Ratio and Proportion

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 ₹2100.

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.

Find the value of x in the following proportions :

(i) 10 : 35 = x : 42

(ii) 3 : x = 24 : 2

(iii) 2.5 : 1.5 = x : 3

(iv) x : 50 :: 3 : 2.

Answer

(i) Given, 10 : 35 = x : 42

$\Rightarrow \dfrac{10}{35} = \dfrac{x}{42} \\[0.5em] \Rightarrow x = \dfrac{10}{35} \times 42 \\[0.5em] \Rightarrow x = \dfrac{420}{35} \\[0.5em] \Rightarrow x = 12.$

Hence, the value of x = 12.

(ii) Given, 3 : x = 24 : 2

$\Rightarrow \dfrac{3}{x} = \dfrac{24}{2} \\[0.5em] \Rightarrow x = \dfrac{3}{24} \times 2 \\[0.5em] \Rightarrow x = \dfrac{6}{24} \\[0.5em] \Rightarrow x = \dfrac{1}{4}.$

Hence, the value of x = $\dfrac{1}{4}$.

(iii) Given, 2.5 : 1.5 = x : 3

$\Rightarrow \dfrac{2.5}{1.5} = \dfrac{x}{3} \\[0.5em] \Rightarrow x = \dfrac{2.5}{1.5} \times 3 \\[0.5em] \Rightarrow x = \dfrac{7.5}{1.5} \\[0.5em] \Rightarrow x = 5.$

Hence, the value of x = 5.

(iv) Given, x : 50 :: 3 : 2

$\Rightarrow \dfrac{x}{50} = \dfrac{3}{2} \\[0.5em] \Rightarrow x = \dfrac{3}{2} \times 50 \\[0.5em] \Rightarrow x = \dfrac{150}{2} \\[0.5em] \Rightarrow x = 75.$

Hence, the value of x = 75.

Find the fourth proportional to :

(i) 3, 12, 15

(ii) $\dfrac{1}{3}, \dfrac{1}{4}, \dfrac{1}{5}$

(iii) 1.5, 2.5, 4.5

(iv) 9.6 kg, 7.2 kg, 28.8 kg.

Answer

(i) Let the fourth proportion be x.

Then 3, 12, 15 and x are in proportion.

$\Rightarrow 3 : 12 :: 15 : x \\[0.5em] \Rightarrow \dfrac{3}{12} = \dfrac{15}{x} \\[0.5em] \Rightarrow x = \dfrac{15}{3} \times 12 \\[0.5em] \Rightarrow x = \dfrac{180}{3} \\[0.5em] \Rightarrow x = 60.$

Hence, the fourth proportion is 60.

(ii) Let the fourth proportion be x.

Then $\dfrac{1}{3}, \dfrac{1}{4}, \dfrac{1}{5}$ and x are in proportion.

$\Rightarrow \dfrac{1}{3} : \dfrac{1}{4} :: \dfrac{1}{5} : x \\[0.5em] \Rightarrow \dfrac{\dfrac{1}{3}}{\dfrac{1}{4}} = \dfrac{\dfrac{1}{5}}{x} \\[0.5em] \Rightarrow x = \dfrac{\dfrac{1}{5}}{\dfrac{1}{3}} \times \dfrac{1}{4} \\[0.5em] \Rightarrow x = \dfrac{3}{5} \times \dfrac{1}{4} \\[0.5em] \Rightarrow x = \dfrac{3}{20}.$

Hence, the fourth proportion is $\dfrac{3}{20}$.

(iii) Let the fourth proportion be x.

Then 1.5, 2.5, 4.5 and x are in proportion.

$\Rightarrow 1.5 : 2.5 :: 4.5 : x \\[0.5em] \Rightarrow \dfrac{1.5}{2.5} = \dfrac{4.5}{x} \\[0.5em] \Rightarrow x = \dfrac{4.5}{1.5} \times 2.5 \\[0.5em] \Rightarrow x = 3 \times 2.5 \\[0.5em] \Rightarrow x = 7.5.$

Hence, the fourth proportion is 7.5.

(iv) Let the fourth proportion be x.

Then 9.6 kg, 7.2 kg, 28.8 kg and x are in proportion.

$\Rightarrow 9.6 : 7.2 :: 28.8 : x \\[0.5em] \Rightarrow \dfrac{9.6}{7.2} = \dfrac{28.8}{x} \\[0.5em] \Rightarrow x = \dfrac{28.8}{9.6} \times 7.2 \\[0.5em] \Rightarrow x = 3 \times 7.2 \\[0.5em] \Rightarrow x = 21.6.$

Hence, the fourth proportion is 21.6 kg.

Find the third proportional to :

(i) 5, 10

(ii) 0.24, 0.6

(iii) ₹3, ₹12

(iv) $5\dfrac{1}{4}$ and 7.

Answer

(i) Let the third proportion be x, then 5, 10, x are in continued proportion.

$\Rightarrow \dfrac{5}{10} = \dfrac{10}{x} \\[0.5em] \Rightarrow x = \dfrac{10}{5} \times 10 \\[0.5em] \Rightarrow x = \dfrac{100}{5} \\[0.5em] \Rightarrow x = 20.$

Hence, the third proportion is 20.

(ii) Let the third proportion be x, then 0.24, 0.6, x are in continued proportion.

$\Rightarrow \dfrac{0.24}{0.6} = \dfrac{0.6}{x} \\[0.5em] \Rightarrow x = \dfrac{0.6}{0.24} \times 0.6 \\[0.5em] \Rightarrow x = \dfrac{0.36}{0.24} \\[0.5em] \Rightarrow x = 1.5.$

Hence, the third proportion is 1.5.

(iii) Let the third proportion be x, then 3, 12, x are in continued proportion.

$\Rightarrow \dfrac{3}{12} = \dfrac{12}{x} \\[0.5em] \Rightarrow x = \dfrac{12}{3} \times 12 \\[0.5em] \Rightarrow x = \dfrac{144}{3} \\[0.5em] \Rightarrow x = 48.$

Hence, the third proportion is ₹ 48.

(iv) Let the third proportion be x, then $5\dfrac{1}{4}$, 7, x are in continued proportion.

$5\dfrac{1}{4} = \dfrac{21}{4} \\[0.5em] \Rightarrow \dfrac{\dfrac{21}{4}}{7} = \dfrac{7}{x} \\[0.5em] \Rightarrow x = \dfrac{7}{\dfrac{21}{4}} \times 7 \\[0.5em] \Rightarrow x = \dfrac{7 \times 4 \times 7}{21} \\[0.5em] \Rightarrow x = \dfrac{28}{3} \\[0.5em] \Rightarrow x = 9\dfrac{1}{3}.$

Hence, the third proportion is $9\dfrac{1}{3}$.

Find the mean proportion of :

(i) 5 and 80

(ii) $\dfrac{1}{12}$ and $\dfrac{1}{75}$

(iii) 8.1 and 2.5

(iv) (a - b) and (a³ - a²b), a > b.

Answer

(i) Let the mean proportion be x.

$\therefore \dfrac{5}{x} = \dfrac{x}{80} \\[0.5em] \Rightarrow x^2 = 5 \times 80 \\[0.5em] \Rightarrow x^2 = 400 \\[0.5em] \Rightarrow x = \sqrt{400} \\[0.5em] \Rightarrow x = 20.$

Hence, the mean proportion is 20.

(ii) Let the mean proportion be x.

$\therefore \dfrac{\dfrac{1}{12}}{x} = \dfrac{x}{\dfrac{1}{75}} \\[0.5em] \Rightarrow x^2 = \dfrac{1}{12} \times \dfrac{1}{75} \\[0.5em] \Rightarrow x^2 = \dfrac{1}{900} \\[0.5em] \Rightarrow x = \sqrt{\dfrac{1}{900}} \\[0.5em] \Rightarrow x = \dfrac{1}{30}$

Hence, the mean proportion is $\dfrac{1}{30}$.

(iii) Let the mean proportion be x.

$\therefore \dfrac{8.1}{x} = \dfrac{x}{2.5} \\[0.5em] \Rightarrow x^2 = 8.1 \times 2.5 \\[0.5em] \Rightarrow x^2 = 20.25 \\[0.5em] \Rightarrow x = \sqrt{20.25} \\[0.5em] \Rightarrow x = 4.5.$

Hence, the mean proportion is 4.5.

(iv) Let the mean proportion be x.

$\therefore \dfrac{(a - b)}{x} = \dfrac{x}{(a^3 - a^2b)} \\[0.5em] \Rightarrow x^2 = (a - b) \times (a^3 - a^2b) \\[0.5em] \Rightarrow x^2 = (a^4 - a^3b - a^3b + a^2b^2) \\[0.5em] \Rightarrow x^2 = (a^4 - 2a^3b + a^2b^2) \\[0.5em] \Rightarrow x^2 = a^2(a^2 - 2ab + b^2) \\[0.5em] \Rightarrow x^2 = a^2(a - b)^2 \\[0.5em] \Rightarrow x = \sqrt{(a^2(a - b)^2)} \\[0.5em] \Rightarrow x = a(a - b).$

Hence, the mean proportion is a(a - b).

If a, 12, 16 and b are in continued proportion, find a and b.

Answer

Given, a, 12, 16 and b are in continued proportion.

$\therefore \dfrac{a}{12} = \dfrac{12}{16} = \dfrac{16}{b} \\[1em] \Rightarrow \dfrac{a}{12} = \dfrac{12}{16} \text{ and } \dfrac{12}{16} = \dfrac{16}{b} \\[1em] \Rightarrow a = \dfrac{12}{16} \times 12 \text{ and } b = \dfrac{16}{12} \times 16 \\[1em] \Rightarrow a = \dfrac{144}{16} \text{ and } b = \dfrac{256}{12} \\[1em] \Rightarrow a = 9 \text{ and } b = \dfrac{64}{3}.$

Hence, the value of a = 9 and b = $\dfrac{64}{3}$.

What number must be added to each of the numbers 5, 11, 19 and 37 so that they are in proportion?

Answer

Let the number to be added be x. So, new numbers are 5 + x, 11 + x, 19 + x, 37 + x

Since, these numbers are in proportion,

$\therefore \dfrac{5 + x}{11 + x} = \dfrac{19 + x}{37 + x} \\[0.5em] \Rightarrow (5 + x)(37 + x) = (19 + x)(11 + x) \\[0.5em] (185 + 5x + 37x + x^2) = 209 + 19x + 11x + x^2 \\[0.5em] \Rightarrow x^2 - x^2 + 42x - 30x + 185 - 209 = 0 \\[0.5em] \Rightarrow 12x - 24 = 0 \\[0.5em] \Rightarrow 12x = 24 \\[0.5em] \Rightarrow x = \dfrac{24}{12} = 2.$

Hence, the number that must be added to make the numbers in proportion is 2.

What numbers should be subtracted from each of the numbers 23, 30, 57 and 78 so that remainders are in proportion?

Answer

Let the number to be subtracted be x. So, new numbers are 23 - x, 30 - x, 57 - x, 78 - x

Since, these numbers are in proportion,

$\therefore \dfrac{23 - x}{30 - x} = \dfrac{57 - x}{78 - x} \\[0.5em] \Rightarrow (23 - x)(78 - x) = (57 - x)(30 - x) \\[0.5em] \Rightarrow (1794 - 23x - 78x + x^2) = (1710 - 57x - 30x + x^2) \\[0.5em] \Rightarrow x^2 - x^2 - 101x + 87x + 1794 - 1710 = 0 \\[0.5em] \Rightarrow -14x + 84 = 0 \\[0.5em] \Rightarrow 14x = 84 \\[0.5em] \Rightarrow x = 6.$

Hence, the number that must be subtracted to make the numbers in proportion is 6.

If k + 3, k + 2, 3k - 7 and 2k - 3 are in proportion, find k.

Answer

Since, k + 3, k + 2, 3k - 7 and 2k - 3 are in proportion,

$\therefore \dfrac{k + 3}{k + 2} = \dfrac{3k - 7}{2k - 3} \\[0.5em] \Rightarrow (k + 3)(2k - 3) = (3k - 7)(k + 2) \\[0.5em] \Rightarrow 2k^2 - 3k + 6k - 9 = 3k^2 + 6k - 7k - 14 \\[0.5em] \Rightarrow 2k^2 + 3k - 9 = 3k^2 - k - 14 \\[0.5em] \Rightarrow 2k^2 - 3k^2 + 3k + k - 9 + 14 = 0 \\[0.5em] \Rightarrow -k^2 + 4k + 5 = 0$

Multiplying equation by -1,

$\Rightarrow k^2 - 4k - 5 = 0 \\[0.5em] \Rightarrow k^2 - 5k + k - 5 = 0 \\[0.5em] \Rightarrow k(k - 5) + 1(k - 5) = 0 \\[0.5em] \Rightarrow (k + 1)(k - 5) = 0 \\[0.5em] \Rightarrow k + 1 = 0 \text{ or } k - 5 = 0 \\[0.5em] \Rightarrow k = -1 \text{ or } k = 5.$

Hence, the value of k is -1 and 5.

If (x + 5) is the mean proportion between x + 2 and x + 9, find the value of x.

Answer

Since, (x + 5) is the mean proportion between x + 2 and x + 9,

$\therefore \dfrac{x + 2}{x + 5} = \dfrac{x + 5}{x + 9} \\[0.5em] \Rightarrow (x + 5)^2 = (x + 2)(x + 9) \\[0.5em] \Rightarrow (x^2 + 25 + 10x) = (x^2 + 9x + 2x + 18) \\[0.5em] \Rightarrow x^2 - x^2 + 10x - 11x + 25 - 18 = 0 \\[0.5em] \Rightarrow -x + 7 = 0 \\[0.5em] \Rightarrow x = 7.$

Hence, the value of x is 7.

What numbers must be added to each of the numbers 16, 26 and 40 so that the resulting numbers must be in continued proportion?

Answer

Let the number to be added to each number be x. So, the new numbers are 16 + x, 26 + x, 40 + x.

Since, new numbers are in continued proportion,

$\therefore \dfrac{16 + x}{26 + x} = \dfrac{26 + x}{40 + x} \\[0.5em] \Rightarrow (16 + x)(40 + x) = (26 + x)(26 + x) \\[0.5em] \Rightarrow 640 + 16x + 40x + x^2 = 676 + 26x + 26x + x^2 \\[0.5em] \Rightarrow x^2 - x^2 + 56x - 52x + 640 - 676 = 0 \\[0.5em] \Rightarrow 4x - 36 = 0 \\[0.5em] \Rightarrow 4x = 36 \Rightarrow x = 9.$

Hence, the number to be added to each number is 9.

Find two numbers such that the mean proportional between them is 28 and the third proportional to them is 224.

Answer

Let the two numbers be x and y.

Given, 28 is the mean proportional between x and y.

$\therefore \dfrac{x}{28} = \dfrac{28}{y} \\[0.5em] \Rightarrow xy = 28 \times 28 \\[0.5em] \Rightarrow xy = 784 \\[0.5em] \Rightarrow x = \dfrac{784}{y} \qquad \text{[....Eq 1]}$

Given, 224 is the third proportional to numbers.

$\therefore \dfrac{x}{y} = \dfrac{y}{224} \\[0.5em] \Rightarrow y^2 = 224x \\[0.5em]$

Putting value of x from equation 1 above:

$\Rightarrow y^2 = 224\Big(\dfrac{784}{y}\Big) \\[0.5em] \Rightarrow y^3 = 175616 \\[0.5em] \Rightarrow y = \sqrt[3]{175616} \\[0.5em] \Rightarrow y = \sqrt[3]{56^3} \\[0.5em] \Rightarrow y = 56 \\[0.5em] \text{ and } x = \dfrac{784}{56} = 14.$

Hence, the two numbers are 14 and 56.

If b is the mean proportional between a and c , prove that a, c, a² + b² and b² + c² are proportional.

Answer

Given, b is the mean proportional between a and c then,

b² = ac.    [....Eq 1]

If a, c, a² + b² and b² + c² are proportional then,

$\Rightarrow \dfrac{a}{c} = \dfrac{a^2 + b^2}{b^2 + c^2} \\[0.5em] \Rightarrow a(b^2 + c^2) = c(a^2 + b^2) \\[0.5em]$

Solving L.H.S first

   a(b² + c²)
= a(ac + c²)     [Putting value of b² from Eq 1]
= ac(a + c)

Solving R.H.S

   c(a² + b²)
= c(a² + ac) [Putting value of b² from Eq 1]
= ac(a + c)

Since, L.H.S. = R.H.S = ac(a + c), hence the numbers,
a, c, a² + b² and b² + c² are in proportion.

If b is the mean proportional between a and c, prove that (ab + bc) is the mean proportional between (a² + b²) and (b² + c²).

Answer

Given, b is the mean proportional between a and c then,

b² = ac.    [....Eq 1]

For (ab + bc) to be the mean proportional between (a² + b²) and (b² + c²) following condition must be satisfied,

(ab + bc)² = (a² + b²)(b² + c²)

Solving L.H.S. first,

$\Rightarrow (ab + bc)^2 \\[0.5em] = a^2b^2 + 2ab^2c + b^2c^2 \\[0.5em]$

Putting value of b² from equation 1:

$= a^2(ac) + 2ac(ac) + c^2(ac) \\[0.5em] = a^3c + 2a^2c^2 + ac^3 \\[0.5em] = ac(a^2 + c^2 + 2ac) \\[0.5em] = ac(a + c)^2$

Now, solving R.H.S. ,

$\Rightarrow (a^2 + b^2)(b^2 + c^2) \\[0.5em] = (a^2b^2 + a^2c^2 + b^4 + b^2c^2) \\[0.5em]$

Putting value of b² from equation 1:

$= (a^2(ac) + a^2c^2 + (ac)^2 + (ac)(c^2) \\[0.5em] = a^3c + a^2c^2 + a^2c^2 + ac^3 \\[0.5em] = a^3c + 2a^2c^2 + ac^3 \\[0.5em] = ac(a^2 + 2ac + c^2) \\[0.5em] = ac(a + c)^2$

Since, L.H.S. = R.H.S. = ac(a + c)² hence,
(ab + bc) is the mean proportional between (a² + b²) and (b² + c²).

If y is the mean proportional between x and z, prove that

xyz(x + y + z)³ = (xy + yz + zx)³

Answer

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

y² = xz    [....Eq 1]

Given, xyz(x + y + z)³ = (xy + yz + zx)³

Solving L.H.S. first,

   xyz(x + y + z)³
= xz.y(x + y + z)³
Putting value of xz as y² from equation 1:
= y².y(x + y + z)³
= y³(x + y + z)³
= (y(x + y + z))³
= (xy + y² + yz)³
= (xy + xz + yz)³ = R.H.S.

L.H.S. = R.H.S. , hence proved, xyz(x + y + z)³ = (xy + yz + zx)³.

If a + c = mb and $\dfrac{1}{b} + \dfrac{1}{d} = \dfrac{m}{c}$, prove that a, b, c and d are in proportion.

Answer

Given,

a + c = mb and $\dfrac{1}{b} + \dfrac{1}{d} = \dfrac{m}{c}$

Solving, a + c = mb

Dividing the equation by b,

$\Rightarrow \dfrac{a}{b} + \dfrac{c}{b} = m$     [....Eq 1]

Now solving,

$\dfrac{1}{b} + \dfrac{1}{d} = \dfrac{m}{c}$

Multiplying the equation by c,

$\Rightarrow \dfrac{c}{b} + \dfrac{c}{d} = m$

Putting the value of m from Equation 1,

$\Rightarrow \dfrac{c}{b} + \dfrac{c}{d} = \dfrac{a}{b} + \dfrac{c}{b} \\[0.5em] \Rightarrow \dfrac{a}{b} + \bcancel{\dfrac{c}{b}} = \bcancel{\dfrac{c}{b}} + \dfrac{c}{d} \\[0.5em] \Rightarrow \dfrac{a}{b} = \dfrac{c}{d}$

Since, $\dfrac{a}{b} = \dfrac{c}{d}$ hence, a, b, c, d are in proportion.

If $\dfrac{x}{a} =\dfrac{y}{b} = \dfrac{z}{c},$ prove that

$\begin{array}{ll} \text{(i)} & \dfrac{x^3}{a^2} + \dfrac{y^3}{b^2} + \dfrac{z^3}{c^2} = \dfrac{(x + y + z)^3}{(a + b + c)^2} \\ \text{(ii)} & \Big(\dfrac{a^2x^2 + b^2y^2 + c^2z^2}{a^3x + b^3y + c^3z}\Big)^3 = \dfrac{xyz}{abc} \\ \text{(iii)} & \dfrac{ax - by}{(a + b)(x - y)} + \dfrac{by - cz}{(b + c)(y - z)} \\ & + \dfrac{cz - ax}{(c + a)(z - x)} = 3. \end{array}$

Answer

(i) Let $\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} = k$

∴ x = ak, y = bk, z = ck.

$\text{L.H.S.} = \dfrac{x^3}{a^2} + \dfrac{y^3}{b^2} + \dfrac{z^3}{c^2} \\[0.5em] = \dfrac{a^3k^3}{a^2} + \dfrac{b^3k^3}{b^2} + \dfrac{c^3k^3}{c^2} \\[0.5em] = ak^3 + bk^3 + ck^3 \\[0.5em] =k^3(a + b + c).$

$\text{R.H.S.} = \dfrac{(x + y + z)^3}{(a + b + c)^2} \\[0.5em] = \dfrac{(ak + bk + ck)^3}{(a + b + c)^2} \\[0.5em] = \dfrac{k^3(a + b + c)^3}{(a + b + c)^2} \\[0.5em] = k^3(a + b + c).$

Since, L.H.S. = R.H.S.,

Hence proved, that