## ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 6 Ratio and Proportion Ex 6.1

Question 1.

Express the following ratios in simplest form:

Solution:

Question 2.

Find the ratio of each of the following in simplest form:

(i) ₹ 5 to 50 paise

(ii) 3 km to 300 m

(iii) 9 m to 27 cm

(iv) 15 kg to 210 g

(v) 25 minutes to 1.5 hours

(vi) 30 days to 36 hours

Solution:

(i) ₹ 5 to 50 paise

= 500 paise : 50 paise

= 10 : 1 (Dividing by 50)

(ii) 3 km to 300 m

= 3000 m to 300 m

= 10 : 1 (Dividing by 300)

(iii) 9 m to 27 cm

= 9 × 100 cm : 27 cm

= 100 : 3 (Dividing by 9)

(iv) 15 kg to 210 g

= 15 × 1000 g : 210 g

= 15000 : 210

= 500 : 7 (Dividing by 30)

(v) 25 minutes to 1.5 hours

= 25 minutes to \(\frac { 3 }{ 2 }\) × 60

= 25 : 90

= 5 : 18

(vi) 30 days to 36 hours

= 30 × 24 hours to 36 hours

= 720 : 36

= 20 : 1 (Dividing by 36)

Question 3.

If A : B = 3 : 4 and B : C = 8 : 9, then find A : C.

Solution:

A : B = 3 : 4 and B : C = 8 : 9

\(\frac { A }{ B }\) = \(\frac { 3 }{ 4 }\)

Question 4.

If A : B = 5 : 8 and B : C = 18 : 25, then find A : B : C.

Solution:

A : B = 5 : 8 and B : C = 18 : 25

Here, In A : B, B = 8

and In B : C, B = 18

LCM of 8, 18 is 72

Question 5.

If 3A = 2B = 5C, then find A : B : C.

Solution:

Let 3A = 2B = 5C = 1

Question 6.

Out of daily income of ₹ 120, a labourer spends ₹ 90 on food and shelter and saves the rest. Find the ratio of his

(i) spending to income

(ii) saving to income

(iii) saving to spending.

Solution:

Daily income = ₹ 120

Expenditure = ₹ 90

Savings = ₹ 120 – ₹ 90 = ₹ 30

(i) Ratio between spending to income

= 90 : 120

= 3 : 4 (Dividing by 30)

(ii) Ratio between saving to income

= 30 : 120

= 1 : 4 (Dividing by 30)

(iii) Ratio between saving to spending

= 30 : 90

= 1 : 3 (Dividing by 30)

Question 7.

5 grams of an alloy contains 3\(\frac { 3 }{ 4 }\) grams copper and the rest is nickel. Find the ratio by weight of nickel to copper.

Solution:

Total weight of an alloy = 5 gms

3 15

Weight of copper = 3\(\frac { 3 }{ 4 }\) gms = \(\frac { 15 }{ 4 }\) gms

Weight of nickel = Total weight of alloy – weight of copper

Question 8.

A pole of height 3 metres is struck by a speeding car and breaks into two pieces such that the first piece is \(\frac { 1 }{ 2 }\) of the second. Find the length of both pieces.

Solution:

Total height of pole = 3 metres

Let length of 2nd piece = x

Length of 1st piece = \(\frac { 1 }{ 2 }\) x

Ratio of lengths of two parts = \(\frac { 1 }{ 2 }\) x : 1x

Length of 1st part = 1 m

Length of 2nd part = 2m

Question 9.

Heights of Anshul and Dhruv are 1.04 m and 78 cm respectively. Divide 35 sweets between them in the ratio of their heights.

Solution:

Height of Anshul : Height of Dhruv

1.4 m : 78 cm

(1.04 × 100) cm : 78 cm

= 104 : 78

= \(\frac { 104 }{ 78 }\) (Dividing both by 2)

= \(\frac { 52 }{ 39 }\) (Dividing boht by 13)

= \(\frac { 4 }{ 3 }\)

= 4 : 3

Ratio of heights of Anshul and Dhruv is 4 : 3

Thus, we are to divide 35 sweets in the ratio 4 : 3

Sum of the terms of the ratios = 4 + 3 = 7

Share of Anshul = \(\frac { 4 }{ 7 }\) of 35 sweets

= \(\frac { 4 }{ 7 }\) × 35

= 20 sweets

Share of Dhruv = \(\frac { 3 }{ 7 }\) of 35 sweets

= \(\frac { 3 }{ 7 }\) × 35

= 15 sweets

Question 10.

₹ 180 are to be divided among three children in the ratio \(\frac { 1 }{ 3 } :\frac { 1 }{ 4 } :\frac { 1 }{ 6 }\) Find the share of each child.

Solution:

First we will simplify the given ratio

Given ratio \(\frac { 1 }{ 3 } :\frac { 1 }{ 4 } :\frac { 1 }{ 6 }\)

Taking L.C.M. of 3, 4 and 6

L.C.M. of 3, 4 and 6 = 12

Question 11.

A natural number has been divided into two parts in the ratio 7 : 11. If the difference between the two parts is 20, find the number and the two parts.

Solution:

Let the first part = 7x

Second part = 11x

According to given statement,

11x – 7x = 20

⇒ 4x = 20

⇒ x = 5

First part = 7x = 7 × 5 = 35

Second part = 11x = 11 × 5 = 55

and number will be 35 + 55 = 90

Question 12.

A certain sum of money has been divided into two parts in the ratio 9 : 13. If the second part is ₹ 260, find the total amount.

Solution:

Let the total amount = ₹ x

The amount has been divided into two parts in the ratio 9 : 13.

Sum of the terms of the ratio = 9 + 13 = 22

First part = \(\frac { 9 }{ 22 }\) of total amount

Second part = \(\frac { 13 }{ 22 }\) of total amount

According to given statement

Question 13.

The ratio of the present ages of Anjali and Ashu is 2 : 3. Five years hence, the ratio of theif ages will be 3 : 4. Find their present ages.

Solution:

Ratio of present ages of Anjali and Ashu = 2 : 3

Let age of Anjali = 2x

and age of Ashu = 3x

5 years hence,

Age of Anjali = 2x + 5

and age of Ashu = 3x + 5

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

9x + 15 = 8x + 20

9x – 8x = 20 – 15

x = 5

Present age of Anjali = 2x = 2 × 5 = 10 years

and age of Ashu = 3x = 3 × 5 = 15 years

Question 14.

The present ages of A and B are in the ratio 5 : 6. Three years ago, their ages were in the ratio 4 : 5. find their present ages.

Solution:

Ratio of the present age of A and B = 5 : 6

Let age of A = 5x

and age of b = 6x

3 years ago,

Age of A was = 5x – 3

and age of B was = 6x – 3

\(\frac { 5x-3 }{ 6x-3 }\) = \(\frac { 4 }{ 5 }\)

⇒ 25x – 15 = 24x – 12

⇒ 25x – 24x = -12 + 15

⇒ x = 3

Present age of A = 5x = 5 × 3 = 15 years

and age of B = 6x = 6 × 3 = 18 years

Question 15.

Two numbers are in the ratio 5 : 6. When 2 is added to first and 3 is added to second, they are in the ratio 4 : 5. Find the numbers.

Solution:

Ratio in two numbers = 5 : 6

Let first number = 5x

Then second number = 6x

Adding 2 in the first and 3 in the second

A = 5x + 2

B = 6x + 3

\(\frac { 5x+2 }{ 6x+3 }\) = \(\frac { 4 }{ 5 }\)

25x + 10 = 24x + 12

25x – 24x = 12- 10

x = 2

First number = 5x = 5 × 2 = 10

and second = 6x = 6 × 2 = 12

Question 16.

The ratio of number of boys to the number of girls in a school of 1430 students is 7 : 6. If 26 new girls are admitted in the school, find how many new boys may be admitted so that the ratio of number of boys to the number of girls may change to 8 : 7.

Solution:

Number of students = 1430

Ratio in number of boys and girls = 7 : 6

Let number of boys = 7x and of girls = 6x

7x + 6x = 1430

⇒ 13x = 1430

⇒ x = 110

Number of boys = 7x = 7 × 110 = 770

and number of girls = 6x = 6 × 110 = 660

Now adding 26 new girls, the number of girls will be = 660 + 26 = 686

Let new boys be added = y

The number of boys = 770 + y

Now new ratio = 8 : 7

\(\frac { 770+y }{ 686 }\) = \(\frac { 8 }{ 7 }\)

5390 + 7y = 5488

7y = 5488 – 5390 = 98

y = 14

Number of new boys admitted = 14

Question 17.

Which ratio is greater?

(i) 5 : 6 or 6 : 7

(ii) 13 : 24 or 17 : 32

Solution:

(i) 5 : 6 or 6 : 7

5 : 6 = \(\frac { 5 }{ 6 }\) and 6 : 7 = \(\frac { 6 }{ 7 }\)

Converting them into equivalent fraction by taking L.C.M. of 6 and 7 = 42