Class 8 Maths Solution | Lesson 4 Ratio and Proportion | Curriculum Development Centre (CDC)

Lesson 4 Ratio and Proportion

Exercise 4.2

1. Check and write whether the following numbers are proportional:

(a) 5, 8, 10, 15

Not proportional. Cross-product: 75 ≠ 80.

(b) 3, 5, 6, 10

Proportional. Cross-product: 30 = 30.

(c) 1kg, 4kg, 6kg, 10kg

Not proportional. Cross-product: 10 ≠ 24.

(d) 5 cm, 8 cm, 10 cm, 16 cm

Proportional. Cross-product: 80 = 80.

(e) 5m, 3m, 25m, 15m

Proportional. Cross-product: 75 = 75.

(f) 3ft, 8ft, 12ft, 22ft

Not proportional. Cross-product: 66 ≠ 96.

1. Find the unknown terms if the following numbers are proportional:

(a) a, 3, 3, 9

Solution:

3/a = 9/3

or, a × 9 = 3 × 3

or, 9a = 9

or, a = 9/9

∴ a = 1 Ans

(b) 3, x, 6, 8

Solution:

3/x = 6/8

or, 6x = 3 × 8

or, 6x = 24

or, x = 24/6

∴ x = 4 Ans

(c) 2, 5, 8, d

Solution:

2/5 = 8/d

or, 2d = 5 × 8

or, 2d = 40

or, d = 40/2

∴ d = 20 Ans

(d) x, 2, 6, 4

Solution:

x/2 = 6/4

or, 4x = 6 × 2

or, 4x = 12

or, x = 12/4

∴ x = 3 Ans

(e) 16, 4, 4, y

Solution:

16/4 = 4/y

or, 16y = 4 × 4

or, 16y = 16

or, y = 16/16

∴ y = 1 Ans

(f) 7, 9, z, 18

Solution:

7/9 = z/18

or, 9z = 7 × 18

or, 9z = 126

or, z = 126/9

∴ z = 14 Ans

3. Find the value of x:

Solution:

(a) x:5 = 10:25

25x = 50

or, x = 50/25

∴ x = 2 Ans

(b) 3:7 = 21:x

Solution:

or, 3x = 7 × 21

or, 3x = 147

or, x = 147/3

∴ x = 49 Ans

(c) 10:x = 2:11

Solution:

10 × 11 = 2x

or, 110 = 2x

or, x = 110/2

∴ x = 55 Ans

(d) 25:15 = x:3

Solution:

25 × 3 = 15x

or, 75 = 15x

or, x = 75/15

∴ x = 5 Ans

4. What should be added or subtracted to make the ratios:

(a) What should be added to 12 and 21 to make the ratio 5:8?

Solution:

Let the number to be added be x.

(12 + x) : (21 + x) = 5 : 8

or, 8(12 + x) = 5(21 + x)

or, 96 + 8x = 105 + 5x

or, 8x - 5x = 105 - 96

or, 3x = 9

∴ x = 3

So, 3 should be added.

(b) What should be added to 15 and 25 to make the ratio 2:3?

Solution:

Let the number to be added be x.

(15 + x) : (25 + x) = 2 : 3

or, 3(15 + x) = 2(25 + x)

or, 45 + 3x = 50 + 2x

or, 3x - 2x = 50 - 45

or, x = 5

So, 5 should be added.

(c) What should be subtracted from 24 and 30 to make the ratio 3:4?

Solution:

Let the number to be subtracted be x.

(24 - x) : (30 - x) = 3 : 4

or, 4(24 - x) = 3(30 - x)

or, 96 - 4x = 90 - 3x

or, -4x + 3x = 90 - 96

or, -x = -6

∴ x = 6

So, 6 should be subtracted.

5. Solve the following ratio problems:

(a) If 12 notebooks are available in Rs. 600, how many notebooks are available in Rs. 900?

Solution:

12 : 600 = x : 900

or, 12 × 900 = 600x

or, x = (12 × 900) / 600

∴ x = 18

So, 18 notebooks are available.

(b) If 5 pens are available in Rs. 50, how many pens are available in Rs. 240?

Solution:

5 : 50 = x : 240

or, 5 × 240 = 50x

or, x = (5 × 240) / 50

∴ x = 24

So, 24 pens are available.

6. Solve the following ratio word problems:

(a) The ratio of the students using pencils and pens in Rastriya Basic School is 10:11. If there are 110 students using pencils, find the number of students using pens.

Solution:

10 : 11 = 110 : y

or, 10y = 11 × 110

or, 10y = 1210

or, y = 1210 / 10

∴ y = 121

So, 121 students are using pens.

(b) The ratio of the students who have lunch at school and bring food from home in Mahendra Secondary School is 3:2. If the number of students who have lunch at school is 321, find the number of students who bring their lunch from home.

Solution:

3 : 2 = 321 : y

or, 3y = 2 × 321

or, 3y = 642

or, y = 642 / 3

∴ y = 214

So, 214 students bring their lunch from home.

(c) Reena and Meena are two sisters. Reena is in class 10 and Meena is in class 8. The ratio of their expenditure to buy books is 3:2. If Meena paid Rs. 824, how much did Reena pay?

Solution:

3 : 2 = R : 824

or, 3 × 824 = 2 × R

or, 2472 = 2R

or, R = 2472 / 2

∴ R = 1236

So, Reena paid Rs. 1236.

(d) The ratio of the milk and sugar in a type of sweet is 5:2. If the milk is 750 gm, what is the part of sugar?

Solution:

5 : 2 = 750 : y

or, 5y = 2 × 750

or, 5y = 1500

or, y = 1500 / 5

∴ y = 300

So, 300 gm of sugar is used.

(e) The ratio of gravity of moon and earth is 1:6. Find the weight of an object on the moon whose weight is 90N on earth.

Solution:

1 : 6 = W : 90

or, 1 × 90 = 6 × W

or, W = 90 / 6

∴ W = 15

So, the weight on the moon is 15N.