(2y – 3)
12 – 8y = 4(3 – 2y)
12 – 8y and 4(2y – 3) are not equivalent.

Math in Focus Course 1A Practice 7.4 Answer Key

Expand each expression.

Question 1.
5(x + 2)
Answer:
5x + 10
Explanation:
5(x + 2)
= 5 X x + 5 X 2
= 5x + 10

Question 2.
7(2x – 3)
Answer:
14x – 21
Explanation:
7(2x – 3)
= 7 X 2x – 7 X 3
= 14x – 21

Question 3.
4(y – 3)
Answer:
4y – 12
Explanation:
4(y – 3)
= 4 X y – 4 X 3
= 4y – 12

Question 4.
8(3y – 4)
Answer:
24y – 32
Explanation:
8(3y – 4)
= 8 X 3y – 8 X 4
= 24y – 32

Question 5.
3(x+ 11)
Answer:
3x + 33
Explanation:
3(x+ 11)
= 3 X x + 3 X 11
= 3x + 33

Question 6.
9(4x – 7)
Answer:
36x – 63
Explanation:
9(4x – 7)
= 9 X 4x – 9 X 7
= 36 x – 63

Factor each expression.

Question 7.
6p + 6
Answer:
6 , (p + 1)
Explanation:
6p + 6
Take 6 as common factor
6 (p + 1)
6, (p + 1) are the factors of  6p + 6

Question 8.
3p + 18
Answer:
3, (p + 6)
Explanation:
3p + 18
Take 3 as common factor
3 (p + 6)
3, (p+6) are the factors of 3p + 18

Question 9.
12 + 3q
Answer:
3, (4 + q)
Explanation:
12 + 3q
Take 3 as common factor
3 (4 + q)
3, (4 + q) are the factors of 12 + 3q

Question 10.
4w – 16
Answer:
4, (w – 4)
Explanation:
4w – 16
take 4 as common factor,
4 (w – 4)
4, (w – 4) are the factors of 4w – 16

Question 11.
14r – 8
Answer:
2, (7r – 4)
Explanation:
14r – 8
take 2 as common factor,
2 (7r – 4)
2, (7r – 4) are the factors of 14r – 8

Question 12.
12r – 12
Answer:
12, (r – 1)
Explanation:
12r – 12
take 12 as common factor,
12 (r – 1)
12, (r – 1) are the factors of 12r – 12

State whether each pair of expressions are equivalent.

Question 13.
4x + 12 and 4(x + 3)
Answer:
YES, pair of expressions are equivalent
Explanation:
4x + 12 and 4(x + 3)
4x + 12 and 4x + 12
both are equal

Question 14.
5(x – 1) and 5x – 1
Answer:
NO, pair of expressions are not equivalent
Explanation:
5(x – 1) and 5x – 1
5x – 5 and 5x – 1
both are not equal

Question 15.
7(5 + y) and 7y + 35
Answer:
YES, pair of expressions are equivalent
Explanation:
7(5 + y) and 7y + 35
35 + 7y and 7y + 35
both are equal

Question 16.
9(y – 2) and 18 – 9y
Answer:
NO, pair of expressions are not equivalent
Explanation:
9(y – 2) and 18 – 9y
9y – 18 and 18 – 9y
both are not equal

Expand and simplify each expression.

Question 17.
3(m + 2) + 4(6 + m)
Answer:
7m + 30
Explanation:
3(m + 2) + 4(6 + m)
= 3m + 6 + 24 +4m
= 7m + 30

Question 18.
5(2p + 5) + 4(2p – 3)
Answer:
18p+13
Explanation:
5(2p + 5) + 4(2p – 3)
= 10p + 25 + 8p – 12
= 18p + 13

Question 19.
4(6k + 7)+ 9 – 14k
Answer:
10k + 37
Explanation:
4(6k + 7)+ 9 – 14k
= 24k + 28 + 9 – 14k
= 10k + 37

Simplify each expression. Then factor the expression.

Question 20.
14x + 13 – 8x – 1
Answer:
6, (x+2)
Explanation:
14x + 13 – 8x – 1
= 14x – 8x + 13 – 1
= 6x + 12
= 6(x + 2)
6, (x + 2) are the factors of 14x + 13 – 8x – 1

Question 21.
8(y + 3) + 6 – 3y
Answer:
5, (y+6)
Explanation:
8(y + 3) + 6 – 3y
= 8y + 24 + 6 – 3y
= 8y – 3y + 24 + 6
= 5y + 30
= 5(y + 6)
5, (y + 6) are the factors of the 8(y + 3) + 6 – 3y

Question 22.
4(3z + 7) + 5(8 + 6z)
Answer:
2 , (21z+34)
Explanation:
4(3z + 7) + 5(8 + 6z)
= 12z + 28 + 40 + 30z
= 12z + 30z + 40 + 28
= 42z + 68
= 2 (21z + 34)
2 , (21z + 34) are the factors of 4(3z + 7) + 5(8 + 6z)

Solve.

Question 23.
Expand and simplify the expression 3(x – 2) + 9(x + 1) + 5(1 + 2x) + 2(3x – 4).
Answer:
28x
Explanation:
3(x – 2) + 9(x + 1) + 5(1 + 2x) + 2(3x – 4).
= 3x – 6 + 9x + 9 + 5 + 10x + 6x – 8.
= 3x + 9x + 10x  + 6x – 6 +9 +5 – 8
= 28x + 0
= 28x

Question 24.
Are the two expressions equivalent? Justify your reasoning.
15(y + 6) + 10(y – 5) + 20(2y + 3)and 5(20 + 13y)
Answer:
Yes, the two expressions equivalent
Explanation:
15(y + 6) + 10(y – 5) + 20(2y + 3) and 5(20 + 13y)
= 15(y + 6) + 10(y – 5) + 20(2y + 3) and 5(20 + 13y)
= 15y + 90 + 10y – 50 +40y + 60  and 100 + 65y
= 65y + 100 and 100 + 65y
So, both are same.

Question 25.
A yard of lace costs w cents and a yard of fabric costs 40c more than the lace. Kimberly wants to buy one yard of lace and 2 yards of fabric. How much money will she need? Express your answer in terms of w.
Answer:
$80 w cents
Explanation:
cost of one yard of lace = w cents
cost of 2 yards of fabric = 2 × 40 = $80
Therefore, the total money needed is $80 + w cents
= $80 w cents

Question 26.
The average weight of 6 packages ¡s (9m + 8) pounds. 2 more packages, with weights of (12m + 1 2) pounds and (14m + 12) pounds, are added to the original 6 packages. Find the average weight of the 8 packages.
Answer:
10m + 9
Explanation:
9m + 8 is the average weight of the 6 packages = 54m + 48
12m + 12 and 14m + 12 is added
12m+12+14m+12 = 26m + 24
26m + 24 is added to the 54m + 48
= 54m + 26m + 48 + 24
= 80m + 72
the average weight of the 8 packages is = (80m + 72)/8 = 10m + 9

Question 27.
The figure shows two rectangles joined to form rectangle ABCD.

a) Write the length of \(\overline{B C}\) in terms of x. Then write an expression for the area of the rectangle ABCD in terms of x.
Answer:
Area =3x + 6 cm
Explanation:
the length of \(\overline{B C}\) in terms of x = x + 2
an expression for the area of the rectangle ABCD in terms of x
= x + 2 X 3
= (x + 2)3
= 3x + 6cm

b) Write an expression for the area of each of the two smaller rectangles.
Answer:
6 cm
Explanation:
the area of each of the two smaller rectangles
= 2 x 3 = 6cm

c) Math Journal Explain how you can use your answers in a) and b) to show that the following expressions are equivalent.  3x + 6 and 3(x + 2)
Answer:
Yes, both are equivalent
Explanation:
3x + 6 and 3(x + 2)
3(x + 2) can be written as 3x + 6
So, both are equal.

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 7 Lesson 7.2 Evaluating Algebraic Expressions to score better marks in the exam.

Math in Focus Grade 6 Course 1 A Chapter 7 Lesson 7.2 Answer Key Evaluating Algebraic Expressions

Math in Focus Grade 6 Chapter 7 Lesson 7.2 Guided Practice Answer Key

Evaluate each algebraic expression for the given value of x.

Question 1.

Answer:

Explanation:
An expression is a term in that describes a group of variables, numbers and operators.
Operators include division, multiplication, addition and subtraction.
Variables in expression are usually denoted as x and y, but it can be and other symbol.
The expressions can be written as verbal phrase or algebraic expression.

Math in Focus Course 1A Practice 7.2 Answer Key

Evaluate each expression for the given value of the x.

Question 1.
x + x + 5 when x = 7
Answer:
19
Explanation:
x = 7
by substituting x value as 7
x + x + 5 = 7 + 7 + 5 = 19

Question 2.
3x + 5 when x = 5
Answer:
20
Explanation:
3x + 5, where as
x = 5
by substituting x value =5
3 x 5 + 5 = 15 + 5 = 20

Question 3.
5y – 8 when y = 3
Answer:
7
Explanation:
5 x y – 8
y = 3
by substituting y value = 3
5 x 3 – 8
15 – 8 = 7

Question 4.
40 – 9y when y = 2
Answer:
22
Explanation:
40 – 9y
y = 2
by substituting y value = 2
40 – 9 x 2 = 40 – 18 = 22

Question 5.
33 – 7w + 6 when w = 4
Answer:
19
Explanation:
33 – 7w + 6
w = 4
by substituting w value = 4
33 – 7 x 4 = 33 – 14 = 19

Question 6.
\(\frac{7w}{2}\) when w = 18
Answer:
63
Explanation:
= \(\frac{7w}{2}\)
w = 18
by substituting w value = 18
= \(\frac{7 x 18}{2}\)
= \(\frac{126}{2}\) = 63

Question 7.
4 + \(\frac{5z}{6}\) when z = 12
Answer:
14
Explanation:
4 + \(\frac{5z}{6}\)
z = 12
by substituting z value = 12
= 4 + \(\frac{5×12}{6}\)
= 4 + \(\frac{60}{6}\)
= 4+10 = 14

Question 8.
\(\frac{4+5z}{2}\) when z = 12
Answer:
32
Explanation:
\(\frac{4+5z}{2}\)
z = 12
by substituting z value =12
= \(\frac{4+5 x 12}{2}\)
= \(\frac{4+60}{2}\)
= \(\frac{64}{2}\) = 32

Question 9.
20 – \(\frac{4r}{5}\) when r = 10
Answer:
16
Explanation:
20 – \(\frac{4r}{5}\)
r = 10
by substituting r value =10
= 20 – \(\frac{4x 10}{5}\)
= 20 – \(\frac{40}{5}\)
= 20 – 8 = 16

Question 10.
\(\frac{8r}{9}\) – 15 when r = 27
Answer:
9
Explanation:
\(\frac{8r}{9}\) – 15
r = 27
by substituting r value =27
= \(\frac{8 x 27}{9}\) – 15
= \(\frac{216}{9}\) – 15
24 – 15 = 9

Question 11.
16 – \(\frac{2 z-4}{3}\) when z = 18
Answer:
\(\frac{16}{3}\)
Explanation:
16 – \(\frac{2 z-4}{3}\) when z = 18
by substituting z value = 18
16 – \(\frac{2 z-4}{3}\)
=16 – \(\frac{2 x 18 – 4}{3}\)
=16 – \(\frac{36-4}{3}\)
=16 – \(\frac{32}{3}\)
=\(\frac{48-32}{3}\)
\(\frac{16}{3}\)

Question 12.
16 – \(\frac{2 z}{3}\) – 4 when z = 18
Answer:
0
Explanation:
16 – \(\frac{2 z}{3}\) – 4 when z = 18
by substituting  z value = 18
16 – \(\frac{2 x 18}{3}\) – 4
= 16 – \(\frac{36}{3}\) – 4
=16 – 12 – 4
=0

Evaluate each expression when x = 3.

Question 13.
\(\frac{x+1}{2}\) + \(\frac{5 x-3}{10}\)
Answer:
\(\frac{16}{5}\)
Explanation:
by substituting  x value = 3 in the given algebraic equation
\(\frac{x+1}{2}\) + \(\frac{5 x-3}{10}\)
= \(\frac{3+1}{2}\) + \(\frac{5 x 3 – 3}{10}\)
= \(\frac{4}{2}\) + \(\frac{12}{10}\)
= 2 + \(\frac{6}{5}\)
= \(\frac{10 + 6}{5}\)
= \(\frac{16}{5}\)

Question 14.
\(\frac{11+x}{2}\) – \(\frac{9 x-3}{4}\)
Answer:
1
Explanation:
by substituting  x value = 3 in the given algebraic equation
\(\frac{11+x}{2}\) – \(\frac{9 x-3}{4}\)
= \(\frac{11+3}{2}\) – \(\frac{9 x 3-3}{4}\)
= \(\frac{14}{2}\) – \(\frac{27-3}{4}\)
= \(\frac{7}{1}\) – \(\frac{24}{4}\)
7 – 6 = 1

Question 15.
\(\frac{7 x-6}{3}\) + 4(8 + 2x)
Answer:
61
Explanation:
by substituting  x value = 3 in the given algebraic equation
\(\frac{7 x-6}{3}\) + 4(8 + 2x)
= \(\frac{7 x 3 -6}{3}\) + 4(8 + 2 x 3)
= \(\frac{21- 6}{3}\) + 4(8 + 6)
= \(\frac{15}{3}\) + 56
= 5 + 56
= 61

Question 16.
13(11 – 3x) – \(\frac{5(16-4 x)}{2}\)
Answer:
16
Explanation:
by substituting  x value = 3 in the given algebraic equation
13(11 – 3x) – \(\frac{5(16-4 x)}{2}\)
13(11 – 3 x 3) – \(\frac{5(16-4 x 3)}{2}\)
13(11 – 9) – \(\frac{5(16 – 12)}{2}\)
13(2) – \(\frac{5(4)}{2}\)
26 – \(\frac{20}{2}\)
= 26 – \(\frac{20}{2}\)
= 26 – 10
= 16

Question 17.
5(x + 2) + 2(6 – x) + \(\frac{2 x+3}{3}\)
Answer:
34
Explanation:
by substituting  x value = 3 in the given algebraic equation
5(x + 2) + 2(6 – x) + \(\frac{2 x+3}{3}\)
5(3 + 2) + 2(6 – 3) + \(\frac{2 x 3+3}{3}\)
5(5) + 2(3) + \(\frac{6+3}{3}\)
25 + 6 + 3 = 34

Question 18.
\(\frac{5 x-3}{4}\) + \(\frac{5(x+5)}{8}\) + 3(13 – 2x)
Answer:
29
Explanation:
by substituting  x value = 3 in the given algebraic equation
\(\frac{5 x-3}{4}\) + \(\frac{5(x+5)}{8}\) + 3(13 – 2x)
= \(\frac{5 x 3 – 3}{4}\) + \(\frac{5(3+5)}{8}\) + 3(13 – 2 x 3)
= \(\frac{15-3}{4}\) + \(\frac{5(8)}{8}\) + 3(13 – 6)
= \(\frac{12}{4}\) + \(\frac{40}{8}\) + 3(7)
= 3 + 5 + 21
= 29

Question 19.
\(\frac{2 x+4}{5}\) – \(\frac{x+1}{4}\) + \(\frac{x}{6}\)
Answer:
\(\frac{3}{2}\)
Explanation:
by substituting  x value = 3 in the given algebraic equation
\(\frac{2 x+4}{5}\) – \(\frac{x+1}{4}\) + \(\frac{x}{6}\)
= \(\frac{2 x 3+4}{5}\) – \(\frac{3+1}{4}\) + \(\frac{3}{6}\)
= \(\frac{10}{5}\) – \(\frac{4}{4}\) + \(\frac{1}{2}\)
= 2-1+\(\frac{1}{2}\)
=1+\(\frac{1}{2}\)
= \(\frac{3}{2}\)

Question 20.
7x – \(\frac{x}{5}\) + \(\frac{7-x}{9}\)
Answer:
21 – \(\frac{3}{5}\) + \(\frac{4}{9}\)
Explanation:
by substituting  x value = 3 in the given algebraic equation
7x – \(\frac{x}{5}\) + \(\frac{7-x}{9}\)
= 7 x 3 – \(\frac{3}{5}\) + \(\frac{7-3}{9}\)
= 21 – \(\frac{3}{5}\) + \(\frac{4}{9}\)
= 21 – \(\frac{3}{5}\) + \(\frac{4}{9}\)

Evaluate each of the following when y = 7.

Question 21.
(5y + 2) minus (2y + 5).
Answer:
18
Explanation:
by substituting  y value = 7 in the given algebraic equation
(5y + 2) minus (2y + 5)
= (5 x 7 + 2) – (2 x 7 + 5).
= (35 + 2) – (14 + 5).
= 37 – 19
= 18

Question 22.
The sum of \(\frac{y}{3}\) and \(\frac{4y}{9}\)
Answer:
\(\frac{49}{9}\)
Explanation:
by substituting  y value = 7 in the given algebraic equation
\(\frac{y}{3}\) + \(\frac{4y}{9}\)
= \(\frac{7}{3}\) + \(\frac{4 x 7}{9}\)
= \(\frac{7}{3}\) + \(\frac{28}{9}\)
= \(\frac{21+28}{9}\)
= \(\frac{49}{9}\)

Question 23.
The product of (y + 1)and(y – 1)
Answer:
\(\frac{4}{3}\)
Explanation:
by substituting  y value = 7 in the given algebraic equation
(y + 1)and(y – 1)
= (7 + 1)and(7 – 1)
= (8)and(6)
= \(\frac{8}{6}\)
= \(\frac{4}{3}\)

Question 24.
8(2y – 1) minus \(\frac{14 y+37}{5}\)
Answer:
8(2y – 1) minus \(\frac{14 y+37}{5}\)
Explanation:
by substituting  y value = 7 in the given algebraic equation
8(2y – 1) minus \(\frac{14 y+37}{5}\)
= 8(2 X 7 – 1) minus \(\frac{14 X 7 + 37}{5}\)
= 8(2y – 1) minus \(\frac{14 y+37}{5}\)

Question 25.
The quotient of 9(7y – 15) and \(\frac{110-6 y}{4}\)
Answer:
18
Explanation:
by substituting  y value = 7 in the given algebraic equation
9(7y – 15) and \(\frac{110-6 y}{4}\)
= 9(7 x 7 – 15) and \(\frac{110 – 6 x 7}{4}\)
= 9(49 – 15) and \(\frac{110-42}{4}\)
= 9(34) and \(\frac{68}{4}\)
= 306/17 = 18

Question 26.
The sum of \(\frac{5 y}{6}\) and 4(\(\frac{3 y}{7}\) + 2y)
Answer:
\(\frac{443}{6}\)
Explanation:
by substituting  y value = 7 in the given algebraic equation
\(\frac{5 y}{6}\) and 4(\(\frac{3 y}{7}\) + 2y)
= \(\frac{5 x 7}{6}\) and 4(\(\frac{3 x 7}{7}\) + 2 x 7)
= \(\frac{35}{6}\) and 4(\(\frac{21}{7}\) + 14)
= \(\frac{35}{6}\) and 4(3 + 14)
= \(\frac{35}{6}\) +68
= \(\frac{35 + 408 }{6}\)
= \(\frac{443}{6}\)

Question 27.
The quotient of (\(\frac{y}{2}+\frac{2 y}{3}\)) and (\(\frac{5 y}{6}-\frac{y}{3}\))
Answer:
\(\frac{7}{3}\)
Explanation:
by substituting  y value = 7 in the given algebraic equation
(\(\frac{y}{2}+\frac{2 y}{3}\)) and (\(\frac{5 y}{6}-\frac{y}{3}\))
= (\(\frac{7}{2}+\frac{2 x 7}{3}\)) and (\(\frac{5 x 7}{6}-\frac{7}{3}\))
= (\(\frac{7}{2}+\frac{14}{3}\)) and (\(\frac{35}{6}-\frac{7}{3}\))
= (\(\frac{21 + 28}{6}\)) and (\(\frac{35 – 14}{6}\))
= (\(\frac{49}{6}\)) and (\(\frac{21}{6}\))
= (\(\frac{49}{6}\)) X (\(\frac{6}{21}\))
= \(\frac{49}{21}\)
= \(\frac{7}{3}\)

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

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

Concepts and Skills

Express each percent as a fraction In simplest form.

Question 1.
46%
Answer:
\(\frac{23}{50}\)
Explanation:
Divide the given percent by 100 to get the decimal number $\frac{percent}{100}$.
\(\frac{46}{100}\)
simplify the fraction with the greatest common factor on both sides.
So, 2 is the greatest common factor.
\(\frac{23}{50}\)

Question 2.
16%
Answer:
\(\frac{1}{5}\)
Explanation:
Divide the given percent by 100 to get the decimal number $\frac{percent}{100}$.
\(\frac{16}{100}\)
Simplify the fraction with the greatest common factor on both sides.
So, 4 is the greatest common factor.
\(\frac{1}{5}\)

Question 3.
88%
Answer:
\(\frac{22}{25}\)
Explanation:
Divide the given percent by 100 to get the decimal number $\frac{percent}{100}$.
\(\frac{88}{100}\)
Simplify the fraction with the greatest common factor on both sides.
So, 4 is the greatest common factor.
\(\frac{22}{25}\)

Express each percent as a decimal.

Question 4.
34%
Answer:
0.34
Explanation:
In order to convert percent to decimal number,
percentage should be divided by 100.
\(\frac{34}{100}\) = 0.34

Question 5.
60%
Answer:
0.60
Explanation:
In order to convert percent to decimal number,
percentage should be divided by 100.
\(\frac{60}{100}\) = 0.60

Question 6.
9%
Answer:
0.09
Explanation:
In order to convert percent to decimal number,
percentage should be divided by 100.
\(\frac{9}{100}\) = 0.09

Express each fraction as a percent.

Question 7.
\(\frac{17}{20}\)
Answer:
85%
Explanation:
Divide the numerator by the denominator
Multiply the resulting decimal of the above division by 100
Express the value using the symbol “%”
\(\frac{17}{20}\) = 0.85
= 0.85 x 100 = 85%

Question 8.
\(\frac{21}{25}\)
Answer:
48%
Explanation:
Divide the numerator by the denominator
Multiply the resulting decimal of the above division by 100
Express the value using the symbol “%”
\(\frac{21}{25}\) = 0.48
= 0.48 x 100 = 48%

Question 9.
\(\frac{270}{300}\)
Answer:
90%
Explanation:
Divide the numerator by the denominator
Multiply the resulting decimal of the above division by 100
Express the value using the symbol “%”
\(\frac{270}{300}\) = 0.9
= 0.9 x 100 = 90%

Express each decimal as a percent.

Question 10.
0.02
Answer:
2%
Explanation:
Convert a decimal to a percentage is done by multiplying the decimal value by 100 and adding %.
0.02 x 100 = 2

Question 11.
0.63
Answer:
63%
Explanation:
Convert a decimal to a percentage is done by multiplying the decimal value by 100 and adding %.
0.63 x 100 = 63

Question 12.
0.9
Answer:
90%
Explanation:
Convert a decimal to a percentage is done by multiplying the decimal value by 100 and adding %.
0.9 x 100 = 90

Find the quantity represented by each percent.

Question 13.
35% of 500 kilograms
Answer:
175 kg
Explanation:
finding the fraction of a number is same as multiplying the fraction with the number, we have.
\(\frac{35}{100}\) x 500
= 35 x 5 = 175

Question 14.
68% of $2,800
Answer:
$1,904
Explanation:
finding the fraction of a number is same as multiplying the fraction with the number, we have.
\(\frac{68}{100}\) x 2,800
=68 x 28
= 1,904

Solve. Show your work.

Question 15.
There were 45 adults on a bus. 60% of them were women. How many women were on the bus?
Answer:
27 women
Explanation:
Number of adults in bus = 40
60% of them were women.
Total women in bus,
\(\frac{60}{100}\) x 45
= \(\frac{6}{10}\) x 45
= \(\frac{270}{10}\) = 27

Question 16.
Iris had $900. She spent 22% of this amount on a mobile phone. How much did she pay for the mobile phone?
Answer:
$198
Explanation:
Iris had $900.
She spent 22% of this amount on a mobile phone.
Amount she paid for the mobile phone,
\(\frac{22}{100}\) x 900
=22 x 9 = 198

Question 17.
22% of a number is 44. Find the number.
Answer:
200 is the number
Explanation:
Let the number be x
44 = x x \(\frac{22}{100}\)
44 = \(\frac{22x}{100}\)
22x = 44 x 100
x = \(\frac{4400}{22}\)
x = 200

Question 18.
125% of a number is 65. Find the number.
Answer:
52 is the number
Explanation:
Let the number be x
65 = x x \(\frac{125}{100}\)
65 = \(\frac{125x}{100}\)
125x = 65 x 100
x = \(\frac{6500}{125}\)
x = 52

Problem Solving

Solve. Show your work.

Question 19.
Harold had 1,400 stamps. He gave 350 of them to his brother and the rest to his sister.
a) What percent of the stamps did he give to his brother?
Answer:
25%
Explanation:
Harold had 1,400 stamps.
He gave 350 of them to his brother
\(\frac{350}{1,400}\) x 100
= \(\frac{350}{14}\)
= 25

b) What percent of the stamps did he give to his sister?
Answer:
75%
Explanation:
Harold had 1,400 stamps.
He gave 350 of them to his brother and rest to his sister
1,400 – 350 = 1,050
\(\frac{1,050}{1,400}\) x 100
= \(\frac{1,050}{14}\)
= 75

Question 20.
There were 15 girls and 25 boys in the Science club. What percent of the members were girls?
Answer:
37.5%
Explanation:
Total students in science club 15 + 25 = 40
Percent of the girls = \(\frac{15}{40}\) x 100
= \(\frac{1,500}{40}\)
= 37.5

Question 21.
Tyrone paid $2,800 plus 7% sales tax on mountain bikes for a bike club. How much did Tyrone pay in total for the mountain bikes?
Answer:
$2996
Explanation:
Tyrone paid $2,800 plus 7% sales tax on mountain bikes for a bike club.
Amount paid by Tyrone for the mountain bikes
2,800 x \(\frac{7}{100}\) = 196
=  2800 + 196
= 2996

Question 22.
Howard opens a savings account with a deposit of $800. The bank will pay him 3% interest per year.
a) How much interest will Howard receive at the end of \(\frac{1}{2}\) year?
Answer:
$120
savings account with a deposit of $800
3% interest per year
SI = \(\frac{P R T}{100}\)
=  \(\frac{800 x 3 x 0.5}{100}\) = 120
for 6 months $120

b) How much interest will he receive at the end of 1 year?
Answer:
$240
Explanation:
SI (simple intrest)= \(\frac{P R T}{100}\)
=  \(\frac{800 x 3 x 1}{100}\) = 240
for 1 year $240

Question 23.
A grocery shop sells walnuts at a 40% markup. The shop pays $4.50 per pound for the walnuts. At what price per pound will the shop sell the walnuts?
Answer:
$6.30
Explanation:
The shop pays $4.50 per pound for the walnuts.
A grocery shop sells walnuts at a 40% markup.
profit = 4.50 x \(\frac{40}{100}\)
p = \(\frac{180}{100}\) = 1.8
selling price = 4.50 + 1.80 = 6.30

Question 24.
Inez earns $160 per month by babysitting. She saves 25% of her monthly salary. If her salary increases by 10%, how much will she now save each month?
Answer:
$44
Explanation:
Initial salary = $160
she gives = \(\frac{25}{100}\) x 160 = 40
New monthly salary = 160 + \(\frac{10}{100}\) x 160 = 176
she saves = \(\frac{25}{100}\) x 176 = 44

Question 25.
Last year, Manuel saved $50 per month. This year, he increases his monthly savings by 25%. He plans to buy a smart phone for $375. At this rate, how long will it take him to save enough money to buy the smart phone if he starts saving for it this year?
Answer:
6 months
Explanation:
interest is increased by 25% or \(\frac{25}{100}\)
= \(\frac{1}{4}\)
50 time of \(\frac{1}{4}\) = \(\frac{50}{4}\) = 12.5
So, \(\frac{50}{12.5}\) = 62.5 per month
\(\frac{375}{62.5}\) = 6 months

Question 26.
The original price of a watch was $550. During a mid-year sale, the selling price of the watch was $440. Find the percent discount.
Answer:
20%
Explanation:
The discount amount is $550 – $440 = $110
Let x represent the unknown percent discount.
550x = 11,000
x = \(\frac{11,000}{550}\)
x = 20%

Question 27.
In January, the price of a kilogram of Brand X rice was $7.60. In May, the price of a kilogram of Brand X rice became $8.80. Find the percent increase in price. Round your answer to 2 decimal places.
Answer:
$15.79
Explanation:
In January, the price of rice was $7.60.
In May, the price of rice became $8.80.
The price increased to $1.80
The percent increase in price = 120%
Original price = \(\frac{120}{7.60}\) = 15.789
Round your answer to 2 decimal places = 15.79

Question 28.
A tank contained 35 gallons of water at first. Due to a leak at the bottom of the tank, the amount of water in the tank decreased to 32 gallons after 1 hour. After another hour, the amount of water in the tank was 26 gallons.
a) Find the percent decrease in the amount of water from 35 gallons to 32 gallons. Round your answer to 1 decimal place.
Answer:
1%
Explanation:
A tank contained 35 gallons of water at first.
Due to a leak at the bottom of the tank, the amount of water in the tank decreased to 32 gallons after 1 hour, 35 – 32 = 3 gallons
\(\frac{3}{35}\)x100 = 1.05% = 1%

b) Find the percent decrease in the amount of water from 32 gallons to 26 gallons. Round your answer to 1 decimal place.
Answer:
19%
Explanation:
At first the water in tank is 32 gallons.
After another hour, the amount of water in the tank was 26 gallons
32 – 26 = 6 gallons
\(\frac{6}{32}\)x100 = 18.75% = 19%

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 6 Lesson 6.4 Real-World Problems: Percent to score better marks in the exam.

Math in Focus Grade 6 Course 1 A Chapter 6 Lesson 6.4 Answer Key Real-World Problems: Percent

Math in Focus Grade 6 Chapter 6 Lesson 6.4 Guided Practice Answer Key

Solve.

Question 1.
Mr. Jefferson is making 80 cups of fruit punch for the grand opening of his bakery. He uses 52 cups of fruit juice and the rest is sparkling water.
a) What percent of the punch is fruit juice?
Method 1
Fraction of fruit juice in punch = \(\frac{?}{?}\)
Percent of fruit juice in punch = \(\frac{?}{?}\) × %
= %
= % of the punch is fruit juice.

Method 2
cups → 100%
1 cup → \(\frac{100}{?}\)%
cups → ? × \(\frac{100}{?}\) = %
% of the punch is fruit juice.
Answer:
65%
Explanation:
Method 1
Fraction of fruit juice in punch = \(\frac{52}{80}\)
Percent of fruit juice in punch = \(\frac{52}{80}\) × 100%
= 65%
= 65% of the punch is fruit juice.

b) What percent of the punch is sparkling water?
100% – % = %
% of the punch is sparkling water.
Answer:
35%
Explanation:
100% – 65% = 35%
35% of the punch is sparkling water.

Solve. Check for reasonableness.

Question 2.
A laptop computer displayed at a shop costs $720. A sales tax of 7% will be added to the price. What is the total cost of the laptop computer?
Method 1

The total cost of the laptop computer is $.

Method 2
Sales tax:

The total cost of the laptop computer is $.
Answer:
$770.40
Explanation:

Question 3.
Dave went to lunch with his friends. The food cost $78.50, and a sales tax of $4.71 was added to the cost of the meal. What was the sales tax rate?
The cost of the food is %.
$ → %
$1 → \(\frac{?}{?}\) %
$ → × \(\frac{?}{?}\) % = %
The sales tax rate was %.
Answer:
6%
Explanation:
The cost of the food is $78.50 %.
$4.71 → 1%
$1 → \(\frac{4.71}{78.50}\) %
$100 → 100× \(\frac{471}{7850}\) % = 0.06%
The sales tax rate was 6%.

Question 4.
Mr. Diaz earns a 3% commission on all the furniture he sells. If he receives $2,880 in commission, what is the dollar amount of his sales?

% → $
1% → $\(\frac{?}{?}\) % = $ %
100% → 100 × $ = $
The dollar amount of his sales is $.
Answer:
$96,000
Explanation:
3% → $2,880
1% → $\(\frac{2880}{3}\) % = $960 %
100% → 100 × $ 960= $96,000
The dollar amount of his sales is $96,000.

Solve.

Question 5.
A firm has $30,000 in a bank account at the beginning of the year. Interest will be paid at a rate of 2% at the end of the year. How much interest will the firm receive for the year?
Interest = imgg 1 % of $ for 1 year
= \(\frac{?}{?}\) × $ × 1
= $
The firm will receive $ in interest for the year.
Answer:
$600
Explanation:
Interest = imgg 1 % of $30,000 for 1 year
= \(\frac{2}{100}\) × $30,000 × 1
= 2 X 300
= $600

Question 6.
A company has $500,000 in a savings account. The interest is 4% per year. How much interest will it earn at the end of \(\frac{1}{2}\) year?
Interest = \(\frac{?}{100}\) × $ × \(\frac{?}{?}\)
= $
The company will earn $ in interest at the end of \(\frac{1}{2}\) year.
Answer:
$10,000
Explanation:
Interest = \(\frac{4}{100}\) × $500,000 × \(\frac{1}{2}\)
= 20,000 X \(\frac{1}{2}\)
= $10,000
The company will earn $10,000 in interest at the end of \(\frac{1}{2}\) year.

Math in Focus Course 1A Practice 6.4 Answer Key

Solve. Show your work. Check that your answers are reasonable.

Question 1.
Ellen had 25 hair clips. 7 of them were blue and the rest were purple.
a) What percent of the hair clips were blue?
Answer:
28%
b) What percent of the hair clips were purple?
Answer:
72%
Explanation:
Ellen had 25 hair clips,
7 of them were blue and the rest were purple.
Number of blue clips = 7
Number of purple clips = 125 – 7 = 18
percentage of blue clips = \(\frac{7}{25}\) × 100 = 28%
percentage of purple clips = \(\frac{18}{25}\) × 100 = 72%

Question 2.
Gabriel had $60. He spent $36 on a pair of shoes and the rest on a shirt.

a) What percent of the money did he spend on the pair of shoes?
Answer:
60%
b) What percent of the money did he spend on the shirt?
Answer:
40%
Explanation:
Gabriel had $60.
He spent $36 on a pair of shoes and the rest on a shirt.
percent of the money did he spend on the pair of shoes
= \(\frac{36}{60}\) × 100
= 60%
Total cost spent on shirt = 60 – 36 = $24
percent of the money did he spend on the shirt
= \(\frac{24}{60}\) × 100
= 40%

Question 3.
Ms. Pierce bought a camera that cost $450. In addition, she had to pay 4% u sales tax. How much did Ms. Pierce pay for the camera?
Answer:
$468
Explanation:
Ms. Pierce bought a camera that cost $450.
In addition, she had to pay 4% u sales tax.
Total price of the camera = \(\frac{4}{100}\) × 450 = $18
Total amount he paid = camera cost + tax
= $450 + $18 = $468

Question 4.
Mr. Osmond bought a computer that cost $2,500. How much did he pay for the computer if the sales tax rate was 7%?
Answer:
$2,675
Explanation:
Mr. Osmond bought a computer that cost $2,500.
Tax paid for the computer was 7%.
Total amount he paid = \(\frac{7}{100}\) × 2,500 = $175
Total amount he paid including tax = cost of computer + tax
= $3,500 + $175 = $2,675

Question 5.
There were 320 girls and 180 boys at a playground. What percent of children were girls?
Answer:
64%
Explanation:
There were 320 girls and 180 boys at a playground.
Total boys and girls = 320 + 180 = 500
percent of children were girls
= \(\frac{320}{500}\) × 100 = 64%

Question 6.
Sally spent $36 and had $12 left. What percent of her money did she spend?
Answer:
75%
Explanation:
Sally spent $36 and had $12 left.
Total amount = spent + left
= $36 + $12 = $48
percent of her money did she spend
= \(\frac{36}{48}\) × 100 = 75%

Question 7.
An artist receives 20% royalty on the retail price of his recordings. If he receives $36,000 in royalties, what is the dollar amount of the retail price of his recordings?
Answer:
$18,000,000
Explanation:
An artist receives 20% royalty on the retail price of his recordings.
If he receives $36,000 in royalties,
Total dollar amount of the retail price of his recordings
\(\frac{20}{100}\) of X = 36,000
100% of X = ?
X = 36000 x \(\frac{100}{20}\) x 100
X = 36,000 x 5 x 100
X = 18,000,000

Question 8.
A club deposited $50,000 in a savings account at the beginning of the year. Interest will be paid at a rate of 3% at the end of the year. How much interest will the club receive for the year?
Answer:
$1,500
Explanation:
A club deposited $50,000 in a savings account at the beginning of the year.
Interest will be paid at a rate of 3% at the end of the year.
Total interest will the club receive for the year
= \(\frac{3}{100}\) × 50,000 = 1,500

Question 9.
Company X has $128,000 in a savings account that pays 6% interest per year. How much interest will it earn at the end of \(\frac{1}{2}\) year?
Answer:
$3,840
Explanation:
Company X has $128,000 in a savings account that pays 6% interest per year.
= \(\frac{6}{100}\) × 128,000 = 7,680
Total interest will it earn at the end of \(\frac{1}{2}\) year
= \(\frac{1}{2}\) × 7,680 = 3,840

 

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 6 Lesson 6.3 Percent of a Quantity to score better marks in the exam.

Math in Focus Grade 6 Course 1 A Chapter 6 Lesson 6.3 Answer Key Percent of a Quantity

Math in Focus Grade 6 Chapter 6 Lesson 6.3 Guided Practice Answer Key

Complete. Use the models to help you.

Question 1.
What is 40% of 720 centimeters?
Method 1

The model shows that:
100% → cm
1% → \(\frac{?}{?}\) = cm
40% → × = cm
40% of 720 centimeters is centimeters.

Method 2
40% of 720 cm = \(\frac{?}{100}\) × 720
= cm
40% of 720 centimeters is centimeters.
Answer:
288centimeters
Explanation:
Method 1
The model shows that:
100% → 720cm
1% → \(\frac{720}{100}\) = 7.2cm
40% → 7.2 × 40 = 288cm
40% of 720 centimeters is 288 centimeters.

Method 2
40% of 720 cm = \(\frac{40}{100}\) × 720
= 288 cm
40% of 720 centimeters is 288centimeters.

Question 2.
What is 75% of 800 kilograms?
Method 1

The model shows that:
100% → kg
1% → \(\frac{?}{?}\) = kg
75% → × = kg
75% of 800 kg is kg.

Method 2
75% of 800 kg = \(\frac{?}{?}\) × 800
= kg
75% of 800 kg is kg.
Answer:
600kg
Explanation:
The model shows that:
100% → 800kg
1% → \(\frac{800}{100}\) = 8kg
75% → 75 × 8 = 600kg
75% of 800 kg is 600kg.

Method 2
75% of 800 kg = \(\frac{75}{100}\) × 800
= 600kg
75% of 800 kg is 600kg.

Find the percent of each whole.

Question 3.
30% of 450
Answer: 135
Explanation:
30% of 450
= \(\frac{30}{100}\) × 450
= 135
30% of 450 is 135.

Question 4.
225% of $60
Answer: 135
Explanation:
225% of $60
= \(\frac{225}{100}\) × 60
= 135
225% of 60 is 135.

Question 5.
55% of 320
Answer: 176
Explanation:
55% of 320
= \(\frac{55}{100}\) × 320
= 176
55% of 320 is 176.

Question 6.
110% of $550
Answer: 605
Explanation:
110% of $550
= \(\frac{110}{100}\) × 550
= 605
110% of 550 is 605.

Solve. Use the model to help you.

Question 7.
27% of the students in a school are in grade 6. This is 540 students. How many students are there in the school?

The model shows that:
% → students
1% → \(\frac{?}{?}\) = students
100% → 100 × = students
There are students in the school.
Answer:
2000 students.
Explanation:
The model shows that:
27% → 540 students
1% → \(\frac{540}{27}\) = 20 students
100% → 100 × 20 = 2000 students
There are 2000 students in the school.

Solve. You may draw a model to help you.

Question 8.
At an amusement park, 60% of the people were adults, and the rest were children. There were 720 adults. How many people were at the amusement park in all?
 % → people
1% → \(\frac{?}{?}\) = people
100% → × = people
There were people at the amusement park in all.
Answer:
1200 people
Explanation:
60% → 720people
1% → \(\frac{720}{60}\) = 12 people
100% → 100× 12 = 1200 people
There were 1200 people at the amusement park in all.

Find the missing value.

Question 9.
20% of is 163.
Answer: 185
Explanation:
20% of is 163.
\(\frac{20}{100}\) x = 163
x = (163 x 100)  ÷ 20
= 16300  ÷ 20
= 815

Question 10.
45% of is 150.
Answer: 333
Explanation:
45% of is 150.
\(\frac{45}{100}\) x = 150
x = (150 x 100)  ÷ 45
= 15000 ÷ 45
= 333

Math in Focus Course 1A Practice 6.3 Answer Key

Solve. Use the models to help you.

Question 1.
What is 15% of 12 meters?

Answer: 1.8 m
Explanation:

Question 2.
A park ranger finds that 35% of the park’s visitors stay at the campground. If 105 visitors stayed at the campground one day, how many visitors did the park have that day?

Answer: 36.75
Explanation:
A park ranger finds that 35% of the park’s visitors stay at the campground.
105 visitors stayed at the campground one day.
number of visitors did the park have that day

Find the quantity represented by each percent.

Question 3.
45% of $360
Answer:
$162
Explanation:
45% of $360
= \(\frac{45}{100}\) × 360
= 162
45% of 360 is 162.

Question 4.
66% of 740 kilometers
Answer:
488.4kg
Explanation:
66% of $740
= \(\frac{66}{100}\) × 740
= 488.4
66% of 740 is 488.4.

Solve. Show your work.

Question 5.
There are 1,500 students in a school. 65% of them are girls. How many girls are there in the school?
Answer:
975 girls.
Explanation:
There are 1,500 students in a school.
65% of them are girls.
Total girls in the school 65% of 1500
= \(\frac{65}{100}\) × 1500
= 975

Question 6.
A school raises $4,000 for its new library. 36% of the money is used to buy reference books. How much money is used to buy reference books?
Answer:
$1,440
Explanation:
A school raises $4,000 for its new library.
36% of the money is used to buy reference books.
Total money is used to buy reference books
36% of $4,000
= \(\frac{36}{100}\) × 4000
= 1440

Question 7.
Ms. Galan spent 55% of her savings on a television that cost $550. How much money did she have in her savings before she bought the television?
Answer:
$1,000
Explanation:
Ms. Galan spent 55% of her savings on a television that cost $550.
Money she have in her savings before she bought the television
550 = 55% of her savings
550 = \(\frac{55}{100}\) x
x = (550 x 100) ÷ 55
x = 1000

Question 8.
75% of a number is 354. Find the number.
Answer:
472 is the number
Explanation:
75% of x = 354
x = (354 x 100)  ÷ 75
x = 472

Question 9.
Ahyoka has 250 CDs. 10% are country, 70% are pop, and the rest are hip-hop. How many CDs are hip-hop?
Answer:
50 CDs
Explanation:
Ahyoka has 250 CDs.
10% are country, 70% are pop = 70 + 10 = 80%
the rest are hip-hop = 100 – 80 = 20%
Total hip-hop CDs = 20% of 250
= \(\frac{20}{100}\) × 250
= 50

Question 10.
There are 820 people at a stadium. 65% of them are adults, 20% of them are boys, and the rest are girls. How many girls are there at the stadium?
Answer:
123 girls
Explanation:
There are 820 people at a stadium.
65% of them are adults, 20% of them are boys = 65 + 20 = 55%
Rest of the girls = 100 – 85 = 15%
Number of girls at the stadium = 15% of 820
= \(\frac{15}{100}\) × 820
= 123

Question 11.
Ms. Stapleton had 2,600 hens, ducks, and goats on her farm. 35% of them were hens, and 25% of them were goats. How many ducks did she have on her farm?
Answer:
Ducks 790
Explanation:
total 2,600 hens, ducks, and goats in farm
Hens = 2,600 x 35%
= (2,600 x 35)/100
= 26 x 35
= 910

Goats = 25% of 2,600
= (2600 x 25)/100
= 26 x 25
= 900

Ducks = 2600 – (Hens + Goats)
Hens + Goats = 910 + 900 = 1,810
Ducks = 2,600 – 1,810
= 790

Question 12.
There are 1,505 fruits for sale at a farmer’s market. 39% of them are apples, 28% are oranges, 13% are honeydew melons, and the rest are watermelons. How many watermelons are there?
Answer:
301 watermelons
Explanation:
There are 1,505 fruits for sale at a farmer’s market.
39% of them are apples, 28% are oranges, 13% are honeydew melons,
Total apples, oranges and honeydew melons = 39 + 28 + 13 = 80
Number of watermelons = 100 – 80 = 20%
Number of watermelons 20% of 1505
= \(\frac{20}{100}\) × 1505
= 301

Question 13.
120% of a number is 45. Find the number.
Answer:
37.5
Explanation:
120% of x = 45
x = (45 x 100)  ÷ 120
x = 37.5

Question 14.
250% of a number is 60. Find the number.
Answer:
24 is the number
Explanation:
250% of x = 60
x = (60 x 100)  ÷ 250
x = 24

Question 15.
400% of a number is 45. Find the number.
Answer:
11.25 is the number
Explanation:
400% of x = 45
x = (45 x 100) ÷ 400
x = 11.25

Question 16.
Last month, Alex spent 40% of his salary on a laptop. He then spent 30% of it on his bills and saved the remaining $1,200. What was his salary?
Answer:
$4,000
Explanation:
He spent 70% of his salary and had $1200 left.
Let ‘x’ be his salary.
100% – 70% = 30% left.
30% of x = 1200
= \(\frac{30}{100}\) x = 1200
0.3x = 1200
x = 1200 ÷ 0.3
x = 4000

Question 17.
20% of the spectators at a tennis match are women. 10% of them are girls, 30% are boys, and the remaining 3,600 spectators are men. Find the total number of spectators at the match.
Answer:
9,000 spectators
Explanation:
20% of the spectators at a tennis match are women.
10% of them are girls, 30% are boys,
Total women, boys and girls = 20 + 10 + 30 = 60%
Men spectators = 3600
The total number of spectators at the match
40 % of x = 3600
\(\frac{40}{100}\) x = 3600
0.4x = 3600
x = 3600 ÷ 0.4
x = 9,000

Question 18.
Jovita made 300 greeting cards. She sold 40% of the cards, gave 85% of the remaining cards to her friends, and kept the rest of the cards for herself. How many greeting cards did she keep for herself?

Answer:
27 greeting cards
Explanation:
Jovita made 300 greeting cards.
She sold 40% of the cards = 300
\(\frac{40}{100}\) x 300 = 120
she sold 120 out of 300 = 300 – 120 = 180
she gave 85% of the remaining cards to her friends,
\(\frac{85}{100}\) x 180
Remaining cards = 180 – 153
Total greeting cards she kept for herself = 27

Question 19.
Patrick saved $500. He received 25% of the money for his birthday, saved 30% of the remainder from his allowance, and earned the rest of it by mowing lawns. How much of his savings did he earn by mowing lawns?
Answer:
$262.50
Explanation:
Patrick saved $500.
He received 25% of the money for his birthday,
saved 30% of the remainder from his allowance = 25 + 30 = 75%
75% of 500
= \(\frac{75}{100}\) x 500
= 75 x 5 = 375
Total savings he earn by mowing lawns
100 – 30 = 70%
70% of x = 375
0.7x = 375
x = 375 x 0.7
x = 262.50

Question 20.
Math Journal Michelle and Michael checked some books out of the library. 20% of the books Michelle checked out were fiction books, and 40% of the books Michael checked out were fiction books. Your friend thinks that Michael checked out more fiction books than Michelle. Explain the error in your friend’s thinking. Use an example to support your reasoning.
Answer:
The error in the assessment is that Michael checked more fiction books than Michelle.
Explanation:
We can determine who has more books only when we know how many books each person has checked out in total,
but not the proportion of fiction of books, (20% and 40%).
So, we can assume that Michael checked more fiction books than Michelle.

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 6 Lesson 6.2 Fractions, Decimals, and Percents to score better marks in the exam.

Math in Focus Grade 6 Course 1 A Chapter 6 Lesson 6.2 Answer Key Fractions, Decimals, and Percents

Hands-On Activity

EXPRESSING FRACTIONS OR MIXED NUMBERS AS PERCENTS

Work in pairs.

Use these fractions and mixed numbers.

STEP 1: Find which of the fractions or mixed numbers can be expressed as percents using the method of writing an equivalent fraction. List the fractions that cannot be expressed as percents using this method.
Example

Answer:

STEP 2: If you want to use the method of writing an equivalent fraction to express a fraction as a percent, what must be true of the fraction or mixed number?
Explanation:
The denominator must me 100 in the fraction,
then the fraction in the numerator is expressed as percentage.
fractions with denominator with 9, 8 , 6, 3 can not be expressed as 100 in the denominator.

Math Journal Explain how you would express 1\(\frac{3}{4}\) as a percent.
Answer:
175%
Explanation:
Express 1\(\frac{3}{4}\) as a percent.
\(\frac{7}{4}\)
To express the fraction with denominator 100,
multiply both the numerator and denominator with 25
= \(\frac{175}{100}\)
= 175%

Math in Focus Grade 6 Chapter 6 Lesson 6.2 Guided Practice Answer Key

Express each fraction or mixed number as a percent.

Question 1.
\(\frac{4}{7}\) = \(\frac{4}{7}\) × %
= \(\frac{?}{?}\) %
= %
Answer:
57%
Explanation:
\(\frac{4}{7}\) = \(\frac{4}{7}\) × 100%
= \(\frac{400}{7}\) %
= 57%

Question 2.
1\(\frac{5}{9}\) = 1\(\frac{5}{9}\) × %
= \(\frac{?}{9}\) × %
= %
Answer:
155%
Explanation:
1\(\frac{5}{9}\) =1 \(\frac{5}{9}\) × 100%
= \(\frac{14}{9}\) x 100%
= 155%

Question 3.
\(\frac{5}{6}\)
Answer:
83%
Explanation:
\(\frac{5}{6}\) = \(\frac{5}{6}\) × 100%
= \(\frac{500}{6}\) %
= 83%

Question 4.
1\(\frac{7}{8}\)
Answer:
187.5%
Explanation:
1\(\frac{7}{8}\) = 1\(\frac{7}{8}\) × 100%
= \(\frac{15}{8}\) x 100 %
= 187.5%

Complete.

Question 5.
Express 0.82 as a percent.
Method 1
0.82 = \(\frac{?}{100}\)
=

Method 2
0.82 = \(\frac{?}{9}\) × 100 %
=
Answer:
82%
Explanation:
Express 0.82 as a percent.
Method 1
Converting from a decimal to a percentage is done by multiplying the decimal value by 100 and adding %.
0.82 = \(\frac{82}{100}\)
= 0.82

Express each decimal as a percent.

Question 6.
0.04
Answer:
4%
Explanation:
Express 0.04 as a percent.
Method 1
Converting from a decimal to a percentage is done by multiplying the decimal value by 100 and adding %.
0.04 = \(\frac{4}{100}\)
= 0.04

Question 7.
0.98
Answer:
98%
Explanation:
Express 0.98 as a percent.
Method 1
Converting from a decimal to a percentage is done by multiplying the decimal value by 100 and adding %.
0.98 = \(\frac{98}{100}\)
= 0.98

Question 8.
0.6
Answer:
60%
Explanation:
Express 0.6 as a percent.
Method 1
Converting from a decimal to a percentage is done by multiplying the decimal value by 100 and adding %.
0.6 = \(\frac{60}{100}\)
= 0.6

Express each percent as a fraction in simplest form.

Question 9.
83\(\frac{1}{3}\) %
Answer:
8333.33
Explanation:
83 \(\frac{1}{3}\)%
= \(\frac{250}{3}\) x 100
= 8333.33

Question 10.
84.4%
Answer:
\(\frac{211}{250}\)
Explanation:
84.4% = \(\frac{84.4}{100}\)
multiply the numerator and denominator with 10,
as numerator has decimal
= \(\frac{844}{1000}\)
Divided numerator and denominator with 2 to simplify
= \(\frac{211}{250}\)

Math in Focus Course 1A Practice 6.2 Answer Key

Express each fraction or mixed number as a percent.

Question 1.
\(\frac{3}{5}\)
Answer:
60%
Explanation:
Simplify the given fraction
\(\frac{3}{5}\) = 0.6
Converting from a decimal to a percentage is done by multiplying the decimal value by 100 and adding %.
0.6 x 100 = 60%

Question 2.
\(\frac{3}{8}\)
Answer:
37.5%
Explanation:
Simplify the given fraction
\(\frac{3}{8}\) = 0.375
Converting from a decimal to a percentage is done by multiplying the decimal value by 100 and adding %.
0.375 x 100 = 37.5%

Question 3.
\(\frac{1}{3}\)
Answer:
33%
Explanation:
Simplify the given fraction
\(\frac{1}{3}\) = 0.33
Converting from a decimal to a percentage is done by multiplying the decimal value by 100 and adding %.
0.33 x 100 = 33%

Question 4.
2\(\frac{1}{5}\)
Answer:
220%
Explanation:
Simplify the given fraction
2\(\frac{1}{5}\)
= \(\frac{11}{5}\) = 2.2
Converting from a decimal to a percentage is done by multiplying the decimal value by 100 and adding %.
2.2 x 100 = 220%

Question 5.
7\(\frac{3}{4}\)
Answer:
442%
Explanation:
Simplify the given fraction
7\(\frac{3}{4}\)
\(\frac{31}{7}\) = 4.42
Converting from a decimal to a percentage is done by multiplying the decimal value by 100 and adding %.
4.42 x 100 = 442%

Question 6.
9\(\frac{7}{8}\)
Answer:
987%
Explanation:
Simplify the given fraction
9\(\frac{7}{8}\)
\(\frac{79}{8}\) = 9.87
Converting from a decimal to a percentage is done by multiplying the decimal value by 100 and adding %.
9.87 x 100 = 987%

Express each decimal as a percent.

Question 7.
0.46
Answer:
46%
Explanation:
Converting from a decimal to a percentage is done by multiplying the decimal value by 100 and adding %.
0.46 x 100 = 46%

Question 8.
0.7
Answer:
70%
Explanation:
Converting from a decimal to a percentage is done by multiplying the decimal value by 100 and adding %.
0.7 x 100 = 70%

Question 9.
0.06
Answer:
6%
Explanation:
Converting from a decimal to a percentage is done by multiplying the decimal value by 100 and adding %.
0.06 x 100 = 6%

Question 10.
1.52
Answer:
152%
Explanation:
Converting from a decimal to a percentage is done by multiplying the decimal value by 100 and adding %.
1.52 x 100 = 152%

Question 11.
6.03
Answer:
603%
Explanation:
Converting from a decimal to a percentage is done by multiplying the decimal value by 100 and adding %.
6.03 x 100 = 603%

Question 12.
8.9
Answer:
890%
Explanation:
Converting from a decimal to a percentage is done by multiplying the decimal value by 100 and adding %.
8.9 x 100 = 890%

Express each percent as a fraction in simplest form.

Question 13.
5.75%
Answer:
\(\frac{23}{400}\)
Explanation:
5.75% can be written as \(\frac{5.75}{100}\)
To convert percent as a fraction in simplest form,
multiply with 100 on both numerator and denominator.
\(\frac{5.75}{100}\) X \(\frac{100}{100}\)
= \(\frac{575}{10000}\)
Simplify the fraction divided with the greatest common factor 5 on both sides.
= \(\frac{23}{400}\)

Question 14.
25.5%
Answer:
\(\frac{51}{200}\)
Explanation:
25.5% can be written as \(\frac{25.5}{100}\)
To convert percent as a fraction in simplest form,
multiply with 10 on both numerator and denominator.
\(\frac{25.5}{100}\) X \(\frac{100}{100}\)
= \(\frac{255}{1000}\)
Simplify the fraction divided with the greatest common factor 5 on both sides.
= \(\frac{51}{200}\)

Question 15.
85.25%
Answer:
\(\frac{341}{400}\)
Explanation:
85.25% can be written as \(\frac{85.25}{100}\)
To convert percent as a fraction in simplest form,
multiply with 100 on both numerator and denominator.
\(\frac{85.25}{100}\) X \(\frac{100}{100}\)
= \(\frac{8525}{10000}\)
Simplify the fraction divided with the greatest common factor 5 on both sides.
= \(\frac{341}{400}\)

Question 16.
16\(\frac{2}{3}\) %
Answer:
\(\frac{1}{6}\)
Explanation:
16\(\frac{2}{3}\) %
= \(\frac{50}{3}\) %
= \(\frac{50}{3}\) X \(\frac{100}{100}\)
= \(\frac{1}{6}\)

Question 17.
42\(\frac{3}{8}\)%
Answer:
\(\frac{339}{800}\)
Explanation:
42\(\frac{3}{8}\) %
= \(\frac{339}{8}\) %
= \(\frac{339}{8}\) X \(\frac{100}{100}\)
= \(\frac{339}{800}\)

Question 18.
79\(\frac{5}{6}\) %
Answer:
\(\frac{479}{600}\)
Explanation:
79\(\frac{5}{6}\) %
= \(\frac{479}{6}\) %
= \(\frac{479}{6}\) X \(\frac{100}{100}\)
= \(\frac{479}{600}\)

Express each fraction as a percent. Round your answer to the nearest whole number.

Question 19.
\(\frac{76}{125}\)
Answer:
61%
Explanation:
\(\frac{76}{125}\) = 0.608
Round 0.608 to the nearest whole number = 61
So, percentage of fraction = 61%

Question 20.
\(\frac{98}{230}\)
Answer:
43%
Explanation:
\(\frac{98}{230}\) = 0.426
Round 0.426 to the nearest whole number = 43
So, percentage of fraction = 43%

Question 21.
\(\frac{102}{350}\)
Answer:
29%
Explanation:
\(\frac{102}{350}\) = 0.291
Round 0.291 to the nearest whole number = 29
So, percentage of fraction = 29%

Find the missing fractions and decimals.

Question 22.

Answer:

Explanation:
The below fractions are missing fractions.
a)12\(\frac{1}{2}\) %
= \(\frac{25}{2}\) %
= \(\frac{25}{2}\) X 100
= \(\frac{25}{200}\)
= \(\frac{1}{8}\)

b) 25%
= \(\frac{25}{100}\) %
= \(\frac{25}{100}\)
= \(\frac{1}{4}\)

c) 37\(\frac{1}{2}\) %
= \(\frac{75}{2}\) %
= \(\frac{75}{2}\) X \(\frac{100}{100}\)
= \(\frac{75}{200}\)
= \(\frac{3}{8}\)

d) 50%
= \(\frac{50}{100}\)
= \(\frac{1}{2}\)

e) 62\(\frac{1}{2}\) %
= \(\frac{125}{2}\) %
= \(\frac{125}{2}\) X \(\frac{100}{1}\)
= \(\frac{125}{200}\)
= \(\frac{5}{8}\)

f) 75%
= \(\frac{75}{100}\)
= \(\frac{3}{4}\)

g) 87\(\frac{1}{2}\) %
= \(\frac{175}{2}\) %
= \(\frac{175}{2}\) X \(\frac{100}{1}\)
= \(\frac{175}{200}\)
= \(\frac{7}{8}\)

Below are the missing decimals.
a) 25%
\(\frac{25}{100}\) = 0.25
b) \(\frac{3}{8}\) = 0.375
c) \(\frac{5}{8}\) = 0.625
d) \(\frac{3}{4}\) = 0.75
e) \(\frac{7}{8}\) = 0.87

Solve. Show your work.

Question 23.
Math Journal School A has 450 seniors, and 432 of them plan to go to college. School B has 380 seniors, and 361 of them plan to go to college. In which school does a greater percent of students plan to go to college? Justify your reasoning.
Answer:
In School A
Explanation:
School A has 450 seniors, and 432 of them plan to go to college.
A = \(\frac{432}{450}\)
percentage of the students = \(\frac{432}{450}\) X 100 = 96%
School B has 380 seniors, and 361 of them plan to go to college.
B = \(\frac{361}{380}\)
percentage of the students = \(\frac{361}{380}\) = 95%
So, greater percent of students plan to go to college are School A.

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Cumulative Review Chapters 3-5 to score better marks in the exam.

Math in Focus Grade 7 Course 2 A Cumulative Review Chapters 3-5 Answer Key

Concepts and Skills

Simplify each expression. (Lessons 3.1, 3.2, 3.3)

Question 1.
3.9p + 0.9p – 1.8p
Answer:
3.9p + 0.9p – 1.8p
= 4.8p – 1.8p
= 3.0p

Question 2.
\(\frac{8}{3}\)x – \(\frac{3}{5}\)y + \(\frac{2}{5}\)x + \(\frac{7}{4}\)y
Answer:
\(\frac{8}{3}\)x – \(\frac{3}{5}\)y + \(\frac{2}{5}\)x + \(\frac{7}{4}\)y

= \(\frac{8}{3}\)x + \(\frac{2}{5}\)x + \(\frac{7}{4}\)y – \(\frac{3}{5}\)y

= \(\frac{(8 x 5) + (2 x 3)}{(3 x 5)}\)x + \(\frac{(7 x 5) – (3 x 4)}{(4 x 5)}\)y

= \(\frac{(40) + (6)}{15}\)x + \(\frac{(35) – (12)}{20}\)y

= \(\frac{46}{15}\)x + \(\frac{23}{20}\)y

Expand and simplify each expression. (Lesson 3.4)

Question 3.
-0.2(0.8q – 4)
Answer:
-0.2(0.8q – 4)
= (-0.2) x (0.8q) + (-0.2) x (-4)
= – 0.16q + 0.8

Question 4.
–\(\frac{2}{3}\)(-\(\frac{3}{4}\)y – \(\frac{1}{2}\))
Answer:
–\(\frac{2}{3}\)(-\(\frac{3}{4}\)y – \(\frac{1}{2}\))

= \(\frac{-2}{3}\)(\(\frac{-3}{4}\)y + \(\frac{-1}{2}\))

=(\(\frac{-2 X -3}{3 X 4}\)y) + (\(\frac{-2 X -1}{3 X 2}\))

=\(\frac{2}{4}\)y + \(\frac{1}{3}\)

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

Question 5.
5(\(\frac{1}{15}\)x – 6y) – \(\frac{1}{3}\)x
Answer:
5(\(\frac{1}{15}\)x – 6y) – \(\frac{1}{3}\)x

=\(\frac{5 X 1}{15}\)x – (5 X 6)y – \(\frac{1}{3}\)x

= \(\frac{5}{15}\)x – 30y – \(\frac{1}{3}\)x

= \(\frac{5}{5 X 3}\)x – 30y – \(\frac{1}{3}\)x

= \(\frac{1}{3}\)x – 30y – \(\frac{1}{3}\)x

= \(\frac{1}{3}\)x – \(\frac{1}{3}\)x – 30y

= -30y

Question 6.
2(m – n) – 6(n – 2m)
Answer:
2(m – n) – 6(n – 2m)
=2m – 2n -6n + 12m
=2m + 12m -2n-6n
=14m-8n

Write an algebraic expression for the shaded area shown in questions 7 and 8. Then expand and simplify the expression. (Lesson 3.4)

Question 7.

Answer:
Length = x in
Base = (y + x) in
Height = 5 m
Area of Trapezium = \(\frac{5}{2}\)(x + y + x) m²
\(\frac{5}{2}\)(x + y + x) m²
= \(\frac{5}{2}\)(2x + y) m² Add like terms
= \(\frac{5}{2}\)(2x) + \(\frac{5}{2}\)(y) m² Use Distributive Property
= (5x + \(\frac{5}{2}\)y)m² Multiply

Question 8.

Answer:
Big Rectangle :
Length = 12 + x + x = (12 + 2x) m
width = 10 in
Area of Big Rectangle = Length × Width = (12 + 2x) ∙ (10)
= 12(10) + 2x(10) = (120 + 20x) m²
Small Rectangle :
Length = 12 m
Width = 2x m
Area of Small Rectangle = Length × Width = 12 ∙ 2x = 24x m²
Area of shaded area = (120 + 20x) – 24x m²
= (120 + 20x) – 24x
= 120 + 20x – 24x
= (120 – 4x) m²
Area = (120 + 20x — 24x) m² = (120 – 4x) m²

Factor each expression. (Lesson 3.5)

Question 9.
-4k – 36
Answer:
-4k – 36
= (-4)k + (-4)(9)
= -4(k + 9)

Question 10.
9 + 15m – 21n
Answer:
9 + 15m – 21n
=(3)(3) + (3)(5)m – (3)(7)n
= 3(3 + 5m – 7n)

Translate each verbal description into an algebraic expression. Simplify the expression when you can. (Lesson 3.6)

Question 11.
50% of one-twentieth of the product of 12 and 5z + 5.
Answer:
50% of one-twentieth of the product of 12 and 5z + 5

=(50%) X (\(\frac{1}{20}\)) X (12) X (5z + 5)

= (\(\frac{50}{100}\)) X (\(\frac{1}{20}\)) X (12) X (5z + 5)

= (\(\frac{1}{2}\)) X (\(\frac{1}{20}\)) X (12) X (5z + 5)

= (\(\frac{1 X 1 X 12}{2 X 20}\)) X (5z + 5)

= (\(\frac{12}{40}\)) X (5z + 5)

= (\(\frac{12}{40}\)) X (5z + 5)

= (\(\frac{3}{10}\)) X (5z + 5)

= (\(\frac{3}{10}\)) X (5)(z + 1)

=(\(\frac{3 X 5}{10}\)) X (z + 1)

=(\(\frac{3}{2}\)) X (z + 1)

=\(\frac{3}{2}\)z +\(\frac{3}{2}\)

 

Question 12.
21 plus 6p minus two-thirds the sum of 14p and 3q.
Answer:
21 plus 6p minus two-thirds the sum of 14p and 3q
= 21 + 6p – \(\frac{2}{3}\)(14p + 3q)

= 21 + 6p – \(\frac{2 X 14p}{3}\) – \(\frac{2 X 3q}{3}\)

= 21 + 6p – \(\frac{2 X 14p}{3}\) – \(\frac{2 X 3q}{3}\)

= 21 + 6p – \(\frac{28p}{3}\) – 2q

= 21 + \(\frac{3 X 6p}{3}\) – \(\frac{28p}{3}\) – 2q

= 21 + \(\frac{18p}{3}\) – \(\frac{28p}{3}\) – 2q

= 21 + \(\frac{18p-28p}{3}\)– 2q

= 21 + \(\frac{-10p}{3}\)– 2q

= 21 – \(\frac{10}{3}\)p- 2q

Tell whether each pair of equations are equivalent. (Lesson 4.1)

Question 13.
x – 7 = 1 and x = 6
Answer:
The above equations are not equivalent
If x = 6, and if we substitute it in
x – 7 = 1
(6) – 7 = 1
-1 = 1  -> False

Hence they are not equivalent.

Question 14.
0.2x = 0.6 and 3x + 1 =10
Answer:
The above equations are equivalent.
Equation 1
0.2x = 0.6
2x = 6
x = 3

Equation 2
3x + 1 = 10
3(3) + 1 = 10
9 + 1 = 10
10 = 10  -> True

Hence they are equivalent.

Solve each equation. (Lesson 4.2)

Question 15.
11 + 4k = 7
Answer:
11 + 4k = 7
4k = 7 – 11
4k = -4
k = -1

Question 16.
5p + \(\frac{2}{15}\) = \(\frac{3}{5}\) + \(\frac{4}{5}\)p
Answer:

5p + \(\frac{2}{15}\) = \(\frac{3}{5}\) + \(\frac{4}{5}\)p

5p – \(\frac{4}{5}\)p = \(\frac{3}{5}\) – \(\frac{2}{15}\)

\(\frac{5 X 5}{5}\)p – \(\frac{4}{5}\)p = \(\frac{3 X 3}{5 X 3}\) – \(\frac{2}{15}\)

\(\frac{25}{5}\)p – \(\frac{4}{5}\)p = \(\frac{9}{15}\) – \(\frac{2}{15}\)

\(\frac{21}{5}\)p = \(\frac{7}{15}\)

\(\frac{7 X 3}{5}\)p = \(\frac{7}{5 X 3}\)

\(\frac{3}{1}\)p = \(\frac{1}{3}\)

3p = \(\frac{1}{3}\)

p = \(\frac{1}{3 X 3}\)

p = \(\frac{1}{9}\)

Question 17.
\(\frac{8}{9}\)(4x – 3) = \(\frac{2}{3}\)
Answer:
\(\frac{8}{9}\)(4x – 3) = \(\frac{2}{3}\)

\(\frac{4 X 2}{3 X 3}\)(4x – 3) = \(\frac{2}{3}\)

\(\frac{4}{3}\)(4x – 3) = \(\frac{2 X 3}{3 X 2}\)

\(\frac{4}{3}\)(4x – 3) = \(\frac{1}{1}\)

\(\frac{4}{3}\)(4x – 3) = 1

4(4x – 3) = 3
(4)(4x) – (4)(3) = 3
16x – 12 = 3
16x = 3 + 12
16x = 15

x = \(\frac{15}{16}\)

Question 18.
7(x + 5) – 3x = x – 7
Answer:
7(x + 5) – 3x = x – 7
7x + 35 – 3x = x – 7
7x  – 3x – x = – 7 – 35
7x – 4x = -42
3x = – 42
x = -14

 

Solve each inequality. Then graph each solution set on a number line. (Lesson 4.4)

Question 19.
8x – 7 > 9
Answer:
8x – 7 > 9
8x > 9 + 7
8x > 16
x > 2

Question 20.
6 + \(\frac{2}{5}\)y ≤ \(\frac{1}{3}\)y + 6\(\frac{1}{3}\)
Answer:

6 + \(\frac{2}{5}\)y ≤ \(\frac{1}{3}\)y + 6\(\frac{1}{3}\)

\(\frac{2}{5}\)y – \(\frac{1}{3}\)y ≤ 6\(\frac{1}{3}\) – 6

\(\frac{2}{5}\)y – \(\frac{1}{3}\)y ≤ \(\frac{1}{3}\)

\(\frac{2 X 3}{5 X 3}\)y – \(\frac{1 X 5}{3 X 5}\)y ≤ \(\frac{1}{3}\)

\(\frac{6}{15}\)y – \(\frac{5}{15}\)y ≤ \(\frac{1}{3}\)

\(\frac{1}{15}\)y ≤ \(\frac{1}{3}\)

y ≤ \(\frac{15}{3}\)

y ≤ \(\frac{5 X 3}{3}\)

y ≤ 5

Question 21.
1.8(x – 5) ≥ 0.2(x – 13)
Answer:
1.8(x – 5) ≥ 0.2(x – 13)
(10)(1.8)(x – 5) ≥ (10)(0.2)(x – 13)
18(x – 5) ≥ 2(x – 13)
9(x – 5) ≥ 1(x – 13)
9x – 45 ≥ x – 13
9x – x ≥ -13 + 45
8x ≥ 32
x ≥ 4

Question 22.
2(3 – r) + 4(1 + 3r) > – 2(3 – 7r)
Answer:
2(3 — r) + 4(1 + 3r) > —2(3 — 7r)
2(3) + 2(—r) + 4(1) + 4(3r) > —2(3) + (—2)(—7r) Use Distributive Property
6 — 2r + 4 + 12r > —6 + 14r Multiply
10r + 10 > —6 + 14r Combine like terms
10r + 10 — 10 > —6 + 14r — 10 Subtract 10 from both sides
10r > -16 + 14r Simplify
10r – 14r > -16 + 14r – 14r Subtract 14r from both sides
-4r > -16 Simplify
-4r > -16
\(\frac{-4}{-4} r\) > \(\frac{-16}{-4}\) Divide both sides by -4 and reverse sign of inequality
r < 4

Use shaded circle for ≥ or ≤ and empty circle for < or >
The number line for r < 4 is shown

Tell whether each table, equation, or graph represents a direct proportion, an inverse proportion, or neither. Find the constant of proportionality for the direct and inverse proportion identified. (Lessons 5.1, 5.2, 5.4)

Question 23.

Answer:

Direct Proportion :
For each pair of values x and y →
\(\frac{y}{x}\) = \(\frac{12}{1.5}\) = 8
\(\frac{y}{x}\) = \(\frac{20}{2.5}\) = 8
\(\frac{y}{x}\) = \(\frac{28}{3.5}\) = 8
\(\frac{y}{x}\) is a constant so x and y is direct proportional
Inverse proportion :
For each pair of values x and y →
xy = 1.5 ∙ 12 = 18
xy = 2.5 ∙ 20 = 50
xy = 3.5 ∙ 28 = 98
xy is not a constant so y is not inversely proportional to x
NOTE:
Direct proportion : y is directly proportional to x then they have a constant of proportionality k, and it can be rewritten as y = kx or \(\frac{y}{x}\) = k
Inverse Proportion : x and y are in Inverse proportion then their product is a constant of proportionality k, and it can be rewritten as xy = k or y = \(\frac{k}{x}\)
Direct Proportion

Question 24.

Answer:

Direct Proportion :
For each pair of values x and y →
\(\frac{y}{x}\) = \(\frac{90}{10}\) = 9
\(\frac{y}{x}\) = \(\frac{45}{20}\) = 2.25
\(\frac{y}{x}\) = \(\frac{30}{30}\) = 1
\(\frac{y}{x}\) is not a constant so x and y is not direct proportional

Inverse Proportion :
For each pair of values x and y →
xy = 10 ∙ 90 = 900
xy = 20 ∙ 45 = 900
xy = 30 ∙ 30 = 900
xy is a constant so y is inversely proportional to x

NOTE :
Direct Proportion : y is directly proportional to x then they have a constant of proportionality k, and it can be rewritten as y = kx or \(\frac{y}{x}\) = k
Inverse Proportion : x and y are in inverse proportion then their product is a constant of proportionality k, and it can be rewritten as xy = k or y = \(\frac{k}{x}\)
Inverse proportion

Question 25.

Answer:

Direct Proportion :
For each pair of valtes x and y →
\(\frac{y}{x}\) = \(\frac{36}{3}\) = 12
\(\frac{y}{x}\) = \(\frac{60}{6}\) = 10
\(\frac{y}{x}\) = \(\frac{84}{9}\) = 9.33
\(\frac{y}{x}\) is not a constant so x and y is not direct proportional

Inverse Proportion :
For each pair of values x and y →
xy = 3 ∙ 36 = 108
xy = 6 ∙ 60 = 360
xy = 9 ∙ 84 = 756
xy is not a constant so y is not inversely proportzonal to x

NOTE :
Direct Proportion : y is directly proportional to x then they have a constant of proportionality k, and it can be rewritten as y = kx or \(\frac{y}{x}\) = k
Inverse Proportion : x and y are in inverse proportion then their product is a constant of proportionality k, and it can be rewritten as xy = k or y = \(\frac{k}{x}\)
Neither

Question 26.
y = 1.8x + 3.6
Answer:
y = 1.8x + 3.6
Direct proportion :
y = 1.8x + 3.6
Because the original equation y = 1.8x + 3.6 cannot be rewritten as an equivalent equation in the form y = kx, it does not represent a direct proportion.

Inverse Proportion :
y = 1.8x + 3.6
Because the original equation y = 1.8x + 3.6 cannot be rewritten as an equivalent equation in the form xy = k, it does not represent an inverse proportion.

NOTE :
Direct Proportion : y is directly proportional to x then they have a constant of proportionality k, and it can be rewritten as y = kx or \(\frac{y}{x}\) = k
Inverse Proportion : x and y are in inverse proportion then their product is a constant of proportionality k, and it can be rewritten as xy = k or y = \(\frac{k}{x}\)
Neither Direct Proportion nor Inverse Proportion

Question 27.
\(\frac{2}{5}\)y = \(\frac{1}{2}\)x
Answer:
\(\frac{2}{5}\)y = \(\frac{1}{2}\)x

Direct proportion :
\(\frac{2}{5}\)y = \(\frac{1}{2}\)
\(\frac{5}{2}\) ∙ \(\frac{2}{5}\)y = \(\frac{5}{2}\) ∙ \(\frac{1}{2}\)x Multiply both sides by \(\frac{5}{2}\)
y = \(\frac{5}{4}\)x
Because the original equation \(\frac{2}{5}\)y = \(\frac{1}{2}\)x can be rewritten as an equivalent equation in the form y = kx, it represent a direct proportion.

Inverse Proportion :
\(\frac{2}{5}\)y = \(\frac{1}{2}\)x
\(\frac{5}{2}\) ∙ \(\frac{2}{5}\)y = \(\frac{5}{2}\) ∙ \(\frac{1}{2}\)x Multiply both sides by \(\frac{5}{2}\)y

Because the original equation \(\frac{2}{5}\)y = \(\frac{1}{2}\)x cannot be rewritten as an equivalent equation in the form xy = k, it does not represent an inverse proportion.

NOTE:
Direct Proportion : y ‘is directly proportional to x then they have a constant of proportionality k, and it can be rewritten as y = kx or \(\frac{y}{x}\) = k
Inverse Proportion : x and y are in inverse proportion then their product is a constant of proportionality k, and it can be rewritten as y = k or y = \(\frac{k}{x}\)
Direct proportion

Question 28.
3y = \(\frac{18}{x}\)
Answer:
3y = \(\frac{18}{x}\)
Direct Proportion :
3y = \(\frac{18}{x}\)
\(\frac{x}{3}\) ∙ 3y = \(\frac{x}{3}\) ∙ \(\frac{18}{3}\) Multiply both sides by \(\frac{x}{3}\)
Because the original equation 3y = \(\frac{18}{3}\) cannot be rewritten as an eqnivalent equation in the form y = kx, it does not represent a direct proportion.

Inverse Proportion :
3y = \(\frac{18}{3}\)
\(\frac{x}{3}\) ∙ 3y = \(\frac{x}{3}\) ∙ \(\frac{18}{3}\) Multiply both sides by \(\frac{x}{3}\)
xy = 6
18
Because the original eqtiation 3y = \(\frac{18}{x}\) can be rewritten as an equivalent equation in the form xy = k, il represent an inverse proportion.

NOTE :
Direct Proportion : y is directly proportional to z then they have a constant of proportionality k, and it can be rewritten as y = kx or \(\frac{y}{x}\) = k
Inverse Proportion : x and y are in inverse proportion then their product is a constant
of proportionality k, and it can be rewritten as xy = k or y = \(\frac{k}{x}\)
Inverse Proportion

Question 29.

Answer:
Direct Proportion :
$From\ the\ graph$
\(\frac{y}{x}\) = \(\frac{5}{1}\) = 5
\(\frac{y}{x}\) = \(\frac{10}{2}\) = 5
\(\frac{y}{x}\) = Constant
The given graph is a straight line that does not lie along the x — axis or y — axis and does pass through the origin.
So, the given graph represents a direct proportion.
Inverse Proportion :
$Use\graph\: $
x ∙ y = 1 ∙ 5 = 5
x ∙ y = 2 ∙ 10 = 20
The product of x and y is not a constant so x and y does not represent an inverse proportion.
Direct proportion

Question 30.

Answer:

Question 31.

Answer:
Direct Proportion:
The given graph is not a straight tine that does tie along the x — axis and don not pass through the origin.
So, the given graph does not represents a direct proportion.

Inverse Proportion:
Use graph
x ∙ y = 1 ∙ 30 = 30
x ∙ y = 3 ∙ 10 = 30
The product of x and y is a constant so x and y represent an inverse proportion.
Inverse Proportion

In each table, y is directly proportional to x. Find the constant of proportionality and then complete the table. (Lesson 5.3)

Question 32.

Answer:

Find Constant of Proportionality :
y is directly proportional to x :
\(\frac{y}{x}\) = \(\frac{4}{20}\)
\(\frac{y}{x}\) = \(\frac{1}{5}\) simplify
Constant of proportionality = \(\frac{1}{5}\)

Find Missing numbers :
Find y when x = 40
\(\frac{y}{x}\) = \(\frac{1}{5}\)
\(\frac{y}{40}\) = \(\frac{1}{5}\) Multiply both sides by 40
40 ∙ \(\frac{y}{40}\) = 40 ∙ \(\frac{y}{40}\)
y = 8
Find x when y = 16
\(\frac{y}{x}\) = \(\frac{1}{5}\)
\(\frac{16}{x}\) = \(\frac{1}{5}\) Evaluate y = 16
5 ∙ 16 = x ∙ 1 Cross Muttipliction
x = 80

Constant of proportionality = \(\frac{1}{5}\)
The missing number are 8 and 80 respectively

Question 33.

Answer:

y is directly proportional lo x :
\(\frac{y}{x}\) = \(\frac{27}{3}\)
\(\frac{y}{x}\) = 9 simplify
Constant of Proportionality = 9
Find Missing Numbers :
Find y when x = 1
\(\frac{y}{x}\) = 9
\(\frac{y}{1}\) = 9 Evaluate x = 1
y = 9
Find x when y = 90
\(\frac{y}{x}\) = 9
\(\frac{90}{x}\) = 9 Evaluate y = 90
\(\frac{x}{90}\) ∙ \(\frac{90}{x}\) = \(\frac{x}{9}\) ∙ 9 Multiply both sides by \(\frac{x}{9}\)
x = 10

Constant of Proportionality = 9
The missing numbers are 9 and 10 respectively

In each table, y is inversely proportional to x. Find the constant of proportionality and then complete the table. (Lesson 5.4)

Question 34.

Answer:

Find Constant of Proportionality
y is Inversely proportional to x:
xy = 12 • 50
xy = 600 Slmplif y
Find missing value in tite table.
when x = 40 find y.
xy = 600
40 • y = 600 Evaluate x = 40
\(\frac{40 y}{40}\) = \(\frac{600}{40}\) Divide both sides by 4
y = 15 Simplify
when y = 10 find x,
x ∙ 10 = 600 Evaluate y = 10
\(\frac{10 x}{10}\) = \(\frac{600}{10}\) Divide both sides by 10
x = 60 Simplify

Constant of Proportionality = 600
The missing number are y = 15 and x = 60

Question 35.

Answer:

Find Constant of Proportionality
y is inversely proportional to x:
xy = 2 ∙ 4.5
xy = 9 Simplify
Find missing value in the table.
When x = 1 find y,
xy = 9
1 ∙ y = 9 Evaluate x = 1
y = 9 Simplify
When y = 3 find x,
x ∙ 3 = 9 Evaltate y = 3
\(\frac{3 x}{3}\) = \(\frac{9}{3}\) Divide both sides by 3
x = 3 Simplify

Constant of Proportionality = 9
The missing number are y = 9 and x = 3

Solve using proportional reasoning. (Lessons 5.3, 5.4)

Question 36.
y varies directly as x, and y = 4.8 when x = 0.2. Find y when x = 0.8.
Answer:
y = 4.8 and x = 0.2
y is directly proportional to x :
\(\frac{y}{x}\) = \(\frac{4.8}{0.2}\)
\(\frac{y}{x}\) = 24
x ∙ \(\frac{y}{x}\) = 24 ∙ x Multiply both sides by x
y = 24x
Find the value of y when x = 0.8
We know y = 24x,
y = 24 ∙ 0.8 Substituting x = 0.8
y = 19.2 Simplify

Question 37.
P is inversely proportional to Q, and P = \(\frac{1}{4}\) when Q = \(\frac{1}{2}\). Find P when Q = 2 \(\frac{1}{2}\).
Answer:
P = \(\frac{1}{4}\) and Q = \(\frac{1}{2}\)
Write an inverse proportion equation :
P is inversely proportional to Q :
P ∙ Q = \(\frac{1}{4}\) ∙ \(\frac{1}{2}\)
PQ = \(\frac{1}{8}\)

Find the value of P when Q = 2\(\frac{1}{2}\)
PQ = \(\frac{1}{8}\)
P ∙ 2\(\frac{1}{2}\) = \(\frac{1}{8}\) Substituting Q = 2\(\frac{1}{2}\)
\(\frac{5}{2}\)P = \(\frac{1}{8}\) Simplifying
\(\frac{2}{5}\) ∙ \(\frac{5}{2}\)P = \(\frac{1}{8}\) ∙ \(\frac{2}{5}\) Multiplying both sides by \(\frac{2}{5}\)
P = \(\frac{1}{20}\)

Problem Solving

Solve. Show your work.

Question 38.
The table shows the relationship between the capacity in quarts, x, and the capacity in cups, y. Graph the relationship between y and x. Use 1 unit on the horizontal axis to represent 1 quart and 1 unit on the vertical axis to represent 4 cups. (Chapter 5)

Answer:


a) Is the number of cups directly proportional to the number of quarts? If so, find the constant of proportionality.
Answer:
The graph is a straight line through the origin, and It does not lie along the x – axis or y – axis.
So, it represents a direct proportion.
Yes the number of cups is directly proportional to the number of quarts
Find the Constant of Proportionality :
\(\frac{y}{x}\) = \(\frac{4}{1}\) = 4
\(\frac{y}{x}\) = \(\frac{8}{2}\) = 4
\(\frac{y}{x}\) = \(\frac{12}{3}\) = 4
\(\frac{y}{x}\) = \(\frac{16}{4}\) = 4
The Constant of Proportionality is 4

b) A soup recipe calls for 5 quarts of stock. How many cups of stock does the soup recipe need?
Answer:
$We \know$
\(\frac{y}{x}\) = 4
\(\frac{y}{5}\) = 4 Substituting x = 5
5 ∙ \(\frac{y}{5}\) = 5 ∙ 4 Multiplying bothsides by 5
y = 20
The recipe will need 20 cups of stocks

c) Convert 24 cups of water to quarts of water.
Answer:
$We\ know$
\(\frac{y}{x}\) = 4
\(\frac{24}{x}\) = 4 Substituting x = 24
\(\frac{x}{4}\) ∙ \(\frac{24}{x}\) = \(\frac{x}{4}\) ∙ 4 Multiplying bothsides by \(\frac{x}{4}\)
x = 6
24 cups of water = 6 quarts of water

Question 39.
A printer takes 2.4 seconds longer to print a page in color than a page in black and white. A page in black and white can be printed in 4 seconds.
There are (-\(\frac{5}{8}\)w + 6) color pages and (1.2w + 5) black and white pages to print. (Chapters 3, 4)
Answer:
The printer takes 4 seconds to print a page in black and white.
So, to print a page in color, the printer wilitake = 4 + 2.4 = 6.4 seconds
Number of pages to print in Blackund White 1.2w+ 5
Number of pages to print in Color = \(\frac{5}{8}\)w + 6

a) How long does it take to print all the pages?
Answer:
4 (1.2w + 5) + 6.4 (\(\frac{5}{8}\)w + 6)
= 4(1.2w) + 4(5) + 6.4 (\(\frac{5}{8}\)w) + 6.4(6) Use Distributive Property
= 4.8w + 20 + 4w + 38.4 Multiply
= 4.8w + 4w + 20 + 38.4 Group like terms
= (8.8w + 58.4) seconds

b) If w = 40, how long does it take to print all the pages?
Answer:
8.8w + 58.4
= 8.8(40) + 58.4 Substituting w = 40
= 352 + 58.4 Multiply
= 410.4 seconds

Question 40.
Jason has to choose between two offers of a gym club membership as shown below. (Chapter 4)

Answer:
Gym A = $55 per month membership fee plus $15 per hour of training with a personal trainer.
${\color{#4257b2} Gym\ B) = $220\ per\ month\ membership\ fee\ and\ training\ with\ a\ personal\ trainer\ at\ no\ extra\ cost $
a) After how many hours of training per month would Gym B be a better deal than Gym A?
Answer:
55 + 15x > 220
55 + 15x – 55 > 220 — 55 Subtract 35 from both sides
15x > 165 Simplify
\(\frac{15 x}{15}\) > \(\frac{165}{15}\) Divide both sides by 15
x > 11
After 11 hours of training per month Gym B would be a better deal than Gym A

b) If Jason plans to train for at least 12 hours per month, which gym club membership should he take up? Explain your answers.
Answer:
Gym A:
55 + 15 • 12
= 55 + 180 Multiply
= 235
Gym B:
220
Jason should take membership of Gym B since it is cheaper if he plans to train for at least 12 hours per month

Question 41.
A commission is an amount of money earned by a sales person, based on the amount of sales the person makes. At a particular home-improvement store, the commission made, C, is directly proportional to the amount of sales, S. A sales person earns $198 when he makes total sales of $3,600. (Chapter 5)
Answer:
Commission (C) = $198
Sales (S) = $3600

a) Write a direct proportion equation that relates C and 5. Then express the store’s commission rate as a percent.
Answer:
C is directly proportional to S :
\(\frac{C}{S}\) = \(\frac{198}{3600}\)
\(\frac{C}{S}\) = \(\frac{11}{200}\)
S ∙ \(\frac{C}{S}\) = \(\frac{11}{200}\) ∙ S Multiply both sides by S
C = \(\frac{11}{200}\)S
Express the store’s commission rate as a percent
C is directly proportional to S :
\(\frac{C}{S}\) = \(\frac{11}{200}\)
\(\frac{C}{S}\) = 0.055
\(\frac{C}{S}\) = 0.055 ∙ 100
\(\frac{C}{S}\) = 5.5%

b) Find the amount of sales made if his commission is $297.
Answer:
We know C = \(\frac{11}{200}\)S,
297 = \(\frac{11}{200}\)S Substituting C = 297
\(\frac{200}{11}\) ∙ 297 = \(\frac{200}{11}\) ∙\(\frac{11}{200}\)S Multiplying both sides by \(\frac{200}{11}\)
S = 5400 Simplify
He made a saie of amount $5400 if his commission is $297

Question 42.
Amy and her friends want to equally share the cost of David’s birthday gift. They plan to buy a baseball glove that costs $79.98. The amount of money that each has to contribute, C, is inversely proportional to the number of people sharing the cost, n. (Chapter 5)
Answer:
Cost of baseball gloves = $79.98
Amount of money that each has to contribute = C
Number of people sharing the cost = n
a) Write an inverse proportion equation that relates C and n.
Answer:
Inverse proportion equation :
C ∙ n = 79.98
Cn = 79.98 Multiplying
\(\frac{C n}{n}\) = \(\frac{79.98}{n}\) Divide both sides by n
C = \(\frac{79.98}{n}\)

b) How much will each person have to pay if 6 people are sharing the cost?
Answer:
n = 6 and C = \(\frac{79.98}{n}\),
C = \(\frac{79.98}{6}\) Substituting n = 6
C = 13.33
Each person has to contribute $13.33 if 6 person are sharing the cost.

Question 43.
The time it takes Tim to cycle to his destination varies inversely as his cycling speed in miles per hour. The graph shows the relationship between y and x. (Chapter 5)

a) Find the constant of proportionality. What does this value represent in this situation?
Answer:
Use the point (1, 12) from the graph to find the constant of proportionality.
x • y = 1 • 12 Choose the point (1, 12)
xy = 12 Multiply
The constant of proportionality is 12
This value represents the distance’ traveled by Tim

b) Write an equation that relates speed and time taken.
Answer:
Use the point (1, 12) from the graph to find the constant of proportionality.
x • y = 1 • 12 Choose the point (1, 12)
xy = 12 Multiply

c) Explain what the point (1, 12) represents in this situation.
Answer:
It means that when Tim cycle at a speed of 12 miles per hour he will reach his destination in 1 hour.

d) How long does it take Tim to reach his destination if he decreases his speed to 6 miles per hour?
Answer:
From the graph :
If Tim cycle at a speed of 6 miles per hour he will reach his destination in 2 hour.

Question 44.
Two-thirds of the capacity of jar A is equivalent to eight-fifteenths of the capacity of jar B. Suppose the full capacity of jar B is (6x – 3) liters. (Chapters 3, 4)
a) What is the full capacity of jar A in terms of x?
Answer:
Let the full capacity of Tank A be A
Let the full capacity of Tank B be B

Two — third of the capacity of Jar A is equivalent to eight — fifteenths of the capacity of jar B :
\(\frac{2}{3}\) ∙ A = \(\frac{8}{15}\) ∙ B Multiplying both sides by \(\frac{3}{2}\)
A = \(\frac{4}{5}\) ∙ B Simplifying
A = \(\frac{4}{5}\) ∙ (6x – 3) Substituting B = (6x – 3)
A = \(\frac{4}{5}\)(6x) – \(\frac{4}{5}\)(3) Using Distributive Property
A = \(\frac{24}{5}\)x – \(\frac{12}{5}\) Multiplying
A = \(\frac{24 x-12}{5}\)

b) If the capacity of jar A is 18 liters, what is the value of x?
Answer:
A = \(\frac{24 x-12}{5}\)
A = \(\frac{24 x-12}{5}\) Substituting A = 18
5 ∙ 18 = 5 ∙ \(\frac{24 x-12}{5}\) Multiplying both sides by 5
90 = 24x – 12 Simplifying
90 + 12 = 24x – 12 + 12 Add 12 to both sides
24x = 102 Simplifying
\(\frac{24 x}{24}\) = \(\frac{102}{24}\) Divide both sides by 24
x = 4.25

Question 45.
A rectangle measures 12 inches by 8 inches. The length of the rectangle is increased by r percent while the width remains unchanged. (Chapters 3, 4)
Answer:
Expanded Length = (12 + \(\frac{12}{100}\)r) inches
Width = 8 inches
a) Write an algebraic expression of the area of the expanded rectangle.
Answer:
Algebraic Expression for the area of the expanded rectangle
Area
= Length ∙ Width
= (12 + \(\frac{12}{100}\)r) ∙ 8
= (12 + 0.12r) ∙ 8
= 8(12) + 8(0.12r)
= (96 + 0.96r) square inches

b) If the area of the expanded rectangle is 120 square inches, find the value of r.
Answer:
Area = 120 square inches
96 + 0.96r = 120
96 + 0.96r – 96 = 120 – 96 Subtract 96 from both suies
0.96r = 24 Simplify
\(\frac{0.96 r}{0.96}\) = \(\frac{24}{0.96}\)
Divide both sides by 0.96
r = 25

This handy Math in Focus Grade 3 Workbook Answer Key Chapter 18 Practice 1 Classifying Polygons provides detailed solutions for the textbook questions.

Math in Focus Grade 3 Chapter 18 Practice 1 Answer Key Classifying Polygons

Circle the polygons.

Question 1.

Answer:

Explanation:
The above figure contains 5 polygons.
A polygon is a closed two-dimensional shape that is formed by enclosing line segments.
A minimum of three line segments are required to make a polygon.

Mark the angles. Label the parts of each polygon.

Question 2.

Answer:

Explanation:
Given figure is a polygon.
The line segments forming a polygon are called its sides.
Sides are AB, BC, CD  and DA .
The meeting point of a pair of sides is called its vertex.

Question 3.

Answer:

Explanation:
Given picture is called a polygon.
The line segments forming a polygon are called its sides.
Sides are AB, BC, CD, DE and EA .
The meeting point of a pair of sides is called its vertex.

Identify each polygon.

Question 4.

Answer: Square

Explanation:
Given figure is a polygon. Square in shape.
The line segments forming a polygon are called its sides.
Sides are AB, BC, CD  and DA .
The meeting point of a pair of sides is called its vertex.

Question 5.

Answer: Pentagon

Explanation:
Given picture is called a polygon. Pentagon in shape.
The line segments forming a polygon are called its sides.
Sides are AB, BC, CD, DE and EA .
The meeting point of a pair of sides is called its vertex.

Question 6.

Answer: Octagon

Explanation:
Given picture is called a polygon. Octagon in shape.
The line segments forming a polygon are called its sides.
What are the sides of polygon ABCDEFGH
Sides are AB, BC, CD, DE , EF and FG, GH and HA .
The meeting point of a pair of sides is called its vertex.

Question 7.

Answer: Square

Explanation:
Given figure is a polygon. Square in shape.
The line segments forming a polygon are called its sides.
Sides are AB, BC, CD  and DA .
The meeting point of a pair of sides is called its vertex.

Question 8.

Answer: Hexagon

Explanation:
Given picture is called a polygon. Hexagon in shape.
The line segments forming a polygon are called its sides.
What are the sides of polygon ABCDEF
Sides are AB, BC, CD, DE , EF and FA .
The meeting point of a pair of sides is called its vertex.

Question 9.

Answer: Trapezoid

Explanation:
Given figure is a polygon. Trapezoid in shape.
The line segments forming a polygon are called its sides.
Sides are AB, BC, CD  and DA .
The meeting point of a pair of sides is called its vertex.

Complete the table. Then answer the question.

Question 10.

Answer:

Explanation:
A Square is a regular quadrilateral or polygon.
which has all the four sides of equal length, four vertices and,
four angles with right angle triangles.
Octagon is a polygon in geometry, which has 8 sides, 8 vertices and 8 angles.
That means the number of vertices is 8 and the number of edges is 8.
So, octagon is a 8-sided polygon and it is a two dimensional plane figure.
All the sides are joined with each other end-to-end to form a shape.
In geometry, a hexagon can be defined as a polygon with six sides.
The two-dimensional shape has 6 sides, 6 vertices and 6 angles.

Answer:

Explanation:
The above Triangle is polygon.
A triangle has three sides, 3 vertices and
3 angles sum to 180 degrees.
The above figure rectangular polygon,
it has four sides, four vertices and four angles, with 90 degrees.
And the angles are right angle triangles.
A pentagon is a five-sided polygon with five straight sides,
five vertices and five interior angles that sum up to 540°540°.
A pentagon shape is a plane figure, or flat (two-dimensional) 5-sided geometric shape.

Answer:

Explanation:
Parallelogram is a four-sided closed figure with opposite sides are equal and opposites angles are equal.
A parallelogram has 4 sides, 4 vertices and 4 angles.
A rhombus is a quadrilateral whose four sides have the same length.
Every rhombus has two diagonals connecting pairs of opposite vertices, and two pairs of parallel sides.
Opposite angles of a rhombus have equal length.
A trapezoid is a quadrilateral with exactly one pair of parallel sides.
The sum of the interior angles of a trapezoid equals 360 degrees,
So, it has four angles and four vertices, also called corners.

Question 11.
Which figures have the same number of sides, vertices and angles?

Answer:
All the above mentioned polygons have sides, vertices and angles.
Explanation:
Polygons are named according to the number of sides and angles they have.
The most familiar polygons are the triangle, the rectangle, and the square.
A regular polygon is one that has equal sides.
Some special polygons have equal sides, vertices and angles.

Are these statements correct? Write true or false.

Question 12.
A hexagon has seven sides and six angles. ____________
Answer: False
Explanation:
hexa means six
In geometry, a hexagon can be defined as a polygon with six sides.
The two-dimensional shape has 6 sides, 6 vertices and 6 angles.

Question 13.
All polygons have four sides. ____________
Answer: False
Explanation:
A plane figure that is described by a finite number of straight line segments connected to form a closed polygon.
Polygons are named according to the number of sides and angles they have.

Question 14.
All parallelograms, squares, and trapezoids have four angles. ____________
Answer: True
Explanation:
All parallelograms, squares, and trapezoids have four angles,
4 sides and 4 vertices as mentioned above.

Question 15.
An octagon has eight vertices and seven sides. ____________
Answer: False
Explanation:
Octagon is a polygon in geometry, which has 8 sides, 8 vertices and 8 angles.
That means the number of vertices is 8 and the number of edges is 8.
So, octagon is a 8-sided polygon and it is a two dimensional plane figure.
All the sides are joined with each other end-to-end to form a shape.

Question 16.
A pentagon has six angles. ____________
Answer: False
Explanation:
A pentagon is a five-sided polygon with five straight sides,
five vertices and five interior angles that sum up to 540°540°.
A pentagon shape is a plane figure, or flat (two-dimensional) 5-sided geometric shape.

Question 17.
A triangle has two vertices. ____________
Answer: False
Explanation:
A triangle has three sides, 3 vertices and
3 angles sum to 180 degrees.

Question 18.
A parallelogram can be separated into 4 triangles. ____________
Answer: True
Explanation:
Parallelogram is a four-sided closed figure with opposite sides are equal and opposites angles are equal.
A parallelogram has 4 sides, 4 vertices and 4 angles.

Question 19.
A rectangle has four right angles. ____________
Answer: True
Explanation:
A rectangle has four sides, four vertices and four angles, with 90 degrees.
And the angles are right angle triangles.

Cut out the tangram below and complete the table.

Question 20.

Answer:

Explanation:
Answer may vary,
We can write all polygons like square, triangle, rectangle and pentagon …. so on.
Polygons are named according to the number of sides and angles they have.
The most familiar polygons are the triangle, the rectangle, and the square.
A regular polygon is one that has equal sides.
Some special polygons have equal sides, vertices and angles.

Use at least 5 polygons to make a figure. Complete the table.

Question 21.

Answer:


Explanation:
All types of polygons be used.
here in the above figure we used 5 polygons
i) Pentagon
ii) Triangle
iii) Rhombus
iv) Trapezoid
v) Rectangle
As we know polygon is a two-dimensional geometric figure that has a finite number of sides and a closed figure.
Polygons are named according to the number of sides and angles they have.
A regular polygon is one that has equal sides.
Some special polygons have equal sides, vertices and angles.

Question 22.
Name the figure that you have made.

Answer:

Explanation:
From the above figure,
we can cut 4 triangles, 1 parallelogram and 1 rectangle.

Solve.

Question 23.
I am a polygon. I have 1 more angle than a rectangle has. What am I? ___________
Answer:
Pentagon
Explanation:
A pentagon is a five-sided polygon with five straight sides,
five vertices and five interior angles that sum up to 540°540°.
A pentagon shape is a plane figure, or flat (two-dimensional) 5-sided geometric shape.

Question 24.
I am a polygon. I have 1 more side than a pentagon has. What am I? ___________
Answer:
Hexagon
Explanation:
hexa means six
In geometry, a hexagon can be defined as a polygon with six sides.
The two-dimensional shape has 6 sides, 6 vertices and 6 angles.

Question 25.
I am a polygon. I have 1 more vertex than a triangle has. What am I? ___________
Answer:
Rectangle
Explanation:
A rectangle has four sides, four vertices and four angles, with 90 degrees.
And the angles are right angle triangles.

Question 26.
Add one more polygon to the shape below to make it a hexagon.

Answer:

Explanation:
The above drawn figure is Hexagon, hexa means six.
In geometry, a hexagon can be defined as a polygon with six sides.
The two-dimensional shape has 6 sides, 6 vertices and 6 angles.

Identify each quadrilateral. Then explain your answer.

Example

This is a rectangle. A rectangle has 2 pairs of opposite sides that are parallel. Only the opposite sides of a rectangle need to be of equal length. All 4 angles of a rectangle are right angles.

Question 27.

This is a ____________________
Answer:
This is a parallelogram.
Explanation:
A parallelogram has 2 pairs of opposite sides that are parallel.
Only the opposite sides of a rectangle need to be of equal length.
All 4 angles of a rectangle are right angles.

Question 28.

This is a ____________________
Answer:
This is a rhombus.
Explanation:
A rhombus has 2 pairs of opposite sides that are parallel.
Only the opposite sides of a rectangle need to be of equal length.
All 4 angles of a rectangle are right angles.

Question 29.

This is a ____________________
Answer:
This is a Trapezoid.
Explanation:
A trapezoid has 2 pairs of opposite sides that are parallel.
Only the opposite sides of a rectangle need to be of equal length.
All 4 angles of a rectangle are right angles.

Write P for a parallelogram, R for rhombus, or T for trapezoid on the shapes.

Question 30.

Answer:

Explanation:
As we all ready know that clearly about Parallelogram, Rhombus and Trapezoid in the above question no.27,
identify the Parallelogram, Rhombus and Trapezoid and mention with their initials.

Question 31.
How is a trapezoid different from a parallelogram?
Answer:

Both are quadrilaterals.
Explanation:
A parallelogram has two pairs of parallel sides.
A trapezoid only has to have one pair of parallel sides.
So a parallelogram is a trapezoid (a special case of one), but a trapezoid is not, in general, a parallelogram.

Question 32.
How is a trapezoid similar to a parallelogram?
Answer:

Explanation:
A trapezoid has one pair of parallel sides and a parallelogram has two pairs of parallel sides.
So a parallelogram is also a trapezoid.

This handy Math in Focus Grade 3 Workbook Answer Key End of Year Review detailed solutions for the textbook questions.

Math in Focus Grade 3 End of Year Review Answer Key

Test Prep

Multiple Choice

Fill in the circle next to the correct answer.

Question 1.
John spends $1.35 on bus fare and $2.50 on food each day. How much does he spend in two days? (Lesson 10.1)
(A) $3.85
(B) $6.60
(C) $6.70
(D) $7.70
Answer:
Option(D)
Explanation:
John spends $1.35 on bus fare and
$2.50 on food each day.
Total amount he spend in two days
$2.50 + $1.35 = 3.85
3.85 x 2 = 7.7

Question 2.
Paige jogs around a 400-meter track 3 times a day. What is the distance she jogs each day? (Lesson 11.2)
(A) 400 m
(B) 1 km 200 m
(C) 1 km 400 m
(D) 10 km 200 m
Answer:
Option(B)
Explanation:
Paige jogs around a 400-meter track 3 times a day.
The distance she jogs each day
400 x 3 = 1200 m

Question 3.
Which mass is not the same as the others? (Lesson 11.3)
(A) 7,220 g
(B) 7,022 g
(C) 7,000 g + 22 g
(D) 7 kg 22 g
Answer:
Option(A)
Explanation:
7,220 g is not same as the other,
other three are 7kg + 22g or 7022g or 7000g + 22g represent same.

Question 4.
Which is incorrect? (Lesson 14.3)

Answer:
Option(C)
Explanation:
A, B and D are correct fractions.
C is incorrect fraction.
As \(\frac{2}{3}\) is not a perfect division.

Question 5.
Look at the measuring cups. (Lesson 11.4)

Which is correct?
(A) There is 500 milliliters more water in X than Y.
(B) There is a total of 1,500 milliliters of water in X and Y.
(C) Z contains 180 milliliters less water than X.
(D) The difference in the volume of water in Y and Z is 170 milliliters.
Answer:
Option(D)
Explanation:
The amount of space in an object is known as volume.
Compare both the volumes in Y and Z,
The difference in the volume of water in Y and Z is 170 milliliters.
V = Y – Z = 200 – 30 = 170 milliliters.

Question 6.
What fraction of the figure is shaded? (Lesson 14.1)

(A) \(\frac{1}{5}\)
(B) \(\frac{2}{5}\)
(C) \(\frac{6}{13}\)
(D) \(\frac{2}{3}\)
Answer:
Option(B)
Explanation:
Total shaded regions are 4 and total regions are 10, and the fraction form is written as follows
\(\frac{4}{10}\) = \(\frac{2}{5}\)

Question 7.
Look at the line segments. (Lesson 17.6)

Which line segments are parallel?
(A) Segments AB and AF
(B) Segments BC and EF
(C) Segments AF and BC
(D) Segments AB and CD
Answer: D
Explanation:
Two lines or line segments can either intersect (cross) each other or be parallel.
Parallel line segments never meet, no matter how far they are extended.
So, Segments AB and CD

Question 8.
Which is a polygon? (Lesson 18.1)

(A) Figure W
(B) Figure X
(C) Figure Y
(D) Figure Z
Answer:
Option(D)
Explanation:
A polygon is defined as a flat or plane, two-dimensional closed shape with straight sides.
It does not have curved sides.
So, figure Z is polygon.

Question 9.
Which tarts weigh the same? (Lesson 15.2)

(A) Lemon and Strawberry
(B) Lemon and Peach
(C) Blueberry and Strawberry
(D) Blueberry and Peach
Answer:
Option(B)
Explanation:
1 pound = 16 ounce
Lemon = 2 pounds
Peach = 32 ounces
2 x 16 = 32
So, Lemon and Peach weigh the same.

Question 10.
8 of the 20 buttons in a box are gray. The rest are white. What fraction of the buttons are white? (Lesson 14.6)

(A) \(\frac{2}{5}\)
(B) \(\frac{3}{5}\)
(C) \(\frac{4}{5}\)
(D) \(\frac{3}{7}\)
Answer:
Option(A)
Explanation:
8 of the 20 buttons in a box are gray.
The rest are white.
\(\frac{8}{20}\)
Divide both numerator and denominator with 4 to simplify.
\(\frac{2}{5}\) fraction of the buttons are white.

Short Answer

Read the questions carefully. Write your answers in the space provided.

Question 11.
Order the fractions from greatest to least. (Lesson 14.4)
\(\frac{1}{4}\), \(\frac{7}{8}\), \(\frac{3}{4}\)
Answer:
\(\frac{7}{8}\), \(\frac{3}{4}\), \(\frac{1}{4}\)
Explanation:
Simplify,
\(\frac{1}{4}\) = 0.25,
\(\frac{7}{8}\) = 0.87,
\(\frac{3}{4}\) = 0.75,
So, when we arrange the fractions greatest to least
\(\frac{7}{8}\), \(\frac{3}{4}\), \(\frac{1}{4}\)

Question 12.
String A is 28 inches long. String B is 4 feet long. Which is longer? (Lesson 15.1)
String ____________
Answer:
String B is longer than A.
Explanation:
String A is 28 inches long.
String B is 4 feet long.
1 feet = 12 inches
4 feet = 12 x 4 = 48 inches
So, string B is longer.

Question 13.
George starts on his science project at 8:25 A.M. He finishes at 10:10 A.M. How long did he take? (Lesson 16.5)
__________ h ___________ min
Answer:
1 hrs 85 min
Explanation:
George starts on his science project at 8:25 A.M.
He finishes at 10:10 A.M.
Total time he took
10:10 A.M – 8:25 A.M = 1 hrs 85 min

Question 14.
Mrs. Freeman puts 3 cups of lemon juice in a punch bowl. She adds 6 pints of water. How many cups of liquid are there in total? (Lesson 15.3)
____________ cups
Answer: 6 cups
Explanation:
Mrs. Freeman puts 3 cups of lemon juice in a punch bowl.
She adds 6 pints of water.
1 cup = 0.5 pints
6 pints = 0.5 x 6 = 3 cups
Already 3 cups are there, so she adds 3 more
Total cups of liquid in bowl 3 + 3 = 6

Question 15.
Which angles in the figure are less than a right angle? (Lesson 17.3)

Angles ___________ and ___________
Answer:
Angles F and D
Explanation:
Angle F and D are less then 90 degrees ,
So F and D are angles in the figure are less than a right angle.

Look at the figures to answer Exercises 16 and 17. (Lesson 19.4)

Question 16.
Which figure has a greater area?
Figure ____________
Answer:
Figure B has a greater area
Explanation:
Figure A has 6 in.²
Figure B 7 in.²,
when compare both the figures the area of B is greater.
Question 17.
How much greater?
__________ in.²
Answer: 1 in.²
Explanation:
Figure A has 6 in.²
Figure B 7 in.²,
when compare both the figures the area of B is greater.
Question 18.
Which figures are congruent? (Lesson 18.2)

Figures __________ and ___________
Answer:
Figures A and C
Explanation:
If two figures or objects have the same shape and size or known as congruent,
or if one has the same shape and size as the mirror image of the other.

Look at the line plot to answer Exercises 19 and 20.

Beth surveyed her friends on the number of books they read last week. She drew a line plot to show her data. (Lesson 13.3)

Question 19.
How many friends did she survey?
Answer:
13 friends
Explanation:
Beth surveyed her friends on the number of books they read last week.
The above line plot tells that she surveyed on 13 friends.

Question 20.
How many friends read more than three books last week?
Answer:
4 friends
Explanation:
Beth surveyed her friends on the number of books they read last week.
The above line plot tells that she surveyed on 13 friends.
Among them 4 of her friends read more than three books last week.

Extended Response

The table and the bar graph show the number of books checked out of a library over five days. Some of the bars on the bar graph were incorrectly drawn.

Look at the table and bar graph to answer Exercises 21 to 26.

Question 21.
Complete the bar graph for Tuesday.
Answer:

Explanation:
The table show the number of books checked out of a library over five days.
On Tuesday the number of books checked out are 25.

Question 22.
One bar on the bar graph was incorrectly drawn for one of the days. On which day is it?
Answer: Tuesday
Explanation:
The table show the number of books checked out of a library over five days.
On Friday the number of books checked out are 40.
But the graph drawn in correctly as 35.

Question 23.
Show the correct number of books checked out for that day in the bar graph.
Answer:
40 b00ks
Explanation:

Question 24.
How many books were checked out during that week?
____________ books
Answer:
150 books
Explanation:
Number of books on Monday = 20
Number of books on Tuesday = 25
Number of books on Wednesday = 30
Number of books on Thursday = 35
Number of books on Friday = 40
Total number of books checked that week
20 + 25 + 30 + 35 + 40 = 150

Question 25.
On which day was the number of books checked out twice as many as Monday?
Answer:
Friday
Explanation:
Information given in the table tells,
Number of books checked on Monday are 20
Number of books checked on Friday are 40
So, on Friday the number of books checked out twice as many as Monday.

Question 26.
Look at the number of books checked out from Monday to Friday. What is the pattern?
Answer:
Ascending order
Explanation:
When we observe the number of books checked on each day,
5 books increased on each day,
following ascending order.
Ascending order means to arrange numbers in increasing order, that is, from smallest to largest.

Solve. Show your work.

Question 27.
After a garage sale, Norman makes $105.50. Julie makes $38.75 more than Norman. Lana makes $19.20 less than Julie. How much does Lana make?
Answer:
$163.45
Explanation:
Norman makes $105.50.
Julie makes $38.75 more than Norman.
$105.50 + 38.75 = $144.25
Lana makes $19.20 less than Julie
$144.25 + $ 19.20 = $163.45

Question 28.
Colin uses a wire to make a square. Each side is 6 centimeters long. He then uses the same wire to make a triangle of three equal sides. How long is each side of the triangle?

Answer: 8 cm
Explanation:
Perimeter of Square P = 4 x s = 4 x 6 = 24 cm
Perimeter of the Triangle = 3s
P = 3s = 24
s = \(\frac{24}{3}\) = 8 cm

Solve. Show your work.

Question 29.
Pauline went to a party. She spends 3 hours 25 minutes there. She goes home at 2:15 P.M. What time did she go to the party?

Answer:
10 : 50 AM

Explanation:
time line diagram as shown above
Pauline went to party and she spends 3hr 25 min and she goes home at 2:15 pm
2 : 15 – 3 :25 = 10 :50
Pauline go to party at 10:50 AM
10:50 + 3:25 = 2:15 PM

Solve. Show your work.

Question 30.
A family has two dogs, a husky and a terrier. The husky’s mass is 23 kilograms. If he gains 7 kilograms, his mass will be five times that of the terrier. What is the mass of the terrier?
Answer: 6kg
Explanation:
Husky’s mass is 23 kilograms
Terrier mass is x
23 + 7 = 5x
30 = 5 x
x = \(\frac{30}{5}\)
x= 6 kg