Concise Ratio and Proportion Solution Chapter-7 Selina Maths for ICSE Board Class 10th Concise Solution Ratio and Proportion Chapter -7 for ICSE Maths Class 10 is available here. All Solution of Concise of Chapter 7 Ratio and Proportion has been solved according instruction given by council. This is the  Solution of Chapter-7 Ratio and Proportion for ICSE Class 10th .ICSE Maths text book of Concise is In series of famous ICSE writer in maths. Concise is most famous among students. With the help of Concise Solution student can achieve their goal in 2020 exam of council.

## Concise Ratio and Proportion Solution Chapter-7 Selina Maths

The Solution of Concise Mathematics Chapter 7 Ratio and Proportion for ICSE Class 10  have been solved by experience teachers from across the globe  to help students of class 10th  ICSE board exams conducted by the ICSE (Indian Council of Secondary Education) board papering in 2020. Therefore the ICSE Class 10th Maths Solution of Concise solve problems of exercise related to various topics which are prescribed in most ICSE Maths textbooks.

### How to Solve Concise Maths Selina Publications Chapter-7 Ratio and Proportion

Note:- Before viewing Solutions of Chapter -7 Ratio and Proportion of Concise Maths  read the Chapter Carefully then solve all example of your text book. The Chapter- 7 Ratio and Proportion is main Chapter in ICSE board .

### Exercise-7(A), Concise Ratio and Proportion Solution Chapter-7

#### Question 1

If a: b = 5: 3, find: ………..

#### Question 2

If x: y = 4: 7, find the value of (3x + 2y): (5x + y).

#### Question 3

If a: b = 3: 8, find the value of ………

#### Question 4

If (a – b): (a + b) = 1: 11, find the ratio (5a + 4b + 15): (5a – 4b + 3).

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

#### Question 5

Find The numbers………………….

#### Question 6

If………………Find………….

#### Question 7

Find   x/y , when x2 + 6y2 = 5xy.

x2 + 6y2 = 5xy

Dividing both sides by y2, we get,

#### Question 8

If the ratio between 8 and 11 is the same as the ratio of 2x – y to x + 2y, find the value of …….. Given,

Devide…………..

#### Question 10

A school has 630 students. The ratio of the number of boys to the number of girls is 3 : 2. This ratio changes to 7 : 5 after the admission of 90 new students. Find the number of newly admitted boys.

#### Question 11

What quantity must be subtracted from each term of the ratio 9: 17 to make it equal to 1: 3?

Let x be subtracted from each term of the ratio 9: 17. Thus, the required number which should be subtracted is 5.

#### Question 12

The monthly pocket money of Ravi and Sanjeev are in the ratio 5 : 7. Their expenditures are in the ratio 3 : 5. If each saves Rs. 80 every month, find their monthly pocket money.

#### Question 13

The work done by (x – 2) men in (4x + 1) days and the work done by (4x + 1) men in (2x – 3) days are in the ratio 3: 8. Find the value of x.

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, #### Question 14

The bus fare between two cities is increased in the ratio 7: 9. Find the increase in the fare, if:

(i) the original fare is Rs 245;

(ii) the increased fare is Rs 207.

According to the given information,

Increased (new) bus fare = 9/7 x  original bus fare

(i) We have:

Increased (new) bus fare =9/7 x Rs 245 = Rs 315

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

(ii) We have:

Rs 207 = 9/7 x original bus fare

Original bus fare =

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

#### Question 15

By increasing the cost of entry ticket to a fair in the ratio 10: 13, the number of visitors to the fair has decreased in the ratio 6: 5. In what ratio has the total collection increased or decreased?

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.

#### Question 16

In a basket, the ratio between the number of oranges and the number of apples is 7: 13. If 8 oranges and 11 apples are eaten, the ratio between the number of oranges and the number of apples becomes 1: 2. Find the original number of oranges and the original number of apples in the basket.

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 x 5 = 35 and 13x  5 = 65 respectively.

#### Question 17

In a mixture of 126 kg of milk and water, milk and water are in ratio 5 : 2. How much water must be added to the mixture to make this ratio 3 : 2?

#### Question 18

(A) If A: B = 3: 4 and B: C = 6: 7, find:

(i) A: B: C (ii) A: C

(B) If A : B = 2 : 5 and A : C = 3 : 4, find

(i) A : B : C

(A)

(i) (ii)

(B)

(i) #### Question 19

If 3A = 4B = 6C; find A: B: C.

3A = 4B = 6C

3A = 4B

4B = 6C Hence, A: B: C = 4: 3: 2

#### Question 20

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

#### Question 21

Find the compound ratio of:

(i) 2: 3, 9: 14 and 14: 27

(ii) 2a: 3b, mn: x2 and x: n.

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

(ii) Required compound ratio = 2a xmn x x: 3b x x2  xn (iii) Required compound ratio = #### Question 22

Find duplicate ratio of:

(i) 3: 4 (ii)

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

(ii) Duplicate ratio of #### Question 23

Find the triplicate ratio of:

(i) 1: 3 (ii)

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

(ii) Triplicate ratio of #### Question 24

Find sub-duplicate ratio of:

(i) 9: 16 (ii) (x – y)4: (x + y)6

(i) Sub-duplicate ratio of 9: 16 = (ii) Sub-duplicate ratio of(x – y)4: (x + y)6

=

#### Question 25

Find the sub-triplicate ratio of:

(i) 64: 27 (ii) x3: 125y3

(i) Sub-triplicate ratio of 64: 27 = (ii) Sub-triplicate ratio of x3: 125y3 =

#### Question 26

Find the reciprocal ratio of:

(i) 5: 8 (ii) (i) Reciprocal ratio of 5: 8 =

(ii) Reciprocal ratio of #### Question 27

If (x + 3) : (4x + 1) is the duplicate ratio of 3 : 5, find the value of x.

#### Question 28

If m: n is the duplicate ratio of m + x: n + x; show that x2 = mn. #### Question 29

If (3x – 9) : (5x + 4) is the triplicate ratio of 3 : 4, find the value of x.

#### Question 30

Find the ratio compounded of the reciprocal ratio of 15: 28, the sub-duplicate ratio of 36: 49 and the triplicate ratio of 5: 4.

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

=

#### Question 31

If r=pq, show that p : q is the duplicate ratio of (p + r) : (q + r). #### Question 32 ### Selina Solution of Chapter 7 – Ratio and Proportion Concise Maths Exercise 7-(B)

#### Question 1

Find the fourth proportional to:

(i) 1.5, 4.5 and 3.5 (ii) 3a, 6a2 and 2ab2

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

⇒ 1.5 : 4.5 = 3.5 : x

so ⇒ 1.5  x = 3.5  4.5

Hence⇒ x = 10.5

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

⇒ 3a : 6a2 = 2ab2 : x

and ⇒ 3a  x = 2ab2 6a2

so ⇒ 3a  x = 12a3b2

Hence ⇒ x = 4a2b2

#### Question 2

Find the third proportional to:

(i) 2 2/3  and 4 (ii) a – b and a2 – b2

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

⇒ 2 2/3 , 4, x are in continued proportion.

⇒ 2 2/3  : 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

#### Question 3

Find the mean proportional between:

(i) 6 + 3 and 8 – 4

(ii) a – b and a3 – a2b

(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 x = (6 +3 ) (8 – 4   )

x= 48 + 24 – 24  – 36

and x= 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)

#### Question 4

If x + 5 is the mean proportional between x + 2 and x + 9; find the value of x.

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

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

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

so (x + 5)2 = (x + 2)(x + 9)

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

25 – 18 = 11x – 10x

Hence x = 7

#### Question 5

If x2, 4 and 9 are in continued proportion, find x.

#### Question 6

What least number must be added to each of the numbers 6, 15, 20 and 43 to make them proportional?

Let the number added be x.

(6 + x) : (15 + x) :: (20 + x) (43 + x) Thus, the required number which should be added is 3.

#### Question 8 #### Question 9 #### Question 10

What least number must be subtracted from each of the numbers 7, 17 and 47 so that the remainders are in continued proportion?

Let the number subtracted be x.

(7 – x) : (17 – x) :: (17 – x) (47 – x) Thus, the required number which should be subtracted is 2.

#### Question 11

If y is the mean proportional between x and z; show that xy + yz is the mean proportional between x2+y2 and y2+z2.

Since y is the mean proportion between x and z

Therefore, y= 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.

#### Question 12

If q is the mean proportional between p and r, show that:

pqr (p + q + r)3 = (pq + qr + rp)3.

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

q2 = pr #### Question 13

If three quantities are in continued proportion; show that the ratio of the first to the third is the duplicate ratio of the first to the second.

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 = x: y2

That is we need to prove that

xy= x2z

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

Hence, proved.

#### Question 14

If y is the mean proportional between x and z, prove that:

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

y2 = xz #### Question 15

Given four quantities a, b, c and d are in proportion. Show that:  LHS = RHS

Hence proved.

#### Question 16

Find two numbers such that the mean mean proportional between them is 12 and the third proportional to them is 96.

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.

#### Question 17

Find the third proportional to

Let the required third proportional be p.

, p are in continued proportion. #### Question 18

If p: q = r: s; then show that:

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

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

#### Question 19

If p + r = mq and ; then prove that p : q = r : s.

Hence, proved.

### Exercise -7 (C) , Ratio and Proportion Concise Maths Selina Publication for ICSE Board Class 10th

#### Question 1

If a : b = c : d, prove that:

(i) 5a + 7b : 5a – 7b = 5c + 7d : 5c – 7d.

(ii) (9a + 13b) (9c – 13d) = (9c + 13d) (9a – 13b).

(iii) xa + yb : xc + yd = b : d.

#### Question 2

If a : b = c : d, prove that:

(6a + 7b) (3c – 4d) = (6c + 7d) (3a – 4b).

#### Question 3

Given, , prove that:

#### Question 4

If ; then prove that:

x: y = u: v.

#### Question 5

If (7a + 8b) (7c – 8d) = (7a – 8b) (7c + 8d), prove that a: b = c: d.

#### Question 6

(i) If x = , find the value of:

.

(ii) If a = , find the value of

:

#### Question 7

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

#### Question 8

If , show that 2ad = 3bc.

#### Question 9

If ; prove that: .

#### Question 10

If a, b and c are in continued proportion, prove that: Given, a, b and c are in continued proportion.

#### Question 11

Using properties of proportion, solve for x:  #### Question 12

If , prove that: 3bx2 – 2ax + 3b = 0.

#### Question 13 #### Question 14

If , express n in terms of x and m.

#### Question 15

If , show that:

nx = my.

### Exercise 7 (D) – Ratio and Proportion Concise Maths Solution

#### Question 1

If a: b = 3: 5, find:

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

Given, #### Question 2

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

#### Question 3

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

(3x – 4y): (2x – 3y) = (5x – 6y): (4x – 5y) #### Question 4

Find the:

(i) duplicate ratio of

(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.

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

(iii) Sub-duplicate ratio of 9x2a: 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 =

#### Question 5

Find the value of x, if:

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

(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

#### Question 6

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

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

Then, we have: #### Question 7

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

#### Question 8

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

15(2x2 – y2) = 7xy

#### Question 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).

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

2xy: x2 = y2: n

2xy x 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) #### Question 10

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

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.

#### Question 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.

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

#### Question 12

If    , find the value of . #### Question 13

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

a: b = c: d. #### Question 14

If     , show that:

(a + b) : (c + d) = #### Question 15

There are 36 members in a student council in a school and the ratio of the number of boys to the number of girls is 3: 1. How any more girls should be added to the council so that the ratio of the number of boys to the number of girls may be 9: 5?

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.

#### Question 16

If 7x – 15y = 4x + y, find the value of x: y. Hence, use componendo and dividend to find the values of: 7x – 15y = 4x + y

7x – 4x = y + 15y

3x = 16y #### Question 17

If  , use properties of proportion to find:

(i) m: n

(ii) #### Question 18

If x, y, z are in continued proportion, prove that x, y, z are in continued proportion,

Therefore, (By alternendo) Hence Proved.

#### Question 19

Given x =. Use componendo and dividendo to prove that b2 =.

x = By componendo and dividendo, Squaring both sides,

By componendo and dividendo,  b2 =

Hence Proved.

#### Question 20

If , find:

#### Question 21

Using componendo and dividendo find the value of x: #### Question 22 #### Question 23 #### Question 24 #### Question 25

If b is the mean proportion between a and c, show that: Given that b is the mean proportion between a and c.

#### Question 26

If , use properties of proportion to find:

i. ii.

From (i), #### Question 27

1. i. If x and y both are positive and (2x2– 5y2): xy = 1: 3, find x: y.
2. Find x, if.

1. i.(2x2– 5y2): xy = 1: 3

ii

.

⇒ x = a/3 or x = 3a

Thanks