# Maths Wiz Solutions Class 7 Chapter 8

## Maths Wiz Class 7 Solutions Chapter 8 Ratio Proportion and Unitary Method

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Maths Wiz Class 7 Math Book, Chapter 8, Ratio Proportion and Unitary Method. Here students can easily find step by step solutions of all the problems for Ratio Proportion and Unitary Method, Exercise 8A, 8B, 8C, 8D, 8E and 8F Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 8 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.

Ratio Proportion and Unitary Method Exercise 8A Solution :

Question no – (1)

Solution :

(a) – (i) roses to lilies,

Roses : daisies : lilies

= 10 : 8 : 12

= 5 : 4 : 6

(ii) daisies to roses,

Roses : lilies :

= 5 : 6

(iii) lilies to daisies,

6 : 4

= 3 : 2

(iv) roses to all flowers,

(b) Roses : flowers

= 10 : (10 + 8 + 12)

= 10 : 30

= 1 : 3

(i) Class 6 students to class 7 students

= Class 6 : class 7 : class 8 = 30 : 25 : 40

= 6 : 5 : 8

(ii) Class 6 students to total number of students

= Class 6 : class 7

= 6 : 5

(iii) Class 7 students to Class 8 students

= Class 7 : class 8 = 5 : 8

(iv) Class 7 and Class 8 students taken together to class 6 students

= Class 7 + class 8) : class 6

= (25 + 40) : 30

= 65 : 30

= 13 : 6

Question no – (2)

Solution :

(a) My dad is twice as tall as I am.

My dad : me = 2 : 1

(b) My lawn is thrice as large as the lawn in Mr. Shukla’s house.

My lawn : Mr. Shukla’s house

= 3 : 1

(c) 2/5 of the candidates who appeared for the test cleared it.

Passed student : present student

= 2 : 5

Question no – (3)

Solution :

(a) Given, 200 : 300

200 : 300

= 2 : 3

(b) Given, 76 : 57

76 : 57

= 4 : 3

(c) Given, 144 : 168

144 : 168

= 6 : 7

(d) One dozen to one score,

12 : 20

= 3 : 5

(e) Ten thousand to one lakh

10000 : 100000

= 1 : 10

Question no – (3)

Solution :

(a) 15 minutes to 1 hour in the simplest form,

15 minutes : 1 hour

= 15 : 60

= 1 : 4

(c) 150 mm to 75 cm in the simplest form.

150 mm to 75 cm

= 150 mm : 75 cm

= 150 : 750

= 1 : 5

(d) 1 hours to 40 seconds

1 hour : 40 seconds

= 3600 : 40

= 1 : 5

(e) 8 hours to 1 day

8 : 24

= 1 : 3

(f) Rs 120 to Rs 300

₹ 120 : ₹ 300

= 2 : 5

Question no – (4)

Solution :

(a) boys to girls

Boys : girls

= 500 : 350

= 10 : 7

(b) girls to total number of students.

Girls : total students

= 350 (500 + 350)

= 350 : 800

= 7 : 17

(c) teachers to total number of students

Teacher : total students

= 50 : (500 + 350)

= 50 : 850

= 1 : 17

Question no – (6)

Solution :

S1 = 48 km/hour

S2 = 120/2

= 60 km/hour

S1 : S2

= 48 : 60

= 4 : 5

Therefore, the ratio of their speed is 4 : 5

Ratio Proportion and Unitary Method Exercise 8B Solution :

Question no – (1)

Solution :

(a) 2/3, 3/2

= 2 × 2/3 × 2

= 4/6

3 × 3/2 × 3

= 9/6

2/3 < 3/2

(b) 3 : 5, 1 : 2

= 3 × 2/5 × 2

= 6/10

1 × 5/2 × 5

= 5/10

3/5 > 1/2

(c) 7 : 10, 11 : 15

7 × 15/10 × 15

= 150/150

11 × 10/15 × 10

= 110/150

7/10 < 11/15

(d) 4 : 7, 42 : 72

4/7 < 42/72

(e) 1/2 : 1/3, 1/4 : 1/9

= 3/2 : 9/4

= 3 × 4/2 × 4

= 12/8

9 × 2/4 × 2

= 18/8

1/2 : 1/3 < 1/4 : 1/9

(f) 5/18, 13/48

= 5 × 48/18 × 48 : 13 × 18/48 × 18

= 240/848

= 5/18 > 13/48

Question no – (2)

Solution :

My school = boys : girls

= 500 : 300

= 5 : 3

= 5 × 5/3 × 5

My sister’s = boys : girls

= 400 : 250

= 8 : 5

= 8 × 3/5 × 3

= 5 × 5/3 × 5

= 25/15

8 × 3/5 × 3

= 24/15

5/3 > 8/5

Therefore, My school has higher ratio of boys to girls.

Question no – (3)

Solution :

My class = 1/10

= 1 × 36/10 × 36

= 36/360

My sister’s class = 1/36

= 5 × 10/36 × 10

= 50/360

Therefore, My sister class > my class.

Ratio Proportion and Unitary Method Exercise 8C Solution :

Question no – (1)

Solution :

(a) Given, ₹ 84 in the ratio 3 : 4.

Sum of ratio,

= 3 + 4

= 7

1st part

= 3/7 × 84

= 36 Rs.

2nd part

= 4/7 × 84

= 48 Rs.

(b) Sum of ratio,

= 11 + 14

= 25

2m = 200 cm

1st part,

= 11/25 × 200

= 88

2nd part,

= 14/25 × 200

= 112

(c) Given sum is = Rs 4800

Sum of ratio,

= 1 + 2 + 3

= 6

1st part,

= 4800 × 1/6

= 800 Rs

2nd part,

= 4800 × 2/6

= 1600 Rs

3rd part,

= 4800 × 3/6

= 2400 Rs

(d) As we know, 5 litres = 5000 m L

Sum of ratio,

= 11 + 17 + 22

= 50

1st part,

= 11/50 × 5000

= 11000

2nd part,

= 17/50 × 5000

= 17000

3rd part,

= 22/50 × 5000

= 22000

Question no – (2)

Solution :

1st partner : 2nd partner : 3rd partner

= 3 : 4 : 8

Profit = 7000 Rs

Total sum,

= 3 + 4 + 8

= 15 Rs

1st partner,

= 3/15 × 75000

= 15000 Rs

2nd partner,

= 4/15 × 75000

= 20000 Rs

3rd partner,

= 8/15 × 75000

= 4000 Rs

Question no – (3)

Solution :

Sum of total students

= 3 + 2

= 5

Boy : girls,

= 3 : 2

Number of boy,

= 3/5 × 600

= 90

Number of girl,

= 2/5 × 600

= 60

Question no – (4)

Solution :

Sum of total side

= 1 + 1.5 + 2

= 4.5

Let, total side = x

According to question,

1.5/4.5 × x

= 8

Or, 1.5x = 8 × 4.5

Or, x = 8 × 4.5/1.5

= 24

2nd part

= 24 × 1/4.5

= 5.33 m

3rd part

= 24 × 2/4.5

= 10.66 m

Question no – (5)

Solution :

Sum of total metre

= 5 + 7

= 12

1st number,

= 132 × 5/12

= 55

2nd number,

= 132 × 7/12

= 77

Therefore, the required two numbers will be 55 and 77.

Ratio Proportion and Unitary Method Exercise 8D Solution :

Question no – (1)

Solution :

As per the question,

(a) 171 in the ratio 10 : 9.

∴ Increase value,

= 171 × 10/9

= 990

(b) 136 L in the ratio 5 : 4.

∴ Increase value,

= 136 × 5/4

= 170

Question no – (2)

Solution :

According to the question,

(a) 115 in the ratio 4 : 5.

= 115 × 4/5

= 92

(b) 126 kg in the ratio 3 : 4.

= 126 × 3/4

= 94.5

Question no – (4)

Solution :

Given, 45 be increased to become 54

Ratio,

= 54 : 45

= 6 : 5

Question no – (5)

Solution :

Given, 130 be decreased to become 104?

Ratio

= 125 : 100

= 5 : 4

Question no – (6)

Solution :

Given, Multiplying factor decreases 72 to 60?

Ratio,

= 60 : 72

= 5 : 6

Question no – (8)

Solution :

Ratio

= 1 : 10

= 1 : 10

Ratio Proportion and Unitary Method Exercise 8E Solution :

Question no – (1)

Solution :

(a) 25 : 16 and 75 : 48

∴ 25/16 = 25 × 3/16 × 3

= 75/48; 75/48

Therefore, Yes, its proportional.

(b) 16 : 24 and 20 : 30

16/24 = 2/3;

= 20/30 = 2/3

= 2/3 = 2/3

Thus, Yes its proportional.

(c) 15 : 20 and 25 : 30

15/20 = 3/4

20/30 = 2/3

Hence, it is not in proportional

(d) 500 men : 400 men and ₹ 240 : ₹ 920

500/400 = 4/5;

240/920 = 6/23

No, not proportional.

(e) 64 kg : 56 kg and ₹ 80 : ₹ 70

= 64/56 = 8/7;

= 80/70 = 8/7

Yes, it is proportional.

Question no – (2)

Solution :

(a) 14, 7, 18, 9

= Yes, it is a proportion

(b) 12, 16, 6, 8

= Yes, it is a proportion

(c) 4, 7, 9, 17

= No, it is not is proportion

(d) 180, 100, 360, 200

= Yes, it is a proportion

(e) 20, 8, 30, 9

= No, it is not is proportion

(f) 34, 51, 68, 102

= No, it is not is proportion.

Question no – (3)

Solution :

(a) Given, 24, x, 18, 15

24/x = 18/15

Or, x = 24 × 15/18

= 20

Or, x = 20

Hence, here x will be 20

(b) Given, 3, 8, x, 128

Or, 3/8 = x/128

Or, x = 128 × 3/8

Or, x = 16 × 3

= 48

Therefore, here x will be 48

(c) Given, 36, x, 16

36/x = 16

Or, x = 36/16

Or, x = 9/4

Thus, here x will be 9/4

(d) Given, 15, 45, x

Or, 15 = 45/x

Or, x = 45/15 = 3

Or, x = 3

Therefore, here x will be 3

Question no – (4)

Solution :

(a) Given, 10/n = 5/18

Or, n = 18 × 10/5

Or, n = 36

Therefore, the missing term will be 36.

(b) Given, 24/20 = n/15

Or, n = 24 × 15/20

Or, n = 18

Thus, the missing term is 18

(c) Given, 6.4 : n = 20 : 8.5

Or, 6.4/n = 20/8.5

Or, 8.5 × 6.4/20

= 85 × 64/20 × 100

= n = 2.72

Hence, the missing term will be 2.72.

Question no – (5)

Solution :

Given numbers, 9, 105 and 1750

9/105 = 105/1750

And 105/9 = 1750/150

∴ [a/b = b/c]

Question no – (6)

Solution :

Length : width = 5 : 3

Common factor,

= 5 + 3

= 8

Let, total ground = L

According to question,

3/8 × L

= 45

Or, L = 45 × 8/3

= 120 m

Length of ground,

= 120 – 45

= 75 m.

Therefore, the length of ground will be 75 m.

Question no – (7)

Solution :

Income : expenditure

= 9 : 7

Common factor,

= (9 + 7)

= 16

Let, total amount = x

According to question,

9x/16 = 18000

Or, x = 18000 × 16/9

= 32000 Rs

Expenditure,

= 32000 – 18000

= 14000 Rs

Saving,

= 18000 – 14000

= 4000 Rs

Therefore, the savings will be 4000 Rs.

Question no – (8)

Solution :

Let, total distance = x

= Saving : smaller

= 25 : 9

Common factor,

= 15 + 9

= 24

According to question,

= 15/24 × x

= 180

Or, x = 8 × 180/5

= 288

Smaller distance,

= 288 – 180

= 108

Therefore, the smaller distance will be 108 km.

Question no – (9)

Solution :

Let actual length = L

According to question,

= 1/100000 × L

= 4.7

Or, L = 4.7 × 100000

= 4.7 km

Therefore, the length of a road will be 4.7 km.

Question no – (10)

Solution :

Let, the dimensions of room = L, W

According to question,

= 1/250 × L

= 2

Or, L = 250 × 2

= 500 cm

= 5 m

Again, 1/25 × w

= 1.5

W = 1.5 × 25

= 375

= 3.7 m

Therefore, the dimensions of the room will be 5 by 3.7 m.

Question no – (11)

Solution :

Boys : girls

= 5 : 3

Common factor,

= 5 + 3

= 8

Let, total number of student = x

According to question,

3/8 × x

= 900

Or, x = 900 × 8/3

= 2400

Number of boys,

= 2400 – 900

= 1500

Therefore, the number of boys in school will be 1500.

Ratio Proportion and Unitary Method Exercise 8F Solution :

Question no – (1)

Solution :

As per the question,

Cost of 20 m of cloth is ₹ 1600

Cost of 28 m of cloth = ?

Cost of 28 m cloth,

= 28 × 1600/20

= 2240 Rs

Hence, the cost of 28 m of cloth will be 2240 Rs.

Question no – (2)

Solution :

According to the given question,

Cost of 8 kg of potatoes is = ₹ 46.56.

Cost of 15 kg of potatoes = ?

Cost of 15 kg of potatoes,

= 15 × 46.56/8

= 87.30 Rs

Thus, the cost of 15 kg of potatoes is 87.30 Rs.

Question no – (3)

Solution :

As per the question,

12 pens cost = ₹ 120.

Cost of 20 pens = ?

Cost of 20 pens,

= 20 × 120/12

= 200

Therefore, the cost of 20 pens will be 200 Rs.

Question no – (4)

Solution :

From the question we get,

Car travels 540 km in = 30 L of petrol.

Car travels in 20 L = ?

Car travel in 20 L,

= 20 × 540/30

= 360 km

Therefore, the car will travel 360 km in 20 L of petrol.

Question no – (5)

Solution :

As per the given question,

factory produced = 15000 light bulbs

faulty light bulbs = 60

found faulty in = 6000

Faulty bulbs,

= 6000 × 60/15000

= 24 bulbs

Hence, 24 bulbs are likely to be found faulty if 6000 are tested.

Question no – (6)

Solution :

(a) 5 kg for ₹ 35.50 or 3 kg for ₹18.60

= 35.50/5 ; 18.60/3

= 7.1 per kg ; 6.2 per kg

3 kg for 18.60

Thus, 3 kg for 18.60 is a better buy

(b) 10 pens for ₹ 158 or 8 pens for ₹ 136

= 158/10 ; 136/8

= 15.8 ; 17

10 pens for Rs 158

Hence, 10 pens for Rs 158 is better buy.

Question no – (7)

Solution :

According to the question,

15 shares in a company worth = Rs 3950

Shares can bought for ₹ 5530 = ?

Shares can bought,

= 5530 × 15/3950

= 21

Therefore, 21 shares can be bought for Rs 5530.

Question no – (8)

Solution :

From the question we know,

Journey takes 6 hours at 50 km an hour

Journey takes travel at 60 km an hour = ?

Time taken travel at 60 km,

= 50 × 6/60

= 5 hours

Therefore, it will take 5 hour to travel at 60 km an hour.

Question no – (9)

Solution :

(a) What will be the cost of 12 kg of wheat?

Cost of 12 kg,

= 12 × 120/8

= 180.84 Rs

(b) What quantity of wheat can be purchased for ₹ 376.75?

∴ Quantity,

= 376.75 × 8/120.56

= 25 Rs

Question no – (10)

Solution :

Sachin run,

= 60/5

= 12

Rohit run,

= 72/8

= 9

Therefore, Sachin performance was better.

Next Chapter Solution :

Updated: June 17, 2023 — 11:23 am