SELINA Solutions for Class 10 Maths Chapter 7 - Ratio and Proportion (Including Properties and Uses)
Complete your revision by practising Selina Solutions for ICSE Class 10 Mathematics Chapter 7 Ratio and Proportion (Including Properties and Uses). Understand the concept of ratio by going through problems involving reciprocal ratio, duplicate ratio, triplicate ratio and compounded ratio.
In addition, revise concepts like mean proportional, continued proportion, componendo, dividendo and alternendo through our Selina textbook solutions. Other self-study materials available at TopperLearning consist of ICSE Class 10 Maths video lessons, online practice tests, revision notes and more. You can use these 24/7 accessible learning resources to boost your skills for scoring excellent marks in your exam.
Chapter 7 - Ratio and Proportion (Including Properties and Uses) Exercise 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(i)
If 3A = 4B = 6C; find A: B: C.
Solution 19(i)
3A = 4B = 6C
3A = 4B 
4B = 6C 
Hence, A: B: C = 4: 3: 2
Question 19(ii)
If 2a = 3b and 4b = 5c, find: a : c.
Solution 19(ii)
Question 20
Find the compound ratio of:
(i) 2: 3, 9: 14 and 14: 27
(ii) 2a: 3b, mn: x2 and x: n.
(iii) 
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 = 
Question 21
Find duplicate ratio of:
(i) 3: 4 (ii) 
Solution 21
(i) Duplicate ratio of 3: 4 = 32: 42 = 9: 16
(ii) Duplicate ratio of 
Question 22
Find the triplicate ratio of:
(i) 1: 3 (ii) 
Solution 22
(i) Triplicate ratio of 1: 3 = 13: 33 = 1: 27
(ii) Triplicate ratio of
Question 23
Find sub-duplicate ratio of:
(i) 9: 16 (ii) (x - y)4: (x + y)6
Solution 23
(i) Sub-duplicate ratio of 9: 16 =
(ii) Sub-duplicate ratio of(x - y)4: (x + y)6
= 
Question 24
Find the sub-triplicate ratio of:
(i) 64: 27 (ii) x3: 125y3
Solution 24
(i) Sub-triplicate ratio of 64: 27
= 
(ii) Sub-triplicate ratio of x3:
125y3 = 
Question 25
Find the reciprocal ratio of:
(i) 5: 8 (ii) 
Solution 25
(i) Reciprocal ratio of 5: 8 = 
(ii) Reciprocal ratio of 
Question 26
If (x + 3) : (4x + 1) is the duplicate ratio of 3 : 5, find the value of x.
Solution 26
Question 27
If m: n is the duplicate ratio of m + x: n + x; show that x2 = mn.
Solution 27
Question 28
If (3x - 9) : (5x + 4) is the triplicate ratio of 3 : 4, find the value of x.
Solution 28
Question 29
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 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
= 
Question 30(a)
If r2 =pq, show that p : q is the duplicate ratio of (p + r) : (q + r).
Solution 30(a)
Question 30(b)
Solution 30(b)
Chapter 7 - Ratio and Proportion (Including Properties and Uses) Exercise 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(i)
Solution 7(i)
Question 7(ii)
Solution 7(ii)
Question 7(iii)
Solution 7(iii)
Question 8
What least number must be subtracted from each of the numbers 7, 17 and 47 so that the remainders are in continued proportion?
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.
Question 9
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 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.
Question 10
If q is the mean proportional between p and r, show that:
pqr (p + q + r)3 = (pq + qr + rp)3.
Solution 10
Given, q is the mean proportional between p and r.
q2 =
pr
Question 11
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 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.
Question 12
If y is the mean proportional between x and z, prove that:
Solution 12
Given, y is the mean proportional between x and z.
y2 = xz
Question 13
Given four quantities a, b, c and d are in proportion. Show that:
Solution 13
LHS = RHS
Hence proved.
Question 14
Find two numbers such that the mean mean proportional between them is 12 and the third proportional to them is 96.
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.
Question 15
Find
the third proportional to 
Solution 15
Let the required third proportional be p.
, p are in
continued proportion.
Question 16
If p: q = r: s; then show that:
mp + nq : q = mr + ns : s.
Solution 16
Hence, mp + nq : q = mr + ns : s.
Question 17
If p + r = mq and 
; then prove that p : q = r : s.
Solution 17
Hence, proved.
Chapter 7 - Ratio and Proportion (Including Properties and Uses) Exercise 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) Exercise 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
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.
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,
Therefore,
(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.
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