Ratio and Proportion — Exercise 6.1

An alloy consists of $27\dfrac{1}{2}$ kg of copper and $2\dfrac{3}{4}$ kg of tin. Find the ratio by weight of tin to the alloy.

Answer

Weight of alloy = Weight of tin + Weight of copper

$\therefore \text{Weight of alloy} = 27\dfrac{1}{2} + 2\dfrac{3}{4} \\[0.5em] = \dfrac{55}{2} + \dfrac{11}{4} \\[0.5em] = \dfrac{110 + 11}{4} \\[0.5em] = \dfrac{121}{4}$

Ratio by weight of tin to alloy = $\dfrac{\text{Weight of tin}}{\text{Weight of alloy}}$

$= \dfrac{\dfrac{11}{4}}{\dfrac{121}{4}} \\[0.5em] = \dfrac{11}{121} \\[0.5em] = \dfrac{1}{11}$

Hence, the ratio by weight of tin to alloy is 1 : 11.

Find the compounded ratio of:

(i) 2 : 3 and 4 : 9

(ii) 4 : 5, 5 : 7 and 9 : 11

(iii) (a - b) : (a + b), (a + b)² : (a² + b²) and (a⁴ - b⁴) : (a² - b²)²

Answer

(i) The compounded ratio of 2 : 3 and 4 : 9 is,

$= \dfrac{2}{3} \times \dfrac{4}{9} \\[0.5em] = \dfrac{8}{27}$

Hence, the compounded ratio is 8 : 27.

(ii) The compounded ratio of 4 : 5, 5 : 7 and 9 : 11 is,

$= \dfrac{4}{5} \times \dfrac{5}{7} \times \dfrac{9}{11} \\[0.5em] = \dfrac{180}{385} \\[0.5em]$

Dividing numerator and denominator by 5, we get:

$\dfrac{\overset{36}{\bcancel{180}}}{\underset{77}{\bcancel{385}}} = \dfrac{36}{77}$

Hence, the compounded ratio is 36 : 77.

(iii) The compounded ratio of (a - b) : (a + b), (a + b)² : (a² + b²) and (a⁴ - b⁴) : (a² - b²)² is,

$= \dfrac{(a - b)}{(a + b)} \times \dfrac{(a + b)^2}{(a^2 + b^2)} \times \dfrac{(a^4 - b^4)}{(a^2 - b^2)^2} \\[0.5em] = \dfrac{(a - b)}{\bcancel{(a + b)}} \times \dfrac{\bcancel{(a + b)}(a + b)}{\bcancel{(a^2 + b^2)}} \times \dfrac{\bcancel{(a^2 - b^2)}\bcancel{(a^2 + b^2)}}{\bcancel{(a^2 - b^2)}(a^2 - b^2)} \\[0.5em] = \dfrac{(a - b)(a + b)}{(a^2 - b^2)} \\[0.5em] = \dfrac{(a^2 - b^2)}{(a^2 - b^2)} \\[0.5em] = \dfrac{1}{1}$

Hence, the compounded ratio is 1 : 1.

Find the duplicate ratio of :

(i) 2 : 3

(ii) $\sqrt{5}$ : 7

(iii) 5a : 6b

Answer

(i) The duplicate ratio of 2 : 3 is,

= 2² : 3² = 4 : 9

Hence, the duplicate ratio is 4 : 9.

(ii) The duplicate ratio of $\sqrt{5}$ : 7 is,

= $(\sqrt{5})$² : 7² = 5 : 49

Hence, the duplicate ratio is 5 : 9.

(iii) The duplicate ratio of 5a : 6b is,

= (5a)² : (6b)² = 25a² : 36b²

Hence, the duplicate ratio is 25a² : 36b².

Find the triplicate ratio of :

(i) 3 : 4

(ii) $\dfrac{1}{2} : \dfrac{1}{3}$

(iii) 1³ : 2³

Answer

(i) The triplicate ratio of 3 : 4 is,

= 3³ : 4³ = 27 : 64

Hence, the triplicate ratio is 27 : 64.

(ii) The triplicate ratio of $\dfrac{1}{2} : \dfrac{1}{3}$ is,

$=\big(\dfrac{1}{2}\big)^3 : \big(\dfrac{1}{3}\big)^3 \\[0.5em] = \big(\dfrac{1}{8}\big) : \big(\dfrac{1}{27}\big) \\[0.5em] = \dfrac{\dfrac{1}{8}}{\dfrac{1}{27}} \\[0.5em] = \dfrac{27}{8} = 27 : 8.$

Hence, the triplicate ratio is 27 : 8.

(iii) The triplicate ratio of 1³ : 2³ is,

= (1³)³ : (2³)³ = 1⁹ : 2⁹ = 1: 512

Hence, the triplicate ratio is 1 : 512.

Find the sub-duplicate ratio of :

(i) 9 : 16

(ii) $\dfrac{1}{4} : \dfrac{1}{9}$

(iii) 9a² : 49b²

Answer

(i) The sub duplicate ratio of 9 : 16 is,

$= \sqrt{9} : \sqrt{16} \\[0.5em] = 3 : 4$

Hence, the sub-duplicate ratio is 3 : 4.

(ii) The sub duplicate ratio of $\dfrac{1}{4} : \dfrac{1}{9}$ is,

$= \sqrt{\dfrac{1}{4}} : \sqrt{\dfrac{1}{9}} \\[0.5em] = \dfrac{1}{2} : \dfrac{1}{3} \\[0.5em] = \dfrac{\dfrac{1}{2}}{\dfrac{1}{3}} \\[0.5em] = \dfrac{3}{2} = 3 : 2$

Hence, the sub-duplicate ratio is 3 : 2.

(iii) The sub duplicate ratio of 9a² : 49b² is,

$= \sqrt{9a^2} : \sqrt{49b^2} \\[0.5em] = 3a : 7b$

Hence, the sub-duplicate ratio is 3a : 7b.

Find the sub-triplicate ratio of :

(i) 1 : 216

(ii) $\dfrac{1}{8} : \dfrac{1}{125}$

(iii) 27a³ : 64b³

Answer

(i) The sub-triplicate ratio of 1 : 216 is,

$= \sqrt[3]{1} : \sqrt[3]{216} \\[0.5em] = 1 : 6$

Hence, the sub-triplicate ratio is 1 : 6.

(ii) The sub-triplicate ratio of $\dfrac{1}{8} : \dfrac{1}{125}$ is,

$= \sqrt[3]{\dfrac{1}{8}} : \sqrt[3]{\dfrac{1}{125}} \\[0.5em] = \dfrac{1}{2} : \dfrac{1}{5} \\[0.5em] = \dfrac{\dfrac{1}{2}}{\dfrac{1}{5}} \\[0.5em] = \dfrac{5}{2} = 5 : 2$

Hence, the sub-triplicate ratio is 5 : 2.

(iii) The sub-triplicate ratio of 27a³ : 64b³ is,

$= \sqrt[3]{27a^3} : \sqrt[3]{64b^3} \\[0.5em] = 3a : 4b$

Hence, the sub-triplicate ratio is 3a : 4b.

Find the reciprocal ratio of :

(i) 4 : 7

(ii) 3² : 4²

(iii) $\dfrac{1}{9} : 2$

Answer

(i) The reciprocal ratio of 4 : 7 is,

$= \dfrac{1}{4} : \dfrac{1}{7} \\[0.5em] = \dfrac{\dfrac{1}{4}}{\dfrac{1}{7}} \\[0.5em] = \dfrac{7}{4} \\[0.5em] = 7 : 4$

Hence, the reciprocal ratio is 7 : 4.

(ii) The reciprocal ratio of 3² : 4² is,

$= \dfrac{1}{3^2} : \dfrac{1}{4^2} \\[0.5em] = \dfrac{\dfrac{1}{9}}{\dfrac{1}{16}} \\[0.5em] = \dfrac{16}{9} \\[0.5em] = 16 : 9$

Hence, the reciprocal ratio is 16 : 9.

(iii) The reciprocal ratio of $\dfrac{1}{9} : 2$ is,

$= \dfrac{1}{\dfrac{1}{9}} : \dfrac{1}{2} \\[0.5em] = \dfrac{9}{\dfrac{1}{2}} \\[0.5em] = \dfrac{18}{1} \\[0.5em] = 18 : 1$

Hence, the reciprocal ratio is 18 : 1.

Arrange the following ratios in ascending order of magnitude :
2 : 3, 17 : 21, 11 : 14 and 5 : 7.

Answer

Given, ratios are $\dfrac{2}{3}, \dfrac{17}{21}, \dfrac{11}{14}, \dfrac{5}{7}.$

We convert them into equivalent like fractions.

L.C.M. of 3, 21, 14, 7 = 42

$\dfrac{2}{3} = \dfrac{2 \times 14}{3 \times 14} = \dfrac{28}{42}, \\[0.5em] \dfrac{17}{21} = \dfrac{17 \times 2}{21 \times 2} = \dfrac{34}{42}, \\[0.5em] \dfrac{11}{14} = \dfrac{11 \times 3}{14 \times 3} = \dfrac{33}{42}, \\[0.5em] \dfrac{5}{7} = \dfrac{5 \times 6}{7 \times 6} = \dfrac{30}{42}. \\[0.5em]$

As, 28 < 30 < 33 < 34,

$\Rightarrow \dfrac{28}{42} \lt \dfrac{30}{42} \lt \dfrac{33}{42} \lt \dfrac{34}{42} \\[1em] \therefore \dfrac{2}{3} \lt \dfrac{5}{7} \lt \dfrac{11}{14} \lt \dfrac{17}{21}$

Hence, the given ratios in ascending order are 2 : 3, 5 : 7, 11 : 14, 17 : 21.

If A : B = 2 : 3, B : C = 4 : 5 and C : D = 6 : 7, find A : D.

Answer

$\dfrac{A}{B} = \dfrac{2}{3} \\[0.5em] \Rightarrow B = \dfrac{3A}{2} \\[0.5em]$

Putting this value of B in B : C

$\dfrac{B}{C} = \dfrac{4}{5} \\[0.5em] \Rightarrow \dfrac{\dfrac{3A}{2}}{C} = \dfrac{4}{5} \\[0.5em] \Rightarrow \dfrac{3A}{2} = \dfrac{4C}{5} \\[0.5em] \Rightarrow C = \dfrac{15A}{8} \\[0.5em]$

Putting this value of C in C : D

$C : D = 6 : 7 \\[0.5em] \Rightarrow \dfrac{\dfrac{15A}{8}}{D} = \dfrac{6}{7} \\[0.5em] \Rightarrow \dfrac{15A}{8D} = \dfrac{6}{7} \\[0.5em] \Rightarrow \dfrac{A}{D} = \dfrac{48}{105} = \dfrac{16}{35} \\[0.5em] \Rightarrow A : D = 16 : 35.$

Hence, the value of A : D is 16 : 35.

If x : y = 2 : 3 and y : z = 4 : 7, find x : y : z.

Answer

Given, x : y = 2 : 3 and y : z = 4 : 7

To find x : y : z, we will make y same in both cases.

Taking L.C.M. of two values of y i.e. 3 and 4 = 12

$\text{So }, \dfrac{x}{y} = \dfrac{2 \times 4}{3 \times 4} = \dfrac{8}{12} = 8 : 12 \\[0.5em] \text{and } \dfrac{y}{z} = \dfrac{4}{7} = \dfrac{4 \times 3}{7 \times 3} = \dfrac{12}{21} = 12 : 21$

∴ x : y : z = 8 : 12 : 21

Hence, the ratio of x : y : z is 8 : 12 : 21.

If A : B = $\dfrac{1}{4} : \dfrac{1}{5}$ and B : C = $\dfrac{1}{7} : \dfrac{1}{6}$, find A : B : C.

Answer

Given, A : B = $\dfrac{1}{4} : \dfrac{1}{5}$ = 5 : 4 and B : C = $\dfrac{1}{7} : \dfrac{1}{6}$ = 6 : 7

To find A : B : C, we will make B same in both cases.

Taking L.C.M. of two values of B i.e. 4 and 6 = 12

$\text{So }, \dfrac{A}{B} = \dfrac{5 \times 3}{4 \times 3} = \dfrac{15}{12} = 15 : 12 \\[0.5em] \text{and } \dfrac{B}{C} = \dfrac{6}{7} = \dfrac{6 \times 2}{7 \times 2} = \dfrac{12}{14} = 12 : 14 \\[0.5em]$

∴ A : B : C = 15 : 12 : 14

Hence, the ratio of A : B : C is 15 : 12 : 14.

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

Answer

$3A = 4B \\[0.5em] \Rightarrow \dfrac{A}{B} = \dfrac{4}{3} \\[0.5em] \Rightarrow A : B = 4 : 3$

Similarly,

$4B = 6C \\[0.5em] \Rightarrow \dfrac{B}{C} = \dfrac{6}{4} = \dfrac{3}{2} \\[0.5em] \Rightarrow B : C = 3 : 2$

So we get,

A : B : C = 4 : 3 : 2

Hence, the ratio of A : B : C is 4 : 3 : 2.

If $\dfrac{3x + 5y}{3x - 5y} = \dfrac{7}{3}$, find x : y.

Answer

Given,

$\dfrac{3x + 5y}{3x - 5y} = \dfrac{7}{3} \\[0.5em] \Rightarrow 3(3x + 5y) = 7(3x - 5y) \\[0.5em] \Rightarrow 9x + 15y = 21x - 35y \\[0.5em] \Rightarrow 15y + 35y = 21x - 9x \\[0.5em] \Rightarrow 50y = 12x \\[0.5em] \Rightarrow x = \dfrac{50y}{12} \\[0.5em] \Rightarrow \dfrac{x}{y} = \dfrac{50}{12} = \dfrac{25}{6} \\[0.5em] \Rightarrow x : y = 25 : 6.$

Hence, the ratio of x : y is 25 : 6.

If a : b = 3 : 11, find (15a - 3b) : (9a + 5b).

Answer

a : b = 3 : 11 or,

$\dfrac{a}{b} = \dfrac{3}{11}$

We need to find $\dfrac{15a - 3b}{9a + 5b}$

Dividing the numerator and denominator by b,

$\Rightarrow \dfrac{\dfrac{15a}{b} - \dfrac{3b}{b}}{\dfrac{9a}{b} + \dfrac{5b}{b}} \\[0.5em] \Rightarrow \dfrac{\dfrac{15a}{ b} - 3}{\dfrac{9a}{b} + 5} \\[0.5em]$

Putting value of $\dfrac{a}{b} = \dfrac{3}{11}$,

$\Rightarrow \dfrac{15 \times \dfrac{3}{11} - 3}{9 \times \dfrac{3}{11} + 5} \\[0.5em] \Rightarrow \dfrac{\dfrac{45 - 33}{11}}{\dfrac{27 + 55}{11} } \\[0.5em] \Rightarrow \dfrac{12}{82} = \dfrac{6}{41} = 6 : 41$

Hence, the value of ratio is 6 : 41.

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

Answer

Given, (4x² + xy) : (3xy - y²) = 12 : 5

$\Rightarrow \dfrac{4x^2 + xy}{3xy - y^2} = \dfrac{12}{5} \\[1em] \Rightarrow 5(4x^2 + xy) = 12(3xy - y^2) \\[1em] \Rightarrow 20x^2 + 5xy = 36xy - 12y^2 \\[1em] \Rightarrow 20x^2 - 31xy + 12y^2 = 0 \\[1em] \Rightarrow 20\dfrac{x^2}{y^2} - 31\dfrac{x}{y} + 12 = 0 \\[1em] \Rightarrow 20(\dfrac{x}{y})^2 - 31(\dfrac{x}{y}) + 12 = 0 \\[1em] \Rightarrow 20(\dfrac{x}{y})^2 - 15(\dfrac{x}{y}) - 16(\dfrac{x}{y})+ 12 = 0 \\[1em] \Rightarrow 5(\dfrac{x}{y})\big(4\big(\dfrac{x}{y}\big) - 3\big) - 4\big(4\big(\dfrac{x}{y}\big) - 3\big) \\[1em] \Rightarrow \big(5\big(\dfrac{x}{y}\big) - 4\big) \big(4\big(\dfrac{x}{y}\big) - 3\big) = 0 \\[1em] \Rightarrow \big(5\big(\dfrac{x}{y}\big) - 4\big) = 0 \text{ or } \big(4\big(\dfrac{x}{y}\big) - 3\big) = 0 \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{4}{5} \text{ or } \dfrac{x}{y} = \dfrac{3}{4}.$

We need to find value of (x + 2y) : (2x + y) or $\dfrac{x + 2y}{2x + y}$

Dividing the numerator and denominator by y,

$\Rightarrow \dfrac{\dfrac{x}{y} + 2}{\dfrac{2x}{y} + 1}$

Putting value of $\dfrac{x}{y} = \dfrac{4}{5}$,

$\Rightarrow \dfrac{\dfrac{4}{5} + 2}{\dfrac{8}{5} + 1} \\[1em] \Rightarrow \dfrac{\dfrac{14}{5}}{\dfrac{13}{5}} \\[1em] \Rightarrow \dfrac{14}{13} = 14 : 13. \\[1em]$

Putting value of $\dfrac{x}{y} = \dfrac{3}{4}$,

$\Rightarrow \dfrac{\dfrac{3}{4} + 2}{\dfrac{6}{4} + 1} \\[1em] \Rightarrow \dfrac{\dfrac{11}{4}}{\dfrac{10}{4}} \\[1em] \Rightarrow \dfrac{11}{10} = 11 : 10$

Hence, the value of ratio (x + 2y) : (2x + y) is 14 : 13 or 11 : 10.

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

Answer

Given, y(3x - y) : x(4x + y) = 5 : 12

$\therefore \dfrac{3xy - y^2}{4x^2 + xy} = \dfrac{5}{12} \\[0.5em] \Rightarrow 12(3xy - y^2) = 5(4x^2 + xy) \\[0.5em] \Rightarrow 36xy - 12y^2 = 20x^2 + 5xy \\[0.5em] \Rightarrow 20x^2 + 12y^2 + 5xy - 36xy = 0 \\[0.5em] \Rightarrow 20x^2 - 31xy + 12y^2 = 0 \\[0.5em]$

Dividing the equation by y²,

$\Rightarrow 20\big(\dfrac{x}{y}\big)^2 - 31\big(\dfrac{x}{y}\big) + 12 = 0 \\[1em] \Rightarrow 20\big(\dfrac{x}{y}\big)^2 - 16\big(\dfrac{x}{y}\big) - 15\big(\dfrac{x}{y}\big) + 12 = 0 \\[1em] \Rightarrow 4\big(\dfrac{x}{y}\big)(5\big(\dfrac{x}{y}\big) - 4) - 3(5\big(\dfrac{x}{y}\big) - 4) = 0 \\[1em] \Rightarrow (5\big(\dfrac{x}{y}\big) - 4)(4\big(\dfrac{x}{y}\big) - 3) = 0 \\[1em] \Rightarrow 5\big(\dfrac{x}{y}\big) - 4 = 0 \text{ or } 4\big(\dfrac{x}{y}\big) - 3 = 0 \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{4}{5} \text{ or } \dfrac{x}{y} = \dfrac{3}{4}.$

We have to find value of (x² + y²) : (x + y)².

$= (x^2 + y^2) : (x^2 + y^2 + 2xy) \\[0.5em] = \dfrac{x^2 + y^2}{x^2 + y^2 + 2xy} \\[0.5em]$

Dividing numerator and denominator by y²,

$= \dfrac{\dfrac{x^2 + y^2}{y^2}}{\dfrac{x^2 + y^2 + 2xy}{y^2}} \\[1em] = \dfrac{\big(\dfrac{x}{y}\big)^2 + 1}{\big(\dfrac{x}{y}\big)^2 + 1 + 2\big(\dfrac{x}{y}\big)} \\[1em]$

Putting value of $\dfrac{x}{y} = \dfrac{4}{5}$,

$\dfrac{\big(\dfrac{4}{5}\big)^2 + 1}{\big(\dfrac{4}{5}\big)^2 + 1 + 2\big(\dfrac{4}{5}\big)} \\[1em] = \dfrac{\big(\dfrac{16}{25}\big) + 1}{\big(\dfrac{16}{25}\big) + 1 + \big(\dfrac{8}{5}\big)} \\[1em] = \dfrac{\dfrac{16 + 25}{25}}{\dfrac{16 + 25 + 40}{25}} \\[1em] = \dfrac{41}{81} \\[1em] = 41 : 81 \\[1em]$

Putting value of $\dfrac{x}{y} = \dfrac{3}{4}$,

$\dfrac{\big(\dfrac{3}{4}\big)^2 + 1}{\big(\dfrac{3}{4}\big)^2 + 1 + 2\big(\dfrac{3}{4}\big)} \\[1em] = \dfrac{\big(\dfrac{9}{16}\big) + 1}{\big(\dfrac{9}{16}\big) + 1 + \big(\dfrac{6}{4}\big)} \\[1em] = \dfrac{\dfrac{9 + 16}{16}}{\dfrac{9 + 16 + 24}{16}} \\[1em] = \dfrac{25}{49} \\[1em] = 25 : 49$

Hence, the value of ratio (x² + y²) : (x + y)² is 41 : 81 or 25 : 49.

If (x - 9) : (3x + 6) is the duplicate ratio of 4 : 9, find the value of x.

Answer

Duplicate ratio of 4 : 9 = 4² : 9² = 16 : 81.

According to question,

(x - 9) : (3x + 6) = 16 : 81

$\Rightarrow \dfrac{x - 9}{3x + 6} = \dfrac{16}{81} \\[0.5em] \Rightarrow 81(x - 9) = 16(3x + 6) \\[0.5em] \Rightarrow 81x - 729 = 48x + 96 \\[0.5em] \Rightarrow 81x - 48x = 96 + 729 \\[0.5em] \Rightarrow 33x = 825 \\[0.5em] \Rightarrow x = \dfrac{825}{33} \\[0.5em] x = 25.$

Hence, the value of x is 25.

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

Answer

Triplicate ratio of 3 : 4 = 3³ : 4³ = 27 : 64.

According to question,

(3x + 1) : (5x + 3) = 27 : 64

$\Rightarrow \dfrac{3x + 1}{5x + 3} = \dfrac{27}{64} \\[0.5em] \Rightarrow 64(3x + 1) = 27(5x + 3) \\[0.5em] \Rightarrow 192x + 64 = 135x + 81 \\[0.5em] \Rightarrow 192x - 135x = 81 - 64 \\[0.5em] \Rightarrow 57x = 17 \\[0.5em] \Rightarrow x = \dfrac{17}{57}.$

Hence, the value of x is $\dfrac{17}{57}$.

If (x + 2y) : (2x - y) is equal to the duplicate ratio of 3 : 2, find x : y.

Answer

Duplicate ratio of 3 : 2 = 3² : 2² = 9 : 4.

According to question,

(x + 2y) : (2x - y) = 9 : 4

$\Rightarrow \dfrac{x + 2y}{2x - y} = \dfrac{9}{4} \\[0.5em] \Rightarrow 4(x + 2y) = 9(2x - y) \\[0.5em] \Rightarrow 4x + 8y = 18x - 9y \\[0.5em] \Rightarrow 4x - 18x = -9y - 8y \\[0.5em] \Rightarrow -14x = -17y \\[0.5em] \Rightarrow \dfrac{x}{y} = \dfrac{-17}{-14} \\[0.5em] \Rightarrow \dfrac{x}{y} = \dfrac{17}{14} \\[0.5em] \Rightarrow x : y = 17 : 14$

Hence, the value of x : y is 17 : 14.

Find two numbers in the ratio of 8 : 7 such that when each is decreased by $12\dfrac{1}{2}$, they are in ratio 11 : 9.

Answer

Since, numbers are in ratio 8 : 7, let the required numbers be 8x and 7x.

According to question,

$\dfrac{8x - 12\dfrac{1}{2}}{7x - 12\dfrac{1}{2}} = \dfrac{11}{9} \\[1em] \Rightarrow \dfrac{8x - \dfrac{25}{2}}{7x - \dfrac{25}{2}} = \dfrac{11}{9} \\[1em] \Rightarrow \dfrac{\dfrac{16x - 25}{2}}{\dfrac{14x - 25}{2}} = \dfrac{11}{9} \\[1em] \Rightarrow \dfrac{16x - 25}{14x - 25} = \dfrac{11}{9} \\[1em] \Rightarrow 9(16x - 25) = 11(14x - 25) \\[1em] \Rightarrow 144x - 225 = 154x - 275 \\[1em] \Rightarrow 144x - 154x = -275 + 225 \\[1em] \Rightarrow -10x = -50 \\[1em] \Rightarrow x = 5.$

∴ x = 5, 8x = 40, 7x = 35.

Hence, the required numbers are 40, 35.

The income of a man is increased in the ratio 10 : 11. If the increase in his income is ₹ 600 per month, find his new income.

Answer

Let the present income = 10x and the new increased income = 11x.

So, the increase per month = 11x - 10x = x

Given, increase in his income is ₹600 per month.

∴ x = 600.

New income = 11x = $11 \times 600$ = ₹6600.

Hence, the new income of man is ₹6600 per month.

A woman reduces her weight in the ratio 7 : 5. What does her weight become if originally it was 91 kg?

Answer

Given, a woman reduces her weight in ratio 7 : 5 and original weight = 91 kg.

$\therefore \dfrac{\text{Original weight}}{\text{Reduced weight}} = \dfrac{7}{5} \\[0.5em] \Rightarrow \text{Reduced weight} = \dfrac{5}{7} \times \text{Original weight} \\[0.5em] \Rightarrow \text{Reduced weight} = \dfrac{5}{7} \times 91 \text{ kg} \\[0.5em] \Rightarrow \text{Reduced weight} = 5 \times 13 = 65 \text{ kg}$

Hence, the reduced weight of woman is 65 kg.

A school collected ₹2100 for charity. It was decided to divide the money between an orphanage and a blind school in the ratio 3 : 4. How much money did each receive?

Answer

Amount collected for charity = ₹2100.

The ratio between orphanage and a blind school = 3 : 4.

Sum of ratio = 3 + 4 = 7

Orphanage share = $\dfrac{3}{7} \times ₹2100 = ₹900.$

Blind school share = $\dfrac{4}{7} \times ₹2100 = ₹1200.$

Hence, the share of orphanage school is ₹900 and the share of blind school is ₹1200.

The sides of a triangle are in the ratio 7 : 5 : 3 and its perimeter is 30 cm. Find the lengths of sides.

Answer

Since the sides of triangle are in the ratio 7 : 5 : 3, let the sides be 7x, 5x and 3x.

Perimeter = Sum of sides of triangle

∴ 7x + 5x + 3x = 30
⇒ 15x = 30
⇒ x = $\dfrac{30}{15}$
⇒ x = 2

∴ x = 2, 7x = 14, 5x = 10, 3x = 6.

Hence, the sides of the triangle are 14cm, 10cm, 6cm.

If the angles of a triangle are in the ratio 2 : 3 : 4, find the angles.

Answer

Since the angles of triangle are in the ratio 2 : 3 : 4, let the angles be 2x, 3x and 4x.

Sum of angle of triangle = 180°

∴ 2x + 3x + 4x = 180
⇒ 9x = 180°
⇒ x = $\dfrac{180°}{9}$
⇒ x = 20°

∴ x = 20°, 2x = 40°, 3x = 60°, 4x = 80°.

Hence, the angles of triangle are 40°, 60°, 80°.

Three numbers are in the ratio $\dfrac{1}{2} : \dfrac{1}{3} : \dfrac{1}{4}$. If the sum of their squares is 244, find the numbers.

Answer

The ratio is $\dfrac{1}{2} : \dfrac{1}{3} : \dfrac{1}{4}$.

L.C.M. of 2, 3, 4 = 12.

$\text{Ratio } = \dfrac{1}{2} \times 12 : \dfrac{1}{3} \times 12 : \dfrac{1}{4} \times 12 \\[0.5em] = 6 : 4 : 3.$

Since the ratio is 6 : 4 : 3, let the numbers be 6x, 4x and 3x.

Given, sum of squares of numbers = 244.

∴ (6x)² + (4x)² + (3x)² = 244
⇒ 36x² + 16x² + 9x² = 244
⇒ 61x² = 244
⇒ x² = $\dfrac{244}{61}$
⇒ x = $\sqrt{4}$ = 2

∴ x = 2, 6x = 12, 4x = 8, 3x = 6.

Hence, the numbers are 12, 8 and 6.

A certain sum was divided among A, B and C in the ratio 7 : 5 : 4. If B got ₹500 more than C, find the total sum divided.

Answer

Since, the ratio of money divided among A : B : C is 7 : 5 : 4. So, the money received by A, B, C be 7x, 5x, 4x respectively.

Given, B receives 500 more than C.

∴ 5x - 4x = 500
x = 500.

Total money divided = 7x + 5x + 4x = 16x = ₹8000.

Hence, the total money divided is ₹8000.

In a business, A invests ₹50000 for 6 months; B ₹60000 for 4 months and C ₹80000 for 5 months. If they together earn ₹18800, find share of each.

Answer

A invests ₹50000 for 6 months, total investment of A = ₹50000 x 6 = ₹300000.

B invests ₹60000 for 4 months, total investment of B = ₹60000 x 4 = ₹240000.

C ₹80000 for 5 months, total investment of C = ₹80000 x 5 = ₹400000.

Ratio of share of A, B and C = 300000 : 240000 : 400000 = 30 : 24 : 40.

Sum of ratios = 94.

Given, total earning = ₹18800.

Share of A = $\dfrac{30}{94} \times 18800$ = $\dfrac{564000}{94}$ = ₹6000.

Share of B = $\dfrac{24}{94} \times 18800$ = $\dfrac{451200}{94}$ = ₹4800.

Share of C = $\dfrac{40}{94} \times 18800$ = $\dfrac{752000}{94}$ = ₹8000.

Hence, the shares of A, B and C are ₹6000, ₹4800 and ₹8000 respectively.

In a mixture of 45 litres, the ratio of milk to water is 13 : 2. How much water must be added to this mixture to make the ratio of milk to water as 3 : 1?

Answer

Ratio of milk to water = 13 : 2.

Total quantity of mixture = 45 litres

Sum of ratio = 13 + 2 = 15.

Quantity of milk = $\dfrac{13}{15}$ x 45 = 13 x 3 = 39 litres.

Quantity of water = $\dfrac{2}{15}$ x 45 = 2 x 3 = 6 litres.

Let the water added be x litres, so quantity of water = (6 + x) litres.

Now ratio = 3 : 1

$\therefore \dfrac{39}{6 + x} = \dfrac{3}{1} \\[0.5em] \Rightarrow 39 = 3(6 + x) \\[0.5em] \Rightarrow 39 = 18 + 3x \\[0.5em] \Rightarrow 3x = 39 - 18 \\[0.5em] \Rightarrow x = \dfrac{21}{3} \\[0.5em] \Rightarrow x = 7.$

The water that must be added is 7 litres.

The ratio of the number of boys to the number of girls in a school of 560 pupils is 5 : 3. If 10 new boys are admitted, find how many new girls may be admitted so that the ratio of number of boys to the number of girls may change to 3 : 2.

Answer

Total students = 560

Ratio of the number of boys to the number of girls = 5 : 3.

Sum of ratio = 5 + 3 = 8.

Number of boys = $\dfrac{5}{8}$ x 560 = 5 x 70 = 350.

Number of girls = $\dfrac{3}{8}$ x 560 = 210.

10 new boys are admitted in school , so total boys now = 350 + 10 = 360.

Let new girls to be admitted be x, so now total girls = (210 + x)

New boys to girls ratio = 3 : 2

$\therefore 360 : (210 + x) = 3 : 2 \\[0.5em] \Rightarrow \dfrac{360}{210 + x} = \dfrac{3}{2} \\[0.5em] \Rightarrow 720 = 3(210 + x) \\[0.5em] \Rightarrow 3x + 630 = 720 \\[0.5em] \Rightarrow 3x = 720 - 630 \\[0.5em] \Rightarrow 3x = 90 \\[0.5em] \Rightarrow x = 30.$

Hence, the number of new girls to be admitted are 30.

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 ₹80 every month, find their monthly pocket money.

Answer

Pocket money ratio of Ravi and Sanjeev = 5 : 7, so let the pocket money be 5x and 7x.

Expenditure ratio of Ravi and Sanjeev = 3 : 5, so let the expenditure be 3y and 5y.

Given, each save ₹80 per month.

∴ For Ravi, savings = 5x - 3y = 80 and for Sanjeev, savings = 7x - 5y = 80.

First solving 5x - 3y = 80.

$\Rightarrow 5x - 3y = 80 \\[0.5em] \Rightarrow 5x = 80 + 3y \\[0.5em] x = \dfrac{80 + 3y}{5}$

Putting above value of x in 7x - 5y = 80.

$\Rightarrow 7\big(\dfrac{80 + 3y}{5}\big) - 5y = 80 \\[1em] \Rightarrow \dfrac{560 + 21y}{5} - 5y = 80 \\[1em] \Rightarrow \dfrac{560 + 21y - 25y}{5} = 80 \\[1em] \Rightarrow \dfrac{560 - 4y}{5} = 80$

On cross multiplying,

$\Rightarrow 560 - 4y = 400 \\[1em] \Rightarrow 560 - 400 = 4y \\[1em] \Rightarrow 4y = 160 \\[1em] y = 40.$

∴ y = 40, x = $\dfrac{80 + 3y}{5} = \dfrac{80 + 3 \times 40}{5} = \dfrac{200}{5} = 40$

∴ 5x = 200, 7x = 280.

Hence, the pocket money of Ravi and Sanjeev is ₹200 and ₹280 respectively.

In class X of a school, the ratio of the number of boys to that of the girls is 4 : 3. If there were 20 more boys and 12 less girls, then the ratio would have been 2 : 1. How many students were there in the class?

Answer

Ratio of the number of boys to girls is 4 : 3, so let the number of boys be 4x and number of girls be 3x.

According to question,

4x + 20 : 3x - 12 = 2 : 1

$\Rightarrow \dfrac{4x + 20}{3x - 12} = \dfrac{2}{1} \\[0.5em] \Rightarrow 4x + 20 = 2(3x - 12) \\[0.5em] \Rightarrow 4x + 20 = 6x - 24 \\[0.5em] \Rightarrow 6x - 4x = 20 + 24 \\[0.5em] \Rightarrow 2x = 44 \\[0.5em] \Rightarrow x = 22.$

∴ Total number of students = 4x + 3x = 7x = $7 \times 22$ = 154.

Hence, the total number of students are 154.

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

Answer

Ratio of passes to failures = 4 : 1. So, the number of students passed = 4x and number of students failed = x.

Total students appeared for examination = 4x + x = 5x.

In second case, number of students appeared = 5x - 30.

Number of students passed in second case = 4x - 20.

So, number of students failed = (5x - 30) - (4x - 20) = 5x - 4x - 30 + 20 = x - 10.

According to question,

$\dfrac{4x - 20}{x - 10} = \dfrac{5}{1} \\[0.5em] \Rightarrow 4x - 20 = 5(x - 10) \\[0.5em] \Rightarrow 4x - 20 = 5x - 50 \\[0.5em] \Rightarrow 4x - 5x = -50 + 20 \\[0.5em] \Rightarrow -x = -30 \\[0.5em] \Rightarrow x = 30.$

∴ 5x = 150

Hence, total number of students appeared for examination were 150.