Selina Chapter 7 Ratio and Proportions ICSE Solutions Class 10 Maths

ICSE Solutions for Selina Concise Chapter 7 Ratio and Proportions Class 10 Maths

Exercise 7(A) 

1. If a: b = 5: 3, find: 5a – 3b/ 5a + 3b

Solution

Given, a: b = 5: 3

So, a/b = 5/3

Now,

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

Solution

Given, x: y = 4: 7

So, x/y = 4/7

3. If a: b = 3: 8, find the value of 4a + 3b/ 6a – b.

Solution

Given, a: b = 3: 8

So, a/b = 3/8

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

Solution

Given,

(a – b)/ (a + b) = 1/ 11

11a – 11b = a + b

10a = 12b

a/b = 12/10 = 6/5

Now, lets take a = 6k and b = 5k

So,

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

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

Solution

Let consider the required number to be x/y

Now, given that

Ratio of 8/21 to 4/9 = (8/21)/ (4/9) = (8/21) × (9/4) = 6/7

Hence, we have

(x/y)/(7/33) = 6/7

x/y = (6/7)×(7/33)

= 2/11

∴ the required number is 2/11.

6. If (m+n)/(m+3n) = 2/3, find : 2n²/(3m²+ mn)

Solution

Given,

3(m + n) = 2(m + 3n)

⇒ 3m + 3n = 2m + 6n

⇒ m = 3n

⇒ m/n = 3/1

Now,

7. Find x/y; when x² + 6y² = 5xy

Solution

Given,

x² + 6y² = 5xy

Dividing by y² both side, we have

Let x/y = a

So,

a² – 5a + 6 = 0

⇒ (a – 2) (a – 3) = 0

⇒ a = 2 or a = 3

∴ x/y = 2 or 3

8. If the ratio between 8 and 11 is the same as the ratio of 2x – y to x + 2y, find the value of 7x/ 9y.

Solution

Given,

(2x – y)/ (x + 2y) = 8/11

On cross multiplying, we get

11(2x – y) = 8(x + 2y)

⇒ 22x – 11y = 8x + 16y

⇒ 14x = 27y

⇒ x/y = 27/14

So, 7x/9y = (7×27)/(9×14) = 3/2

9. Divide Rs 1290 into A, B and C such that A is 2/5 of B and B: C = 4:3.

Solution

Given,

B: C = 4: 3

So, B/C = 4/3

⇒ C = (3/4) B

And, A = (2/5) B

We know that,

A + B + C = Rs 1290

(2/5) B + B + (3/4) B = 1290

Taking L.C.M,

(8B + 20B + 15B)/20 = 1290

⇒ 43B = 1290×20

⇒ B = 1290× 20/43 = 600

So, A = (2/5)×600 = 240

And, C = (3/4)×600 = 450

∴ A gets Rs 600, B gets Rs 240 and C gets Rs 450

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

Let’s consider the number of boys be 3x.

Then, the number of girls = 2x

⇒ 3x + 2x = 630

5x = 630

x = 126

So, the number of boys = 3x = 3×126 = 378

And, number of girls = 2x = 2×126 = 252

After admission of 90 new students,

Total number of students = 630 + 90 = 720

Here, let take the number of boys to be 7x

And, the number of girls = 5x

⇒ 7x + 5x = 720

⇒ 12x = 720

⇒ x = 720/12

⇒ x = 60

So, the number of boys = 7x = 7×60 = 420

And, the number of girls = 5x = 5×60 = 300

∴ the number of newly admitted boys = 420 – 378 = 42

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

Solution

Let x be subtracted from each term of the ratio 9: 17.

⇒ 27 – 3x = 17 – x

⇒ 10 = 2x

⇒ x = 5

∴ the required number which should be subtracted is 5.

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

Given,

The pocket money of Ravi and Sanjeev are in the ratio 5: 7

So, we can assume the pocket money of Ravi as 5k and that of Sanjeev as 7k.

Also, give that

The expenditure of Ravi and Snajeev are in the ratio 3: 5

So, it can be taken as the expenditure of Ravi as 3m and that of Sanjeev as 5m.

And, each of them saves Rs 80

This can be expressed as below:

5k – 3m = 80 ... (a)

7k – 5m = 80 … (b)

Solving equations (a) and (b), we have

k = 40 and m = 40

∴ the monthly pocket money of Ravi is Rs 5k = Rs 5×40 = Rs 200 and that of Sanjeev is Rs 7k = Rs 7×40 = Rs 280.

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

On assuming that the same amount of work is done one day by all the men 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, we have

Amount of work done by (4x + 1) men in (2x – 3) days

= (4x + 1)×(2x – 3) units of work

Then according to the question, we have

⇒ 8x – 16 = 6x – 9

⇒ 2x = 7

⇒ x = 7/2

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

From the question we have,

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

(i) We have,

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

Thus, the increase in fare = Rs 315 – Rs 245 = Rs 70

(ii) Here we have,

Rs 207 = (9/7) × original bus fare

Original bus fare = Rs 207 × 7/9 = Rs 161

Thus, the increase in fare = Rs 207 – Rs 161 = Rs 46

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

Let’s take the cost of the entry ticket initially and at present to be 10x and 13x respectively.

And let the number of visitors initially and at present be 6y and 5y respectively.

So,

Initially, the total collection = 10x × 6y = 60 xy

And at present, the total collection = 13x × 5y = 65 xy

Hence,

The ratio of total collection = 60 xy: 65 xy = 12: 13

∴ it’s seen that the total collection has been increased in the ratio 12: 13.

Exercise 7(B) 

1. Find the fourth proportional to:

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

Solution

(i) Let’s assume 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 = (3.5 × 4.5)/ 1.5

⇒ x = 10.5

(ii) Let’s assume the fourth proportional to 3a, 6a² and 2ab² be x.

3a: 6a² = 2ab²: x

3a × x = 2ab² x 6a²

3a × x = 12a³b²

x = 4a²b²

2. Find the third proportional to:

(i) 2(2/3) and 4

(ii) a – b and a² – b²

Solution

(i) Let’s take the third proportional to

and 4 be x.

So,, 4, x are in continued proportion.

⇒ 8/3: 4 = 4: x

⇒ (8/3)/4 = 4/x

⇒ x = 16 x 3/8 = 6

(ii) Let’s take the third proportional to a – b and a² – b² be x.

So, a – b, a² – b², x are in continued proportion.

a – b: a² – b² = a² – b²: x

3. Find the mean proportional between:

(i) 6 + 3√3 and 8 – 4√3

(ii) a – b and a³ – a²b

Solution

(i) Let the mean proportional between 6 + 3√3 and 8 – 4√3 be x.

So, 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)

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

Solution

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

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

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

⇒ (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

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

Solution

Given, x², 4 and 9 are in continued proportion

So, we have

x²/4 = 4/9

⇒ x² = 16/9

Thus, x = 4/3

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

Solution

Let assume the number added to be x.

So, (6 + x): (15 + x) :: (20 + x): (43 + x)

⇒ (6 + x)/ (15 + x) = (20 + x)/(43 + x)

⇒ (6 + x) (4 + x) = (20 + x) (43 + x)

⇒ 258 + 6x + 43x + x² = 300 + 20x = 15x + x²

⇒ 49x – 35x = 300 – 258

⇒ 14x = 42

⇒ x = 3

∴ the required number which should be added is 3.

7. (i) If a, b, c are in continued proportion,

Show that: (a² + b²)/b(a+c) = b(a+c)/(b² + c²)

Solution

Given,

a, b, c are in continued proportion.

So, we have

a/b = b/c

⇒ b² = ac

Now,

(a² + b²) (b² + c²) = (a² + ac) (ac + c²) [As b² = ac]

= a(a + c) c(a + c)

= ac(a + c)²

= b²(a + c)²

(a² + b²) (b² + c²) = [b(a + c)][b(a + c)]

Thus, L.H.S = R.H.S

Hence, Proved

(ii) If a, b, c are in continued proportion and a(b – c) = 2b, prove that: a – c = 2(a + b)/ a

Solution

Given,

a, b, c are in continued proportion.

So, we have

a/b = b/c

⇒ b² = ac

And, given a(b – c) = 2b

ab – ac = 2b

⇒ ab – b² = 2b

⇒ ab = 2b + b²

⇒ ab = b(2 + b)

⇒ a = b + 2

⇒ a – b = 2

Now, taking the L.H.S we have

L.H.S = a – c

= a(a – c)/ a [Multiply and divide by a]

= a² – ac/ a

= a² – b²/ a

= (a – b) (a + b)/a

= 2(a + b)/a

= R.H.S

Hence, Proved

(iii) If a/b = c/d, show that: (a³c + ac³)/(b³d + bd³) = (a+c)⁴/(b+d)⁴

Solution

Let’s take a/b = c/d = k

So, a = bk and c = dk

Taking L.H.S,

Now, taking the R.H.S

Thus, L.H.S = R.H.S

Hence, Proved

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

Let’s assume the number subtracted to be x.

So, we have

(7 – x): (17 – x):: (17 – x): (47 – x)

⇒ (7 – x)(47 – x) = (17 – x)²

⇒ 329 – 47x – 7x + x² = 289 – 34x + x²

⇒ 329 – 289 = -34x + 54x

⇒ 20x = 40

⇒ x = 2

∴ the required number which must be subtracted is 2.

Exercise 7(C) 

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

(i) Given, a/b = c/d

(ii) Given, a/b = c/d

On cross-multiplication we have,

⇒ (9a + 13b)(9c – 13d) = (9c + 13d)(9a – 13b)

(iii) Given, a/b = c/d

Hence, Proved

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

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

Solution

Given, a/b = c/d

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

Hence, Proved

3. Given, a/b = c/d, prove that:

(3a – 5b)/(3a + 5b) = (3c – 5d)(3c + 5d)

Solution

4. If  (5x+ 6y)/(5u + 6v) = (5x -6y)/(5u-6v);

Then prove that x: y = u: v

Solution

⇒ 10x/12y = 10u/12v

Thus, x/y = u/v ⇒ x: y = u: v

5. If (7a + 8b) (7c – 8d) = (7a – 8b) (7c + 8d);

Prove that a: b = c: d

Solution

The given can the rewritten as,

6. (i) If x = 6ab/ (a + b), find the value of:

(x+ 3a)/(x - 3a) + (x +3b)/(x - 3b)

Solution

Given, x = 6ab/ (a + b)

⇒ x/3a = 2b/ a + b

Now, applying componendo and dividendo we have

Again, x = 6ab/ (a + b)

⇒ x/3b = 2a/ a + b

Now, applying componendo and dividendo we have

From (1) and (2), we get

(ii) If a = 4√6/ (√2 + √3), find the value of:

(a + 2√2)/(a-2√2) + (a+2√3)/(a-2√3)

Solution

Given, a = 4√6/(√2 + √3)

a/2√2 = 2√3/(√2 + √3)

Now, applying componendo and dividendo we have

Again, a = 4√6/(√2 + √3)

a/2√3 = 2√2/(√2 + √3)

Now, applying componendo and dividendo we have

From (1) and (2), we have

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

Rewriting the given, we have

Now, applying componendo and dividendo

Applying componendo and dividendo again, we get

Hence, Proved

Exercise 7(D) 

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

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

Solution

Given, a/b = 3/5

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

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

Solution

Given, 

On cross multiplying, we get

45x + 54y = 64x + 40y

⇒ 14y = 19x

Thus, x/y = 14/19

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

Solution

Given, (3x – 4y): (2x – 3y) = (5x – 6y): (4x – 5y)

This can be rewritten as,

⇒ 5x – 7y = 9x – 11y

⇒ 4y = 4x

⇒ x/y = 1/1

Thus, x: y = 1: 1

4. Find the:

(i) duplicate ratio of 2√2: 3√5

(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

(i) Duplicate ratio of 2√2: 3√5 = (2√2)²: (3√5)² = 8: 45

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

(iii) Sub-duplicate ratio of 9x²a⁴: 25y⁶b² = √(9x²a⁴): √(25y⁶b²) = 3xa²: 5y³b

(iv) Sub-triplicate ratio of 216: 343 = (216)^1/3: (343)^1/3 = 6: 7

(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 = 

5. Find the value of x, if:

(i) (2x + 3): (5x – 38) is the duplicate ratio of √5: √6. 

(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

(i) (2x + 3): (5x – 38) is the duplicate ratio of √5: √6

And, the duplicate ratio of √5: √6 = 5: 6

So, (2x + 3)/(5x – 38) = 5/6

⇒ 12x + 18 = 25x – 190

⇒ 25x – 12x = 190 + 18

⇒ 13x = 208

⇒ x = 208/13 = 16

(ii) (2x + 1): (3x + 13) is the sub-duplicate ratio of 9: 25

Then the sub-duplicate ratio of 9: 25 = 3: 5

⇒ (2x + 1)/(3x + 13) = 3/5

⇒ 10x + 5 = 9x + 39

⇒ x = 34

(iii) (3x – 7): (4x + 3) is the sub-triplicate ratio of 8: 27

And the sub-triplicate ratio of 8: 27 = 2: 3

(3x – 7)/(4x + 3) = 2/3

⇒ 9x – 8x = 6 + 21

⇒ x = 27

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

Solution

Let’s assume the required quantity which has to be added be p.

So, we have

⇒ dx + pd = cy + cp

⇒ pd – cp = cy – dx

⇒ p(d – c) = cy – dx

⇒ p = cy –dx/ (d – c)

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

Solution

Let’s consider the woman’s reduced weight as x.

Given, the original weight = 84 kg

So, we have

84: x = 7: 5

⇒ 84/x = 7/5

⇒ 84 x 5 = 7x

⇒ x = (84 x 5)/ 7

⇒ x = 60

∴ the reduced weight of the woman is 60 kg.

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

Solution

15(2x² – y²) = 7xy

Let’s take the substitution as 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

So, 6a – 5 = 0 or 5a + 3 = 0

a = 5/6 or a = -3/5

As, a cannot be taken negative (ratio)

Thus, a = 5/6

x/y = 5/6

Hence, x: y = 5: 6

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

(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)

⇒ n² = (x -y) (x³ – x²y)

⇒ n² = (x -y) x²(x – y)

⇒ n² = x² (x – y)²

⇒ n = x(x – y)

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

Solution

Let’s assume the required numbers be a and b.

Given, 14 is the mean proportional between a and b.

a: 14 = 14: b

⇒ ab = 196

⇒ a = 196/b …. (1)

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

a: b = b: 112

⇒ b² = 112a …. (2)

Using (1), we have:

b² = 112 × (196/b)

⇒ b³ = 14³ x 2³

⇒ b = 28

From (1),

a = 196/ 28 = 7

∴ the two numbers are 7 and 28.

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

Given,

⇒ x(y² + z² + 2yz) = y(x² + z² + 2xz)

⇒ xy² + xz² + 2yzx = x²y + z²y + 2xzy

⇒ xy² + xz² = x²y + z²y

⇒ xy(y – x) = z²(y – x)

⇒ xy = z²

∴ z is mean proportional between x and y.

12. If  x = 2ab/a+b , find the value of (x+a)/(x-a) + (x+b)/(x-b) .

Solution

x = 2ab/(a + b)

⇒ x/a = 2b/(a + b)

Applying componendo and dividendo,

Also, x = 2ab/ (a + b)

x/b = 2a/ (a + b)

Applying componendo and dividendo, we have

Now, comparing (1) and (2) we have

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

a: b = c: d.

Solution

Given,

Applying componendo and dividendo, we get

⇒ 8a/18b = 8c/18d

⇒a/b = c/d

Hence, Proved