# Class 10 SELINA Solutions Maths Chapter 7 - Ratio and Proportion (Including Properties and Uses)

## Ratio and Proportion (Including Properties and Uses) Exercise Ex. 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

x^{2} + 6y^{2} =
5xy

Dividing both sides by y^{2},
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(i)

3A = 4B = 6C

3A = 4B

4B = 6C

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

### Solution 19(ii)

### Solution 20

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

(ii) Required compound ratio = 2a mn x: 3b x^{2} n

(iii) Required compound ratio =

### Solution 21

(i) Duplicate ratio of 3: 4 = 3^{2}: 4^{2} = 9: 16

(ii) Duplicate ratio of

### Solution 22

(i) Triplicate ratio of 1: 3 = 1^{3}:
3^{3} = 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 x^{3}:
125y^{3} =

### 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 = 5^{3}:
4^{3} = 125: 64

Required compounded ratio

=

### Solution 30(a)

### Solution 30(b)

## Ratio and Proportion (Including Properties and Uses) Exercise Ex. 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

(ii) Let the fourth proportional to 3a, 6a^{2} and 2ab^{2} be x.

3a : 6a^{2} = 2ab^{2} : x

3a x = 2ab^{2} 6a^{2}

3a x = 12a^{3}b^{2}

x = 4a^{2}b^{2}

### 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 a^{2} - b^{2} be x.

a - b, a^{2}
- b^{2}, x are in continued proportion.

a - b : a^{2}
- b^{2} = a^{2} - b^{2} : 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)

x^{2 }= 48 + 24- 24 - 36

x^{2 }= 12

x= 2

(ii) Let the mean proportional between a - b and a^{3} - a^{2}b be x.

a - b, x, a^{3} - a^{2}b are in continued proportion.

a - b : x = x : a^{3} - a^{2}b

x x = (a - b) (a^{3} - a^{2}b)

x^{2} = (a - b) a^{2}(a - b) = [a(a - b)]^{2}

x = a(a - b)

### 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(i)

### Solution 7(ii)

### Solution 7(iii)

### 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, y^{2 }= xz

Now,
we have to prove that xy+yz is the mean proportional between x^{2}+y^{2}
and y^{2}+z^{2}, i.e.,

LHS = RHS

Hence, proved.

### Solution 10

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

q^{2} =
pr

### Solution 11

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

Then, x : y :: y : z y^{2} = xz ....(1)

Now, we have to prove that

x : z = x^{2 }: y^{2}

That is we need to prove that

xy^{2 }= x^{2}z

LHS = xy^{2} = x(xz) = x^{2}z = RHS [Using (1)]

Hence, proved.

### Solution 12

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

y^{2} = xz

### Solution 13

LHS = RHS

Hence proved.

### 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 4

Given, x + 5 is the mean proportion 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)

x^{2} + 25 + 10x = x^{2} + 2x + 9x + 18

25 - 18 = 11x - 10x

x = 7

### 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.

## Ratio and Proportion (Including Properties and Uses) Exercise Ex. 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,

3bx^{2} + 3b = 2ax

3bx^{2} - 2ax + 3b = 0.

### Solution 13

### Solution 14

Applying componendo and dividendo,

### Solution 15

Applying componendo and dividendo,

## Ratio and Proportion (Including Properties and Uses) Exercise Ex. 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} = 8a^{3}
: 27b^{3}

(iii)
Sub-duplicate ratio of 9x^{2}a^{4 }: 25y^{6}b^{2}
=

(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(2x^{2}
- y^{2}) = 7xy

### 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

### Solution 13

### Solution 14

### Solution 15

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 17

### Solution 18

x, y, z are in continued proportion,

Therefore,

(By alternendo)

Hence Proved.

### Solution 19

x =

By componendo and dividendo,

Squaring both sides,

By componendo and dividendo,

b^{2} =

Hence Proved.

### Solution 20

### Solution 21

### Solution 23

### Solution 24

### Solution 25

Given that b is the mean proportion between a and c.

### Solution 26

i.

ii.

From (i),

### Solution 27

i. (2x^{2} - 5y^{2}): xy = 1: 3

ii.

### Solution 9

(i) Let the fourth proportional to 2xy, x^{2} and y^{2} be n.

2xy: x^{2} = y^{2}: n

2xy n = x^{2} y^{2}

n =

(ii) Let the third proportional to a^{2} - b^{2} and a + b be n.

a^{2} - b^{2}, a + b and n are in continued proportion.

a^{2} - b^{2} : a + b = a + b : n

n =

(iii) Let the mean proportion to (x - y) and (x^{3} - x^{2}y) be n.

(x - y), n, (x^{3} - x^{2}y) are in continued proportion

(x - y) : n = n : (x^{3} - x^{2}y)

### Solution 16

7x - 15y = 4x + y

7x - 4x = y + 15y

3x = 16y

### Solution 22