# 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 a**

^{3}

**– a**

^{2}

**b**

**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)

⇒ x

^{2}= 48 + 24√3 - 24√3 - 36

⇒ x

^{2}= 12

⇒ x = 2√3

(ii) Let the mean proportional between a - b and a

⇒ a – b, x, a

⇒ a – b : x = x: a

⇒ x×x = (a - b)(a

⇒ x

⇒ x = a(a - b)

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

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.

∴ (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 + x

^{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)

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

**Solution 4:**(a – b)/(a + b) = 1/1111a –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 + x

^{2}= 300 + 20x + 15x + x^{2}
49x - 35x = 300 - 258

14x = 42

x = 3

Thus, the required number which should be added is 3.

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

(2x

2x/y – y/x = 7/15

Let x/y = a

∴ 2a – 1/a = 7/15

(2a

30a

30a

30a

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

Therefore, y

Now, we have to prove that xy + yz is the mean proportional between x

(xy + yz)

L.H.S = (xy + yz)

= [y(x + z)]

= y

= xz(x + z)

R.H.S. = (x

= (x

= x(x + z)z(x + z)

= xz(x + z)

L.H.S. = R.H.S.

Hence, proved.

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

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(2**x^{2}**– y**^{2}**) = 7xy, find x: y, if x and y both are positive.**

**Solution 8:**15(2x^{2}- y2) = 7xy(2x

^{2}- y^{2})/xy = 7/152x/y – y/x = 7/15

Let x/y = a

∴ 2a – 1/a = 7/15

(2a

^{2}– 1)/a = 7/1530a

^{2}– 15 = 7a30a

^{2}– 7a – 15 = 030a

^{2}– 25a + 18a – 15 = 05a(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**x^{2}**+ y**^{2}**and y**^{2}**+ z**^{2}

**Solution 9:**Since y is the mean proportion between x and z.Therefore, y

^{2}= xzNow, we have to prove that xy + yz is the mean proportional between x

^{2}+ y^{2}and y^{2 }+ z^{2}, i.e.,(xy + yz)

^{2}= (x^{2}+ y^{2})(y^{2}+ z^{2})L.H.S = (xy + yz)

^{2}= [y(x + z)]

^{2}= y

^{2}(x + z)^{2}= xz(x + z)

^{2}R.H.S. = (x

^{2}+ y^{2})(y^{2}+ z^{2})= (x

^{2}+ xz)(xz + z^{2})= 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.****Let the number of boys be 3x.**

Solution 10:Solution 10:

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