Ratio and Proportion Class- 7th RS Aggarwal Exe-7 A Goyal Brothers ICSE Maths Solution

Ratio and Proportion Class- 7th RS Aggarwal Exe-7 A Goyal Brothers ICSE Math Solution . We provide step by step Solutions of lesson-7 Ratio and Proportion for ICSE Class-7 Foundation RS Aggarwal Mathematics of Goyal Brothers Prakashan . Our Solutions contain all type Questions of Exe-7 A to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-7 Mathematics we would learn this article What is a ratio and a proportion? What is ratio and proportion Class 7?  What are the 3 types of proportion? What is the formula of ratio?

Ratio and Proportion Class- 7th RS Aggarwal Exe-7 A Goyal Brothers ICSE Maths Solution

Board	ICSE
Publications	Goyal brothers Prakashan
Subject	Maths
Class	7th
Chapter-7	Ratio and Proportion
Writer	RS Aggrawal
Book Name	Foundation
Topics	Solution of Exe-7 A
Academic Session	2023 – 2024

What is a ratio and a proportion?

A ratio is an ordered pair of numbers a and b, written a / b where b does not equal 0. A proportion is an equation in which two ratios are set equal to each other.

What are the 3 types of proportion?

Exercise – 7 A

Ratio and Proportion Class- 7th RS Aggarwal Goyal Brothers ICSE Maths Solution

1. Express each of the following rations in simplest form :

(i) 20 : 35

Solution: 20 : 35

20 : 35 = (20/35)

= (20 ÷ 5)/(35 ÷ 5)

= (4/7)

= 4 : 7

(ii) 95 : 57

Solution: 95 : 57

95 : 57 = (95/57)

= (95 ÷ 19)/(57 ÷ 19)

= (5/3)

= 5 : 3

(iii) 2.1 : 1.2

Solution: 2.1 : 1.2

2.1 : 1.2 = (2.1/1.2) = (21/12)

= (21 ÷ 3)/(12 ÷ 3)

= (7/4)

= 7 : 4

(iv) 3(1/4) : 6(1/2)

Solution: 3(1/4) : 6(1/2)

3(1/4) : 6(1/2) = (13/4) ÷ (13/2)

= (13/4)/(13/2)

= (13/4) × (2/13)

= (2/4)

= (2 ÷ 2)/(4 ÷ 2)

= (1/2)

= 1 : 2

(v) 6a : 9b

Solution: 6a : 9b

= 6a : 9b = (6a/9b)

= (6a ÷ 3)/(9b ÷ 3)

= (2a/3b)

= 2a : 3b

(vi) 8ab² : 6a²b

Solution: 8ab² : 6a²b

8ab² : 6a²b = (8ab²/6a²b)

= (8ab² ÷ 2ab)/(6a²b ÷ 2ab)

= (4b/3a)

= 4b : 3a

2. Express each of the following rations in simples form :

(i) (1/8) : (1/12)

Solution: (1/8) : (1/12)

L.C.M. of 8 and 12

= 24

= (1/8) × 24 : (1/12) × 24

= 3 : 2

(ii) (1/9) : (1/6)

Solution: (1/9) : (1/6)

L.C.M. of 9 and 6

= 18

= (1/9) × 18 : (1/6) × 18

= 2 : 3

(iii) (2/3) : (5/6)

Solution: (2/3) : (5/6)

L.C.M. of 3 and 6

= 6

= (2/3) × 6 : (5/6) × 6

= 4 : 5

(iv) (1/4) : (1/6) : (1/8)

Solution: (1/4) : (1/6) : (1/8)

L.C.M. of 4, 6 and 8

= 24

= (1/4) × 24 : (1/6) × 24 : (1/8) × 24

= 6 : 4 : 3

(v) 1(2/3) : 2(1/2) : 1(3/4)

Solution: 1(2/3) : 2(1/2) : 1(3/4)

= (5/3) : (5/2) : (7/4)

L.C.M. of 3, 2 and 4

= 12

= (5/3) × 12 : (5/2) × 12 : (7/4) × 12

= 20 : 30: 21

(vi) 2(1/3) : 3(1/4) : 1(1/6)

Solution: 2(1/3) : 3(1/4) : 1(1/6)

= (7/3) : (13/4) : (7/6)

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

= 12

= (7/3) × 12 : (13/4) × 12 : (7/6) × 12

= 28 : 39 : 14

3. Express each of the following rations in simplest form :

(i) 35 paise : ₹ 1

Solution: 35 paise : ₹ 1

= 35 paise : 100 paise

= 35 : 100

= 35 ÷ 5 : 100 ÷ 5

= 7 : 20

(ii) 1 kg : 375 gm

Solution: 1 kg : 375 gm

= (1 kg)/(375gm)

= (1 × 1000 gm)/(375gm)

= (1000 gm ÷ 125)/(375gm ÷ 125)

= 8 : 3

(iii) 60 cm : 1m

Solution: 60 cm : 1m

= (60 cm/1m)

= (60 cm ÷ 20)/(60 cm ÷ 20)

= (3/5)

= 3 : 5

(iv) 400 m : 1 km

Solution: 400 m : 1 km

= (400 m)/1 km = (400 m)/(1000 m)

= (4 ÷ 2)/(10 ÷ 2)

= (2/5)

= 2 : 5

(v) 8 months : 1(2/3) years

Solution: 8 months : 1(2/3) years

= 8 months : (5/3) years

= (8 months)/((5/3) years)

= (8 months)/((5/3) × 12 months

= (8/20) = (8 ÷ 4)/(20 ÷ 4)

= (2/5)

= 2 : 5

(vi) 6(2/3) hours : 1 day

Solution: 6(2/3) hours : 1 day

= 20 hours : 1 day

= (3/1 day)

= (20 hours)/(3 × 24 hours)

= (20 hours)/(72 hours)

= (20 ÷ 4)/(72 ÷ 4)

= (5/18)

= 5 : 18

4. The angles of a triangle are in the ration 5 : 6 : 7. Find each angle.

Solution: We know that the sum of the angles of a straight line = 180°

sum of ratio = 5 + 6 + 7

= 18

1st angle = (5/18) × 180°

= 5 × 10

= 50°

2nd angle = (6/18) × 180°

= 6 × 10

= 60°

3rd angle = (7/18) × 180°

= 7 × 10

= 70°

5. A line segment of length 84 cm has been divided into two parts in the ratio 4 : 3. Find the length of each part.

Solution: Length of line of segment = 84 cm

sum of ratio = 4 + 3

= 7

1st part (4/7) × 48 cm

= 4 × 12 cm

= 48 cm

2nd part (3/7) × 84 cm

= 3 × 12 cm

= 36 cm

6. Divide ₹ 3900 between Kunal and Kirti in the ratio 15 : 11.

Solution: Total amount = ₹ 3900

Ratio of the money between kunal and kirti = 15 : 11

sum of ratio’s = 15 + 11

= 26

Kunal’s share = Rs. (3900 × 15)/26

= Rs. 2250

and Kirti’s share = Rs. (3900 × 11)/26

= Rs. 1650

7. Divide ₹ 16250 among A, B, C in the ratio 5 : 7 : 13.

Solution: Sum of ratio = 5 + 7 + 13

= 25

A share’s = Rs. (5/25) × 16250

= Rs. 5 × 650

= Rs. 3250

B share’s = Rs. (7/25) × 16250

= Rs. 7 × 650

= Rs. 4550

C share’s = Rs. (13/25) × 16250

= Rs. 13 × 650

= Rs. 8450

8. Divide 35 sweets between Sudha and Tanvy in the ratio (1/2) : (1/3).

Solution: Ratio (1/2) : (1/3)

L.C.M. of 2 and 3 = 6

ratio = (1/2) × 6 : (1/3) × 6

= 3 : 2

Sum of ratio = 3 + 2 = 5

Sudha’s share = (3/5) × 35

= 3 × 7

= 21 sweets

Tanvy’s share = (2/5) × 35

= 2 × 7

= 14 sweets

9. Divide 468 g of rice into three heaps containing  the quantities in the ratio (1/6) : (1/8) : (1/12).

[Hint : (1/6) : (1/8) : (1/12) = (24/6) : (24/8) : (24/12) = 4 : 3 : 2]

Solution: Ratio = (1/6) : (1/8) : (1/12)

L.C.M. of 6, 8 and 12 = 24

ratio = (1/6) × 24 : (1/8) × 24 : (1/12) × 24

= 4 : 3 : 2

Sum of ratio = 4 + 3 + 2 = 9

1st heaps contains = (4/9) × 468

= 4 × 52

= 208 gm

2nd heaps contains = (3/9) × 468

= 3 × 52

= 156 gm

3rd heaps contains = (2/9) × 468

= 2 × 52

= 104 gm

10. Divide ₹715 among A, B, C in such a way that B gets three times as much as A gets and C gets half of as much as B gets.

[Hint : Suppose A gets ₹ x, then, B gets ₹ 3x and C gets ₹(3x/2).

∴ A : B : C = x : 3x : (3x/2) = 2 : 6 : 3]

Solution: B = 3A

⇒ (1/3) = (A/B)

⇒ A : B = 1 : 3

C = (1/2)B

⇒ (2/1) = (B/C)

⇒ B : C = 2 : 1

A : B : C

1  : 3  :

2  : 1

2 : 6 : 3

sum of ratio = 2 + 6 + 3 = 11

A share’s = (2/11) × 715

= 2 × 65

= Rs. 130

B share’s = (6/11) × 715

= 6 × 65

= Rs. 390

C share’s = (3/11) × 715

= 3 × 65

= Rs. 195

11. Divide ₹ 760 among A, B, C such that A gets (5/6) of what B gets and the ratio between the shares of B and C is 3 : 4.

[Hint : Let B’s share be ₹ 3x and C’s share be ₹ 4x.

Then, A’s share = ₹[(5/6) × 3x] = ₹(5x/2).

∴ A : B : C = (5x/2) : 3x : 4x = 5 : 6 : 8].

Solution: Let the there ratio of A , B and C be a, b, c then

a = (5/6) – b

(a/b) = (5/6)

⇒ a : b, 5 : 6

and b : c = 3 : 4

a : b : c

5 : 6

3 : 4

⇒ 15 : 18 : 24

⇒ 5 : 6 : 8

Sum of ratio = 5 + 6 + 8 = 19

A share = (5/19) × 760

= 5 × 40

= Rs. 200

B share = (6/19) × 760

= 6 × 40

= Rs. 240

C share = (8/19) × 760

= 8 × 40

= Rs. 320

12. The boys and girls in a school are in the ratio 8 : 3. If the number of girls is 405, how many boys are there in the school?

Solution: Let x be the no. of boy’s

x : 405 = 8 : 3

⇒ (x/405) = (8/3)

⇒ x = (8/3) × 405

⇒ x = 8 × 135

⇒ x = 1080

Number of boy’s in the school = 1080

13. An alloy contains copper and zinc in the ratio 7 : 3. If it contains 12.6 g of copper, how much does this alloy weigh?

Solution: Let weight of zinc = x

∴ 12.6 : x = 7 : 3

⇒ (12.6/x) = (7/3)

⇒ 7x = 3 × 12.6

⇒ x = (3 × 12.6)/7

⇒ x = (3 × 1.8)

⇒ x = 54

∴ weight of zinc = 5.4 gm

∴ weight of alloy = 12.6 + 5.4 = 18 gm

14. Reena weighed 63 kg. She reduced her weight in the ratio 9 : 8. Find her new weight.

Solution: Weight of reena = 63 kg

or reducing the weight, the ratio will be 9 : 8

∴ Reduced weight = 63 × (8/9)

= 56 kg

15. Two numbers are in the ratio 9 : 13 and their sum is 176. Find the numbers.

Solution: Ratio in two numbers = 9 : 13

Let first number = 9x

then second number = 13x

But their sum = 176

∴ 9x + 13x = 176

⇒ 22x = 176

⇒ x = (176/22)

⇒ x = 8

∴ First number = 9x = 9 × 8 = 72

second number = 13x = 13 × 8 = 104

Hence number are 72 and 1o4.

16. Two numbers are in the ratio 5 : 8 and their difference is 12. Find the numbers.

Solution: Ratio in two numbers = 5 : 8

Let First number = 5x

then second number = 8x

and their deference

∴ 8x – 5x = 12

⇒ 3x = 12

⇒ x = (12/3)

⇒ x = 4

⇒ First number = 5x = 5 × 4 = 20

Now second number = 8x = 8 × 4 = 32

Hence number are 20 and 32.

17. If a : b = 3 : 2 and b : c = 4 : 5, find a : c.

Solution: a : b = 3 : 2 and b : c = 4 : 5

⇒ (a/b) = (3/2)    and    (b/c) = (4/5)

⇒ a : c = (a/c)

= (a/b) × (b/c)

= (3/2) × (4/5)

= (6/5)

= 6 : 5

18. If a : b = 5 : 6 and b : c = 2.8 : 3.5, find a : c.

Solution: a : b = 5 : 6

(a/b) = (5/6)

b : c = 2.8 : 3.5

(b/c) = (2.8/3.5)

a : c = (a/c)

= (a/b) × (b/c)

= (5/6) × (2.8/3.5)

= (5/6) × (28/35)

= (4/6) = (2/3) = 2 : 3

⇒ a : c = 2 : 3

19. If p : q = 1(1/3) : 1(1/2) and q : r = (1/2) : (1/3), find p : r.

[Hint : ∴ p : q = (4/3) : (3/2) = 8 : 9;        q : r = (1/2) : (1/3) = 3 :2         p : r = (p/r) = (p/q) × (q/r)]

Solution: p : q = 1(1/3) : 1(1/2)

p : q = (4/3) : (3/2)

p : q = (4/3)/(3/2)

p : q = (4 × 2)/(3 × 3)

⇒ p : q = (8/9)

⇒ p : q = 8 : 9

q : r = (1/2) : (1/3)

q : r = (1/2)/(1/3)

q : r = (1 × 3)/(1 × 2)

⇒ q : r = (3/2)

⇒ q : r = 3 : 2

p : r = (p/r) = (p/q) × (q/r)

p : r = (8/9) × (3/2)

p : r = (4/3)

p : r = 4 : 3

20. If a : b = 3 : 4 and b : c = 8 : 9, find a : b : c.

Solution:  a      :     b      :      c

3      :     4

8      :      9

⇒ 3 × 8 : 8 × 4 : 4 × 9

⇒ 24 : 32 : 36

⇒ 6 : 8 : 9

21. If l : m = 2(1/2) : 1(2/3) and m : n = 1(1/4) : 3(1/2), find l : m : n.

[Hint : l : m = (5/2) : (5/3) = 3 : 2 = 15 : 10, m : n = (5/4) : (7/2) = 5 : 14 = 10 :28]

Solution: l : m : n

2(1/2) : 1(2/3)

1(1/4) : 3(1/2)

⇒ (5/2) × (5/4) : (5/4) × (5/3) : (5/3) × (7/2)

⇒ (25/8) : (25/12) : (35/6)

⇒ (5/8) : (5/12) : (7/6)

L.C.M. of 8, 12 and 6 = 24

⇒ (5/8) × 24 : (5/12) × 24 : (7/6) × 24

⇒ 15 : 10 : 28

22. Which ratio is greater ?

(i) (3 : 4) or (5 : 7)

Solution: (3 : 4) or (5 : 7)

L.C.M. of 4 and 7 = 28

⇒ (3/4) = (3 × 7)/(4 × 7)

= (21/28)

⇒ (5/7) = (5 × 4)/(7 × 4)

= (20/28)

Now its clear that (21/28) is greater than (20/28).

3 : 4 is greater than 5 : 7

(ii) (11 : 21) 0r (19 : 28)

Solution: (11 : 21) 0r (19 : 28)

L.C.M. of 21 and 28 = 84

(57/84) > (44/84)

Hence its clear that

(19/28) > (11/21)

⇒ 19 : 28 > 11 : 21

— : end of Ratio and Proportion Class- 7th RS Aggarwal Exe-7 A Goyal Brothers ICSE Maths Solution:–

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