## Selina Concise Mathematics Class 10 ICSE Solutions Ratio and Proportion (Including Properties and Uses)

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

**Exercise 7(A)**

**Question 1.**

**Solution:**

**Question 2.**

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

**Solution:**

**Question 3.**

**Solution:**

**Question 4.**

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

**Solution:**

**Question 5.**

**Solution:**

**Question 6.**

**Solution:**

**Question 7.**

**Solution:**

**Question 8.**

**Solution:**

**Question 9.**

**Solution:**

**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:**

**Question 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.

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:**

**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:**

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:**

**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:**

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:**

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:**

**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:**

**Question 19(i).**

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

**Solution:**

**Question 19(ii).**

If 2a = 3b and 4b = 5c, find: a : c.

**Solution:**

**Question 20.**

**Solution:**

**Question 21.**

Find duplicate ratio of:

(i) 3: 4 (ii) 3√3 : 2√5

**Solution:**

(i) Duplicate ratio of 3 : 4 = 32 : 42 = 9 : 16

(ii) Duplicate ratio of 3√3 : 2√5 = (3√3)² : (2√5)² = 27 : 20

**Question 22.**

**Solution:**

**Question 23.**

Find sub-duplicate ratio of:

(i) 9: 16 (ii) (x – y)^{4}: (x + y)^{6}

**Solution:**

**Question 24.**

Find the sub-triplicate ratio of:

(i) 64: 27 (ii) x^{3}: 125y^{3}

**Solution:**

(i) Sub-triplicate ratio of 64 : 27 = ∛64 : ∛27 = 4 : 3

(ii) Sub-triplicate ratio of x³ : 125y³ = ∛x³ : ∛125y³ = x : 5y

**Question 25.**

**Solution:**

**Question 26.**

If (x + 3) : (4x + 1) is the duplicate ratio of 3 : 5, find the value of x.

**Solution:**

**Question 27.**

If m: n is the duplicate ratio of m + x: n + x; show that x^{2} = mn.

**Solution:**

**Question 28.**

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

**Solution:**

**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:**

**Question 30(a).**

If r^{2 }_{=}pq, show that p : q is the duplicate ratio of (p + r) : (q + r).

**Solution:**

**Question 30(b).**

**Solution:**

**Exercise 7(B)**

**Question 1.**

Find the fourth proportional to:

(i) 1.5, 4.5 and 3.5 (ii) 3a, 6a^{2} and 2ab^{2}

**Solution:**

(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

(ii) Let the fourth proportional to 3a, 6a^{2} and 2ab^{2} be x.

⇒ 3a : 6a^{2} = 2ab^{2} : x

⇒ 3a × x = 2ab^{2} 6a^{2}

⇒ 3a × x = 12a^{3}b^{2}

⇒ x = 4a^{2}b^{2}

**Question 2.**

**Solution:**

**Question 3.**

**Solution:**

(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^{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 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.

⇒ (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)

⇒ x^{2} + 25 + 10x = x^{2} + 2x + 9x + 18

⇒ 25 – 18 = 11x – 10x

⇒ x = 7

**Question 5.**

If x^{2}, 4 and 9 are in continued proportion, find x.

**Solution:**

**Question 6.**

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

**Solution:**

**Question 7(i).**

**Solution:**

**Question 7(ii).**

**Solution:**

**Question 7(iii).**

**Solution:**

**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:**

**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:**

Since y is the mean proportion between x and z

Therefore, y^{2 }= xz

Now, we have to prove that xy+yz is the mean proportional between x^{2}+y^{2} and y^{2}+z^{2}, 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:**

Given, q is the mean proportional between p and r.

⇒ q^{2} = 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:**

Let x, y and z be the three quantities which are in continued proportion.

Then, x : y :: y : z ⇒ y^{2} = xz ….(1)

Now, we have to prove that

x : z = x^{2 }: y^{2}

That is we need to prove that

xy^{2 }= x^{2}z

LHS = xy^{2} = x(xz) = x^{2}z = RHS [Using (1)]

Hence, proved.

**Question 12.**

**Solution:**

Given, y is the mean proportional between x and z.

⇒ y^{2} = xz

**Question 13.**

**Solution:**

**Question 14.**

Find two numbers such that the mean mean proportional between them is 12 and the third proportional to them is 96.

**Solution:**

**Question 15.**

**Solution:**

**Question 16.**

If p: q = r: s; then show that:

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

**Solution:**

**Question 17.**

**Solution:**

**Question 18.**

**Solution:**

**Question 19.**

**Solution:**

**Question 20.**

**Solution:**

**Exercise 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:**

**Question 2.**

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

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

**Solution:**

**Question 3.**

**Solution:**

**Question 4.**

**Solution:**

**Question 5.**

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

**Solution:**

**Question 6.**

**Solution:**

**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:**

**Question 8.**

**Solution:**

**Question 9.**

**Solution:**

**Question 10.**

**Solution:**

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

**Question 11.**

**Solution:**

**Question 12.**

**Solution:**

**Question 13.**

**Solution:**

**Question 14.**

**Solution:**

**Question 15.**

**Solution:**

**Exercise 7(D)**

**Question 1.**

If a: b = 3: 5, find:

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

**Solution:**

**Question 2.**

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

**Solution:**

**Question 3.**

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

**Solution:**

**Question 4.**

Find the:

(i) duplicate ratio of 2√2 : 3√5

(ii) triplicate ratio of 2a: 3b

(iii) sub-duplicate ratio of 9x^{2}a^{4 }: 25y^{6}b^{2}

(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:**

**Question 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:**

**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:**

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:**

**Question 8.**

If 15(2x^{2} – y^{2}) = 7xy, find x: y; if x and y both are positive.

**Solution:**

**Question 9.**

Find the:

(i) fourth proportional to 2xy, x^{2} and y^{2}.

(ii) third proportional to a^{2} – b^{2} and a + b.

(iii) mean proportional to (x – y) and (x^{3} – x^{2}y).

**Solution:**

**Question 10.**

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

**Solution:**

**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:**

**Question 12.**

**Solution:**

Since, q is the mean proportional between p and r,

q^{2} = pr

**Question 13.**

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

a: c = (a^{2} + b^{2}) : (b^{2} + c^{2})

**Solution:**

**Question 14.**

**Solution:**

**Question 15.**

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

a: b = c: d.

**Solution:**

**Question 16.**

**Solution:**

**Question 17.**

**Solution:**

**Question 18.**

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:**

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 19.**

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

**Solution:**

**Question 20.**

**Solution:**

**Question 21.**

**Solution:**

**Question 22.**

**Solution:**

**Question 23.**

**Solution:**

**Question 24.**

**Solution:**

**Question 25.**

**Solution:**

**Question 26.**

**Solution:**

**More Resources for Selina Concise Class 10 ICSE Solutions**

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