Chapter 7 : Ratio and Proportion (Including Properties and Uses) - Selina Solutions for Class 10 Maths ICSE

Mathematics is one of the crucial and most scoring subjects in ICSE Class 10. TopperLearning presents study materials for ICSE Class 10 Mathematics which will enable students to score well in the board examination. The syllabus includes certain challenging concepts like Value Added Tax, Ratio and Proportion, Quadric Equations among others which require effective study materials. Our study materials for ICSE Class 10 consist of video lessons, question banks, sample papers, revision notes and past year papers which improve the quality of learning.

Read  more
Page / Exercise

Chapter 7 - Ratio and Proportion (Including Properties and Uses) Excercise Ex. 7(A)

Question 1

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

Solution 1

Question 2

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

Solution 2

Question 3

If a: b = 3: 8, find the value of .

Solution 3

Question 4

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

Solution 4

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

Question 5

 

Solution 5

 

Question 6

If.

Solution 6

Question 7

Find, when x2 + 6y2 = 5xy.

Solution 7

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

Solution 8

Given,

Question 9

 

Solution 9

 

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.

Solution 10

Question 11

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

Solution 11

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.

Solution 12

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.

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,

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.

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

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?

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.

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.

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.

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?

Solution 17

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

Solution 18

(A)

(i)

 

(ii)


(B)

(i)

Question 19

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

Solution 19

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.

Solution 20

Question 21

Find the compound ratio of:

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

(ii) 2a: 3b, mn: x2 and x: n.

(iii)

Solution 21

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

 

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

 

(iii) Required compound ratio =

Question 22

Find duplicate ratio of:

(i) 3: 4 (ii)

Solution 22

(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)

Solution 23

(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

Solution 24

(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

Solution 25

(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)

Solution 26

(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. 

Solution 27

Question 28

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

Solution 28

Question 29

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

Solution 29

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.

Solution 30

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 r2 =pq, show that p : q is the duplicate ratio of (p + r) : (q + r).

Solution 31

Question 32

Solution 32

Chapter 7 - Ratio and Proportion (Including Properties and Uses) Excercise Ex. 7(B)

Question 1

Find the fourth proportional to:

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

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

Question 2

Find the third proportional to:

(i) 2 and 4 (ii) a - b and a2 - b2

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

Question 3

Find the mean proportional between:

(i) 6 + 3 and 8 - 4

(ii) a - b and a3 - a2b

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)

Question 4

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

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

Question 5

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

Solution 5

Question 6

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

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.

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 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?

Solution 10

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.

Solution 11

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.

Question 12

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

pqr (p + q + r)3 = (pq + qr + rp)3.

Solution 12

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.

Solution 13

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.

Question 14

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

Solution 14

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:

Solution 15

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.

Solution 16

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

Solution 17

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.

Solution 18

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

Question 19

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

Solution 19

Hence, proved.

Chapter 7 - Ratio and Proportion (Including Properties and Uses) Excercise Ex. 7(C)

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.

Solution 1

Question 2

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

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

Solution 2

Question 3

Given, , prove that:

Solution 3

Question 4

If ; then prove that:

x: y = u: v.

Solution 4

Question 5

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

Solution 5

Given,

Applying componendo and dividendo,

Hence, a: b = c: d.

Question 6

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

.

(ii) If a = , find the value of:

Solution 6

(i) x =

(ii)

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.

Solution 7

Question 8

If , show that 2ad = 3bc.

Solution 8

Question 9

If ; prove that: .

Solution 9

Given,

Question 10

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

Solution 10

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

Question 11

Using properties of proportion, solve for x:

Solution 11

Question 12

If , prove that: 3bx2 - 2ax + 3b = 0.

Solution 12

Since,

Applying componendo and dividendo, we get,

Squaring both sides,

Again applying componendo and dividendo,

3bx2 + 3b = 2ax

3bx2 - 2ax + 3b = 0.

Question 13

Solution 13

Question 14

If , express n in terms of x and m.

Solution 14

Applying componendo and dividendo,

Question 15

If , show that:

nx = my.

Solution 15

Applying componendo and dividendo,

Chapter 7 - Ratio and Proportion (Including Properties and Uses) Excercise Ex. 7(D)

Question 1

If a: b = 3: 5, find:

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

Solution 1

Given,

Question 2

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

Solution 2

Question 3

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

Solution 3

(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 9x2a4 : 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 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 =

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.

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

Question 6

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

Solution 6

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?

Solution 7

Question 8

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

Solution 8

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).

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)

Question 10

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

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

rightwards double arrow a equals 196 over b... left parenthesis 1 right parenthesis

 

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

a: b = b: 112

rightwards double arrow b squared equals 112 a... left parenthesis 2 right parenthesis

 

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.

Solution 11

Given,

Hence, z is mean proportional between x and y.

Question 12

If , find the value of .

Solution 12

Question 13

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

a: b = c: d.

Solution 13

Question 14

If , show that:

(a + b) : (c + d) =

Solution 14

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?

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.

Question 16

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

Solution 16

7x - 15y = 4x + y

7x - 4x = y + 15y

3x = 16y

 

Question 17

If , use properties of proportion to find:

(i) m: n

(ii)

Solution 17

Question 18

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

Solution 18

x, y, z are in continued proportion,

x over y equals y over z rightwards double arrow y squared equals z x.... left parenthesis 1 right parenthesis

Therefore,

fraction numerator x plus y over denominator y end fraction equals fraction numerator y plus z over denominator z end fraction space space space space open parentheses B y space c o m p o n e n d o close parentheses

(By alternendo)

Hence Proved.

Question 19

Given x =.

Use componendo and dividendo to prove that b2 =.

Solution 19

x =

By componendo and dividendo,

Squaring both sides,

By componendo and dividendo,

b2 =

Hence Proved.

Question 20

If , find:

Solution 20

Question 21

Using componendo and dividendo find the value of x:

Solution 21

Question 22

Solution 22

 

 

 

 

 

Question 23

 

Solution 23

 

Question 24

Solution 24

 

Question 25

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

Solution 25

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

 

Question 26

 

If , use properties of proportion to find: 

 

Solution 26

i.

ii.

From (i),

 

Question 27

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

ii. Find x, if .

 

 

Solution 27

i. (2x2 - 5y2): xy = 1: 3

ii.

 

Why to choose our study materials for ICSE Class 10 Mathematics?

  • Designed in simple and comprehensive language
  • Contain 285 video lessons, 37 revision notes, 3709+ questions in bank, set of 6 sample papers and 9 past year papers
  • Prepared by subject experts
  • Revised from time to time
  • Useful for doing quick revision
  • Helpful when doing homework