## ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 6 Ratio and Proportion Objective Type Questions

**Mental Maths**

Question 1.

Fill in the blanks:

(i) The simplest form of the ratio \(\frac { 1 }{ 6 }\) : \(\frac { 1 }{ 4 }\) is ………

(ii) 75 cm : 1.25 m = ……..

(iii) If two ratios are equivalent, then the four quantities are said to be in ………

(iv) If 8, x, 48 and 18 are in proportion then the value of x is ………

(v) If the cost of 10 pencils is ₹ 15, then the cost of 6 pencils is ……..

(vi) If a cyclist is travelling at a speed of 15 km/h, then the distance covered by him in 20 minutes is ……..

Solution:

(i) The simplest form of the ratio \(\frac { 1 }{ 6 }\) : \(\frac { 1 }{ 4 }\) is 2 : 3.

(ii) 75 cm : 1.25 m = 3 : 5

(iii) If two ratios are equivalent,

then the four quantities are said to be in proportion.

(iv) If 8, x, 48 and 18 are in proportion then the value of x is

8 : x :: 48 : 18

x = 3

(v) If the cost of 10 pencils is ₹ 15, then the cost of 6 pencils is

Cost of 10 pencils = ₹ 15

Let cot of 6 pencils = ₹ x

10 : 6 15 : x

10x = 6 × 5

x = 3

Cost of 6 pencils = ₹ 3

(vi) If a cyclist is travelling at a speed of 15 km/ h,

then the distance covered by him in 20 minutes is ……

Speed of a cyclist = 15 km/h

Distance travelled in 20 minutes = \(\frac { 20 }{ 60 }\) × 15 km = 5 km

Question 2.

State whether the following statements are true (T) or false (F):

(i) A ratio is always greater than 1.

(ii) Ratio of half an hour to 20 seconds is 30 : 20.

(iii) The ratio 5 : 7 is greater than the ratio 5 : 6.

(iv) If the numbers 3, 5, 12 and x are in proportion then the value of x is 20.

(v) The ratio 3 : 2 and 4 : 5 are equivalent.

Solution:

(i) A ratio is always greater than 1. (False)

Correct:

Ratio can be less than 1 or equal to 1.

It is not necessary that it is greater than 1.

(ii) Ratio of half an hour to 20 seconds is 30 : 20. (False)

Correct:

= 30 × 60 : 20 = 1800 : 20 = 90 : 1

(iii) The ratio 5 : 7 is greater than the ratio 5 : 6. (False)

Correct:

\(\frac { 5 }{ 7 }\), \(\frac { 5 }{ 6 }\)

\(\frac { 30:35 }{ 42 }\)

30 < 35

5 : 7 is not greater than the ratio 5 : 6

(iv) If the numbers 3, 5, 12 and x are in proportion

then the value of x is 20. (True)

3, 5, 12 and x are in proportion then

x × 3 = 5 × 12

x = 20

(v) The ratio 3 : 2 and 4 : 5 are equivalent. (False)

Correct:

\(\frac { 3 }{ 2 }\) ≠ \(\frac { 4 }{ 5 }\) as 15 ≠ 8

**Multiple Choice Questions**

Choose the correct answer from the given four options (3 to 14):

Question 3.

A ratio equivalent to 6 : 10 is

(a) 3 : 4

(b) 18 : 30

(c) 12 : 40

(d) 5 : 3

Solution:

6 : 10 = 3 : 5

18 : 30 = 3 : 5

6 : 40 is equivalent to 18 : 30 (b)

Question 4.

A ratio equivalent to the ratio \(\frac { 2 }{ 3 }\) : \(\frac { 3 }{ 4 }\) is

(a) 4 : 6

(b) 8 : 9

(c) 6 : 8

(d) 9 : 8

Solution:

\(\frac { 2 }{ 3 }\) : \(\frac { 3 }{ 4 }\)

\(\frac { 8:9 }{ 12 }\)

= 8 : 9 (b)

Question 5.

The ratio of 75 mL to 3 litres is

(a) 25 : 1

(b) 40 : 1

(c) 1 : 40

(d) 3 : 200

Solution:

75 mL to 3 litres

= 75 : 3000

= 1 : 40 (c)

Question 6.

The ratio of the number of sides of a rectangle to the number of edges of a cuboid is

(a) 1 : 2

(b) 1 : 3

(c) 2 : 3

(d) none of these

Solution:

Ratio of number of sides of a rectangle

to the number of edges of a cuboid = 4 : 12 = 1 : 3 (b)

Question 7.

In a class of 35 students, the number of girls is 20. The ratio of number of boys to the number of girls in the class is

(a) 3 : 4

(b) 4 : 3

(c) 7 : 4

(d) 7 : 3

Solution:

Total number of students = 35

No. of girls = 20

No. of boys = 35 – 20 = 15

Ratio in boys and girls = 15 : 20 = 3 : 4 (a)

Question 8.

The ratio of number of girls to the number of boys in a class is 6 : 7. If there are 21 boys in the class, then the number of girls in the class is

(a) 39

(b) 24

(c) 18

(d) 13

Solution:

Girls : boys = 6 : 7

No. of boys = 21

Let no. of girls = x

6 : 7 :: x : 21

7 × x = 6 × 21

x = 18

No. of boys = 18 (c)

Question 9.

Two numbers are in the ratio 3 : 5. If the sum of the numbers is 144, then the largest number is

(a) 48

(b) 54

(c) 72

(d) 90

Solution:

Ratio between two numbers = 3 : 5

Sum of numbers 144

Then larger number = \(\frac { 144 }{ 3+5 }\) × 5

= \(\frac { 144 }{ 8 }\) × 5 = 90 (d)

Question 10.

If x, 12, 8 and 32 are in proportion, then x is

(a) 6

(b) 4

(c) 3

(d) 2

Solution:

x, 12, 8 and 32 are in proportion, then

x × 32 = 12 × 8

x = 3 (c)

Question 11.

If 3, 12 and x are in continued proportion, then x is

(a) 4

(b) 6

(c) 16

(d) 48

Solution:

3, 12 and x are in continued proportion, then

3 : 12 :: 12 : x

3x = 12 × 12

x = 48 (d)

Question 12.

If the weight of 5 bags of sugar is 27 kg, then the weight of one bag of sugar is

(a) 5.4 kg

(b) 5.2 kg

(c) 5.4 kg

(d) 5.6 kg

Solution:

Weight of 5 bags of sugar = 27 kg

Weight of one bag of sugar

= 5 : 1 :: 27 : x

5x = 1 × 27

x = \(\frac { 27 }{ 5 }\) = 5.4 kg (c)

Question 13.

Sonali bought one dozen notebooks for ₹ 66. What did she pay for one notebook?

(a) ₹ 6.50

(b) ₹ 6.60

(c) ₹ 5.60

(d) ₹ 5.50

Solution:

Cost of 12 books = ₹ 66

Let cost of one book = x

12 : 1 = 66 : x

12 × x = 1 × 66

x = 5.5

Cost of 1 book = ₹ 5.50 (d)

Question 14.

The speed of 90 km/h is equal to

(a) 10 m/sec

(b) 18 m/sec

(c) 25 m/sec

(d) none of these

Solution:

Speed of 90 km/h = \(\frac { 90\times 5 }{ 18 }\) m/sec

= 25 m/sec (c)

**Value Based Questions**

Question 1.

Sudhanshu divided his property into two parts in the ratio 8 : 5. If the first part is ₹ 1,60,000 and second part is donated to an orphanage, find the amount donated to the orphanage. What values are being promoted?

Solution:

Ratio in two parts of a property = 8 : 5

First part = ₹ 1,60,000

Second part = \(\frac { 160000\times 5 }{ 8 }\) = ₹ 10,00,000

It is a good to donate the needy people and support them.

**Higher Order Thinking Skills (HOTS)**

Question 1.

Present ages of Rohit and Mayank are in the ratio 11 : 8. 8 years hence the ratio of their ages will be 5 : 4. Find their present ages.

Solution:

Ratio in the present ages of Rohit and Mayank = 11 : 8

Let age of Rohit = 11x, then that of Mayank = 8x

8 years hence, their ages will be

Age of Rohit = 11x + 8 years

and age of Mayank = 8x + 8 years

and the ratio of their ages after 8 years = 5 : 4

\(\frac { 11x+8 }{ 8x+8 }\) = \(\frac { 5 }{ 4 }\)

⇒ 44x + 32 = 40x + 40

⇒ 44x – 40x = 40 – 32

⇒ 4x = 8

⇒ x = 2

Present age of Rohit = 11x = 11 × 2 = 22 years

and age of Mayank = 8x = 8 × 2 = 16 years

Question 2.

Ratio of length and breadth of a rectangle is 3 : 2. If the length of the rectangle is 5 m more than the breadth, find the perimeter of the rectangle.

Solution

Ratio in length and breadth of a rectangle = 3 : 2

Let length = 3x m and Breadth = 2x

Also, l of rectangle is 5 m more than the breadth

i.e. 3x = 2x + 5

⇒ 3x – 2x = 5

⇒ x = 5

Length = 3x = 3 × 5 = 15m

Breadth = 2x = 2 × 5 = 10m

Perimeter = 2(Length + Breadth) = 2 (15 + 10) m = 2 × 25 = 50 m