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# Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 7 - Ratio and Proportion (Including Properties and Uses)

## Selina Textbook Solutions Chapter 7 - Ratio and Proportion (Including Properties and Uses)

Selina Textbook Solutions are a perfect way to ace your examination with high marks. These Textbook Solutions are extremely helpful for solving difficult questions in the ICSE Class 10 Mathematics exam. Our Selina Textbook Solutions are written by our subject experts. Find all the answers to the Selina textbook questions of Chapter 7 - Ratio and Proportion (Including Properties and Uses).

All Selina textbook questions of Chapter 7 - Ratio and Proportion (Including Properties and Uses) solutions are created in accordance with the latest ICSE syllabus. These free Textbook Solutions for ICSE Class 10 Selina Concise Mathematics will give you a deeper insight on the fundamentals in this chapter and will help you to score more marks in the final examination. ICSE Class 10 students can refer to these solutions while doing their homework and while studying and revising for the Mathematics exam.

Exercise/Page

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 7 - Ratio and Proportion (Including Properties and Uses) Page/Excercise 7(C)

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Given,

Applying componendo and dividendo,

Hence, a: b = c: d.

Solution 6

(i) x =

(ii)

Solution 7

Solution 8

Solution 9

Given,

Solution 10

Given, a, b and c are in continued proportion.

Solution 11

Solution 12

Since,

Applying componendo and dividendo, we get,

Squaring both sides,

Again applying componendo and dividendo,

3bx2 + 3b = 2ax

3bx2 - 2ax + 3b = 0.

Solution 13

Solution 14

Applying componendo and dividendo,

Solution 15

Applying componendo and dividendo,

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 7 - Ratio and Proportion (Including Properties and Uses) Page/Excercise 7(D)

Solution 1

Given,

Solution 2

Solution 3

(3x - 4y): (2x - 3y) = (5x - 6y): (4x - 5y)

Solution 4

(i) Duplicate ratio of

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

(iii) Sub-duplicate ratio of 9x2a4 : 25y6b2 =

(iv) Sub-triplicate ratio of 216: 343 =

(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 =

Solution 5

(i) (2x + 3): (5x - 38) is the duplicate ratio of

Duplicate ratio of

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

Sub-duplicate ratio of 9: 25 = 3: 5

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

Sub-triplicate ratio of 8: 27 = 2: 3

Solution 6

Let the required quantity which is to be added be p.

Then, we have:

Solution 7

Solution 8

15(2x2 - y2) = 7xy

Solution 9

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

2xy: x2 = y2: n

2xy n = x2 y2

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 =

(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)

Solution 10

Let the required numbers be a and b.

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

a: 14 = 14: b

ab = 196

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

a: b = b: 112

Using (1), we have:

Thus, the two numbers are 7 and 28.

Solution 11

Given,

Hence, z is mean proportional between x and y.

Solution 12

Since, q is the mean proportional between p and r,

q2 = pr

Solution 13

Given, a, b and c are in continued proportion.

a: b = b: c

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Ratio of number of boys to the number of girls = 3: 1

Let the number of boys be 3x and number of girls be x.

3x + x = 36

4x = 36

x = 9

Number of boys = 27

Number of girls = 9

Le n number of girls be added to the council.

From given information, we have:

Thus, 6 girls are added to the council.

Solution 19

7x - 15y = 4x + y

7x - 4x = y + 15y

3x = 16y

Solution 20

Solution 21

x, y, z are in continued proportion,

Therefore,

(By alternendo)

Hence Proved.

Solution 22

x =

By componendo and dividendo,

Squaring both sides,

By componendo and dividendo,

b2 =

Hence Proved.

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 7 - Ratio and Proportion (Including Properties and Uses) Page/Excercise 7(A)

Solution 1

Solution 2

Solution 3

Solution 4

Hence, (5a + 4b + 15): (5a - 4b + 3) = 5: 1

Solution 5

Solution 6

Solution 7

x2 + 6y2 = 5xy

Dividing both sides by y2, we get,

Solution 8

Given,

Solution 9

Solution 10

Solution 11

Let x be subtracted from each term of the ratio 9: 17.

Thus, the required number which should be subtracted is 5.

Solution 12

Solution 13

Assuming that all the men do the same amount of work in one day and one day work of each man = 1 units, we have,

Amount of work done by (x - 2) men in (4x + 1) days

= Amount of work done by (x - 2)(4x + 1) men in one day

= (x - 2)(4x + 1) units of work

Similarly,

Amount of work done by (4x + 1) men in (2x - 3) days

= (4x + 1)(2x - 3) units of work

According to the given information,

Solution 14

According to the given information,

Increased (new) bus fare = original bus fare

(i) We have:

Increased (new) bus fare = Rs 245 = Rs 315

Increase in fare = Rs 315 - Rs 245 = Rs 70

(ii) We have:

Rs 207 = original bus fare

Original bus fare =

Increase in fare = Rs 207 - Rs 161 = Rs 46

Solution 15

Let the cost of the entry ticket initially and at present be 10 x and 13x respectively.

Let the number of visitors initially and at present be 6y and 5y respectively.

Initially, total collection = 10x 6y = 60 xy

At present, total collection = 13x 5y = 65 xy

Ratio of total collection = 60 xy: 65 xy = 12: 13

Thus, the total collection has increased in the ratio 12: 13.

Solution 16

Let the original number of oranges and apples be 7x and 13x.

According to the given information,

Thus, the original number of oranges and apples are 7 5 = 35 and 13 5 = 65 respectively.

Solution 17

Solution 18

(A)

(i)

(ii)

(B)

(i)

Solution 19(ii)

Solution 19(i)

3A = 4B = 6C

3A = 4B

4B = 6C

Hence, A: B: C = 4: 3: 2

Solution 20

(i) Required compound ratio = 2 9 14: 3 14 27

(ii) Required compound ratio = 2a mn x: 3b x2 n

(iii) Required compound ratio =

Solution 21

(i) Duplicate ratio of 3: 4 = 32: 42 = 9: 16

(ii) Duplicate ratio of

Solution 22

(i) Triplicate ratio of 1: 3 = 13: 33 = 1: 27

(ii) Triplicate ratio of

Solution 23

(i) Sub-duplicate ratio of 9: 16 =

(ii) Sub-duplicate ratio of(x - y)4: (x + y)6

=

Solution 24

(i) Sub-triplicate ratio of 64: 27 =

(ii) Sub-triplicate ratio of x3: 125y3 =

Solution 25

(i) Reciprocal ratio of 5: 8 =

(ii) Reciprocal ratio of

Solution 26

Solution 27

Solution 28

Solution 29

Reciprocal ratio of 15: 28 = 28: 15

Sub-duplicate ratio of 36: 49 =

Triplicate ratio of 5: 4 = 53: 43 = 125: 64

Required compounded ratio

=

Solution 30(b)

Solution 30(a)

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 7 - Ratio and Proportion (Including Properties and Uses) Page/Excercise 7(B)

Solution 1

(i) Let the fourth proportional to 1.5, 4.5 and 3.5 be x.

1.5 : 4.5 = 3.5 : x

1.5 x = 3.5 4.5

x = 10.5

(i) Let the fourth proportional to 3a, 6a2 and 2ab2 be x.

3a : 6a2 = 2ab2 : x

3a x = 2ab2 6a2

3a x = 12a3b2

x = 4a2b2

Solution 2

(i) Let the third proportional to 2 and 4 be x.

2, 4, x are in continued proportion.

2 : 4 = 4 : x

(ii) Let the third proportional to a - b and a2 - b2 be x.

a - b, a2 - b2, x are in continued proportion.

a - b : a2 - b2 = a2 - b2 : x

Solution 3

(i) Let the mean proportional between 6 + 3 and 8 - 4 be x.

6 + 3, x and 8 - 4 are in continued proportion.

6 + 3 : x = x : 8 - 4

x x = (6 + 3) (8 - 4)

x2 = 48 + 24- 24 - 36

x2 = 12

x= 2

(ii) Let the mean proportional between a - b and a3 - a2b be x.

a - b, x, a3 - a2b are in continued proportion.

a - b : x = x : a3 - a2b

x x = (a - b) (a3 - a2b)

x2 = (a - b) a2(a - b) = [a(a - b)]2

x = a(a - b)

Solution 4

Given, x + 5 is the mean proportional between x + 2 and x + 9.

(x + 2), (x + 5) and (x + 9) are in continued proportion.

(x + 2) : (x + 5) = (x + 5) : (x + 9)

(x + 5)2 = (x + 2)(x + 9)

x2 + 25 + 10x = x2 + 2x + 9x + 18

25 - 18 = 11x - 10x

x = 7

Solution 5

Solution 6

Let the number added be x.

(6 + x) : (15 + x) :: (20 + x) (43 + x)

Thus, the required number which should be added is 3.

Solution 7(iii)

Solution 7(ii)

Solution 7(i)

Solution 8

Let the number subtracted be x.

(7 - x) : (17 - x) :: (17 - x) (47 - x)

Thus, the required number which should be subtracted is 2.

Solution 9

Since y is the mean proportion between x and z

Therefore, y2 = xz

Now, we have to prove that xy+yz is the mean proportional between x2+y2 and y2+z2, i.e.,

LHS = RHS

Hence, proved.

Solution 10

Given, q is the mean proportional between p and r.

q2 = pr

Solution 11

Let x, y and z be the three quantities which are in continued proportion.

Then, x : y :: y : z y2 = xz ....(1)

Now, we have to prove that

x : z = x2 : y2

That is we need to prove that

xy2 = x2z

LHS = xy2 = x(xz) = x2z = RHS [Using (1)]

Hence, proved.

Solution 12

Given, y is the mean proportional between x and z.

y2 = xz

Solution 13

LHS = RHS

Hence proved.

Solution 14

Let a and b be the two numbers, whose mean proportional is 12.

Now, third proportional is 96

Therefore, the numbers are 6 and 24.

Solution 15

Let the required third proportional be p.

, p are in continued proportion.

Solution 16

Hence, mp + nq : q = mr + ns : s.

Solution 17

Hence, proved.

Solution 18

Then, a = bk and c = dk

Solution 19

a, b, c and d are in proportion

Then, a = bk and c = dk

Solution 20

Then, x = ak, y = bk and z = ck

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TopperLearning provides step-by-step solutions for each question in each chapter. Access Chapter 7 - Ratio and Proportion (Including Properties and Uses)  for ICSE Class 10 Mathematics free of cost. The solutions are provided by our subject matter experts. Refer to our solutions for the Selina Concise Mathematics textbook to revise the whole chapter and clear your fundamentals before the examination. By referring to the solutions for this chapter and the others, we hope that you are able to write your exams well.

# Text Book Solutions

ICSE X - Mathematics

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