# ICSE Solutions for Chapter 7 Ratio and Proportion Class 10 Mathematics

Question 1: If a : b = 3 : 5, find;
(10a + 3b) : (5a + 2b)

Solution 1:
Given, a/b = 3/5
(10a + 3b)/(5a + 2b)
= {10(a/b) + 3}/{5(a/b) + 2}
= {10(3/5) + 3}/{5(3/5) + 2}
= (6 + 3)/(3 + 2)
= 9/5

Question 2: If a : b = c : d, prove that:
(6a +7b) (3c - 4d) = (6c + 7d) (3a - 4b).

Solution 2: Given, a/b = c/d
⇒ 6a/7b = 6c/7d (Mutiplying each side by 6/7)
⇒ (6a + 7b)/7b = (6c + 7d)/7d (By componendo)
⇒ (6a + 7b)/(6c + 7d) = 7b/7d = b/d  ---(1)
Also, a/b = c/d
⇒ 3a/4b = 3c/4d (Mutiplying each side by ¾)
⇒ (3a – 4b)/4b = (3c - 4d)/4d  (By dividendo)
⇒ (3a – 4b)/(3c – 4d) = 4b/4d = b/d ...(2)
From (1) and (2),
(6a + 7b)/(6c + 7d) = (3a – 4b)/(3c – 4d)
(6a + 7b)(3c - 4d) =  (6c + 7d)(3a – 4b)

Question 3: Find the mean proportional between:
(i) 6 + 3√3 and 8 – 4√3
(ii) a – b and a3 – a2

Solution 3:
(i) Let the mean proportional between 6 + 3√3 and 8 - 4√3 be x.
⇒ 6 + 3√3, x and 8 - 4√3 are in continued proportion.
⇒ 6 + 3√3:  x = x : 8 - 4√3
⇒ x × x = (6 + 3√3) (8 - 4√3)
⇒ x2 = 48 + 24√3 - 24√3 - 36
⇒ x2 = 12
⇒ x = 2√3

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

Solution 4:  (a – b)/(a + b)  = 1/11
11a –11b = a + b
10a = 12b
a/b = 12/10 = 6/5
So, let a = 6k and b = 5k
(5a + 4b + 15)(5a – 4b + 3) = {5(6k) + 4(5k) + 15}/{5(6k) – 4(5k) + 3}
= (30k + 20k + 15)/(30k  - 20k + 3)
= (50k + 15)(10k + 3)
= 5(10k + 3)/(10k + 3)
= 5
Hence, (5a + 4b + 15);(5a - 4b + 3) = 5:1

Question 5: Find the number which bears the same ratio to 7/33 that 8/21 does to 4/9.

Solution 5: Let the required number be x/y.
Now, Ratio of 8/21 to 4/9 = (8/21)/(4/9) = 8/21 × 9/4 = 6/7
Thus, we have
(x/y)/(7/33) = 6/7
⇒ x/y = (6/7)/(7/33)
⇒ x/y = 6/7 × 7/33
⇒ x/y = 2/11
Hence, the required number is 2/11.

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)
(6 + x)/(15 + x) = (20 + x)/(43 + x)
(6 + x)(43 + x) = (20 + x)(15 + x)
258 + 6x + 43x + x2 = 300 + 20x + 15x + x2
49x - 35x = 300 - 258
14x = 42
x = 3
Thus, the required number which should be added is 3.

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: Given, (a+b+c+d)/(a–b+c–d) = (a-b+c-d)/(a-b-c+d)
Applying componenedo and dividendo,
(a + b + c + d) + (a – b + c – d)/(a + b + c + d) – (a + b – c – d) = (a – b + c – d) + (a – b – c + d)/(a – b + c – d) – (a – b – c + d)
{2(a + b)}/{2(c + d)} = {2(a – b)/2(c – d)}
(a + b)/(c + d) = (a – b)/(c – d)
(a + b)/(a – b) = (c + d)/(c – d)
Applying componendo and dividendo,
(a + b + a - b)/(a + b – a + b)  = (c + d + c – d)/(c + d – c + d)
2a/2b = 2c/2d
a/b = c/d

Question 8: If 15(2x2  – y2) = 7xy, find x: y, if x and y both are positive.

Solution 8: 15(2x2  - y2) = 7xy
(2x2 - y2)/xy = 7/15
2x/y – y/x = 7/15
Let  x/y = a
∴ 2a – 1/a = 7/15
(2a2 – 1)/a = 7/15
30a2 – 15 = 7a
30a2  – 7a – 15 = 0
30a2 – 25a + 18a – 15 = 0
5a(6a – 5) + 3(6a – 5) = 0
(6a – 5)(5a + 3) = 0
a = 5/6, -3/5
But, a cannot be negative.
∴ a = 5/6
⇒ x/y = 5/6
⇒ x : y = 5 : 6

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 y+ z2, i.e.,
(xy + yz)2 = (x2 + y2)(y2 + z2)
L.H.S = (xy + yz)2
= [y(x + z)]2
= y2(x + z)2
= xz(x + z)2
R.H.S. = (x2 + y2)(y2 + z2)
= (x2 + xz)(xz + z2)
= x(x + z)z(x + z)
= xz(x + z)2
L.H.S. = R.H.S.
Hence, proved.

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:
Let the number of boys be 3x.
Then, number of girls = 2x
∴ 3x + 2x = 630
⇒ 5x = 630
⇒ x = 126
⇒ Number of boys = 3x = 3×126 = 378
And, Number of girls = 2x = 2×126 = 252

After admission of 90 new students, we have
total number of students = 630 + 90 = 720
Now, let the number of boys be 7x.
Then, number of girls = 5x
∴ 7x + 5x = 720
⇒ 12x = 720
⇒ x =  60
⇒ Number of boys = 7x = 7×60 = 420
And, Number of girls = 5x = 5×60 = 300
∴ Number of newly admitted boys = 420 - 378 = 42