Selina Solutions Concise Maths Class 10 Chapter 7 Ratio and Proportion Exercise 7(D)

Different models of problems based on the concepts of the entire chapter are listed in this exercise. Additionally, duplicate, triplicate, sub-duplicate and sub-triplicate ratio are some of the key insights in the exercise. The Selina Solutions for Class 10 Maths can be used by students for any quick references as well as a guide for students in order to help their exam preparation. The solutions to this exercise can be accessed in the Concise Selina Solutions for Class 10 Maths Chapter 7 Ratio and Proportion Exercise 7(D) PDF, which is provided in the links below.

Selina Solutions Concise Maths Class 10 Chapter 7 Ratio and Proportion Exercise 7(D) Download PDF

 

selina solutions concise maths class 10 chapter 7d
selina solutions concise maths class 10 chapter 7d
selina solutions concise maths class 10 chapter 7d
selina solutions concise maths class 10 chapter 7d
selina solutions concise maths class 10 chapter 7d
selina solutions concise maths class 10 chapter 7d

 

Access other exercises of Selina Solutions Concise Maths Class 10 Chapter 7 Ratio and Proportion

Exercise 7(A) Solutions

Exercise 7(B) Solutions

Exercise 7(C) Solutions

Access Selina Solutions Concise Maths Class 10 Chapter 7 Ratio and Proportion Exercise 7(D)

1. If a: b = 3: 5, find:

(10a + 3b): (5a + 2b)

Solution:

Given, a/b = 3/5

(10a + 3b)/ (5a + 2b)

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(D) - 1

2. If 5x + 6y: 8x + 5y = 8: 9, find x: y.

Solution:

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(D) - 2Given,

On cross multiplying, we get

45x + 54y = 64x + 40y

14y = 19x

Thus,

x/y = 14/19

3. If (3x – 4y): (2x – 3y) = (5x – 6y): (4x – 5y), find x: y.

Solution:

Given, (3x – 4y): (2x – 3y) = (5x – 6y): (4x – 5y)

This can be rewritten as,

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(D) - 3

5x – 7y = 9x – 11y

4y = 4x

x/y = 1/1

Thus,

x: y = 1: 1

4. Find the:

(i) duplicate ratio of 2√2: 3√5

(ii) triplicate ratio of 2a: 3b

(iii) sub-duplicate ratio of 9x2a: 25y6b2

(iv) sub-triplicate ratio of 216: 343

(v) reciprocal ratio of 3: 5

(vi) ratio compounded of the duplicate ratio of 5: 6, the reciprocal ratio of 25: 42 and the sub-duplicate ratio of 36: 49.

Solution:

(i) Duplicate ratio of 2√2: 3√5 = (2√2)2: (3√5)2 = 8: 45

(ii) Triplicate ratio of 2a: 3b = (2a)3: (3b)3 = 8a3: 27b3

(iii) Sub-duplicate ratio of 9x2a4: 25y6b2 = √(9x2a4): √(25y6b2) = 3xa2: 5y3b

(iv) Sub-triplicate ratio of 216: 343 = (216)1/3: (343)1/3 = 6: 7

(v) Reciprocal ratio of 3: 5 = 5: 3

(vi) Duplicate ratio of 5: 6 = 25: 36

Reciprocal ratio of 25: 42 = 42: 25

Sub-duplicate ratio of 36: 49 = 6: 7

Required compound ratio =
Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(D) - 4

5. Find the value of x, if:

(i) (2x + 3): (5x – 38) is the duplicate ratio of √5: √6. 

(ii) (2x + 1): (3x + 13) is the sub-duplicate ratio of 9: 25.

(iii) (3x – 7): (4x + 3) is the sub-triplicate ratio of 8: 27.

Solution:

(i) (2x + 3): (5x – 38) is the duplicate ratio of √5: √6

And, the duplicate ratio of √5: √6 = 5: 6

So,

(2x + 3)/ (5x – 38) = 5/6

12x + 18 = 25x – 190

25x – 12x = 190 + 18

13x = 208

x = 208/13 = 16

(ii) (2x + 1): (3x + 13) is the sub-duplicate ratio of 9: 25

Then the sub-duplicate ratio of 9: 25 = 3: 5

(2x + 1)/ (3x + 13) = 3/5

10x + 5 = 9x + 39

x = 34

(iii) (3x – 7): (4x + 3) is the sub-triplicate ratio of 8: 27

And the sub-triplicate ratio of 8: 27 = 2: 3

(3x – 7)/ (4x + 3) = 2/3

9x – 8x = 6 + 21

x = 27

6. What quantity must be added to each term of the ratio x: y so that it may become equal to c: d?

Solution:

Let’s assume the required quantity which has to be added be p.

So, we have

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(D) - 5

dx + pd = cy + cp

pd – cp = cy – dx

p(d – c) = cy – dx

p = cy –dx/ (d – c)

7. A woman reduces her weight in the ratio 7: 5. What does her weight become if originally it was 84 kg?

Solution:

Let’s consider the woman’s reduced weight as x.

Given, the original weight = 84 kg

So, we have

84: x = 7: 5

84/x = 7/5

84 x 5 = 7x

x = (84 x 5)/ 7

x = 60

Therefore, the reduced weight of the woman is 60 kg.

8. If 15(2x2 – y2) = 7xy, find x: y; if x and y both are positive.

Solution:

15(2x2 – y2) = 7xy

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(D) - 6

Let’s take the substitution as x/y = a

2a – 1/a = 7/15

(2a2 – 1)/ a = 7/15

30a2 – 15 = 7a

30a2 – 7a – 15 = 0

30a2 – 25a + 18a – 15 = 0

5a(6a – 5) + 3(6a – 5) = 0

(6a – 5) (5a + 3) = 0

So, 6a – 5 = 0 or 5a + 3 = 0

a = 5/6 or a = -3/5

As, a cannot be taken negative (ratio)

Thus, a = 5/6

x/y = 5/6

Hence, x: y = 5: 6

9. Find the:

(i) fourth proportional to 2xy, x2 and y2.

(ii) third proportional to a2 – b2 and a + b.

(iii) mean proportional to (x – y) and (x3 – x2y).

Solution:

(i) Let the fourth proportional to 2xy, x2 and y2 be n.

2xy: x2 = y2: n

2xy × n = x2 × y2

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(D) - 7

n =

(ii) Let the third proportional to a2 – b2 and a + b be n.

a2 – b2, a + b and n are in continued proportion.

a2 – b2: a + b = a + b: n

n =
Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(D) - 8

(iii) Let the mean proportional to (x – y) and (x3 – x2y) be n.

(x – y), n, (x3 – x2y) are in continued proportion

(x – y): n = n: (x3 – x2y)

n2 = (x -y) (x3 – x2y)

n2 = (x -y) x2(x – y)

n2 = x2 (x – y)2

n = x(x – y)

10. Find two numbers such that the mean proportional between them is 14 and third proportional to them is 112.

Solution:

Let’s assume the required numbers be a and b.

Given, 14 is the mean proportional between a and b.

a: 14 = 14: b

ab = 196

a = 196/b …. (1)

Also, given, third proportional to a and b is 112.

a: b = b: 112

b2 = 112a …. (2)

Using (1), we have:

b2 = 112 × (196/b)

b3 = 143 x 23

b = 28

From (1),

a = 196/ 28 = 7

Therefore, the two numbers are 7 and 28.

11. If x and y be unequal and x: y is the duplicate ratio of x + z and y + z, prove that z is mean proportional between x and y.

Solution:

Given,
Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(D) - 9

x(y2 + z2 + 2yz) = y(x2 + z2 + 2xz)

xy2 + xz2 + 2yzx = x2y + z2y + 2xzy

xy2 + xz2 = x2y + z2y

xy(y – x) = z2(y – x)

xy = z2

Therefore, z is mean proportional between x and y.

12. If Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(D) - 10, find the value of Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(D) - 11.

Solution:

x = 2ab/ (a + b)

x/a = 2b/(a + b)

Applying componendo and dividendo,

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(D) - 12

Also, x = 2ab/ (a + b)

x/b = 2a/ (a + b)

Applying componendo and dividendo, we have

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(D) - 13

Now, comparing (1) and (2) we have

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(D) - 14

13. If (4a + 9b) (4c – 9d) = (4a – 9b) (4c + 9d), prove that:

a: b = c: d.

Solution:

Given,
Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(D) - 15

Applying componendo and dividendo, we get

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(D) - 16

8a/18b = 8c/18d

a/b = c/d

– Hence Proved

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