Ratio & Proportion — Exercise 7(A)

Find the ratio between :

(i) 60 paise and ₹ 1.35

(ii) 1.8 m and 75 cm

(iii) 1.5 kg and 600 g

(iv) 35 min and $1\dfrac{3}{4}$ hrs

Answer

(i) Given,

60 paise and ₹ 1.35

₹ 1 = 100 paise

₹ 1.35 = 135 paise

Required ratio = $\dfrac{60}{135} = \dfrac{4}{9}$

= 4 : 9

Hence, ratio = 4 : 9.

(ii) Given,

1.8 m and 75 cm

1 m = 100 cm

1.8 m = 180 cm

Required ratio = $\dfrac{180}{75} = \dfrac{12}{5}$

= 12 : 5.

Hence, ratio = 12 : 5.

(iii) Given,

1.5 kg and 600 g

1 kg = 1000 g

1.5 kg = 1500 g

Required ratio = $\dfrac{1500}{600} = \dfrac{5}{2}$

= 5 : 2.

Hence, ratio = 5 : 2.

(iv) Given,

35 min and $1\dfrac{3}{4}$ hrs

1 hour = 60 min

$\Rightarrow 1\dfrac{3}{4} = \dfrac{7}{4}$ hrs

= $\dfrac{7}{4} \times 60$ = 105 mins.

Required ratio = $\dfrac{35}{105} = \dfrac{1}{3}$.

Hence, ratio = 1 : 3.

If A : B = 5 : 6 and B : C = 9 : 11, find (i) A : C (ii) A : B : C.

Answer

Given,

A : B = 5 : 6

B : C = 9 : 11

Taking L.C.M of two values of B i.e. 6 and 9 = 18.

So, $\dfrac{A}{B} = \dfrac{5 \times 3}{6 \times 3} = \dfrac{15}{18}$ = 15 : 18

and $\dfrac{B}{C} = \dfrac{9 \times 2}{11 \times 2} = \dfrac{18}{22}$ = 18 : 22

∴ A : B : C = 15 : 18 : 22

⇒ A : C = 15 : 22.

(i)

Hence, A : C = 15 : 22.

(ii)

Hence, A : B : C = 15 : 18 : 22.

If P : Q = 7 : 4 and Q : R = 5 : 14, find (i) R : P (ii) P : Q : R.

Answer

Given,

P : Q = 7 : 4

Q : R = 5 : 14

To find R : P and P : Q : R, we will make Q same in both cases.

Taking L.C.M of two values of Q i.e. 4 and 5 = 20

So, $\dfrac{P}{Q} = \dfrac{7 \times 5}{4 \times 5} = \dfrac{35}{20}$ = 35 : 20

and $\dfrac{Q}{R} = \dfrac{5 \times 4}{14 \times 4} = \dfrac{20}{56}$ = 20 : 56

$\Rightarrow \dfrac{P}{Q} \times \dfrac{Q}{R} = \dfrac{35}{20} \times \dfrac{20}{56} \\[1em] \Rightarrow \dfrac{P}{R} = \dfrac{35}{56} \\[1em] \Rightarrow \dfrac{P}{R} = \dfrac{5}{8} \\[1em] \Rightarrow \dfrac{R}{P} = \dfrac{8}{5}.$

Hence, R : P = 8 : 5.

∴ P : Q : R = 35 : 20 : 56

Hence, ratio of P : Q : R = 35 : 20 : 56.

If 3A = 5B = 6C, find A : B : C.

Answer

Let 3A = 5B = 6C = k.

$A = \dfrac{k}{3}, B = \dfrac{k}{5}, C = \dfrac{k}{6}$

$\Rightarrow A : B : C = \dfrac{k}{3} : \dfrac{k}{5} : \dfrac{k}{6} \\[1em] \Rightarrow A : B : C = \dfrac{1}{3} : \dfrac{1}{5} : \dfrac{1}{6}$

Taking L.C.M of values of 3, 5, 6 = 30.

$\Rightarrow A : B : C = \dfrac{1}{3} \times 30 : \dfrac{1}{5} \times 30 : \dfrac{1}{6} \times 30 \\[1em] \Rightarrow A : B : C = 10 : 6 : 5.$

Hence, A : B : C = 10 : 6 : 5.

If $\dfrac{A}{2} = \dfrac{B}{3} = \dfrac{C}{6}$, find A : B : C.

Answer

Let, $\dfrac{A}{2} = \dfrac{B}{3} = \dfrac{C}{6}$ = k.

∴ A = 2k, B = 3k, C = 6k.

A : B : C = 2k : 3k : 6k

A : B : C = 2 : 3 : 6.

Hence, A : B : C = 2 : 3 : 6.

If a : b = 8 : 5, find (7a + 5b) : (8a − 9b).

Answer

Given,

a : b = 8 : 5

Let a = 8x and b = 5x.

Substituting values in $\dfrac{7a + 5b}{8a − 9b}$ we get,

$\Rightarrow \dfrac{7(8x) + 5(5x)}{8(8x) - 9(5x)} \\[1em] \Rightarrow \dfrac{56x + 25x}{64x - 45x} \\[1em] \Rightarrow \dfrac{81x}{19x} \\[1em] \Rightarrow \dfrac{81}{19}.$

Hence, 7a + 5b : 8a − 9b = 81 : 19.

If x : y = 3 : 2, find (5x − 3y) : (7x + 2y).

Answer

Given,

x : y = 3 : 2

Let x = 3a and y = 2a.

Substituting values in $\dfrac{5x − 3y}{7x + 2y}$ we get,

$\Rightarrow \dfrac{5(3a) - 3(2a)}{7(3a) + 2(2a)} \\[1em] \Rightarrow \dfrac{15a - 6a}{21a + 4a} \\[1em] \Rightarrow \dfrac{9a}{25a} \\[1em] \Rightarrow \dfrac{9}{25}.$

Hence, 5x − 3y : 7x + 2y = 9 : 25.

If x : y = 10 : 3, find (3x² + 2y²) : (3x² − 2y²).

Answer

Given,

x : y = 10 : 3

Let x = 10a and y = 3a.

Substituting values in $\dfrac{3x^2 + 2y^2}{3x^2 − 2y^2}$ we get,

$\Rightarrow \dfrac{3(10a)^2 + 2(3a)^2}{3(10a)^2 - 2(3a)^2} \\[1em] \Rightarrow \dfrac{3(100a^2) + 2(9a^2)}{3(100a^2) - 2(9a^2)} \\[1em] \Rightarrow \dfrac{300a^2 + 18a^2}{300a^2 - 18a^2} \\[1em] \Rightarrow \dfrac{318a^2}{282a^2} \\[1em] \Rightarrow \dfrac{53}{47}.$

Hence, (3x² + 2y²) : (3x² − 2y²) = 53 : 47.

If a : b = 2 : 5, find (3a² − 2ab + 5b²) : (a² + 7ab − 2b²).

Answer

Given,

a : b = 2 : 5

Let a = 2x, then b = 5x.

Substituting values in $\dfrac{3a^2 − 2ab + 5b^2}{a^2 + 7ab − 2b^2}$ we get,

$\Rightarrow \dfrac{3 \times (2x)^2 − 2 \times (2x) \times (5x) + 5 \times (5x)^2}{(2x)^2 + 7 \times (2x) \times (5x) − 2 \times (5x)^2} \\[1em] \Rightarrow \dfrac{12x^2 − 20x^2 + 125x^2}{4x^2 + 70x^2 − 50x^2} \\[1em] \Rightarrow \dfrac{137x^2 − 20x^2}{74x^2 − 50x^2} \\[1em] \Rightarrow \dfrac{117x^2}{24x^2} \\[1em] \Rightarrow \dfrac{39}{8}.$

Hence, 3a² − 2ab + 5b² : a² + 7ab − 2b² = 39 : 8 .

If (5x + 2y) : (7x + 4y) = 13 : 20, find x : y.

Answer

Given,

(5x + 2y) : (7x + 4y) = 13 : 20,

$\Rightarrow \dfrac{5x + 2y}{7x + 4y} = \dfrac{13}{20} \\[1em] \Rightarrow 20(5x + 2y) = 13(7x + 4y) \\[1em] \Rightarrow 100x + 40y = 91x + 52y \\[1em] \Rightarrow 100x - 91x = 52y - 40y \\[1em] \Rightarrow 9x = 12y \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{12}{9} \\[1em] \Rightarrow \dfrac{x}{y} =\dfrac{4}{3}.$

Hence, x : y = 4 : 3.

If (3x + 5y) : (3x − 5y) = 7 : 3, find x : y.

Answer

Given,

(3x + 5y) : (3x − 5y) = 7 : 3,

$\Rightarrow \dfrac{3x + 5y}{3x - 5y} = \dfrac{7}{3} \\[1em] \Rightarrow 3(3x + 5y) = 7(3x - 5y) \\[1em] \Rightarrow 9x + 15y = 21x - 35y \\[1em] \Rightarrow 15y + 35y = 21x - 9x \\[1em] \Rightarrow 50y = 12x \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{50}{12} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{25}{6}.$

Hence, x : y = 25 : 6.

If (6x² − xy) : (2xy − y²) = 6 : 1, find x : y.

Answer

Given,

(6x² − xy) : (2xy − y²) = 6 : 1,

Dividing numerator and denominator by y², we get :

$\Rightarrow \dfrac{\dfrac{6x^2 − xy}{y^2}}{\dfrac{2xy − y^2}{y^2}} = \dfrac{6}{1} \\[1em] \Rightarrow \dfrac{6\Big(\dfrac{x}{y}\Big)^2 - \dfrac{x}{y}}{2\dfrac{x}{y} - 1} = 6$

Let, $\dfrac{x}{y}$ = t.

$\Rightarrow \dfrac{6t^2 - t}{2t - 1} = 6 \\[1em] \Rightarrow 6t^2 - t = (2t - 1) \times 6 \\[1em] \Rightarrow 6t^2 - t = 12t - 6 \\[1em] \Rightarrow 6t^2 - t - 12t + 6 = 0 \\[1em] \Rightarrow 6t^2 - 13t + 6 = 0 \\[1em] \Rightarrow 6t^2 - 4t - 9t + 6 = 0 \\[1em] \Rightarrow 2t(3t - 2) - 3(3t - 2) = 0 \\[1em] \Rightarrow (2t - 3)(3t - 2) = 0 \\[1em] \Rightarrow (2t - 3) = 0 \text{ or }(3t - 2) = 0 \text{ [Using zero - product rule] } \\[1em] \Rightarrow 2t = 3 \text{ or } 3t = 2 \\[1em] \Rightarrow t = \dfrac{3}{2} \text{ or } t = \dfrac{2}{3} .$

Thus,

$\dfrac{x}{y} = \dfrac{3}{2} \text{ or } \dfrac{x}{y} = \dfrac{2}{3}$.

Hence, x : y = 3 : 2 or 2 : 3.

If (4x² − 3y²) : (2x² + 5y²) = 12 : 19, find x : y.

Answer

Given,

(4x² − 3y²) : (2x² + 5y²) = 12 : 19,

Dividing numerator and denominator both by y² we get,

$\Rightarrow \dfrac{\dfrac{4x^2 − 3y^2}{y^2}}{\dfrac{2x^2 + 5y^2}{y^2}} = \dfrac{12}{19} \\[1em] \Rightarrow \dfrac{4\Big(\dfrac{x}{y}\Big)^2 - 3\dfrac{y^2}{y^2}}{2\Big(\dfrac{x}{y}\Big)^2 + 5\dfrac{y^2}{y^2}} = \dfrac{12}{19} \\[1em] \Rightarrow \dfrac{4\Big(\dfrac{x}{y}\Big)^2 - 3}{2\Big(\dfrac{x}{y}\Big)^2 + 5} = \dfrac{12}{19}$

Let, $\dfrac{x}{y}$ = t

$\Rightarrow \dfrac{4t^2 - 3}{2t^2 + 5} = \dfrac{12}{19} \\[1em] \Rightarrow (4t^2 - 3) \times 19 = (2t^2 + 5) \times 12 \\[1em] \Rightarrow 76t^2 - 57 = 24t^2 + 60 \\[1em] \Rightarrow 76t^2 - 24t^2 = 60 + 57 \\[1em] \Rightarrow 52t^2 = 117 \\[1em] \Rightarrow t^2 = \dfrac{117}{52} \\[1em] \Rightarrow t^2 = \dfrac{9}{4} \\[1em] \Rightarrow t = \sqrt{\dfrac{9}{4}} \\[1em] \Rightarrow t = \dfrac{3}{2}.$

Hence, x : y = 3 : 2.

If x² + 4y² = 4xy, find x : y.

Answer

Given,

x² + 4y² = 4xy

Dividing both sides by xy we get,

$\Rightarrow \dfrac{x^2 + 4y^2}{xy} = \dfrac{4xy}{xy} \\[1em] \Rightarrow \dfrac{x}{y} + 4\dfrac{y}{x} = 4$

Let $\dfrac{x}{y}$ = t

$\Rightarrow t + 4\dfrac{1}{t} = 4 \\[1em] \Rightarrow \dfrac{t^2 + 4}{t} = 4 \\[1em] \Rightarrow t^2 + 4 = 4t \\[1em] \Rightarrow t^2 - 4t + 4 = 0 \\[1em] \Rightarrow t^2 - 2t - 2t + 4 = 0 \\[1em] \Rightarrow t(t - 2) - 2(t - 2) = 0 \\[1em] \Rightarrow (t - 2)(t - 2) = 0 \\[1em] \Rightarrow t - 2 = 0 \text{ or } t - 2 = 0 \\[1em] \Rightarrow t = 2 \\[1em] \Rightarrow \dfrac{x}{y} = 2.$

Hence, x : y = 2 : 1.

If 10x² − 23xy + 9y² = 0, find x : y.

Answer

Given,

10x² − 23xy + 9y² = 0

Dividing both sides by xy we get,

$\Rightarrow \dfrac{10x^2 − 23xy + 9y^2}{xy} = 0 \\[1em] \Rightarrow 10\dfrac{x}{y} - 23 + 9\dfrac{y}{x} = 0 \\[1em] \Rightarrow 10\dfrac{x}{y} + 9\dfrac{y}{x} = 23$

Let, $\dfrac{x}{y}$ = t

$\Rightarrow 10t + 9\dfrac{1}{t} = 23 \\[1em] \Rightarrow \dfrac{10t^2 + 9}{t} = 23 \\[1em] \Rightarrow 10t^2 + 9 = 23t \\[1em] \Rightarrow 10t^2 - 23t + 9 = 0 \\[1em] \Rightarrow 10t^2 - 5t - 18t + 9 = 0 \\[1em] \Rightarrow 5t(2t - 1) - 9(2t - 1) = 0 \\[1em] \Rightarrow (5t - 9)(2t - 1) = 0 \\[1em] \Rightarrow 5t - 9 = 0 \text{ or } 2t - 1 = 0 \\[1em] \Rightarrow 5t = 9 \text{ or } 2t = 1 \\[1em] \Rightarrow t = \dfrac{9}{5} \text{ or } t = \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{9}{5} \text{ or } \dfrac{x}{y} = \dfrac{1}{2}$

Hence, x : y = 9 : 5 or 1 : 2.

A ratio is equal to 3 : 4. If its consequent is 144, what is its antecedent?

Answer

Let the antecedent be x. Then,

3 : 4 = x : 144

$\Rightarrow \dfrac{3}{4} = \dfrac{x}{144} \\[1em] \Rightarrow 3 \times 144 = 4x \\[1em] \Rightarrow 432 = 4x \\[1em] \Rightarrow x = \dfrac{432}{4} \\[1em] \Rightarrow x = 108$

Hence, the antecedent = 108.

Two numbers are in the ratio 8 : 13. If 14 is added to each of the numbers, the ratio becomes 2 : 3. Find the numbers.

Answer

Let the numbers be 8x and 13x.

Given,

If 14 is added to each of the numbers, the ratio becomes 2 : 3.

$\Rightarrow \dfrac{8x + 14}{13x + 14} = \dfrac{2}{3} \\[1em] \Rightarrow (8x + 14) \times 3 = (13x + 14) \times 2 \\[1em] \Rightarrow 24x + 42 = 26x + 28 \\[1em] \Rightarrow 26x - 24x = 42 - 28 \\[1em] \Rightarrow 2x = 14 \\[1em] \Rightarrow x = \dfrac{14}{2} \\[1em] \Rightarrow x = 7.$

The two numbers are 8x and 13x

= 8 × 7, 13 × 7

= 56, 91.

Hence, the numbers are 56 and 91.

Two numbers are in the ratio 5 : 7. If 8 is subtracted from each, the ratio becomes 3 : 5. Find the numbers.

Answer

Let the numbers be 5x and 7x.

Given,

If 8 is subtracted from each, the ratio becomes 3 : 5.

$\Rightarrow \dfrac{5x - 8}{7x - 8} = \dfrac{3}{5} \\[1em] \Rightarrow (5x - 8) \times 5 = (7x - 8) \times 3 \\[1em] \Rightarrow 25x - 40 = 21x - 24\\[1em] \Rightarrow 25x - 21x = 40 - 24 \\[1em] \Rightarrow 4x = 16 \\[1em] \Rightarrow x = \dfrac{16}{4} \\[1em] \Rightarrow x = 4.$

The two numbers are 5x and 7x

= 5 × 4, 7 × 4

= 20, 28.

Hence, the numbers are 20 and 28.

What least number must be added to each term of the ratio 5 : 7 to make it 8 : 9?

Answer

Let x be the least number must be added to each term of the ratio 5 : 7 to make it 8 : 9.

$\Rightarrow \dfrac{5 + x}{7 + x} = \dfrac{8}{9} \\[1em] \Rightarrow (5 + x) \times 9 = (7 + x) \times 8 \\[1em] \Rightarrow 45 + 9x = 56 + 8x \\[1em] \Rightarrow 9x - 8x = 56 - 45 \\[1em] \Rightarrow x = 11.$

Hence, the least number to be added is 11.

Out of the monthly income of ₹ 45,000, Rahul spends ₹ 31,500 and the rest he saves. Find the ratio of his

(i) income to expenditure

(ii) income to savings

(iii) savings to expenditure.

Answer

Given,

Income = ₹ 45,000

Expenditure = ₹ 31,500

Total savings = Income - Expenditure = ₹ 45,000 - 31,500

= ₹ 13,500.

(i) Required ratio,

Income : Expenditure = 45000 : 31500

$= \dfrac{45000}{31500} = \dfrac{10}{7}$

= 10 : 7.

Hence, required ratio = 10 : 7.

(ii) Required ratio,

Income : Savings = 45000 : 13500

$= \dfrac{45000}{13500} = \dfrac{10}{3}$

= 10 : 3.

Hence, required ratio = 10 : 3.

(iii) Required ratio is savings to expenditure.

Savings : expenditure = 13500 : 31500

$=\dfrac{13500}{31500} = \dfrac{3}{7}$

= 3 : 7.

Hence, required ratio = 3 : 7.

The cost of making an umbrella is divided between material, labour and overheads in the ratio 6 : 4 : 1. If the material costs ₹ 132, find the cost of production of an umbrella.

Answer

Given,

The ratio of material : labour : overheads = 6 : 4 : 1

Let the cost of material be 6x, labour be 4x and overheads be x.

Given,

Cost of material = ₹ 132

⇒ 6x = 132

⇒ x = $\dfrac{132}{6}$

⇒ x = 22.

The total cost of production of an umbrella = 6x + 4x + 1x = 11x

= 11 × 22

= ₹ 242.

Hence, cost of production of an umbrella = ₹ 242.

Divide ₹ 6,720 in the ratio 5 : 3.

Answer

Given,

Let ₹ 6,720 be divided in two parts A and B.

A = 5a and B = 3a

To find A's part,

$\Rightarrow \dfrac{5a}{5a + 3a} \times 6,720 \\[1em] \Rightarrow \dfrac{5a}{8a} \times 6,720 \\[1em] \Rightarrow \dfrac{5}{8} \times 6,720 \\[1em] \Rightarrow ₹ 4,200.$

To find B's part,

$\Rightarrow \dfrac{3a}{5a + 3a} \times 6,720 \\[1em] \Rightarrow \dfrac{3a}{8a} \times 6,720 \\[1em] \Rightarrow \dfrac{3}{8} \times 6,720 \\[1em] \Rightarrow ₹ 2,520$

Hence, ₹ 6,720 can be divided into ₹ 4,200 and ₹ 2,520.

Divide ₹ 11,620 among A, B and C in the ratio 35 : 28 : 20.

Answer

Given,

Let A = 35a and B = 28a and C = 20a,

To find A's part,

$\Rightarrow \dfrac{35a}{35a + 28a + 20a} \times 11,620 \\[1em] \Rightarrow \dfrac{35a}{83a} \times 11,620 \\[1em] \Rightarrow \dfrac{35}{83} \times 11,620 \\[1em] \Rightarrow ₹ 4,900.$

To find B's part,

$\Rightarrow \dfrac{28a}{35a + 28a + 20a} \times 11,620 \\[1em] \Rightarrow \dfrac{28a}{83a} \times 11,620 \\[1em] \Rightarrow \dfrac{28}{83} \times 11,620 \\[1em] \Rightarrow ₹ 3,920.$

To find C's part,

$\Rightarrow \dfrac{20a}{35a + 28a + 20a} \times 11,620 \\[1em] \Rightarrow \dfrac{20a}{83a} \times 11,620 \\[1em] \Rightarrow \dfrac{20}{83} \times 11,620 \\[1em] \Rightarrow ₹ 2,800$

Hence, A = ₹ 4,900, B = ₹ 3,920 and C = ₹ 2,800.

Divide ₹ 782 among P, Q and R in the ratio $\dfrac{1}{2} : \dfrac{2}{3} : \dfrac{3}{4}$.

Answer

Given,

P : Q : R = $\dfrac{1}{2} : \dfrac{2}{3} : \dfrac{3}{4}$.

Multiply each ratio by 12 (LCM of denominators) to clear fractions :

= $\dfrac{1}{2} \times 12 : \dfrac{2}{3} \times 12 : \dfrac{3}{4} \times 12$

= 6 : 4 : 9.

Let P = 6a and Q = 8a and R = 9a.

To find P's part,

$\Rightarrow \dfrac{6a}{6a + 8a + 9a} \times 782 \\[1em] \Rightarrow \dfrac{6a}{23a} \times 782 \\[1em] \Rightarrow \dfrac{6}{23} \times 782 \\[1em] \Rightarrow 204$

To find Q's part,

$\Rightarrow \dfrac{8a}{6a + 8a + 9a} \times 782 \\[1em] \Rightarrow \dfrac{8a}{23a} \times 782 \\[1em] \Rightarrow \dfrac{8}{23} \times 782 \\[1em] \Rightarrow 272$

To find Q's part,

$\Rightarrow \dfrac{9a}{6a + 8a + 9a} \times 782 \\[1em] \Rightarrow \dfrac{9a}{23a} \times 782 \\[1em] \Rightarrow \dfrac{9}{23} \times 782 \\[1em] \Rightarrow 306$

Hence, ₹ 782 can be divided into P = ₹ 204, Q = ₹ 272 and R = ₹ 306.

If ₹ 5,100 be divided among A, B, C in such a way that A gets $\dfrac{2}{3}$ of what B gets and B gets $\dfrac{1}{4}$ of what C gets, find their respective shares.

Answer

Given,

A = $\dfrac{2}{3}$ of B and B = $\dfrac{1}{4}$ of C

Express all in terms of B:

A = $\dfrac{2}{3}B$, B = B and C = 4B

Total amount divided among A, B and C = ₹ 5,100

So,

$\Rightarrow \dfrac{2}{3}B + B + 4B = 5100 \\[1em] \Rightarrow \dfrac{2B + 3B + 12B}{3} = 5100 \\[1em] \Rightarrow \dfrac{17B}{3} = 5100 \\[1em] \Rightarrow B = \dfrac{5100 \times 3}{17} \\[1em] \Rightarrow B = ₹ 900.$

B's share = ₹ 900

Therefore,

A's share = $\dfrac{2}{3} \times 900$ = ₹ 600

C's share = 4 × 900 = ₹ 3,600

Hence, A = ₹ 600, B = ₹ 900 and C = ₹ 3,600.

Divide ₹ 8,300 among A, B and C such that 4 times A’s share, 5 times B’s share and 7 times C’s share may all be equal.

Answer

Given,

Let, 4A = 5B = 7C = k

Then, A = $\dfrac{k}{4}$, B = $\dfrac{k}{5}$, C = $\dfrac{k}{7}$

Total amount divided among A, B and C = ₹ 8,300

So,

$\Rightarrow \dfrac{k}{4} + \dfrac{k}{5} + \dfrac{k}{7} = 8300 \\[1em] \Rightarrow k\Big(\dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{7}\Big) = 8300 \\[1em] \Rightarrow k\Big(\dfrac{35 + 28 + 20}{140}\Big) = 8300 \\[1em] \Rightarrow k\Big(\dfrac{83}{140}\Big) = 8300 \\[1em] \Rightarrow k = \dfrac{8300 \times 140}{83} \\[1em] \Rightarrow k = 14000$

Therefore,

A' share = $\dfrac{14000}{4}$ = ₹ 3,500

B's share = $\dfrac{14000}{5}$ = ₹ 2,800

C"s share = $\dfrac{14000}{7}$ = ₹ 2,000

Hence, A = ₹ 3,500, B = ₹ 2,800 and C = ₹ 2,000.

A sum of money is divided between A and B in the ratio 6 : 11. If B’s share is ₹ 7,315, find (i) A’s share (ii) the total amount of money.

Answer

(i) Let A's share be ₹ x.

Then,

$\Rightarrow \dfrac{6}{11} = \dfrac{x}{7315} \\[1em] \Rightarrow x = \dfrac{6}{11} \times 7315 \\[1em] \Rightarrow x = ₹ 3,990.$

Hence, A' share of money = ₹ 3,990.

(ii) Total sum of money = A's share + B's share

= ₹ 3,990 + ₹ 7,315

= ₹ 11,305.

Hence, total sum of money = ₹ 11,305.

The ages of Tanvy and Divya are in the ratio 5 : 7. Five years hence, their ages will be in the ratio 3 : 4. Find their present ages.

Answer

Given,

The ages of Tanvy and Divya are in the ratio 5 : 7.

Let the present age of Tanvy be 5x and the present age of Divya be 7x.

Five years from now :

Tanvy's age will be 5x + 5.

Divya's age will be 7x + 5.

Given,

The ratio of their ages in five years will be 3 : 4.

$\Rightarrow \dfrac{5x + 5}{7x + 5} = \dfrac{3}{4}$

⇒ 4(5x + 5) = 3(7x + 5)

⇒ 20x + 20 = 21x + 15

⇒ 21x - 20x = 20 - 15

⇒ x = 20 - 15

⇒ x = 5.

Tanvy's present age = 5x = 5(5) = 25 years.

Divya's present age = 7x = 7(5) = 35 years.

Hence, present age of Tnavy and Divya are 25 years and 35 years respectively.

One year ago, the ratio of Amit’s and Arun’s ages was 6 : 7 respectively. Four years hence, their ages will be in the ratio 7 : 8. How old is Amit?

Answer

Given,

The ratio of Amit’s and Arun’s ages was 6 : 7.

Let Amit's age one year ago be 6x and Arun's age one year ago be 7x.

Their present ages are :

Amit's present age = 6x + 1

Arun's present age = 7x + 1

After four years their ages will be:

Amit's age will be (6x + 1) + 4 = 6x + 5

Arun's age will be (7x + 1) + 4 = 7x + 5

Given,

The ratio of their ages four years hence will be 7 : 8.

$\therefore \dfrac{6x + 5}{7x + 5} = \dfrac{7}{8}$

⇒ 8(6x + 5) = 7(7x + 5)

⇒ 48x + 40 = 49x + 35

⇒ 40 - 35 = 49x - 48x

⇒ x = 5

Amit's present age is 6x + 1.

= 6(5) + 1 = 30 + 1

= 31 years.

Hence, Amit's present age = 31 years.

Reena reduces her weight in the ratio 5 : 4. What is her weight now, if originally it was 70 kg?

Answer

Given,

Reena reduces her weight in the ratio 5 : 4

Let her orignal weight be 5x and new weight be 4x.

Reena's orignal weight = 70 kg

⇒ 5x = 70

⇒ x = $\dfrac{70}{5}$

⇒ x = 14 kg

Reena's new weight = 4x

= 4 × 14

= 56 kg.

Hence, Reena's present weight = 56 kg.

68 kg of a mixture contains milk and water in the ratio 27 : 7. How much more water is to be added to this mixture to get a new mixture containing milk and water in the ratio 3 : 1?

Answer

Given,

Milk : Water = 27 : 7 and total = 68 kg

Milk = $\dfrac{27}{27 + 7} \times 68 = \dfrac{27}{34} \times 68 = 54\text{ kg}$

Water = $\dfrac{7}{27 + 7} \times 68 = \dfrac{7}{34} \times 68 = 14\text{ kg}$

Let x kg of water be added to make the ratio of milk to water as 3 : 1.

$\Rightarrow \dfrac{54}{14 + x} = \dfrac{3}{1} \\[1em] \Rightarrow 54 = 3(14 + x) \\[1em] \Rightarrow 54 = 42 + 3x \\[1em] \Rightarrow 54 - 42 = 3x \\[1em] \Rightarrow 12 = 3x \\[1em] \Rightarrow x = 4.$

Hence, 4 kg of water must be added.

A mixture contains milk and water in the ratio 5 : 1. On adding 5 litres of water, the ratio of milk to water becomes 5 : 2. Find the quantity of milk in the original mixture.

Answer

Given,

Milk : Water = 5 : 1

Let Milk = 5x and Water = x

After adding 5 litres of water, the ratio of milk to water becomes 5 : 2.

$\Rightarrow \dfrac{5x}{x + 5} = \dfrac{5}{2} \\[1em] \Rightarrow 2(5x) = 5(x + 5) \\[1em] \Rightarrow 10x = 5x + 25 \\[1em] \Rightarrow 10x - 5x = 25 \\[1em] \Rightarrow 5x = 25 \\[1em] \Rightarrow x = 5.$

Milk in the original mixture = 5x = 5 × 5 = 25 litres.

Hence, the quantity of milk in the original mixture = 25 litres.

In an examination, the ratio of passes to failures was 4 : 1. Had 30 less appeared and 20 less passed, the ratio of passes to failures would have been 5 : 1. How many students appeared for the examination?

Answer

Given,

Ratio of Passes to Failures = 4 : 1

Let students who Pass = 4k and Fail = k

Total students appeared for examination = 4k + k = 5k

Given,

30 students didn't appear for examination.

Now total students appeared for exam = 5k − 30

If 20 less students passed the exam, then students that pass the exam now = 4k − 20

Number of students that fail in exam = (5k − 30) − (4k − 20) = k − 10.

The new ratio of passes to failures = 5 : 1

$\therefore \dfrac{4k - 20}{k - 10} = \dfrac{5}{1} \\[1em] \Rightarrow 4k - 20 = 5(k - 10) \\[1em] \Rightarrow 4k - 20 = 5k - 50 \\[1em] \Rightarrow -20 + 50 = 5k - 4k \\[1em] \Rightarrow k = 30.$

Total students who appeared exam = 5k = 5(30) = 150.

Hence, total number of students appeared for the examination = 150.

Find the angles of a triangle which are in the ratio 5 : 4 : 3.

Answer

Given,

Let the angles of triangle be 5x, 4x and 3x.

We know that,

Sum of angles of a triangle = 180°

$\Rightarrow 5x + 4x + 3x = 180 \\[1em] \Rightarrow 12x = 180 \\[1em] \Rightarrow x = \dfrac{180}{12} = 15.$

⇒ 5x = 5(15) = 75°

⇒ 4x = 4(15) = 60°

⇒ 3x = 3(15) = 45°

Hence, the angles of the triangle are 75°, 60° and 45°.

The sides of a triangle are in the ratio $\dfrac{1}{2} : \dfrac{1}{3} : \dfrac{1}{4}$ and its perimeter is 91 cm. Find the lengths of the sides of the triangle.

Answer

Given,

Side₁ : Side₂ : Side₃ = $\dfrac{1}{2} : \dfrac{1}{3} : \dfrac{1}{4}$

= $\dfrac{1}{2} \times 12 : \dfrac{1}{3} \times 12 : \dfrac{1}{4} \times 12$

= 6 : 4 : 3.

Let Side₁ = 6a and Side₂ = 4a and Side₃ = 3a

Sum of sides of triangle = 6a + 4a + 3a = 13a

Given,

Perimeter of triangle = 91 cm

⇒ 13a = 91

⇒ a = $\dfrac{91}{13}$

⇒ a = 7.

Therefore,

Length of Side₁ = 6a = 6(7) = 42 cm

Length of Side₂ = 4a = 4(7) = 28 cm

Length of Side₃ = 3a = 3(7) = 21 cm

Hence, the lengths of the sides are 42 cm, 28 cm and 21 cm.

In a school, the boys and girls are in the ratio 9 : 5. If there are 425 girls, what is the total number of students in the school?

Answer

Given,

Boys : Girls = 9 : 5

Let number of Boys = 9x and Girls = 5x.

Total number of girls in school = 425

So,

⇒ 5x = 425

⇒ x = $\dfrac{425}{5}$

⇒ x = 85.

Therefore,

Number of Boys in school = 9x = 9(85) = 765.

Total number of students in school = 765 + 425 = 1190.

Hence, the total number of students in the school = 1190.

Compare the following ratios :

(i) (7 : 9) and (11 : 16)

(ii) (19 : 25) and (17 : 20)

(iii) $\Big(\dfrac{1}{2} : \dfrac{1}{5}\Big)$ and (5 : 2)

Answer

(i) To compare 2 ratios, the consequent of the first ratio and 2nd ratio must be made equal.

Given,

A : B = 7 : 9 and C : D = 11: 16

L.C.M. of 9 and 16 is 144.

$\Rightarrow \dfrac{A}{B} = \dfrac{7 \times 16}{9 \times 16} = \dfrac{112}{144} \\[1em] \Rightarrow \dfrac{C}{D} = \dfrac{11 \times 9}{16 \times 9} = \dfrac{99}{144} \\[1em] \Rightarrow \dfrac{112}{144} \gt \dfrac{99}{144} \\[1em] \Rightarrow \dfrac{7}{9} \gt \dfrac{11}{16}$

Hence, 7 : 9 > 11 : 16.

(ii) To compare 2 ratios, the consequent of the first ratio and 2nd ratio must be made equal.

Let A : B = 19 : 25 and C : D = 17 : 20.

L.C.M. of 25 and 20 is 100.

$\Rightarrow \dfrac{A}{B} = \dfrac{19 \times 4}{25 \times 4} = \dfrac{76}{100} \\[1em] \Rightarrow \dfrac{C}{D} = \dfrac{17 \times 5}{20 \times 5} = \dfrac{85}{100} \\[1em] \Rightarrow \dfrac{76}{100} \lt \dfrac{85}{100} \\[1em] \Rightarrow \dfrac{19}{25} \lt \dfrac{17}{20}$

Hence, 19 : 25 < 17 : 20.

(ii) To compare 2 ratios, the consequent of the first ratio and 2nd ratio must be made equal.

Let A : B = $\Big(\dfrac{1}{2} : \dfrac{1}{5}\Big)$ and C : D = 5:2

First simplifying A : B,

L.C.M of 2 and 5 is 10.

$\Rightarrow \Big(\dfrac{1}{2} \times 10 : \dfrac{1}{5} \times 10\Big) \\[1em] \Rightarrow 5:2$

Since both ratios are same.

Hence, $\Big(\dfrac{1}{2} : \dfrac{1}{5}\Big)$ = 5 : 2.

Arrange the following ratios in descending order of magnitudes :

(i) (5 : 6), (8 : 9), (13 : 18) and (19 : 24)

(ii) (6 : 7), (13 : 14), (19 : 21) and (23 : 28)

(iii) (7 : 12), (9 : 16), (13 : 20) and (5 : 8)

Answer

(i) Given,

(5 : 6), (8 : 9), (13 : 18) and (19 : 24)

We convert them into equivalent like fractions.

L.C.M of 6, 9, 18, 24 is 72.

$\Rightarrow \dfrac{5 \times 12}{6 \times 12} = \dfrac{60}{72} \\[1em] \Rightarrow \dfrac{8 \times 8}{9 \times 8} = \dfrac{64}{72} \\[1em] \Rightarrow \dfrac{13 \times 4}{18 \times 4} = \dfrac{52}{72} \\[1em] \Rightarrow \dfrac{19 \times 3}{24 \times 3} = \dfrac{57}{72} \\[1em] \Rightarrow \dfrac{64}{72} \gt \dfrac{60}{72} \gt \dfrac{57}{72} \gt \dfrac{52}{72}$

8 : 9 > 5 : 6 > 19 : 24 > 13 : 18

Hence, the ratios in descending order are (8 : 9) > (5 : 6) > (19 : 24) > (13 : 18).

(ii) Given,

(6 : 7), (13 : 14), (19 : 21) and (23 : 28)

We convert them into equivalent like fractions.

L.C.M of 7, 14, 21, 28 is 84

$\Rightarrow \dfrac{6 \times 12}{7 \times 12} = \dfrac{72}{84} \\[1em] \Rightarrow \dfrac{13 \times 6}{14 \times 6} = \dfrac{78}{84} \\[1em] \Rightarrow \dfrac{19 \times 4}{21 \times 4} = \dfrac{76}{84} \\[1em] \Rightarrow \dfrac{23 \times 3}{28 \times 3} = \dfrac{69}{84} \\[1em] \Rightarrow \dfrac{78}{84} \gt \dfrac{76}{84} \gt \dfrac{72}{84} \gt \dfrac{69}{84}$

(13 : 14) > (19 : 21) > (6 : 7) > (23 : 28).

Hence, the ratios in descending order are (13 : 14) > (19 : 21) > (6 : 7) > (23 : 28).

(iii) Given,

(7 : 12), (9 : 16), (13 : 20) and (5 : 8)

We convert them into equivalent like fractions.

L.C.M of 12, 16, 20, 8 is 240.

$\Rightarrow \dfrac{7 \times 20}{12 \times 20} = \dfrac{140}{240} \\[1em] \Rightarrow \dfrac{9 \times 15}{16 \times 15} = \dfrac{135}{240} \\[1em] \Rightarrow \dfrac{13 \times 12}{20 \times 12} = \dfrac{156}{240} \\[1em] \Rightarrow \dfrac{5 \times 30}{8 \times 30} = \dfrac{150}{240} \\[1em] \Rightarrow \dfrac{156}{240} \gt \dfrac{150}{240} \gt \dfrac{140}{240} \gt \dfrac{135}{240}$

(13 : 20) > (5 : 8) > (7 : 12) > (9 : 16).

Hence, the ratios in descending order are (13 : 20) > (5 : 8) > (7 : 12) > (9 : 16).

Arrange the following ratios in ascending order of magnitudes :

(i) (4 : 9), (6 : 11), (7 : 13) and (27 : 50)

(ii) (2 : 3), (8 : 15), (11 : 12) and (7 : 16)

(iii) (3 : 5), (4 : 9), (5 : 11) and (10 : 17)

Answer

(i) Given,

(4 : 9), (6 : 11), (7 : 13) and (27 : 50)

We convert them into decimals.

⇒ $\dfrac{4}{9}$ ≈ 0.444

⇒ $\dfrac{6}{11}$ ≈ 0.5455

⇒ $\dfrac{7}{13}$ ≈ 0.5385

⇒ $\dfrac{27}{50}$ ≈ 0.5400

⇒ 0.444 < 0.5385 < 0.5400 < 0.5455

⇒ (4 : 9) < (7 : 13) < (27 : 50) < (6 : 11).

Hence, the ratios in ascending order are (4 : 9), (7 : 13), (27 : 50), (6 : 11).

(ii) Given,

(2 : 3), (8 : 15), (11 : 12) and (7 : 16)

We convert them into decimals.

$\dfrac{2}{3}$ = 0.667

$\dfrac{8}{15}$ ≈ 0.533

$\dfrac{11}{12}$ ≈ 0.917

$\dfrac{7}{16}$ = 0.4375

0.4375 < 0.533 < 0.667 < 0.917

(7 : 16) < (8 : 15) < (2 : 3) < (11 : 12)

Hence, the ratios in ascending order are (7 : 16), (8 : 15), (2 : 3), (11 : 12).

(iii) Given,

(3 : 5), (4 : 9), (5 : 11) and (10 : 17)

We convert them into decimals.

$\dfrac{3}{5}$ = 0.600

$\dfrac{4}{9}$ ≈ 0.444

$\dfrac{5}{11}$ ≈ 0.455

$\dfrac{10}{17}$ = 0.588

⇒ 0.444 < 0.455 < 0.588 < 0.600

⇒ (4 : 9) < (5 : 11) < (10 : 17) < (3 : 5)

Hence, the ratios in ascending order are (4 : 9), (5 : 11), (10 : 17), (3 : 5).

If (3a + 2b) : (5a + 3b) = 18 : 29, find (a : b).

Answer

Given,

(3a + 2b) : (5a + 3b) = 18 : 29

Solving,

$\dfrac{3a + 2b}{5a + 3b} = \dfrac{18}{29}$

⇒ 29(3a + 2b) = 18(5a + 3b)

⇒ 87a + 58b = 90a + 54b

⇒ 58b - 54b = 90a - 87a

⇒ 4b = 3a

⇒ $\dfrac{a}{b} = \dfrac{4}{3}$

⇒ a : b = 4 : 3

Hence, a : b = 4 : 3.