Ratio and Proportion

Weight of alloy = Weight of tin + Weight of copper

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

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

$$\begin{aligned}= \frac{\frac{11}{4}}{\frac{121}{4}} \\[0.5em] = \frac{11}{121} \\[0.5em] = \frac{1}{11}\end{aligned}$$

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

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

$$\begin{aligned}= \frac{2}{3} \times \frac{4}{9} \\[0.5em] = \frac{8}{27}\end{aligned}$$

Hence, the compounded ratio is 8 : 27.

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

$$\begin{aligned}= \frac{4}{5} \times \frac{5}{7} \times \frac{9}{11} \\[0.5em] = \frac{180}{385} \\[0.5em]\end{aligned}$$

Dividing numerator and denominator by 5, we get:

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

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,

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

Hence, the compounded ratio is 1 : 1.

(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.

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

= 3³ : 4³ = 27 : 64

Hence, the triplicate ratio is 27 : 64.

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

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

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.

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

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

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

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

$$\begin{aligned}= \sqrt{\frac{1}{4}} : \sqrt{\frac{1}{9}} \\[0.5em] = \frac{1}{2} : \frac{1}{3} \\[0.5em] = \frac{\frac{1}{2}}{\frac{1}{3}} \\[0.5em] = \frac{3}{2} = 3 : 2\end{aligned}$$

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

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

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

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

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

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

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

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

$$\begin{aligned}= \sqrt[3]{\frac{1}{8}} : \sqrt[3]{\frac{1}{125}} \\[0.5em] = \frac{1}{2} : \frac{1}{5} \\[0.5em] = \frac{\frac{1}{2}}{\frac{1}{5}} \\[0.5em] = \frac{5}{2} = 5 : 2\end{aligned}$$

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

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

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

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

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

$$\begin{aligned}= \frac{1}{4} : \frac{1}{7} \\[0.5em] = \frac{\frac{1}{4}}{\frac{1}{7}} \\[0.5em] = \frac{7}{4} \\[0.5em] = 7 : 4\end{aligned}$$

Hence, the reciprocal ratio is 7 : 4.

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

$$\begin{aligned}= \frac{1}{3^2} : \frac{1}{4^2} \\[0.5em] = \frac{\frac{1}{9}}{\frac{1}{16}} \\[0.5em] = \frac{16}{9} \\[0.5em] = 16 : 9\end{aligned}$$

Hence, the reciprocal ratio is 16 : 9.

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

$$\begin{aligned}= \frac{1}{\frac{1}{9}} : \frac{1}{2} \\[0.5em] = \frac{9}{\frac{1}{2}} \\[0.5em] = \frac{18}{1} \\[0.5em] = 18 : 1\end{aligned}$$

Hence, the reciprocal ratio is 18 : 1.

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

We convert them into equivalent like fractions.

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

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

As, 28 < 30 < 33 < 34,

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

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

$$\begin{aligned}\frac{A}{B} = \frac{2}{3} \\[0.5em] \Rightarrow B = \frac{3A}{2} \\[0.5em]\end{aligned}$$

Putting this value of B in B : C

$$\begin{aligned}\frac{B}{C} = \frac{4}{5} \\[0.5em] \Rightarrow \frac{\frac{3A}{2}}{C} = \frac{4}{5} \\[0.5em] \Rightarrow \frac{3A}{2} = \frac{4C}{5} \\[0.5em] \Rightarrow C = \frac{15A}{8} \\[0.5em]\end{aligned}$$

Putting this value of C in C : D

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

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

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

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

∴ x : y : z = 8 : 12 : 21

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

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

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

∴ A : B : C = 15 : 12 : 14

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

$$\begin{aligned}3A = 4B \\[0.5em] \Rightarrow \frac{A}{B} = \frac{4}{3} \\[0.5em] \Rightarrow A : B = 4 : 3\end{aligned}$$

Similarly,

$$\begin{aligned}4B = 6C \\[0.5em] \Rightarrow \frac{B}{C} = \frac{6}{4} = \frac{3}{2} \\[0.5em] \Rightarrow B : C = 3 : 2\end{aligned}$$

So we get,

A : B : C = 4 : 3 : 2

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

Given,

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

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

a : b = 3 : 11 or,

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

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

Dividing the numerator and denominator by b,

$$\begin{aligned}\Rightarrow \frac{\frac{15a}{b} - \frac{3b}{b}}{\frac{9a}{b} + \frac{5b}{b}} \\[0.5em] \Rightarrow \frac{\frac{15a}{ b} - 3}{\frac{9a}{b} + 5} \\[0.5em]\end{aligned}$$

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

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

Hence, the value of ratio is 6 : 41.

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

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

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

Dividing the numerator and denominator by y,

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

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

$$\begin{aligned}\Rightarrow \frac{\frac{4}{5} + 2}{\frac{8}{5} + 1} \\[1em] \Rightarrow \frac{\frac{14}{5}}{\frac{13}{5}} \\[1em] \Rightarrow \frac{14}{13} = 14 : 13. \\[1em]\end{aligned}$$

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

$$\begin{aligned}\Rightarrow \frac{\frac{3}{4} + 2}{\frac{6}{4} + 1} \\[1em] \Rightarrow \frac{\frac{11}{4}}{\frac{10}{4}} \\[1em] \Rightarrow \frac{11}{10} = 11 : 10\end{aligned}$$

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

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

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

Dividing the equation by y²,

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

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

$$\begin{aligned}= (x^2 + y^2) : (x^2 + y^2 + 2xy) \\[0.5em] = \frac{x^2 + y^2}{x^2 + y^2 + 2xy} \\[0.5em]\end{aligned}$$

Dividing numerator and denominator by y²,

$$\begin{aligned}= \frac{\frac{x^2 + y^2}{y^2}}{\frac{x^2 + y^2 + 2xy}{y^2}} \\[1em] = \frac{\big(\frac{x}{y}\big)^2 + 1}{\big(\frac{x}{y}\big)^2 + 1 + 2\big(\frac{x}{y}\big)} \\[1em]\end{aligned}$$

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

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

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

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

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

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

According to question,

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

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

Hence, the value of x is 25.

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

According to question,

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

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

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

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

According to question,

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

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

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

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

According to question,

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

∴ x = 5, 8x = 40, 7x = 35.

Hence, the required numbers are 40, 35.

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.

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

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

Hence, the reduced weight of woman is 65 kg.

Amount collected for charity = ₹2100.

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

Sum of ratio = 3 + 4 = 7

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

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

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

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 = $\frac{30}{15}$
⇒ x = 2

∴ x = 2, 7x = 14, 5x = 10, 3x = 6.

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

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 = $\frac{180°}{9}$
⇒ x = 20°

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

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

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

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

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

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 = $\frac{244}{61}$
⇒ x = $\sqrt{4}$ = 2

∴ x = 2, 6x = 12, 4x = 8, 3x = 6.

Hence, the numbers are 12, 8 and 6.

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.

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 = $\frac{30}{94} \times 18800$ = $\frac{564000}{94}$ = ₹6000.

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

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

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

Ratio of milk to water = 13 : 2.

Total quantity of mixture = 45 litres

Sum of ratio = 13 + 2 = 15.

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

Quantity of water = $\frac{2}{15}$ 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

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

The water that must be added is 7 litres.

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 = $\frac{5}{8}$ x 560 = 5 x 70 = 350.

Number of girls = $\frac{3}{8}$ 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

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

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

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.

$$\begin{aligned}\Rightarrow 5x - 3y = 80 \\[0.5em] \Rightarrow 5x = 80 + 3y \\[0.5em] x = \frac{80 + 3y}{5}\end{aligned}$$

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

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

On cross multiplying,

$$\begin{aligned}\Rightarrow 560 - 4y = 400 \\[1em] \Rightarrow 560 - 400 = 4y \\[1em] \Rightarrow 4y = 160 \\[1em] y = 40.\end{aligned}$$

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

∴ 5x = 200, 7x = 280.

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

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

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

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

Hence, the total number of students are 154.

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,

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

∴ 5x = 150

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

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

$$\begin{aligned}\Rightarrow \frac{10}{35} = \frac{x}{42} \\[0.5em] \Rightarrow x = \frac{10}{35} \times 42 \\[0.5em] \Rightarrow x = \frac{420}{35} \\[0.5em] \Rightarrow x = 12.\end{aligned}$$

Hence, the value of x = 12.

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

$$\begin{aligned}\Rightarrow \frac{3}{x} = \frac{24}{2} \\[0.5em] \Rightarrow x = \frac{3}{24} \times 2 \\[0.5em] \Rightarrow x = \frac{6}{24} \\[0.5em] \Rightarrow x = \frac{1}{4}.\end{aligned}$$

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

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

$$\begin{aligned}\Rightarrow \frac{2.5}{1.5} = \frac{x}{3} \\[0.5em] \Rightarrow x = \frac{2.5}{1.5} \times 3 \\[0.5em] \Rightarrow x = \frac{7.5}{1.5} \\[0.5em] \Rightarrow x = 5.\end{aligned}$$

Hence, the value of x = 5.

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

$$\begin{aligned}\Rightarrow \frac{x}{50} = \frac{3}{2} \\[0.5em] \Rightarrow x = \frac{3}{2} \times 50 \\[0.5em] \Rightarrow x = \frac{150}{2} \\[0.5em] \Rightarrow x = 75.\end{aligned}$$

Hence, the value of x = 75.

(i) Let the fourth proportion be x.

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

$$\begin{aligned}\Rightarrow 3 : 12 :: 15 : x \\[0.5em] \Rightarrow \frac{3}{12} = \frac{15}{x} \\[0.5em] \Rightarrow x = \frac{15}{3} \times 12 \\[0.5em] \Rightarrow x = \frac{180}{3} \\[0.5em] \Rightarrow x = 60.\end{aligned}$$

Hence, the fourth proportion is 60.

(ii) Let the fourth proportion be x.

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

$$\begin{aligned}\Rightarrow \frac{1}{3} : \frac{1}{4} :: \frac{1}{5} : x \\[0.5em] \Rightarrow \frac{\frac{1}{3}}{\frac{1}{4}} = \frac{\frac{1}{5}}{x} \\[0.5em] \Rightarrow x = \frac{\frac{1}{5}}{\frac{1}{3}} \times \frac{1}{4} \\[0.5em] \Rightarrow x = \frac{3}{5} \times \frac{1}{4} \\[0.5em] \Rightarrow x = \frac{3}{20}.\end{aligned}$$

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

(iii) Let the fourth proportion be x.

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

$$\begin{aligned}\Rightarrow 1.5 : 2.5 :: 4.5 : x \\[0.5em] \Rightarrow \frac{1.5}{2.5} = \frac{4.5}{x} \\[0.5em] \Rightarrow x = \frac{4.5}{1.5} \times 2.5 \\[0.5em] \Rightarrow x = 3 \times 2.5 \\[0.5em] \Rightarrow x = 7.5.\end{aligned}$$

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.

$$\begin{aligned}\Rightarrow 9.6 : 7.2 :: 28.8 : x \\[0.5em] \Rightarrow \frac{9.6}{7.2} = \frac{28.8}{x} \\[0.5em] \Rightarrow x = \frac{28.8}{9.6} \times 7.2 \\[0.5em] \Rightarrow x = 3 \times 7.2 \\[0.5em] \Rightarrow x = 21.6.\end{aligned}$$

Hence, the fourth proportion is 21.6 kg.

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

$$\begin{aligned}\Rightarrow \frac{5}{10} = \frac{10}{x} \\[0.5em] \Rightarrow x = \frac{10}{5} \times 10 \\[0.5em] \Rightarrow x = \frac{100}{5} \\[0.5em] \Rightarrow x = 20.\end{aligned}$$

Hence, the third proportion is 20.

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

$$\begin{aligned}\Rightarrow \frac{0.24}{0.6} = \frac{0.6}{x} \\[0.5em] \Rightarrow x = \frac{0.6}{0.24} \times 0.6 \\[0.5em] \Rightarrow x = \frac{0.36}{0.24} \\[0.5em] \Rightarrow x = 1.5.\end{aligned}$$

Hence, the third proportion is 1.5.

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

$$\begin{aligned}\Rightarrow \frac{3}{12} = \frac{12}{x} \\[0.5em] \Rightarrow x = \frac{12}{3} \times 12 \\[0.5em] \Rightarrow x = \frac{144}{3} \\[0.5em] \Rightarrow x = 48.\end{aligned}$$

Hence, the third proportion is ₹ 48.

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

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

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

(i) Let the mean proportion be x.

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

Hence, the mean proportion is 20.

(ii) Let the mean proportion be x.

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

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

(iii) Let the mean proportion be x.

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

Hence, the mean proportion is 4.5.

(iv) Let the mean proportion be x.

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

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

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

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

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

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,

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

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

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,

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

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

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

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

Multiplying equation by -1,

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

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

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

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

Hence, the value of x is 7.

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,

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

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

Let the two numbers be x and y.

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

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

Given, 224 is the third proportional to numbers.

$$\begin{aligned}\therefore \frac{x}{y} = \frac{y}{224} \\[0.5em] \Rightarrow y^2 = 224x \\[0.5em]\end{aligned}$$

Putting value of x from equation 1 above:

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

Hence, the two numbers are 14 and 56.

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,

$$\begin{aligned}\Rightarrow \frac{a}{c} = \frac{a^2 + b^2}{b^2 + c^2} \\[0.5em] \Rightarrow a(b^2 + c^2) = c(a^2 + b^2) \\[0.5em]\end{aligned}$$

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.

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,

$$\begin{aligned}\Rightarrow (ab + bc)^2 \\[0.5em] = a^2b^2 + 2ab^2c + b^2c^2 \\[0.5em]\end{aligned}$$

Putting value of b² from equation 1:

$$\begin{aligned}= a^2(ac) + 2ac(ac) + c^2(ac) \\[0.5em] = a^3c + 2a^2c^2 + ac^3 \\[0.5em] = ac(a^2 + c^2 + 2ac) \\[0.5em] = ac(a + c)^2\end{aligned}$$

Now, solving R.H.S. ,

$$\begin{aligned}\Rightarrow (a^2 + b^2)(b^2 + c^2) \\[0.5em] = (a^2b^2 + a^2c^2 + b^4 + b^2c^2) \\[0.5em]\end{aligned}$$

Putting value of b² from equation 1:

$$\begin{aligned}= (a^2(ac) + a^2c^2 + (ac)^2 + (ac)(c^2) \\[0.5em] = a^3c + a^2c^2 + a^2c^2 + ac^3 \\[0.5em] = a^3c + 2a^2c^2 + ac^3 \\[0.5em] = ac(a^2 + 2ac + c^2) \\[0.5em] = ac(a + c)^2\end{aligned}$$

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

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.

Given,

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

Solving, a + c = mb

Dividing the equation by b,

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

Now solving,

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

Multiplying the equation by c,

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

Putting the value of m from Equation 1,

$$\begin{aligned}\Rightarrow \frac{c}{b} + \frac{c}{d} = \frac{a}{b} + \frac{c}{b} \\[0.5em] \Rightarrow \frac{a}{b} + \bcancel{\frac{c}{b}} = \bcancel{\frac{c}{b}} + \frac{c}{d} \\[0.5em] \Rightarrow \frac{a}{b} = \frac{c}{d}\end{aligned}$$

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

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

∴ x = ak, y = bk, z = ck.

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

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

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

Hence proved, that

$\frac{x^3}{a^2} + \frac{y^3}{b^2} + \frac{z^3}{c^2} = \frac{(x + y + z)^3}{(a + b + c)^2}$

(ii) Let $\frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k$

∴ x = ak, y = bk, z = ck.

$$\begin{aligned}\text{L.H.S.} = \Big(\frac{a^2x^2 + b^2y^2 + c^2z^2}{a^3x + b^3y + c^3}\Big)^3 \\[1em] = \Big(\frac{a^2(ak)x + b^2(bk)y + c^2(ck)z}{a^3x + b^3y + c^3z}\Big)^3 \\[1em] = \Big(\frac{a^3xk + b^3yk + c^3zk}{a^3x + b^3y + c^3z}\Big)^3 \\[1em] = \frac{k^3(a^3x + b^3y + c^3z)^3}{(a^3x + b^3y + c^3z)^3} \\[1em] = k^3 \\[1em] = k \times k \times k \\[1em] = \frac{x}{a} \times \frac{y}{b} \times \frac{z}{c} \\[1em] = \frac{xyz}{abc} = \text{R.H.S.}\end{aligned}$$

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

Hence proved, that $\Big(\frac{a^2x^2 + b^2y^2 + c^2z^2}{a^3x + b^3y + c^3z}\Big)^3 = \frac{xyz}{abc}$.

(iii) Let $\frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k$

∴ x = ak, y = bk, z = ck.

$$\begin{aligned}\text{L.H.S.} = \frac{ax - by}{(a + b)(x - y)} + \frac{by - cz}{(b + c)(y - z)} + \frac{cz - ax}{(c + a)(z - x)} \\[1em] = \frac{a(ak) - b(bk)}{(a + b)((ak) - (bk))} + \frac{b(bk) - c(ck)}{(b + c)(bk - ck)} + \frac{c(ck) - a(ak)}{(c + a)(ck - ak)} \\[1em] = \frac{a^2k - b^2k}{k(a + b)(a - b)} + \frac{b^2k - c^2k}{k(b + c)(b - c)} + \frac{c^2k - a^2k}{k(c + a)(c - a)} \\[1em] = \frac{k(a^2 - b^2)}{k(a^2 - b^2)} + \frac{k(b^2 - c^2)}{k(b^2 - c^2)} + \frac{k(c^2 - a^2)}{k(c^2 - a^2)} \\[1em] = 1 + 1 + 1 \\[1em] = 3 = \text{R.H.S.}\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that,

$\frac{ax - by}{(a + b)(x - y)} + \frac{by - cz}{(b + c)(y - z)} + \frac{cz - ax}{(c + a)(z - x)} = 3.$

(i) Let $\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k$

$\therefore a = bk, c = dk, e = fk.$

$$\begin{aligned}\text{L.H.S.} = (b^2 + d^2 + f^2)(a^2 + c^2 + e^2) \\[0.5em] = (b^2 + d^2 + f^2)(b^2k^2 + d^2k^2 + f^2k^2) \\[0.5em] = k^2(b^2 + d^2 + f^2)(b^2 + d^2 + f^2) \\[0.5em] = k^2(b^2 + d^2 + f^2)^2. \\[1em] \text{R.H.S.} = (ab + cd + ef)^2 \\[0.5em] = (bk.b + dk.d + fk .f)^2 \\[0.5em] = (b^2k + d^2k + f^2k)^2 \\[0.5em] = k^2(b^2 + d^2 + f^2)^2.\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that, (b² + d² + f²)(a² + c² + e²) = (ab + cd + ef)².

(ii) Let $\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k$

$\therefore a = bk, c = dk, e = fk.$

$$\begin{aligned}\text{L.H.S.} = \frac{(a^3 + c^3)^2}{(b^3 + d^3)^2} \\[1em] = \frac{(b^3k^3 + d^3k^3)^2}{(b^3 + d^3)^2} \\[1em] = \frac{[k^3(b^3 + d^3)]^2}{(b^3 + d^3)^2} \\[1em] = \frac{k^6(b^3 + d^3)^2}{(b^3 + d^3)^2} \\[1em] = k^6 \\[1em] \text{R.H.S.} = \frac{e^6}{f^6} \\[1em] = \frac{f^6k^6}{f^6} \\[1em] = k^6\end{aligned}$$

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

(iii) Let $\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k$

$\therefore a = bk, c = dk, e = fk.$

$$\begin{aligned}\text{L.H.S.} = \frac{a^2}{b^2} + \frac{c^2}{d^2} + \frac{e^2}{f^2} \\[1em] = \frac{b^2k^2}{b^2} + \frac{d^2k^2}{d^2} + \frac{f^2k^2}{f^2} \\[1em] = k^2 + k^2 + k^2 \\[1em] = 3k^2 \\[1em] \text{R.H.S.} = \frac{ac}{bd} + \frac{ce}{df} + \frac{ae}{bf} \\[1em] = \frac{(bk)dk}{bd} + \frac{(dk)fk}{df} + \frac{(bk)fk}{bf} \\[1em] = k^2 + k^2 + k^2 \\[1em] = 3k^2\end{aligned}$$

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

(iv) Let $\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k$

$\therefore a = bk, c = dk, e = fk.$

$$\begin{aligned}\text{L.H.S.} = bdf\Big(\frac{a + b}{b} + \frac{c + d}{d} + \frac{e + f}{f}\Big)^3 \\[1em] = bdf\Big(\frac{bk + b}{b} + \frac{dk + d}{d} + \frac{fk + f}{f}\Big)^3 \\[1em] = bdf(k + 1 + k + 1 + k + 1)^3 \\[1em] = bdf(3k + 3)^3 \\[1em] = bdf(3)^3(k + 1)^3 \\[1em] = 27bdf(k + 1)^3. \\[1em] \text{R.H.S.} = 27(a + b)(c + d)(e + f) \\[1em] = 27(bk + b)(dk + d)(fk + f) \\[1em] = 27b(k + 1)d(k + 1)f(k + 1) \\[1em] = 27bdf(k + 1)^3.\end{aligned}$$

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

$$\begin{aligned}\text{Let } ax = by = cz = k \\[1em] \therefore x = \frac{k}{a}, y = \frac{k}{b}, z = \frac{k}{c} \\[1em] \text{L.H.S.} = \frac{x^2}{yz} + \frac{y^2}{zx} + \frac{z^2}{xy} \\[1em] = \frac{\frac{k^2}{a^2}}{\frac{k}{b} \times \frac{k}{c}} + \frac{\frac{k^2}{b^2}}{\frac{k}{c} \times \frac{k}{a}} + \frac{\frac{k^2}{c^2}}{\frac{k}{a} \times \frac{k}{b}} \\[1em] = \frac{k^2 \times bc}{k^2 \times a^2} + \frac{k^2 \times ac}{k^2 \times b^2} + \frac{k^2 \times ab}{k^2 \times c^2} \\[1em] = \frac{bc}{a^2} + \frac{ac}{b^2} + \frac{ab}{c^2} = \text{R.H.S.} \\[1em]\end{aligned}$$

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

(i) a, b, c, d are in proportion

$\therefore \frac{a}{b} = \frac{c}{d} = k$

Hence, a = bk, c = dk.

$$\begin{aligned}\text{L.H.S.} = (5a + 7b)(2c - 3d) \\[0.5em] = (5bk + 7b)(2dk - 3d) \\[0.5em] = bd(5k + 7)(2k - 3) \\[1em] \text{R.H.S.} = (5c + 7d)(2a - 3b) \\[0.5em] = (5dk + 7d)(2bk - 3b) \\[0.5em] = bd(5k + 7)(2k - 3).\end{aligned}$$

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

(ii) a, b, c, d are in proportion

$\therefore \frac{a}{b} = \frac{c}{d} = k$

Hence, a = bk, c = dk.

$$\begin{aligned}\text{L.H.S.} = (ma + nb) : b \\[0.5em] = \frac{ma + nb}{b} \\[0.5em]\end{aligned}$$

Putting value of a = bk,

$$\begin{aligned}= \frac{mbk + nb}{b} \\[0.5em] = mk + n. \\[1em] \text{R.H.S.} = (mc + nd) : d \\[0.5em]\end{aligned}$$

Putting value of c = dk,

$$\begin{aligned}= \frac{mdk + nd}{d} \\[0.5em] = mk + n.\end{aligned}$$

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

(iii) a, b, c, d are in proportion

$\therefore \frac{a}{b} = \frac{c}{d} = k$

Hence, a = bk, c = dk.

$$\begin{aligned}\text{L.H.S.} = (a^4 + c^4) : (b^4 + d^4) \\[1em] = \frac{a^4 + c^4}{b^4 + d^4} \\[1em] = \frac{k^4b^4 + k^4d^4}{b^4 + d^4} \\[1em] = k^4. \\[1em] \text{R.H.S.} = a^2c^2 : b^2d^2 \\[1em] = \frac{a^2c^2}{b^2d^2} \\[1em] = \frac{(bk)^2(dk)^2}{b^2d^2} \\[1em] = \frac{k^4b^2d^2}{b^2d^2} \\[1em] = k^4.\end{aligned}$$

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

(iv) a, b, c, d are in proportion

$\therefore \frac{a}{b} = \frac{c}{d} = k$

Hence, a = bk, c = dk.

$$\begin{aligned}\text{L.H.S.} = \frac{a^2 + ab}{c^2 + cd} \\[1em] = \frac{b^2k^2 + b^2k}{d^2k^2 + d^2k} \\[1em] = \frac{b^2k(k + 1)}{d^2k(k + 1)} \\[1em] = \frac{b^2}{d^2} \\[1em] \text{R.H.S.} = \frac{b^2 - 2ab}{d^2 - 2cd} \\[1em] = \frac{b^2 - 2b^2k}{d^2 - 2d^2k} \\[1em] = \frac{b^2(1 - 2k)}{d^2(1 - 2k)} \\[1em] = \frac{b^2}{d^2}.\end{aligned}$$

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

(v) a, b, c, d are in proportion

$\therefore \frac{a}{b} = \frac{c}{d} = k$

Hence, a = bk, c = dk.

$$\begin{aligned}\text{L.H.S.} = \frac{(a + c)^3}{(b + d)^3} \\[1em] = \frac{(bk + dk)^3}{(b + d)^3} \\[1em] = \frac{k^3(b + d)^3}{(b + d)^3} \\[1em] = k^3 \\[1em] \text{R.H.S.} = \frac{a(a - c)^2}{b(b - d)^2} \\[1em] = \frac{bk(bk - dk)^2}{b(b - d)^2} \\[1em] = \frac{bk(k^2(b - d)^2)}{b(b - d)^2} \\[1em] = \frac{bk^3(b - d)^2}{b(b - d)^2} \\[1em] = k^3.\end{aligned}$$

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

(vi) a, b, c, d are in proportion

$\therefore \frac{a}{b} = \frac{c}{d} = k$

Hence, a = bk, c = dk.

$$\begin{aligned}\text{L.H.S.} = \frac{a^2 + ab + b^2}{a^2 - ab + b^2} \\[1em] = \frac{b^2k^2 + (bk)b + b^2}{b^2k^2 - (bk)b + b^2} \\[1em] = \frac{b^2k^2 + b^2k + b^2}{b^2k^2 - b^2k + b^2} \\[1em] = \frac{b^2(k^2 + k + 1)}{b^2(k^2 - k + 1)} \\[1em] = \frac{k^2 + k + 1}{k^2 - k + 1} \\[1em] \text{R.H.S.} = \frac{c^2 + cd + d^2}{c^2 - cd + d^2} \\[1em] = \frac{d^2k^2 + (dk)d + d^2}{d^2k^2 - (dk)d + d^2} \\[1em] = \frac{d^2(k^2 + k + 1)}{d^2(k^2 - k + 1)} \\[1em] = \frac{k^2 + k + 1}{k^2 - k + 1}.\end{aligned}$$

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

(vii) a, b, c, d are in proportion

$\therefore \frac{a}{b} = \frac{c}{d} = k$

Hence, a = bk, c = dk.

$$\begin{aligned}\text{L.H.S.} = \frac{a^2 + b^2}{c^2 + d^2} \\[1em] = \frac{b^2k^2 + b^2}{d^2k^2 + d^2} \\[1em] = \frac{b^2(k^2 + 1)}{d^2(k^2 + 1)} \\[1em] = \frac{b^2}{d^2} \\[1em] \text{R.H.S.} = \frac{ab + ad - bc}{bc + cd - ad} \\[1em] = \frac{(bk)b + (bk)d - b(dk)}{b(dk) + (dk)d - (bk)d} \\[1em] = \frac{b^2k + bdk - bdk}{bdk + d^2k - bdk} \\[1em] = \frac{b^2k}{d^2k} \\[1em] = \frac{b^2}{d^2}.\end{aligned}$$

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

(viii) a, b, c, d are in proportion

$\therefore \frac{a}{b} = \frac{c}{d} = k$

Hence, a = bk, c = dk.

$$\begin{aligned}\text{L.H.S.} = abcd\Big(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + \frac{1}{d^2} \Big) \\[1em] = (bk)b(dk)d\Big(\frac{1}{b^2k^2} + \frac{1}{b^2} + \frac{1}{d^2k^2} + \frac{1}{d^2}\Big) \\[1em] = b^2d^2k^2\Big(\frac{d^2 + d^2k^2 + b^2 + b^2k^2}{b^2d^2k^2}\Big) \\[1em] = d^2(1 + k^2) + b^2(1 + k^2) \\[1em] = (d^2 + b^2)(1 + k^2). \\[1em] \text{R.H.S.} = a^2 + b^2 + c^2 + d^2 \\[1em] = b^2k^2 + b^2 + d^2k^2 + d^2 \\[1em] = b^2(k^2 + 1) + d^2(k^2 + 1) \\[1em] = (d^2 + b^2)(1 + k^2).\end{aligned}$$

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

Since, x, y, z are in continued proportion

$$\begin{aligned}\therefore \frac{x}{y} = \frac{y}{z} = k \\[1em] \Rightarrow y = zk \text{ and } x = yk = zk^2 \\[1em] \text{L.H.S.} = \frac{(x + y)^2}{(y + z)^2} \\[1em] = \frac{(zk^2 + zk)^2}{(zk + z)^2} \\[1em] = \frac{z^2k^4 + z^2k^2 + 2z^2k^3}{z^2k^2 + z^2 + 2z^2k} \\[1em] = \frac{z^2k^2(k^2 + 1 + 2k)}{z^2(k^2 + 1 + 2k)} \\[1em] = k^2. \\[1em] \text{R.H.S.} = \frac{x}{z} \\[1em] = \frac{zk^2}{z} = k^2.\end{aligned}$$

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

Since, a, b, c are in continued proportion

$$\begin{aligned}\therefore \frac{a}{b} = \frac{b}{c} = k \\[1em] \Rightarrow b = ck \text{ and } a = bk = ck^2 \\[1em] \text{L.H.S.} = \frac{pa^2 + qab + rb^2}{pb^2 + qbc + rc^2} \\[1em] = \frac{p(ck^2)^2 + q(ck^2)(ck) + r(ck)^2}{p(ck)^2 + q(ck)c + rc^2} \\[1em] = \frac{pc^2k^4 + qc^2k^3 + rc^2k^2}{pc^2k^2 + qc^2k + rc^2} \\[1em] = \frac{c^2k^2(pk^2 + qk + r)}{c^2(pk^2 + qk + r)} \\[1em] = k^2 \\[1em] \text{R.H.S.} = \frac{a}{c} \\[1em] = \frac{ck^2}{c} \\[1em] = k^2.\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that,

$\frac{pa^2 + qab + rb^2}{pb^2 + qbc + rc^2} = \frac{a}{c}.$

(i) Since, a, b, c are in continued proportion

$$\begin{aligned}\therefore \frac{a}{b} = \frac{b}{c} = k \\[1em] \Rightarrow b = ck \text{ and } a = bk = ck^2 \\[1em] \text{L.H.S.} = \frac{a + b}{b + c} \\[1em] = \frac{ck^2 + ck}{ck + c} \\[1em] = \frac{ck(k + 1)}{c(k + 1)} \\[1em] = k. \\[1em] \text{R.H.S.} = \frac{a^2(b - c)}{b^2(a - b)} \\[1em] = \frac{(ck^2)^2(ck - c)}{(ck)^2(ck^2 - ck)} \\[1em] = \frac{c^2k^4c(k - 1)}{c^2k^2ck(k - 1)} \\[1em] = \frac{c^3k^4(k - 1)}{c^3k^3(k - 1)} \\[1em] = k.\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that,

$\frac{a + b}{b + c} = \frac{a^2(b - c)}{b^2(a - b)}$.

(ii) Since, a, b, c are in continued proportion

$$\begin{aligned}\therefore \frac{a}{b} = \frac{b}{c} = k \\[1em] \Rightarrow b = ck \text{ and } a = bk = ck^2 \\[1em] \text{L.H.S.} = \frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3} \\[1em] = \frac{1}{(ck^2)^3} + \frac{1}{(ck)^3} + \frac{1}{c^3} \\[1em] = \frac{1}{c^3k^6} + \frac{1}{c^3k^3} + \frac{1}{c^3} \\[1em] = \frac{1}{c^3}\Big(\frac{1}{k^6} + \frac{1}{k^3} + 1\Big) \\[1em] \text{R.H.S.} = \frac{a}{b^2c^2} + \frac{b}{c^2a^2} + \frac{c}{a^2b^2} \\[1em] = \frac{ck^2}{(ck)^2c^2} + \frac{ck}{c^2(ck^2)^2} + \frac{c}{(ck^2)^2(ck)^2} \\[1em] = \frac{ck^2}{c^4k^2} + \frac{ck}{c^4k^4} + \frac{c}{c^4k^6} \\[1em] = \frac{1}{c^3} + \frac{1}{c^3k^3} + \frac{1}{c^3k^6} \\[1em] = \frac{1}{c^3}\Big(1 + \frac{1}{k^3} + \frac{1}{k^6}\Big) \\[1em]\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that,

$\frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3} = \frac{a}{b^2c^2} + \frac{b}{c^2a^2} + \frac{c}{a^2b^2}$.

(iii) Since, a, b, c are in continued proportion

$$\begin{aligned}\therefore \frac{a}{b} = \frac{b}{c} = k \\[1em] \Rightarrow b = ck \text{ and } a = bk = ck^2 \\[1em] \text{L.H.S.} = a : c \\[1em] = \frac{a}{c} \\[1em] = \frac{ck^2}{c} \\[1em] = k^2. \\[1em] \text{R.H.S.} = (a^2 + b^2) : (b^2 + c^2) \\[1em] = \frac{a^2 + b^2}{b^2 + c^2} \\[1em] = \frac{((ck^2)^2 + (ck)^2)}{((ck)^2 + c^2)} \\[1em] = \frac{c^2k^4 + c^2k^2}{c^2k^2 + c^2} \\[1em] = \frac{c^2k^2(k^2 + 1)}{c^2(k^2 + 1)} \\[1em] = k^2. \\[1em]\end{aligned}$$

Since, L.H.S = R.H.S hence proved that,

a : c = (a² + b²) : (b² + c²).

(iv) Since, a, b, c are in continued proportion

$$\begin{aligned}\therefore \frac{a}{b} = \frac{b}{c} = k \\[1em] \Rightarrow b = ck \text{ and } a = bk = ck^2 \\[1em] \text{L.H.S.} = a^2b^2c^2(a^{-4} + b^{-4} + c^{-4}) \\[1em] = (ck^2)^2(ck)^2c^2((ck^2)^{-4} + (ck)^{-4} + c^{-4}) \\[1em] = c^2k^4c^2k^2c^2(c^{-4}k^{-8} + c^{-4}k^{-4} + c^{-4}) \\[1em] = c^6k^6c^{-4}(k^{-8} + k^{-4} + 1) \\[1em] = c^2k^6\big(\frac{1}{k^8} + \frac{1}{k^4} + 1\big) \\[1em] = c^2k^6\big(\frac{1 + k^4 + k^8}{k^8}\big) \\[1em] = \frac{c^2k^6}{k^8}(1 + k^4 + k^8) \\[1em] = \frac{c^2}{k^2}(1 + k^4 + k^8) \\[1em] \text{R.H.S.} = b^{-2}(a^4 + b^4 + c^4) \\[1em] = (ck)^{-2}((ck^2)^4 + (ck)^4 + c^4) \\[1em] = c^{-2}k^{-2}(c^4k^8 + c^4k^4 + c^4) \\[1em] = c^{-2}k^{-2}c^4(k^8 + k^4 + 1) \\[1em] = c^2k^{-2}(k^8 + k^4 + 1) \\[1em] = \frac{c^2}{k^2}(1 + k^4 + k^8). \\[1em]\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that,

a²b²c²(a^-4 + b^-4 + c^-4) = b^-2(a⁴ + b⁴ + c⁴).

(v) Since, a, b, c are in continued proportion

$$\begin{aligned}\therefore \frac{a}{b} = \frac{b}{c} = k \\[1em] \Rightarrow b = ck \text{ and } a = bk = ck^2 \\[1em] \text{L.H.S.} = abc(a + b + c)^3 \\[1em] = (ck^2)(ck)c(ck^2 + ck + c)^3 \\[1em] = c^3k^3(c^3(k^2 + k + 1)^3) \\[1em] = c^6k^3(k^2 + k + 1)^3 \\[1em] \text{R.H.S.} = (ab + bc + ca)^3 \\[1em] = ((ck^2)(ck) + (ck)(c) + (c)(ck^2))^3 \\[1em] = (c^2k^3 + c^2k + c^2k^2)^3 \\[1em] = ((c^2k)^3(k^2 + 1 + k)^3) \\[1em] = c^6k^3(k^2 + k + 1)^3. \\[1em]\end{aligned}$$

Since L.H.S. = R.H.S. hence proved that,

abc(a + b + c)³ = (ab + bc + ca)³.

(vi) Since, a, b, c are in continued proportion

$$\begin{aligned}\therefore \frac{a}{b} = \frac{b}{c} = k \\[1em] \Rightarrow b = ck \text{ and } a = bk = ck^2 \\[1em] \text{L.H.S.} = (a + b + c)(a - b + c) \\[1em] = (ck^2 + ck + c)(ck^2 - ck + c) \\[1em] = c(k^2 + k + 1)c(k^2 - k + 1) \\[1em] = c^2(k^4 - k^3 + k^2 + k^3 - k^2 + k + k^2 - k + 1) \\[1em] = c^2(k^4 - \bcancel{k^3} + \bcancel{k^2} + \bcancel{k^3} - \bcancel{k^2} + \bcancel{k} + k^2 - \bcancel{k} + 1) \\[1em] = c^2(k^4 + k^2 + 1) \\[1em] \text{R.H.S.} = a^2 + b^2 + c^2 \\[1em] = (ck^2)^2 + (ck)^2 + c^2 \\[1em] = c^2k^4 + c^2k^2 + c^2 \\[1em] = c^2(k^4 + k^2 + 1). \\[1em]\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that,

(a + b + c)(a - b + c) = a² + b² + c².

(i) Since, a, b, c, d are in continued proportion

$$\begin{aligned}\therefore \frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k \\[1em] \therefore c = dk, b = ck = (dk)k = dk^2 \text{ and } a = bk = (ck)k = (dk)k^2 = dk^3. \\[1em] \text{L.H.S.} = \frac{a^3 + b^3 + c^3}{b^3 + c^3 + d^3} \\[1em] = \frac{(dk^3)^3 + (dk^2)^3 + (dk)^3}{(dk^2)^3 + (dk)^3 + d^3} \\[1em] = \frac{d^3k^9 + d^3k^6 + d^3k^3}{d^3k^6 + d^3k^3 + d^3} \\[1em] = \frac{d^3k^3(k^6 + k^3 + 1)}{d^3(k^6 + k^3 + 1)} \\[1em] = k^3. \\[1em] \text{R.H.S.} = \frac{a}{d} \\[1em] = \frac{dk^3}{d} \\[1em] = k^3. \\[1em]\end{aligned}$$

Since, L.H.S. = R.H.S. hence, proved that,

$\frac{a^3 + b^3 + c^3}{b^3 + c^3 + d^3} = \frac{a}{d}$.

(ii) Since, a, b, c, d are in continued proportion

$$\begin{aligned}\therefore \frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k \\[1em] \therefore c = dk, b = ck = (dk)k = dk^2 \text{ and } a = bk = (ck)k = (dk)k^2 = dk^3. \\[1em] \text{L.H.S.} = (a^2 - b^2)(c^2 - d^2) \\[1em] = [(dk^3)^2 - (dk^2)^2][(dk)^2 - (d^2)] \\[1em] = (d^2k^6 - d^2k^4)(d^2k^2 - d^2) \\[1em] = d^2k^4(k^2 - 1)d^2(k^2 - 1) \\[1em] = d^4k^4(k^2 - 1)^2. \\[1em] \text{R.H.S.} = (b^2 - c^2)^2 \\[1em] = ((dk^2)^2 - (dk)^2)^2 \\[1em] = (d^2k^4 - d^2k^2)^2 \\[1em] = (d^2k^4- d^2k^2)(d^2k^4 - d^2k^2) \\[1em] = d^2k^2(k^2 - 1)d^2k^2(k^2 - 1) \\[1em] = d^4k^4(k^2 - 1)^2. \\[1em]\end{aligned}$$

Since, L.H.S. = R.H.S. hence, proved that,

(a² - b²)(c² - d²) = (b² - c²)².

(iii) Since, a, b, c, d are in continued proportion

$$\begin{aligned}\therefore \frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k \\[1em] \therefore c = dk, b = ck = (dk)k = dk^2 \text{ and } a = bk = (ck)k = (dk)k^2 = dk^3. \\[1em] \text{L.H.S.} = (a + d)(b + c) - (a + c)(b + d) \\[1em] = (dk^3 + d)(dk^2 + dk) - (dk^3 + dk)(dk^2 + d) \\[1em] = d(k^3 + 1)dk(k + 1) - dk(k^2 + 1)d(k^2 + 1) \\[1em] = d^2k[(k^3 + 1)(k + 1) - (k^2 + 1)^2] \\[1em] = d^2k[(k^4 + k^3 + k + 1) - (k^4 + 1 + 2k^2)] \\[1em] = d^2k(k^4 - k^4 + k^3 - 2k^2 + k + 1 - 1) \\[1em] = d^2k(k^3 - 2k^2 + k) \\[1em] = d^2k(k(k^2 - 2k + 1)) \\[1em] = d^2k^2(k^2 - 2k + 1) \\[1em] = d^2k^2(k - 1)^2. \\[1em] \text{R.H.S.} = (b - c)^2 \\[1em] = (dk^2 - dk)^2 \\[1em] = [(dk)^2(k - 1)^2] \\[1em] = d^2k^2(k - 1)^2.\end{aligned}$$

Since, L.H.S = R.H.S, hence proved that,

(a + d)(b + c) - (a + c)(b + d) = (b - c)².

(iv) Since, a, b, c, d are in continued proportion

$$\begin{aligned}\therefore \frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k \\[1em] \therefore c = dk, b = ck = (dk)k = dk^2 \text{ and } a = bk = (ck)k = (dk)k^2 = dk^3. \\[1em] \text{L.H.S.} = a : d \\[1em] = \frac{a}{d} \\[1em] = \frac{dk^3}{d} \\[1em] = k^3. \\[1em]\end{aligned}$$

R.H.S. = triplicate ratio of (a - b) : (b - c)

$$\begin{aligned}= (a - b)^3 : (b - c)^3 \\[1em] = \frac{(a - b)^3}{(b - c)^3} \\[1em] = \frac{(dk^3 - dk^2)^3}{(dk^2 - dk)^3} \\[1em] = \frac{[k(dk^2 - dk)]^3}{(dk^2 - dk)^3} \\[1em] = \frac{k^3(dk^2 - dk)^3}{(dk^2 - dk)^3} \\[1em] = k^3.\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that,

a : d = triplicate ratio of (a - b) : (b - c).

(v) Since, a, b, c, d are in continued proportion

$$\begin{aligned}\therefore \frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k \\[1em] \therefore c = dk, b = ck = (dk)k = dk^2 \text{ and } a = bk = (ck)k = (dk)k^2 = dk^3. \\[1em] \text{L.H.S.} = \big(\frac{a - b}{c} + \frac{a - c}{b}\big)^2 - \big(\frac{d - b}{c} + \frac{d - c}{b}\big)^2. \\[1em] = \big(\frac{dk^3 - dk^2}{dk} + \frac{dk^3 - dk}{dk^2}\big)^2 - \big(\frac{d - dk^2}{dk} + \frac{d - dk}{dk^2}\big)^2. \\[1em] = \big(\frac{k(dk^3 - dk^2) + dk^3 - dk}{dk^2} \big)^2 - \big(\frac{k(d - dk^2) + d - dk}{dk^2}\big)^2. \\[1em] = \big(\frac{dk^4 - dk^3 + dk^3 - dk}{dk^2} \big)^2 - \big(\frac{kd - dk^3 + d - dk}{dk^2}\big)^2. \\[1em] = \big(\frac{dk^4 - dk}{dk^2} \big)^2 - \big(\frac{d - dk^3}{dk^2}\big)^2. \\[1em] = \big(\frac{dk(k^3 - 1)}{dk^2}\big)^2 - \big(\frac{d(1 - k^3)}{dk^2}\big)^2 \\[1em] = \frac{d^2k^2(k^3 - 1)^2}{d^2k^4} - \frac{d^2(1 - k^3)^2}{d^2k^4} \\[1em] = \frac{(k^3 - 1)^2}{k^2} - \frac{(1 - k^3)^2}{k^4} \\[1em] = \frac{k^6 + 1 -2k^3}{k^2} - \frac{1 + k^6 - 2k^3}{k^4} \\[1em] = \frac{k^2(k^6 + 1 - 2k^3) - (1 + k^6 - 2k^3)}{k^4} \\[1em] = \frac{k^8 + k^2 - 2k^5 - 1 - k^6 + 2k^3}{k^4} \\[1em] \text{R.H.S.} = (a - d)^2\big(\frac{1}{c^2} - \frac{1}{b^2}\big) \\[1em] = (dk^3 - d)^2\big(\frac{1}{(dk)^2} - \frac{1}{(dk^2)^2}\big) \\[1em] = d^2(k^3 - 1)^2 \big(\frac{1}{d^2k^2} - \frac{1}{d^2k^4}\big) \\[1em] = \frac{d^2}{d^2k^2}(k^3 - 1)^2\big(1 - \frac{1}{k^2}\big) \\[1em] = \frac{(k^3 - 1)^2(k^2 - 1)}{k^4} \\[1em] = \frac{(k^6 + 1 - 2k^3)(k^2 - 1)}{k^4} \\[1em] = \frac{k^8 - k^6 + k^2 - 1 + 2k^3 - 2k^5}{k^4}\end{aligned}$$

Since, L.H.S = R.H.S hence proved that,

$\big(\frac{a - b}{c} + \frac{a - c}{b}\big)^2 - \big(\frac{d - b}{c} + \frac{d - c}{b}\big)^2 = (a - d)^2\big(\frac{1}{c^2} - \frac{1}{b^2}\big).$

(i) Given, a : b : : c : d,

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{c}{d} \\[0.5em]\end{aligned}$$

Multiplying the equation by $\frac{2}{5}$,

$$\begin{aligned}\Rightarrow \frac{2a}{5b} = \frac{2c}{5d} \\[0.5em]\end{aligned}$$

By componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{2a + 5b}{2a - 5b} = \frac{2c + 5d}{2c - 5d} \\[0.5em]\end{aligned}$$

Hence, proved that $\frac{2a + 5b}{2a - 5b} = \frac{2c + 5d}{2c - 5d}.$

(ii) Given, a : b : : c : d,

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{c}{d} \\[0.5em]\end{aligned}$$

On multiplying the equation by $\frac{5}{11}$,

$$\begin{aligned}\Rightarrow \frac{5a}{11b} = \frac{5c}{11d} \\[0.5em]\end{aligned}$$

By componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{5a + 11b}{5a - 11b} = \frac{5c + 11d}{5c - 11d} \\[0.5em]\end{aligned}$$

By alternendo,

$\Rightarrow \frac{5a + 11b}{5c + 11d} = \frac{5a - 11b}{5c - 11d}$

Hence, proved that $\frac{5a + 11b}{5c + 11d} = \frac{5a - 11b}{5c - 11d}.$

(iii) Given, a : b : : c : d,

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{c}{d} \\[0.5em]\end{aligned}$$

On multiplying equation by $\frac{2}{3}$,

$$\begin{aligned}\Rightarrow \frac{2a}{3b} = \frac{2c}{3d} \\[0.5em]\end{aligned}$$

By componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{2a + 3b}{2a - 3b} = \frac{2c + 3d}{2c - 3d} \\[0.5em]\end{aligned}$$

On cross multiplication,

$\Rightarrow (2a + 3b)(2c - 3d) = (2c + 3d)(2a - 3b).$

Hence, proved that (2a + 3b)(2c - 3d) = (2a - 3b)(2c + 3d).

(iv) Given, a : b : : c : d,

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{c}{d} \\[0.5em]\end{aligned}$$

On multiplying the equation by $\frac{l}{m}$,

$$\begin{aligned}\Rightarrow \frac{la}{mb} = \frac{lc}{md} \\[0.5em]\end{aligned}$$

By componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{la + mb}{la - mb} = \frac{lc + md}{lc - md} \\[0.5em]\end{aligned}$$

By alternendo,

$$\begin{aligned}\Rightarrow \frac{la + mb}{lc + md} = \frac{la - mb}{lc - md} \\[0.5em] \Rightarrow (la + mb) : (lc + md) : : (la - mb) : (lc - md).\end{aligned}$$

Hence, proved that (la + mb) : (lc + md) : : (la - mb) : (lc - md).

Given,

$\frac{5x + 7y}{5u + 7v} = \frac{5x - 7y}{5u - 7v}$

By alternendo,

$$\begin{aligned}\Rightarrow \frac{5x + 7y}{5x - 7y} = \frac{5u + 7v}{5u - 7v} \\[0.5em]\end{aligned}$$

By componendo & dividendo,

$$\begin{aligned}\Rightarrow \frac{5x + 7y + 5x - 7y}{5x + 7y - 5x + 7y} = \frac{5u + 7v + 5u - 7v}{5u + 7v - 5u + 7v} \\[0.5em] \Rightarrow \frac{10x}{14y} = \frac{10u}{14v} \\[0.5em]\end{aligned}$$

On dividing equation by $\frac{10}{14}$,

$$\begin{aligned}\Rightarrow \frac{x}{y} = \frac{u}{v} \\[0.5em]\end{aligned}$$

Hence, proved that $\frac{x}{y} = \frac{u}{v}.$

Given,

$$\begin{aligned}\frac{8a - 5b}{8c - 5d} = \frac{8a + 5b}{8c + 5d} \\[0.5em]\end{aligned}$$

By alternendo,

$$\begin{aligned}\Rightarrow \frac{8a - 5b}{8a + 5b} = \frac{8c - 5d}{8c + 5d} \\[0.5em]\end{aligned}$$

By componendo & dividendo,

$$\begin{aligned}\Rightarrow \frac{8a - 5b + 8a + 5b}{8a - 5b - 8a - 5b} = \frac{8c - 5d + 8c + 5d}{8c - 5d - 8c - 5d} \\[1em] \Rightarrow -\frac{16a}{10b} = -\frac{16c}{10d}\end{aligned}$$

On dividing the equation by $-\frac{16}{10}$,

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{c}{d} \\[0.5em]\end{aligned}$$

Hence, proved that $\frac{a}{b} = \frac{c}{d}.$

Given, (4a + 5b)(4c - 5d) = (4a - 5b)(4c + 5d).

On cross-multiplication,

$$\begin{aligned}\Rightarrow \frac{4a + 5b}{4a - 5b} = \frac{4c + 5d}{4c - 5d} \\[0.5em]\end{aligned}$$

By componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{4a + 5b + 4a - 5b}{4a + 5b - 4a + 5b} = \frac{4c + 5d + 4c - 5d}{4c + 5d - 4c + 5d} \\[0.5em] \Rightarrow \frac{8a}{10b} = \frac{8c}{10d} \\[0.5em]\end{aligned}$$

On dividing the equation by $\frac{8}{10}$,

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{c}{d} \\[0.5em] \Rightarrow a : b : : c : d.\end{aligned}$$

Hence, proved that a, b, c, d are in proportion.

Given, (pa + qb) : (pc + qd) : : (pa - qb) : (pc - qd).

$$\begin{aligned}\Rightarrow \frac{pa + qb}{pc + qd} = \frac{pa - qb}{pc - qd} \\[0.5em]\end{aligned}$$

By alternendo,

$$\begin{aligned}\Rightarrow \frac{pa + qb}{pa - qb} = \frac{pc + qd}{pc - qd} \\[0.5em]\end{aligned}$$

By componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{pa + qb + pa - qb}{pa + qb - pa + qb} = \frac{pc + qd + pc - qd}{pc + qd - pc + qd} \\[0.5em] \Rightarrow \frac{2pa}{2qb} = \frac{2pc}{2qd} \\[0.5em]\end{aligned}$$

On dividing the equation by $\frac{2p}{2q}$,

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{c}{d} \\[0.5em] \Rightarrow a : b : : c : d.\end{aligned}$$

Hence, proved that a, b, c, d are in proportion.

Given, (ma + nb) : b : : (mc + nd) : d.

$$\begin{aligned}\Rightarrow \frac{(ma + nb)}{b} = \frac{(mc + nd)}{d} \\[0.5em] \Rightarrow d(ma + nb) = b(mc + nd) \\[0.5em] \Rightarrow mad + nbd = bmc + bnd \\[0.5em] \Rightarrow mad = bmc \\[0.5em]\end{aligned}$$

On dividing equation by m,

$$\begin{aligned}\Rightarrow ad = bc \\[0.5em] \Rightarrow \frac{a}{b} = \frac{c}{d} \\[0.5em] \Rightarrow a : b : : c : d.\end{aligned}$$

Hence, proved that a, b, c, d are in proportion.

Given, (11a² + 13b²)(11c² - 13d²) = (11a² - 13b²)(11c² + 13d²).

On cross-multiplication,

$$\begin{aligned}\Rightarrow \frac{11a^2 + 13b^2}{11a^2 - 13b^2} = \frac{11c^2 + 13d^2}{11c^2 - 13d^2} \\[0.5em]\end{aligned}$$

By componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{11a^2 + 13b^2 + 11a^2 - 13b^2}{11a^2 + 13b^2 - 11a^2 + 13b^2} = \frac{11c^2 + 13d^2 + 11c^2 - 13d^2}{11c^2 + 13d^2 - 11c^2 + 13d^2} \\[0.5em] \Rightarrow \frac{22a^2}{26b^2} = \frac{22c^2}{26d^2} \\[0.5em]\end{aligned}$$

On dividing the equation by $\frac{22}{26}$,

$$\begin{aligned}\Rightarrow \frac{a^2}{b^2} = \frac{c^2}{d^2} \\[0.5em] \Rightarrow \sqrt{\big(\frac{a^2}{b^2}\big)} = \sqrt{\big(\frac{c^2}{d^2}\big)} \\[0.5em] \Rightarrow \frac{a}{b} = \frac{c}{d} \\[0.5em] \Rightarrow a : b : : c : d.\end{aligned}$$

Hence, proved that a, b, c, d are in proportion.

Given,

$$\begin{aligned}x = \frac{2ab}{a + b} \\[1em] \frac{x}{a} = \frac{\frac{2ab}{a + b}}{a} \\[1em] \therefore \frac{x}{a} = \frac{2b}{a + b}\end{aligned}$$

By componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{x + a}{x - a} = \frac{2b + a + b}{2b - a - b } \\[0.5em] \Rightarrow \frac{x + a}{x - a} = \frac{3b + a}{b - a} \qquad \text{[....Eq 1]} \\[1.5em] \frac{x}{b} = \frac{\frac{2ab}{a + b}}{b} \\[1em] \therefore \frac{x}{b} = \frac{2a}{a + b} \\[1em]\end{aligned}$$

By componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{x + b}{x - b} = \frac{2a + a + b}{2a - a - b} \\[0.5em] \Rightarrow \frac{x + b}{x - b} = \frac{3a + b}{a - b} \qquad \text{[....Eq 2]}\end{aligned}$$

Adding Eq 1 and 2,

$$\begin{aligned}\Rightarrow \frac{x + a}{x - a} + \frac{x + b}{x - b} = \frac{3b + a}{b - a} + \frac{3a + b}{a - b} \\[1em] \Rightarrow \frac{x + a}{x - a} + \frac{x + b}{x - b} = \frac{3b + a}{b - a} - \frac{3a + b}{b - a} \\[1em] \Rightarrow \frac{x + a}{x - a} + \frac{x + b}{x - b} = \frac{3b - b + a - 3a}{b - a} \\[1em] \Rightarrow \frac{x + a}{x - a} + \frac{x + b}{x - b} = \frac{2b - 2a}{b - a} \\[1em] \Rightarrow \frac{x + a}{x - a} + \frac{x + b}{x - b} = \frac{2(b - a)}{b - a} = 2 \\[1em]\end{aligned}$$

Hence, the required value is 2.

Given,

$$\begin{aligned}x = \frac{8ab}{a + b} \\[1em] \frac{x}{4a} = \frac{\frac{8ab}{a + b}}{4a} \\[1em] \therefore \frac{x}{4a} = \frac{2b}{a + b}\end{aligned}$$

By componendo and dividendo, $$\begin{aligned}\Rightarrow \frac{x + 4a}{x - 4a} = \frac{2b + a + b}{2b - a - b } \\[0.5em] \Rightarrow \frac{x + 4a}{x - 4a} = \frac{3b + a}{b - a} \qquad \text{[....Eq 1]} \\[1.5em] \frac{x}{4b} = \frac{\frac{8ab}{a + b}}{4b} \\[1em] \therefore \frac{x}{4b} = \frac{2a}{a + b}\end{aligned}$$

By componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{x + 4b}{x - 4b} = \frac{2a + a + b}{2a - a - b} \\[0.5em] \Rightarrow \frac{x + 4b}{x - 4b} = \frac{3a + b}{a - b} \qquad \text{[....Eq 2]}\end{aligned}$$

Adding Eq 1 and 2,

$$\begin{aligned}\Rightarrow \frac{x + 4a}{x - 4a} + \frac{x + 4b}{x - 4b} = \frac{3b + a}{b - a} + \frac{3a + b}{a - b} \\[1em] \Rightarrow \frac{x + 4a}{x - 4a} + \frac{x + 4b}{x - 4b} = \frac{3b + a}{b - a} - \frac{3a + b}{b - a} \\[1em] \Rightarrow \frac{x + 4a}{x - 4a} + \frac{x + 4b}{x - 4b} = \frac{3b - b + a - 3a}{b - a} \\[1em] \Rightarrow \frac{x + 4a}{x - 4a} + \frac{x + 4b}{x - 4b} = \frac{2b - 2a}{b - a} \\[1em] \Rightarrow \frac{x + 4a}{x - 4a} + \frac{x + 4b}{x - 4b} = \frac{2(b - a)}{b - a} = 2 \\[1em]\end{aligned}$$

Hence, the required value is 2.

Given,

$$\begin{aligned}x = \frac{4\sqrt{6}}{\sqrt{2} + \sqrt{3}} \\[1em] \frac{x}{2\sqrt{2}} = \frac{\frac{4\sqrt{6}}{\sqrt{2} + \sqrt{3}}}{2\sqrt{2}} \\[1em] \therefore \frac{x}{2\sqrt{2}} = \frac{2\sqrt{3}}{\sqrt{2} + \sqrt{3}}\end{aligned}$$

By componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{x + 2\sqrt{2}}{x - 2\sqrt{2}} = \frac{2\sqrt{3} + \sqrt{2} + \sqrt{3}}{2\sqrt{3} - \sqrt{2} - \sqrt{3}} \\[1em] \Rightarrow \frac{x + 2\sqrt{2}}{x - 2\sqrt{2}} = \frac{3\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \qquad \text{[....Eq 1]} \\[1.5em] \frac{x}{2\sqrt{3}} = \frac{\frac{4\sqrt{6}}{\sqrt{2} + \sqrt{3}}}{2\sqrt{3}} \\[1em] \therefore \frac{x}{2\sqrt{3}} = \frac{2\sqrt{2}}{\sqrt{2} + \sqrt{3}}\end{aligned}$$

By componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{x + 2\sqrt{3}}{x - 2\sqrt{3}} = \frac{2\sqrt{2} + \sqrt{2} + \sqrt{3}}{2\sqrt{2} - \sqrt{2} - \sqrt{3}} \\[1em] \Rightarrow \frac{x + 2\sqrt{3}}{x - 2\sqrt{3}} = \frac{3\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} \qquad \text{[....Eq 2]}\end{aligned}$$

Adding Eq 1 and 2,

$$\begin{aligned}\Rightarrow \frac{x + 2\sqrt{2}}{x - 2\sqrt{2}} + \frac{x + 2\sqrt{3}}{x - 2\sqrt{3}} = \frac{3\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} + \frac{3\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} \\[1em] \Rightarrow \frac{x + 2\sqrt{2}}{x - 2\sqrt{2}} + \frac{x + 2\sqrt{3}}{x - 2\sqrt{3}} = \frac{3\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} - \frac{3\sqrt{2} + \sqrt{3}}{\sqrt{3} - \sqrt{2}} \\[1em] \Rightarrow \frac{x + 2\sqrt{2}}{x - 2\sqrt{2}} + \frac{x + 2\sqrt{3}}{x - 2\sqrt{3}} = \frac{3\sqrt{3} + \sqrt{2} - 3\sqrt{2} - \sqrt{3}}{\sqrt{3} - \sqrt{2}} \\[1em] \Rightarrow \frac{x + 2\sqrt{2}}{x - 2\sqrt{2}} + \frac{x + 2\sqrt{3}}{x - 2\sqrt{3}} = \frac{2\sqrt{3} - 2\sqrt{2}}{\sqrt{3} - \sqrt{2}} \\[1em] \Rightarrow \frac{x + 2\sqrt{2}}{x - 2\sqrt{2}} + \frac{x + 2\sqrt{3}}{x - 2\sqrt{3}} = \frac{2(\sqrt{3} - \sqrt{2})}{\sqrt{3} - \sqrt{2}} = 2. \\[1em]\end{aligned}$$

Hence, the required value is 2.

(i) Given,

$\frac{\sqrt{2 - x} + \sqrt{2 + x}}{\sqrt{2 - x} - \sqrt{2 + x}} = \frac{3}{1}$

Applying componendo and dividendo,

$$\begin{aligned}\Rightarrow\frac{\sqrt{2 - x} + \sqrt{2 + x} + \sqrt{2 - x} - \sqrt{2 + x}}{\sqrt{2 - x} + \sqrt{2 + x} - \sqrt{2 - x} + \sqrt{2 + x}} = \frac{3 + 1}{3 - 1} \\[1em] \Rightarrow \frac{2\sqrt{2 - x}}{2\sqrt{2 + x}} = \frac{4}{2} \\[1em] \Rightarrow \frac{\sqrt{2 - x}}{\sqrt{2 + x}} = \frac{2}{1} \\[1em]\end{aligned}$$

Squaring both sides we get,

$$\begin{aligned}\Rightarrow \frac{2 - x}{2 + x} = \frac{4}{1} \\[0.5em] \Rightarrow (2 - x) = 4(2 + x) \\[0.5em] \Rightarrow 2 - x = 8 + 4x \\[0.5em] \Rightarrow 5x = -6 \\[0.5em] \Rightarrow x = -\frac{6}{5}.\end{aligned}$$

Hence, the value of x = $-\frac{6}{5}.$

(ii) Given,

$\frac{\sqrt{x + 4} + \sqrt{x - 10}}{\sqrt{x + 4} - \sqrt{x - 10}} = \frac{5}{2}$

Applying componendo and dividendo,

$$\begin{aligned}\Rightarrow\frac{\sqrt{x + 4} + \sqrt{x - 10} + \sqrt{x + 4} - \sqrt{x - 10}}{\sqrt{x + 4} + \sqrt{x - 10} - \sqrt{x - 4} + \sqrt{x - 10}} = \frac{5 + 2}{5 - 2} \\[1em] \Rightarrow \frac{2\sqrt{x + 4}}{2\sqrt{x - 10}} = \frac{7}{3} \\[1em] \Rightarrow \frac{\sqrt{x + 4}}{\sqrt{x - 10}} = \frac{7}{3}\end{aligned}$$

Squaring both sides,

$$\begin{aligned}\Rightarrow \frac{x + 4}{x - 10} = \frac{49}{9} \\[1em] \Rightarrow 9(x + 4) = 49(x - 10) \\[1em] \Rightarrow 9x + 36 = 49x - 490 \\[1em] \Rightarrow 40x = 526 \\[1em] \Rightarrow x = \frac{526}{40} \\[1em] \Rightarrow x = \frac{263}{20}.\end{aligned}$$

Hence, the value of x = $\frac{263}{20}.$

(iii) Given,

$\frac{\sqrt{1 + x} + \sqrt{1 - x}}{\sqrt{1 + x} - \sqrt{1 - x}} = \frac{a}{b}$

Applying componendo and dividendo,

$$\begin{aligned}\Rightarrow\frac{\sqrt{1 + x} + \sqrt{1 - x} + \sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x} - \sqrt{1 + x} + \sqrt{1 - x}} = \frac{a + b}{a - b} \\[1em] \Rightarrow \frac{2\sqrt{1 + x}}{2\sqrt{1 - x}} = \frac{a + b}{a - b} \\[1em] \Rightarrow \frac{\sqrt{1 + x}}{{\sqrt{1 - x}}} = \frac{a + b}{a - b}\end{aligned}$$

Squaring both sides we get,

$$\begin{aligned}\Rightarrow \frac{1 + x}{1 - x} = \frac{(a + b)^2}{(a - b)^2} \\[0.5em]\end{aligned}$$

By componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{1 + x + 1 - x}{1 + x - 1 + x} = \frac{(a + b)^2 + (a - b)^2}{(a + b)^2 - (a - b)^2} \\[1em] \Rightarrow \frac{2}{2x} = \frac{a^2 + 2ab + b^2 + a^2 - 2ab + b^2}{a^2 + 2ab + b^2 - a^2 + 2ab - b^2} \\[1em] \Rightarrow \frac{2}{2x} = \frac{2a^2 + 2b^2 }{2ab + 2ab} \\[1em] \Rightarrow \frac{2}{2x} = \frac{2(a^2 + b^2)}{4ab} \\[1em] \Rightarrow \frac{1}{x} = \frac{a^2 + b^2}{2ab} \\[1em] \Rightarrow x = \frac{2ab}{a^2 + b^2}.\end{aligned}$$

Hence, the value of x = $\frac{2ab}{a^2 + b^2}.$

(iv) Given,

$\frac{\sqrt{5x} + \sqrt{2x - 6}}{\sqrt{5x} - \sqrt{2x - 6}} = 4.$

By componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{\sqrt{5x} + \sqrt{2x - 6} + \sqrt{5x} - \sqrt{2x - 6}}{\sqrt{5x} + \sqrt{2x - 6} - \sqrt{5x} + \sqrt{2x - 6}} = \frac{4 + 1}{4 - 1} \\[1em] \Rightarrow \frac{2\sqrt{5x}}{2\sqrt{2x - 6}} = \frac{5}{3} \\[1em] \Rightarrow \frac{\sqrt{5x}}{\sqrt{2x - 6}} = \frac{5}{3}\end{aligned}$$

Squaring both sides,

$$\begin{aligned}\Rightarrow \Big(\frac{\sqrt{5x}}{\sqrt{2x - 6}}\Big)^2 = \Big(\frac{5}{3}\Big)^2 \\[1em] \Rightarrow \frac{5x}{2x - 6} = \frac{25}{9} \\[1em] \Rightarrow 5x \times 9 = 25(2x - 6) \\[1em] \Rightarrow 45x = 50x - 150 \\[1em] \Rightarrow 5x = 150 \\[1em] \Rightarrow x = 30.\end{aligned}$$

Hence, the value of x is 30.

(v) Given,

$\frac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x} - \sqrt{a - x}} = \frac{c}{d}$

By componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{\sqrt{a + x} + \sqrt{a - x} + \sqrt{a + x} - \sqrt{a - x}}{\sqrt{a + x} + \sqrt{a - x} - \sqrt{a + x} + \sqrt{a - x}} = \frac{c + d}{c - d} \\[1em] \Rightarrow \frac{2\sqrt{a + x}}{2\sqrt{a - x}} = \frac{c + d}{c - d} \\[1em] \Rightarrow \frac{\sqrt{a + x}}{\sqrt{a - x}} = \frac{c + d}{c - d}\end{aligned}$$

Squaring both sides,

$$\begin{aligned}\Rightarrow \frac{a + x}{a - x} = \Big(\frac{c + d}{c - d}\Big)^2 \\[1em] \Rightarrow \frac{a + x}{a - x} = \frac{c^2 + d^2 + 2cd}{c^2 + d^2 - 2cd} \\[1em]\end{aligned}$$

Again applying componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{a + x + a - x}{a + x - a + x} = \frac{c^2 + d^2 + 2cd + c^2 + d^2 - 2cd}{c^2 + d^2 + 2cd - c^2 - d^2 + 2cd} \\[1em] \Rightarrow \frac{2a}{2x} = \frac{2(c^2 + d^2)}{4cd} \\[1em] \Rightarrow \frac{a}{x} = \frac{c^2 + d^2}{2cd} \\[1em] \Rightarrow x = \frac{2acd}{c^2 + d^2}\end{aligned}$$

Hence, the value of x is $\frac{2acd}{c^2 + d^2}.$

(vi) Given,

$\frac{a + \sqrt{a^2 - 2ax}}{a - \sqrt{a^2 - 2ax}} = \frac{b}{1}.$

By componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{a + \sqrt{a^2 - 2ax} + a - \sqrt{a^2 - 2ax}}{a + \sqrt{a^2 - 2ax} - a + \sqrt{a^2 - 2ax}} = \frac{b + 1}{b - 1} \\[1em] \Rightarrow \frac{2a}{2\sqrt{a^2 - 2ax}} = \frac{b + 1}{b - 1} \\[1em] \Rightarrow \frac{a}{\sqrt{a^2 - 2ax}} = \frac{b + 1}{b - 1} \\[1em]\end{aligned}$$

Squaring both sides,

$$\begin{aligned}\Rightarrow \frac{a^2}{a^2 - 2ax} = \Big(\frac{b + 1}{b - 1}\Big)^2 \\[1em] \Rightarrow \frac{a^2}{a^2 - 2ax} = \frac{b^2 + 1 + 2b}{b^2 + 1 - 2b} \\[1em]\end{aligned}$$

Applying componendo and dividendo again,

$$\begin{aligned}\Rightarrow \frac{a^2 + a^2 - 2ax}{a^2 - a^2 + 2ax} = \frac{b^2 + 1 + 2b + b^2 + 1 - 2b}{b^2 + 1 + 2b - b^2 - 1 + 2b} \\[1em] \Rightarrow \frac{2a(a - x)}{2ax} = \frac{2(b^2 + 1)}{4b} \\[1em] \Rightarrow \frac{a - x}{x} = \frac{b^2 + 1}{2b}\end{aligned}$$

On cross-multiplication,

$$\begin{aligned}\Rightarrow 2b(a - x) = x(b^2 + 1) \\[0.5em] \Rightarrow 2ab - 2bx = b^2x + x \\[0.5em] \Rightarrow b^2x + x + 2bx = 2ab \\[0.5em] \Rightarrow x(b^2 + 1 + 2b) = 2ab \\[0.5em] \Rightarrow x(b + 1)^2 = 2ab \\[0.5em] \Rightarrow x = \frac{2ab}{(b + 1)^2}.\end{aligned}$$

Hence, the value of x is $\frac{2ab}{(b + 1)^2}.$

(i) Given,

$\frac{3x + \sqrt{9x^2 - 5}}{3x - \sqrt{9x^2 - 5}} = \frac{5}{1}.$

By componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{3x + \sqrt{9x^2 - 5} + 3x - \sqrt{9x^2 - 5}}{3x + \sqrt{9x^2 - 5} - 3x + \sqrt{9x^2 - 5}} = \frac{5 + 1}{5 - 1} \\[1em] \Rightarrow \frac{6x}{2\sqrt{9x^2 - 5}} = \frac{6}{4} \\[1em] \Rightarrow \frac{x}{\sqrt{9x^2 - 5}} = \frac{1}{2}\end{aligned}$$

Squaring both sides we get,

$$\begin{aligned}\frac{x^2}{9x^2 - 5} = \frac{1}{4} \\[0.5em] \Rightarrow 4x^2 = 9x^2 - 5 \\[0.5em] \Rightarrow 5x^2 - 5 = 0 \\[0.5em] \Rightarrow 5(x^2 - 1) = 0 \\[0.5em] \Rightarrow (x + 1)(x - 1) = 0 \\[0.5em] \Rightarrow x = 1, -1.\end{aligned}$$

Since, x is positive, hence, x ≠ -1.

Hence, the required value of x is 1.

(ii) Given,

$\frac{2x + \sqrt{4x^2 - 1}}{2x - \sqrt{4x^2 - 1}} = \frac{4}{1}.$

By componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{2x + \sqrt{4x^2 - 1} + 2x - \sqrt{4x^2 - 1}}{2x + \sqrt{4x^2 - 1} - 2x + \sqrt{4x^2 - 1}} = \frac{4 + 1}{4 - 1} \\[1em] \Rightarrow \frac{4x}{2\sqrt{4x^2 - 1}} = \frac{5}{3} \\[1em] \Rightarrow \frac{2x}{\sqrt{4x^2 - 1}} = \frac{5}{3}\end{aligned}$$

Squaring both sides we get,

$$\begin{aligned}\frac{4x^2}{4x^2 - 1} = \frac{25}{9} \\[1em] \Rightarrow 4x^2 \times 9 = 25(4x^2 - 1) \\[1em] \Rightarrow 36x^2 = 100x^2 - 25 \\[1em] \Rightarrow 64x^2 = 25 \\[1em] \Rightarrow x^2 = \frac{25}{64} \\[1em] \Rightarrow x = \sqrt{\frac{25}{64}} \\[1em] \Rightarrow x = \frac{5}{8} \text{ or } -\frac{5}{8} \\[1em]\end{aligned}$$

Since, x is positive, hence, x ≠ $-\frac{5}{8}.$

Hence, the required value of x is $\frac{5}{8}$.

Given,

$\frac{1 + x + x^2}{1 - x + x^2} = \frac{62(1 + x)}{63(1 - x)}$

$$\begin{aligned}\Rightarrow \frac{(1 + x + x^2)(1 - x)}{(1 - x + x^2)(1 + x)} = \frac{62}{63} \\[1em] \Rightarrow \frac{1 + x + x^2 -x -x^2 - x^3}{1 - x + x^2 + x - x^2 + x^3} = \frac{62}{63} \\[1em] \Rightarrow \frac{1 - \cancel{x} + \cancel{x} - \cancel{x^2} + \cancel{x^2} - x^3}{1 + \cancel{x} - \cancel{x} - \cancel{x^2} + \cancel{x^2} + x^3} = \frac{62}{63} \\[1em] \Rightarrow \frac{1 - x^3}{1 + x^3} = \frac{62}{63}\end{aligned}$$

Again applying componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{1 - x^3 + 1 + x^3}{1 - x^3 - 1 -x^3} = \frac{62 + 63}{62 - 63} \\[1em] \Rightarrow \frac{2}{-2x^3} = \frac{125}{-1} \\[1em] \Rightarrow -\frac{1}{x^3} = -125 \\[1em] \Rightarrow x^3 = \frac{1}{125} \\[1em] \Rightarrow x = \frac{1}{\sqrt[3]{125}} \\[1em] \Rightarrow x = \frac{1}{5}.\end{aligned}$$

Hence, the required value is $\frac{1}{5}.$

Given,

$16\Big(\frac{a - x}{a + x}\Big)^3 =\frac{a + x}{a - x}.$

$$\begin{aligned}\Rightarrow \frac{(a - x)^3}{(a + x)^3} \times \frac{(a - x)}{(a + x)} = \frac{1}{16} \\[1em] \Rightarrow \Big(\frac{a - x}{a + x}\Big)^4 = \Big(\frac{1}{2}\Big)^4 \text{ or } \Big(-\frac{1}{2}\Big)^4 \\[1em] \Rightarrow \frac{a - x}{a + x} = \frac{1}{2} \text{ or } -\frac{1}{2}\end{aligned}$$

First Solving,

$\frac{a - x}{a + x} = \frac{1}{2}$

By componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{a - x + a + x}{a - x - a - x} = \frac{1 + 2}{1 - 2} \\[1em] \Rightarrow -\frac{2a}{2x} = -3 \\[1em] \Rightarrow \frac{a}{x} = 3 \\[1em] \Rightarrow x = \frac{a}{3}.\end{aligned}$$

Now Solving,

$\frac{a - x}{a + x} = -\frac{1}{2}$

By componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{a - x + a + x}{a - x - a - x} = \frac{1 - 2}{1 + 2}\\[1em] \Rightarrow -\frac{2a}{2x} = -\frac{1}{3} \\[1em] \Rightarrow \frac{a}{x} = \frac{1}{3} \\[1em] \Rightarrow x = 3a. \\[1em]\end{aligned}$$

Hence, the value of x is $\frac{a}{3}$ and 3a.

Given,

$\frac{x}{1} = \frac{\sqrt{a + 1} + \sqrt{a - 1}}{\sqrt{a + 1} - \sqrt{a - 1}}.$

Applying componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{x + 1}{x - 1} = \frac{\sqrt{a + 1} + \sqrt{a - 1} + \sqrt{a + 1} - \sqrt{a - 1}}{\sqrt{a + 1} + \sqrt{a - 1} - \sqrt{a + 1} + \sqrt{a - 1}} \\[1em] \Rightarrow \frac{x + 1}{x - 1} = \frac{2\sqrt{a + 1}}{2\sqrt{a - 1}} \\[1em] \Rightarrow \frac{x + 1}{x - 1} = \frac{\sqrt{a + 1}}{\sqrt{a - 1}}\end{aligned}$$

Squaring both sides we get,

$$\begin{aligned}\Rightarrow \Big(\frac{x + 1}{x - 1}\Big)^2 = \Big(\frac{\sqrt{a + 1}}{\sqrt{a - 1}}\Big)^2 \\[1em] \Rightarrow \frac{x^2 + 1 + 2x}{x^2 + 1 - 2x} = \frac{a + 1}{a - 1}\end{aligned}$$

Again applying componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{x^2 + 1 + \cancel{2x} + x^2 + 1 - \cancel{2x}}{\cancel{x^2} + \cancel{1} + 2x - \cancel{x^2} - \cancel{1} + 2x} = \frac{a + \cancel{1} + a - \cancel{1}}{\cancel{a} + 1 - \cancel{a} + 1} \\[1em] \Rightarrow \frac{2(x^2 + 1)}{4x} = \frac{2a}{2} \\[1em] \Rightarrow \frac{x^2 + 1}{2x} = a \\[1em] \Rightarrow x^2 + 1 = 2ax \\[1em] \Rightarrow x^2 - 2ax + 1 = 0.\end{aligned}$$

Hence proved that, x² - 2ax + 1 = 0.

Given; x = $\frac{\sqrt{2a + 1} + \sqrt{2a - 1}}{\sqrt{2a + 1} - \sqrt{2a - 1}}$

Applying componendo and dividendo on both sides we get :

$$\begin{aligned}\Rightarrow \frac{x + 1}{x - 1} = \frac{\sqrt{2a + 1} + \sqrt{2a - 1} + \sqrt{2a + 1} - \sqrt{2a - 1}}{\sqrt{2a + 1} + \sqrt{2a - 1} - (\sqrt{2a + 1} - \sqrt{2a - 1})}\\[1em] \Rightarrow \frac{x + 1}{x - 1} = \frac{2\sqrt{2a + 1}}{2\sqrt{2a - 1}}\\[1em] \Rightarrow \frac{x + 1}{x - 1} = \frac{\sqrt{2a + 1}}{\sqrt{2a - 1}}\\[1em] \text{Squaring both sides we get :}\\[1em] \Rightarrow \frac{(x + 1)^2}{(x - 1)^2} = \frac{2a + 1}{2a - 1}\\[1em] \Rightarrow \frac{x^2 + 1 + 2x}{x^2 + 1 - 2x} = \frac{2a + 1}{2a - 1}\\[1em] \text{Applying componendo and dividendo on both sides we get :}\\[1em] \Rightarrow \frac{x^2 + 1 + 2x + x^2 + 1 - 2x}{x^2 + 1 + 2x - (x^2 + 1 - 2x)} = \frac{2a + 1 + 2a - 1}{2a + 1 - (2a - 1)}\\[1em] \Rightarrow \frac{2x^2 + 2}{4x} = \frac{4a}{2}\\[1em] \Rightarrow \frac{x^2 + 1}{2x} = 2a\\[1em] \Rightarrow x^2 + 1 = 4ax\\[1em] \Rightarrow x^2 + 1 - 4ax = 0\end{aligned}$$

Hence, proved that x² - 4ax + 1 = 0.

Given,

$x = \frac{\sqrt{a^2 + b^2} + \sqrt{a^2 - b^2}}{\sqrt{a^2 + b^2} - \sqrt{a^2 - b^2}}$

By componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{x + 1}{x - 1} = \frac{\sqrt{a^2 + b^2} + \sqrt{a^2 - b^2} + \sqrt{a^2 + b^2} - \sqrt{a^2 - b^2}}{\sqrt{a^2 + b^2} + \sqrt{a^2 - b^2} - \sqrt{a^2 + b^2} + \sqrt{a^2 - b^2}} \\[1em] \Rightarrow \frac{x + 1}{x - 1} = \frac{2\sqrt{a^2 + b^2}}{2\sqrt{a^2 - b^2}} \\[1em] \Rightarrow \frac{x + 1}{x - 1} = \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 - b^2}}\end{aligned}$$

On squaring both sides,

$$\begin{aligned}\Rightarrow \Big(\frac{x + 1}{x - 1}\Big)^2 = \frac{a^2 + b^2}{a^2 - b^2} \\[1em] \Rightarrow \frac{x^2 + 1 + 2x}{x^2 + 1 - 2x} = \frac{a^2 + b^2}{a^2 - b^2} \\[1em]\end{aligned}$$

Again applying componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{x^2 + 1 + \cancel{2x} + x^2 + 1 - \cancel{2x}}{\cancel{x^2} + \cancel{1} + 2x - \cancel{x^2} - \cancel{1} + 2x} = \frac{a^2 + \cancel{b^2} + a^2 - \cancel{b^2}}{\cancel{a^2} + b^2 - \cancel{a^2} + b^2} \\[1em] \Rightarrow \frac{2x^2 + 2}{4x} = \frac{2a^2}{2b^2} \\[1em] \Rightarrow \frac{x^2 + 1}{2x} = \frac{a^2}{b^2}\\[1em] \Rightarrow b^2 = \frac{2a^2x}{x^2 + 1}.\end{aligned}$$

Hence, proved that $b^2 = \frac{2a^2x}{x^2 + 1}.$

Given,

$\frac{a^3 + 3ab^2}{b^3 + 3a^2b} = \frac{63}{62}$

By componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{a^3 + 3ab^2 + b^3 + 3a^2b}{a^3 + 3ab^2 - b^3 - 3a^2b} = \frac{63 + 62}{63 - 62} \\[1em] \Rightarrow \Big(\frac{a + b}{a - b}\Big)^3 = 125 \\[1em] \Rightarrow \Big(\frac{a + b}{a - b}\Big)^3 = (5)^3 \\[1em] \Rightarrow \frac{a + b}{a - b} = 5 \\[1em] \Rightarrow a + b = 5a - 5b \\[1em] \Rightarrow 5a - a = b + 5b \\[1em] \Rightarrow 4a = 6b \\[1em] \Rightarrow \frac{a}{b} = \frac{6}{4} = \frac{2}{3} \\[1em] \Rightarrow a : b = 3 : 2.\end{aligned}$$

Hence, the value of a : b = 3 : 2.

Given,

$\frac{x^3 + 12x}{6x^2 + 8} = \frac{y^3 + 27y}{9y^2 + 27}$

By componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{x^3 + 12x + 6x^2 + 8}{x^3 + 12x - 6x^2 - 8} = \frac{y^3 + 27y + 9y^2 + 27}{y^3 + 27y - 9y^2 - 27} \\[1em] \Rightarrow \frac{x^3 + (3 \times x \times 2^2 ) + (3 \times x^2 \times 2 ) + 2^3}{x^3 + (3 \times x \times 2^2 ) - (3 \times x^2 \times 2 ) - 2^3} \\[1em] = \frac{y^3 + (3 \times y \times 3^2) + (3 \times y^2 \times 3) + 3^3}{y^3 + (3 \times y \times 3^2) - (3 \times y^2 \times 3) - 3^3} \\[1em] \Rightarrow \Big(\frac{x + 2}{x - 2}\Big)^3 = \Big(\frac{y + 3}{y - 3}\Big)^3 \\[1em] \Rightarrow \frac{x + 2}{x - 2} = \frac{y + 3}{y - 3}\end{aligned}$$

Again applying componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{x + \cancel{2} + x - \cancel{2}}{\cancel{x} + 2 - \cancel{x} + 2} = \frac{y + \cancel{3} + y - \cancel{3}}{\cancel{y} + 3 - \cancel{y} + 3} \\[1em] \Rightarrow \frac{2x}{4} = \frac{2y}{6} \\[1em] \Rightarrow \frac{x}{y} = \frac{2}{6} \times \frac{4}{2} \\[1em] \Rightarrow \frac{x}{y} = \frac{2}{3} \\[1em] \Rightarrow x : y = 2 : 3.\end{aligned}$$

Hence, the value of ratio x : y is 2 : 3.

Given,

$\frac{x^3 + 3x}{3x^2 + 1} = \frac{341}{91}$

Applying componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{x^3 + 3x + 3x^2 + 1}{x^3 + 3x - 3x^2 - 1} = \frac{341 + 91}{341 - 91} \\[1em] \Rightarrow \frac{(x + 1)^3}{(x - 1)^3} = \frac{432}{250} \\[1em] \Rightarrow \frac{(x + 1)^3}{(x - 1)^3} = \frac{216}{125} \\[1em] \Rightarrow \Big(\frac{x + 1}{x - 1}\Big)^3 = \Big(\frac{6}{5}\Big)^3 \\[1em] \Rightarrow \frac{x + 1}{x - 1} = \frac{6}{5}\end{aligned}$$

Again applying componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{x + \cancel{1} + x - \cancel{1}}{\cancel{x} + 1 - \cancel{x} + 1} = \frac{6 + 5}{6 - 5} \\[1em] \Rightarrow \frac{2x}{2} = \frac{11}{1} \\[1em] \Rightarrow x = 11.\end{aligned}$$

Hence, the value of x is 11.

We know that if $\frac{a}{b} = \frac{c}{d} = \frac{e}{f}$, then each ratio

$= \frac{a + c + e}{b + d + f} = \frac{\text{sum of antecedents}}{\text{sum of consequents}}$

$$\begin{aligned}\therefore \frac{x + y}{ax + by} = \frac{y + z}{ay + bz} = \frac{z + x}{az + bx} \\[1em] = \frac{x + y + y + z + z + x}{ax + by + ay + bz + az + bx} \\[1em] = \frac{2(x + y + z)}{a(x + y + z) + b(x + y + z)} \\[1em] = \frac{2(x + y + z)}{(a + b)(x + y + z)} \\[1em] = \frac{2}{a + b}.\end{aligned}$$

Hence, proved that,

$\frac{x + y}{ax + by} = \frac{y + z}{ay + bz} = \frac{z + x}{az + bx} = \frac{2}{a + b}.$

4 litres = 4 x 1000 ml = 4000 ml

Ratio = $\frac{4000}{900} = \frac{40}{9}$ = 40 : 9.

∴ Option 2 is the correct option.

Let new number be x. Since, 210 is increased in the ratio 5 : 7,

$$\begin{aligned}\therefore \frac{210}{x} = \frac{5}{7} \\[0.5em] \Rightarrow x = 210 \times \frac{7}{5} \\[0.5em] \Rightarrow x = 294.\end{aligned}$$

∴ Option 3 is the correct option.

Since, two numbers are in the ratio 7 : 9, let the numbers be 7x, 9x.

Given, the sum of two numbers = 288.

∴ 7x + 9x = 288
⇒ 16x = 288
⇒ x = $\frac{288}{16}$
⇒ x = 18.

Smaller number = 7x = 126.

∴ Option 1 is the correct option.

Number of edges in a cube = 12

Number of faces in cube = 6

Ratio of number of edges to the number of faces = $\frac{12}{6} = \frac{2}{1}$ = 2 : 1.

∴ Option 1 is the correct option.

Given, x : 12 : : 8 : 32

$$\begin{aligned}\Rightarrow \frac{x}{12} = \frac{8}{32} \\[0.5em] \Rightarrow x = 12 \times \frac{8}{32} \\[0.5em] \Rightarrow x = 3.\end{aligned}$$

∴ Option 3 is the correct option.

Let the fourth proportional be x.

So, the numbers 3, 4, 5, x are in proportion.

$$\begin{aligned}\therefore 3 : 4 : : 5 : x \\[0.5em] \Rightarrow \frac{3}{4} = \frac{5}{x} \\[0.5em] \Rightarrow x = 5 \times \frac{4}{3} \\[0.5em] \Rightarrow x = \frac{20}{3}.\end{aligned}$$

∴ Option 2 is the correct option.

Let the third proportional be x.

Hence, the numbers $6\frac{1}{4}$, 5, x are in continued in proportion.

$$\begin{aligned}\therefore \frac{25}{4} : 5 : : 5 : x \\[0.5em] \Rightarrow \frac{\frac{25}{4}}{5} = \frac{5}{x} \\[0.5em] \Rightarrow \frac{25}{20} = \frac{5}{x} \\[0.5em] \Rightarrow x = 5 \times \frac{20}{25} \\[0.5em] \Rightarrow x = 4.\end{aligned}$$

∴ Option 1 is the correct option.

Let the mean proportional be x.

Hence, the numbers are $\frac{1}{2}$, x and 128 are in continued in proportion.

$$\begin{aligned}\therefore \frac{1}{2} : x : : x : 128 \\[0.5em] \Rightarrow \frac{\frac{1}{2}}{x} = \frac{x}{128} \\[0.5em] \Rightarrow x^2 = \frac{1}{2} \times 128 \\[0.5em] \Rightarrow x^2 = 64 \\[0.5em] \Rightarrow x = \sqrt{64} \\[0.5em] \Rightarrow x = 8.\end{aligned}$$

∴ Option 4 is the correct option.

Given, x is proportional to y.

$$\begin{aligned}\Rightarrow \frac{6}{M} = \frac{12}{18} \\[1em] \Rightarrow M = \frac{6 \times 18}{12} \\[1em] \Rightarrow M = \frac{108}{12} \\[1em] \Rightarrow M = 9. \\[1.5em] \frac{12}{18} = \frac{N}{6} \\[1em] \Rightarrow N = \frac{12 \times 6}{18} \\[1em] \Rightarrow N = \frac{72}{18} \\[1em] \Rightarrow N = 4.\end{aligned}$$

Hence, Option 3 is the correct option.

Speed = $\frac{\text{Distance}}{\text{Time}}$

Average speed = $\frac{60}{2}$ = 30 km/hr

$$\begin{aligned}\Rightarrow \frac{90}{x} = 30 \\[1em] \Rightarrow x = \frac{90}{30} = 3. \\[1em] \phantom{\Rightarrow} \frac{y}{5} = 30 \\[1em] \Rightarrow y = 30 \times 5 = 150.\end{aligned}$$

Hence, Option 4 is the correct option.

Given,

2 parts of red pigments is mixed with 5 parts of the base.

$$\begin{aligned}\therefore \frac{4}{a} = \frac{2}{5} \\[1em] \Rightarrow a = \frac{4 \times 5}{2} = \frac{20}{2} = 10. \\[1em] \therefore \frac{b}{12.5} = \frac{2}{5} \\[1em] \Rightarrow b = \frac{2}{5} \times 12.5 = 2 \times 2.5 = 5 \\[1em] \therefore \frac{6}{c} = \frac{2}{5} \\[1em] \Rightarrow c = \frac{6 \times 5}{2} = 3 \times 5 = 15.\end{aligned}$$

Hence, Option 4 is the correct option.

The mean proportion between two numbers a and b is defined as $\sqrt{ab}$

So, reason (R) is true.

When two numbers are $\frac{2}{3} \text{ and } \frac{15}{2}$

$$\begin{aligned}\text{Mean proportion} = \sqrt{\frac{2}{3} \times \frac{15}{2}}\\[1em] = \sqrt{\frac{30}{6}}\\[1em] = \sqrt{5}\end{aligned}$$

So, assertion (A) is false.

Thus, Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

Three numbers a, b, c are in continued proportion if:

$\frac{a}{b} = \frac{b}{c}$

⇒ b² = ac

Here, a is called the first proportion, b is the mean (or geometric mean), c is the third proportion.

So, assertion (A) is true.

Lets take an example, a = 9, b = 6 and c = 4.

When three numbers a, b, c are in continued proportion if:

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{b}{c}\\[1em] \Rightarrow \frac{9}{6} = \frac{6}{4}\\[1em] \Rightarrow\frac{3}{2} = \frac{3}{2}\end{aligned}$$

Here, a = 9 is first proportion but it is not the smallest number among three.

So, reason (R) is false.

Thus, Assertion (A) is true, but Reason (R) is false.

Hence, option 1 is the correct option.

Given,

a, b, c and d are in proportion.

$\therefore \frac{a}{b} = \frac{c}{d}$

So, reason (R) is true.

Applying componendo and dividendo, we get

$\Rightarrow \frac{a + b}{a - b} = \frac{c + d}{c - d}$

Therefore, a + b, a - b, c + d and c - d are in proportion.

So, assertion (A) is true and reason (R) correctly explains assertion (A).

Thus, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Given,

a, b, c and d are in proportion.

$$\begin{aligned}\therefore \frac{a}{b} = \frac{c}{d} \\[1em] \Rightarrow \frac{a}{c} = \frac{b}{d}.\end{aligned}$$

∴ a, c, b and d are in proportion.

So, reason (R) is true.

$$\begin{aligned}\Rightarrow \frac{a}{c} = \frac{b}{d}\\[1em] \Rightarrow \frac{a}{c} - 1 = \frac{b}{d} - 1\\[1em] \Rightarrow \frac{a - c}{c} = \frac{b - d}{d}\end{aligned}$$

We can say that a - c, c, b - d, d are in proportion.

So, assertion (A) is true and reason (R) correctly explains assertion (A).

Thus, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

The compounded ratio is :

$$\begin{aligned}\frac{(a + b)^2}{(a - b)^2} \times \frac{(a^2 - b^2)}{(a^2 + b^2)} \times \frac{(a^4 - b^4)}{(a + b)^4} \\[1em] = \frac{(a + b)^2}{(a - b)^2} \times \frac{(a - b)(a + b)}{(a^2 + b^2)} \times \frac{(a^2 - b^2)(a^2 + b^2)}{(a + b)^4} \\[1em] = \frac{(a + b)^2}{(a - b)^2} \times \frac{(a - b)(a + b)}{(a^2 + b^2)} \times \frac{(a - b)(a + b)(a^2 + b^2)}{(a + b)^4} \\[1em] = \frac{(a + b)^4(a - b)^2(a^2 + b^2)}{(a - b)^2(a^2 + b^2)(a + b)^4} \\[1em] = \frac{1}{1} = 1 : 1 .\end{aligned}$$

Hence, the compounded ratio is 1 : 1.

Given, (7p + 3q) : (3p - 2q) = 43 : 2.

$$\begin{aligned}\Rightarrow \frac{7p + 3q}{3p - 2q} = \frac{43}{2} \\[1em] \Rightarrow 2(7p + 3q) = 43(3p - 2q) \\[1em] \Rightarrow 14p + 6q = 129p - 86q \\[1em] \Rightarrow 92q = 115p \\[1em] \Rightarrow \frac{p}{q} = \frac{92}{115} = \frac{4}{5}. \\[1em]\end{aligned}$$

Hence, the value of ratio p : q is 4 : 5.

Given,

$$\begin{aligned}\frac{a}{b} = \frac{3}{5} \\[1em] \therefore a = \frac{3b}{5}\end{aligned}$$

Putting value of a in the ratio (3a + 5b) : (7a - 2b),

$$\begin{aligned}= \frac{3 \times \frac{3b}{5} + 5b}{7 \times \frac{3b}{5} - 2b} \\[1em] = \frac{\frac{9b}{5} + 5b}{\frac{21b}{5} - 2b} \\[1em] = \frac{\frac{9b + 25b}{5}}{\frac{21b - 10b}{5}} \\[1em] = \frac{34b}{11b} \\[1em] = \frac{34}{11} = 34 : 11.\end{aligned}$$

Hence, the value of ratio (3a + 5b) : (7a - 2b) is 34 : 11.

Let the length of the shorter side be 5x cm and 12x cm.

Length of hypotenuse = $\sqrt{(5x)^2 + (12x)^2} \text{ cm}$

$$\begin{aligned}= \sqrt{25x^2 + 144x^2} \text{ cm} \\[0.75em] = \sqrt{169x^2} \text{ cm} \\[0.75em] = 13x \text{ cm} \\[0.75em]\end{aligned}$$

According to given,

$$\begin{aligned}5x + 12x + 13x = 360 \\[0.5em] 30x = 360 \\[0.5em] x = 12.\end{aligned}$$

∴ x = 12, 13x = 13 $\times$ 12 = 156.

Hence, the length of longest side is 156 cm.

Let the savings of Lokesh and his sister are 5x and 6x.

Let Lokesh save ₹ y more then according to question,

$$\begin{aligned}\Rightarrow \frac{5x + y}{6x + 30} = \frac{5}{6} \\[0.5em] \Rightarrow 6(5x + y) = 5(6x + 30) \\[0.5em] \Rightarrow 30x + 6y = 30x + 150 \\[0.5em] \Rightarrow 6y = 150 \\[0.5em] \Rightarrow y = 25.\end{aligned}$$

Hence, Lokesh should save ₹ 25 more.

In first case,
Let the number of students passed = 3x and failed = x.
Total candidates appeared = 3x + x = 4x.

In second case,
Total candidates appeared = 4x + 8
Number of passed students = 3x - 6
Number of failed students = (4x + 8) - (3x - 6)
= 4x + 8 - 3x + 6
= x + 14.

According to question, ratio of number of passed to failed student in this case = 2 : 1.

$$\begin{aligned}\therefore \frac{3x - 6}{x + 14} = \frac{2}{1} \\[0.5em] \Rightarrow (3x - 6) = 2(x + 14) \\[0.5em] \Rightarrow 3x - 6 = 2x + 28 \\[0.5em] \Rightarrow 3x - 2x = 28 + 6 \\[0.5em] \Rightarrow x = 34.\end{aligned}$$

∴ x = 34, 4x = 136.

Hence, the number of students who appeared were 136.

Let x be added to each number.

So, 4 + x, 6 + x, 8 + x and 11 + x must be in proportion.

$$\begin{aligned}\therefore \frac{4 + x}{6 + x} = \frac{8 + x}{11 + x} \\[1em] \Rightarrow (4 + x) \times (11 + x) = (8 + x) \times (6 + x) \\[1em] \Rightarrow 44 + 4x + 11x + x^2 = 48 + 8x + 6x + x^2\\[1em] \Rightarrow 44 + 15x + x^2 = 48 + 14x + x^2 \\[1em] \Rightarrow 15x - 14x = 48 - 44 \\[1em] \Rightarrow x = 4.\end{aligned}$$

Hence, required number is 4.

Since, the numbers are in continued proportion,

$$\begin{aligned}\therefore \frac{(a + 2b + c)}{(a - c)} = \frac{(a - c)}{(a - 2b + c)} \\[1em] \Rightarrow (a + 2b + c)(a - 2b + c) = (a - c)^2 \\[1em] \Rightarrow (a^2 - 2ab + ac + 2ab - 4b^2 + 2bc + ac - 2bc + c^2) = (a^2 + c^2 - 2ac) \\[1em] \Rightarrow a^2 + c^2 + 2ac - 4b^2 = a^2 + c^2 - 2ac \\[1em] \Rightarrow 4b^2 = 4ac \\[1em] \Rightarrow b^2 = ac.\end{aligned}$$

Since, b² = 4ac, hence proved that b is the mean proportional between a and c.

2, 6, p, 54 and q are in continued proportion then,

$$\begin{aligned}\Rightarrow \frac{2}{6} = \frac{6}{p} = \frac{p}{54} = \frac{54}{q}.\\[1em] \text{Solving, } \frac{2}{6} = \frac{6}{p} \text{ for p,} \\[1em] \Rightarrow p = \frac{6}{2} \times 6 \\[1em] \Rightarrow p = 18. \\[1em] \text{Now solving, } \frac{p}{54} = \frac{54}{q} \text{ for q,} \\[1em] \Rightarrow q = \frac{54}{p} \times 54 \\[1em] \Rightarrow q = \frac{54}{18} \times 54 \\[1em] \Rightarrow q = 3 \times 54 \\[1em] \Rightarrow q = 162.\end{aligned}$$

Hence, the value of p = 18 and q = 162.

Since, a, b, c, d, e are in continued proportion.

Let, $\frac{a}{b} = \frac{b}{c} = \frac{c}{d} = \frac{d}{e} = k.$

∴ d = ek, c = ek², b = ek³, a = ek⁴.

Now,

$$\begin{aligned}\text{L.H.S. } = \frac{a}{e} \\[1em] = \frac{ek^4}{e} = k^4 \\[1em] \text{R.H.S. } = \frac{(a^4)}{(b^4)} \\[1em] = \frac{(ek^4)^4}{(ek^3)^4} \\[1em] = \frac{e^4k^{16}}{e^4k^{12}} \\[1em] = k^4\end{aligned}$$

Since, L.H.S. = R.H.S. hence, proved that a : e = a⁴ : b⁴.

Let the two numbers be a and b.

Given, mean proportional between a and b is 16.

$\therefore \sqrt{ab} = 16 \qquad \text{[....Eq 1]}$

Given, third proportional is 128.

$$\begin{aligned}\therefore \frac{a}{b} = \frac{b}{128} \\[0.5em] \Rightarrow a = \frac{b^2}{128} \qquad \text{[....Eq 2]}\end{aligned}$$

Putting this value of a in Eq 1,

$$\begin{aligned}\Rightarrow \sqrt{\Big(\frac{b^2}{128}\Big)b} = 16 \\[0.5em] \Rightarrow \sqrt{\Big(\frac{b^3}{128}\Big)} = 16\end{aligned}$$

Squaring both sides,

$$\begin{aligned}\Rightarrow \frac{b^3}{128} = 256 \\[1em] \Rightarrow b^3 = 256 \times 128 \\[1em] \Rightarrow b^3 = 32768 \\[1em] \Rightarrow b = \sqrt[3]{32768} \\[1em] \Rightarrow b = 32\end{aligned}$$

Putting value of b in Eq 2, $$\begin{aligned}a = \frac{b^2}{128} \qquad \text{[....Eq 2]} \\[1em] \Rightarrow a = \frac{(32)^2}{128} \\[1em] \Rightarrow a = \frac{1024}{128} \\[1em] \Rightarrow a = 8\end{aligned}$$

Hence, the value of a = 8 and b = 32.

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

$\therefore q^2 = pr$

Given,

$$\begin{aligned}\Rightarrow p^2 - 3q^2 + r^2 = q^4\Big(\frac{1}{p^2} - \frac{3}{q^2} + \frac{1}{r^2}\Big). \\[1em] \text{L.H.S.} = p^2 - 3q^2 + r^2 \\[1em] = p^2 + r^2 - 3pr. \\[1em] \text{R.H.S.} = q^4\Big(\frac{1}{p^2} - \frac{3}{q^2} + \frac{1}{r^2}\Big) \\[1em] \Rightarrow p^2r^2\Big(\frac{1}{p^2} - \frac{3}{pr} + \frac{1}{r^2}\Big) \\[1em] \Rightarrow p^2r^2\Big(\frac{r^2 - 3pr + p^2}{p^2r^2}\Big) \\[1em] \Rightarrow p^2 + r^2 - 3pr.\end{aligned}$$

Since, L.H.S. = R.H.S. hence proved that,

$p^2 - 3q^2 + r^2 = q^4\big(\frac{1}{p^2} - \frac{3}{q^2} + \frac{1}{r^2}\big).$

(i) Let, $\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k,$ then,

a = bk, c = dk, e = fk.

Given,

$\sqrt{\frac{3a^2 - 5c^2 + 7e^2}{3b^2 - 5d^2 + 7f^2}}$

Putting values of a, c, e in above equation,

$$\begin{aligned}= \sqrt{\frac{3(bk)^2 - 5(dk)^2 + 7(fk)^2}{3b^2 - 5d^2 + 7f^2}} \\[1em] = \sqrt{\frac{3b^2k^2 - 5d^2k^2 + 7f^2k^2}{3b^2 - 5d^2 + 7f^2}} \\[1em] = \sqrt{k^2\Big(\frac{3b^2 - 5d^2 + 7f^2}{3b^2 - 5d^2 + 7f^2}\Big)} \\[1em] = k.\end{aligned}$$

Since, the values of all ratios = k. Hence, proved.

(ii) Let, $\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k,$ then,

a = bk, c = dk, e = fk.

Given,

$\Big(\frac{2a^3 + 5c^3 + 7e^3}{2b^3 + 5d^3 + 7f^3}\Big)^{1/3}$

Putting values of a, c, e in above equation, $$\begin{aligned}= \Big(\frac{2(bk)^3 + 5(dk)^3 + 7(fk)^3}{2b^3 + 5d^3 + 7f^3}\Big)^{1/3} \\[1em] = \Big(\frac{2b^3k^3 + 5d^3k^3 + 7f^3k^3}{2b^3 + 5d^3 + 7f^3}\Big)^{1/3} \\[1em] = \Big[k^3\Big(\frac{2b^3 + 5d^3 + 7f^3}{2b^3 + 5d^3 + 7f^3}\Big)\Big]^{1/3} \\[1em] = k^{3 \times 1/3} \\[1em] = k.\end{aligned}$$

Since, the values of all ratios = k. Hence, proved.

Let $\frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k,$ then,

x = ak, y = bk, z = ck.

Given,

$\frac{3x^3 - 5y^3 + 4z^3}{3a^3 - 5b^3 + 4c^3} = \Big(\frac{3x - 5y + 4z}{3a - 5b + 4c}\Big)^3$

Putting values of x, y, z in above equation and solving L.H.S.,

$$\begin{aligned}= \frac{3(ak)^3 - 5(bk)^3 + 4(ck)^3}{3a^3 - 5b^3 + 4c^3} \\[1em] = \frac{3a^3k^3 - 5b^3k^3 + 4c^3k^3}{3a^3 - 5b^3 + 4c^3} \\[1em] = k^3\Big(\frac{3a^3 - 5b^3 + 4c^3}{3a^3 - 5b^3 + 4c^3}\Big) \\[1em] = k^3.\end{aligned}$$

Solving, R.H.S. now, $$\begin{aligned}= \Big(\frac{3(ak) - 5(bk) + 4(ck)}{3a - 5b + 4c}\Big)^3 \\[1em] = \Big[k\Big(\frac{3a - 5b + 4c}{3a - 5b + 4c}\Big)\Big]^3 \\[1em] = k^3.\end{aligned}$$

Since, L.H.S. = R.H.S. Hence, proved that,

$\frac{3x^3 - 5y^3 + 4z^3}{3a^3 - 5b^3 + 4c^3} = \Big(\frac{3x - 5y + 4z}{3a - 5b + 4c}\Big)^3$.

Given, x : a = y : b,

Let, $\frac{x}{a}$ = $\frac{y}{b}$ = k

∴ x = ak, y = bk

Putting values of x and y in L.H.S. first,

$$\begin{aligned}= \frac{(ak)^4 + a^4}{(ak)^3 + a^3} + \frac{(bk)^4 + b^4}{(bk)^3 + b^3} \\[1em] = \frac{a^4(k^4 + 1)}{a^3(k^3 + 1)} + \frac{b^4(k^4 + 1)}{b^3(k^3 + 1)} \\[1em] = \frac{a(k^4 + 1) + b(k^4 + 1)}{k^3 + 1} \\[1em] = \frac{(a + b)(k^4 + 1)}{k^3 + 1}\end{aligned}$$

Now, putting values in R.H.S., $$\begin{aligned}= \frac{(x + y)^4 + (a + b)^4}{(x + y)^3 + (a + b)^3} \\[1em] = \frac{(ak + bk)^4 + (a + b)^4}{(ak + bk)^3 + (a + b)^3} \\[1em] = \frac{k^4(a + b)^4 + (a + b)^4}{k^3(a + b)^3 + (a + b)^3} \\[1em] = \frac{(a + b)^4(k^4 + 1)}{(a + b)^3(k^3 + 1)} \\[1em] = \frac{(a + b)(k^4 + 1)}{k^3 + 1}.\end{aligned}$$

Since, L.H.S. = R.H.S. Hence, proved that,

$\frac{x^4 + a^4}{x^3 + a^3} + \frac{y^4 + b^4}{y^3 + b^3} = \frac{(x + y)^4 + (a + b)^4}{(x + y)^3 + (a + b)^3}.$

Let, $\frac{x}{b + c - a} = \frac{y}{c + a - b} = \frac{z}{a + b - c} = k.$

∴ x = k(b + c - a), y = k(c + a - b), z = k(a + b - c).

Putting values of x, y and z in $\frac{x + y + z}{a + b + c}$ we get,

$$\begin{aligned}\frac{k(b + c - a) + k(c + a - b) + k(a + b - c)}{a + b + c} \\[1em] = \frac{kb + kc - \cancel{ak} + \cancel{kc} + \cancel{ak} - \cancel{bk} + ak + \cancel{bk} - \cancel{kc}}{a + b + c} \\[1em] = \frac{k(a + b + c)}{(a + b + c)} \\[1em] = k.\end{aligned}$$

Since, the value of all ratios = k, hence, each ratio = $\frac{x + y + z}{a + b + c}.$

(i) Given,

$$\begin{aligned}\frac{a}{b} = \frac{9}{10} \\[0.5em] \Rightarrow a = \frac{9b}{10}\end{aligned}$$

Putting $a = \frac{9b}{10}$ in $\frac{5a + 3b}{5a - 3b}$, we get,

$$\begin{aligned}\frac{5 \times \frac{9b}{10} + 3b}{5 \times \frac{9b}{10} - 3b} \\[1em] = \frac{\frac{9b}{2} + 3b}{\frac{9b}{2} - 3b} \\[1em] = \frac{\frac{9b + 6b}{2}}{\frac{9b - 6b}{2}} \\[1em] = \frac{15b}{3b} \\[1em] = 5.\end{aligned}$$

Hence, the value of $\frac{5a + 3b}{5a - 3b}$ = 5.

(ii) Given,

$$\begin{aligned}\frac{a}{b} = \frac{9}{10} \\[0.5em] \Rightarrow a = \frac{9b}{10}\end{aligned}$$

Putting $a = \frac{9b}{10}$ in $\frac{2a^2 - 3b^2}{2a^2 + 3b^2}$, we get,

$$\begin{aligned}\frac{2 \times \Big(\frac{9b}{10}\Big)^2 - 3b^2}{2 \times \Big(\frac{9b}{10}\Big)^2 + 3b^2} \\[1em] = \frac{2 \times \frac{81b^2}{100} - 3b^2}{2 \times \frac{81b^2}{100} + 3b^2} \\[1em] = \frac{\frac{81b^2 - 150b^2}{50}}{\frac{81b^2 + 150b^2}{50}} \\[1em] = -\frac{69b^2}{231b^2} \\[1em] = -\frac{23}{77}.\end{aligned}$$

Hence, the value of $\frac{2a^2 - 3b^2}{2a^2 + 3b^2} = -\frac{23}{77}$.

Given,

$\frac{3x^2 + 2y^2}{3x^2 - 2y^2} = \frac{11}{9}$

Applying componendo and dividendo to above equation,

$$\begin{aligned}\Rightarrow \frac{3x^2 + 2y^2 + 3x^2 - 2y^2}{3x^2 + 2y^2 - 3x^2 + 2y^2} = \frac{11 + 9}{11 - 9} \\[1em] \Rightarrow \frac{6x^2}{4y^2} = \frac{20}{2} \\[1em] \Rightarrow \frac{3x^2}{2y^2} = 10 \\[1em] \Rightarrow \frac{x^2}{y^2} = \frac{20}{3}\end{aligned}$$

Putting value of $\frac{x^2}{y^2}$ = $\frac{20}{3}$ in $\frac{3x^4 + 25y^4}{3x^4 - 25y^4}$,

$$\begin{aligned}\Rightarrow \frac{3\big(\frac{x^2}{y^2}\big)^2 + 25}{3\big(\frac{x^2}{y^2}\big)^2 - 25} \\[1em] \Rightarrow \frac{3\big(\frac{400}{9}\big) + 25}{3\big(\frac{400}{9}\big) - 25} \\[1em] \Rightarrow \frac{\frac{400}{3} + 25}{\frac{400}{3} - 25} \\[1em] \Rightarrow \frac{\frac{400 + 75}{3}}{\frac{400 - 75}{3}} \\[1em] \Rightarrow \frac{475}{325} \\[1em] \Rightarrow \frac{19}{13}.\end{aligned}$$

Hence, the value of $\frac{3x^4 + 25y^4}{3x^4 - 25y^4} \text{ is } \frac{19}{13}.$

$x = \frac{2mab}{a + b}$

Putting this value of x in

$$\begin{aligned}\phantom{= }\frac{x + ma}{x - ma} + \frac{x + mb}{x - mb} \\[1.5em] = \frac{\frac{2mab}{a + b} + ma}{\frac{2mab}{a + b} - ma} + \frac{\frac{2mab}{a + b} + mb}{\frac{2mab}{a + b} - mb} \\[1em] = \frac{\frac{2mab + ma^2 + mab}{a + b}}{\frac{2mab - ma^2 - mab}{a + b}} + \frac{\frac{2mab + mb^2 + mab}{a + b}}{\frac{2mab - mb^2 - mab}{a + b}} \\[1em] = \frac{2mab + ma^2 + mab}{2mab - ma^2 - mab} + \frac{2mab + mb^2 + mab}{2mab - mb^2 - mab} \\[1em] = \frac{ma(2b + a + b)}{ma(2b - a - b)} + \frac{mb(2a + b + a)}{mb(2a - b - a)} \\[1em] = \frac{3b + a}{b - a} + \frac{3a + b}{a - b} \\[1em] = \frac{3b + a}{b - a} - \frac{3a + b}{b - a} \\[1em] = \frac{3b + a - 3a - b}{b - a} \\[1em] = \frac{2b - 2a}{b - a} \\[1em] = \frac{2(b - a)}{(b - a)} \\[1em] = 2.\end{aligned}$$

Hence, the value of $\frac{x + ma}{x - ma} + \frac{x + mb}{x - mb} = 2.$

Given,

$$\begin{aligned}x = \frac{pab}{a + b} \\[1em] \Rightarrow \frac{x}{pa} = \frac{b}{a + b} \text{ and } \frac{x}{pb} = \frac{a}{a + b}\end{aligned}$$

Applying componendo and dividendo on both equations,

$$\begin{aligned}\Rightarrow \frac{x + pa}{x - pa} = \frac{b + a + b}{b - a - b} \text{ and } \frac{x + pb}{x - pb} = \frac{a + a + b}{a - a - b} \\[1em] \Rightarrow \frac{x + pa}{x - pa} = -\frac{2b + a}{a} \text{ and } \frac{x + pb}{x - pb} = -\frac{2a + b}{b}\end{aligned}$$

Subtracting both the equations,

$$\begin{aligned}\Rightarrow \frac{x + pa}{x - pa} - \frac{x + pb}{x - pb} = -\frac{2b + a}{a} - \Big(-\frac{2a + b}{b}\Big) \\[1em] = -\frac{2b + a}{a} + \frac{2a + b}{b} \\[1em] = \frac{-2b^2 - \cancel{ab} + 2a^2 + \cancel{ab}}{ab} \\[1em] = \frac{2(a^2 - b^2)}{ab} = \text{R.H.S.} \\[1em]\end{aligned}$$

Since, L.H.S. = R.H.S. , hence, proved that,

$\frac{x + pa}{x - pa} - \frac{x + pb}{x - pb} = \frac{2(a^2 - b^2)}{ab}.$

Given,

$\frac{a + x + \sqrt{a^2 - x^2}}{a + x - \sqrt{a^2 - x^2}} = \frac{b}{x}$

Applying componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{a + x + \sqrt{a^2 - x^2} + a + x - \sqrt{a^2 - x^2}}{a + x + \sqrt{a^2 - x^2} - a - x + \sqrt{a^2 - x^2}} = \frac{b + x}{b - x} \\[1em] \Rightarrow \frac{2(a + x)}{2\sqrt{a^2 - x^2}} = \frac{b + x}{b - x} \\[1em] \Rightarrow \frac{(a + x)}{\sqrt{a^2 - x^2}} = \frac{b + x}{b - x} \\[1em]\end{aligned}$$

Squaring both sides,

$$\begin{aligned}\Rightarrow \frac{(a + x)^2}{a^2 - x^2} = \frac{(b + x)^2}{(b - x)^2} \\[1em] \Rightarrow \frac{(a + x)^2}{(a + x)(a - x)} = \frac{(b + x)^2}{(b - x)^2} \\[1em] \Rightarrow \frac{(a + x)}{(a - x)} = \frac{(b + x)^2}{(b - x)^2} \\[1em]\end{aligned}$$

Again applying componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{a + x + a - x}{a + x - a + x} = \frac{(b + x)^2 + (b - x)^2}{(b + x)^2 - (b - x)^2} \\[1em] \Rightarrow \frac{2a}{2x} = \frac{b^2 + x^2 + 2bx + b^2 + x^2 - 2bx}{b^2 + x^2 + 2bx - (b^2 + x^2 - 2bx)} \\[1em] \Rightarrow \frac{2a}{2x} = \frac{b^2 + b^2 + x^2 + x^2 + 2bx - 2bx}{b^2 - b^2 + x^2 - x^2 + 2bx - (-2bx)} \\[1em] \Rightarrow \frac{2a}{2x} = \frac{2(b^2 + x^2)}{4bx} \\[1em] \Rightarrow \frac{a}{x} = \frac{b^2 + x^2}{2bx}\end{aligned}$$

Multiplying both sides by x we get,

$\Rightarrow a = \frac{b^2 + x^2}{2b}$

On cross-multiplication,

$$\begin{aligned}\Rightarrow 2ab = b^2 + x^2 \\[1em] \Rightarrow x^2 = 2ab - b^2 \\[1em] \Rightarrow x = \sqrt{2ab - b^2}.\end{aligned}$$

Hence, the value of x is $\sqrt{2ab - b^2}.$

Given,

$\frac{x}{1} = \frac{\sqrt[3]{a + 1} + \sqrt[3]{a - 1}}{\sqrt[3]{a + 1} - \sqrt[3]{a - 1}}$

Applying componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{x + 1}{x - 1} = \frac{\sqrt[3]{a + 1} + \sqrt[3]{a - 1} + \sqrt[3]{a + 1} - \sqrt[3]{a - 1}}{\sqrt[3]{a + 1} + \sqrt[3]{a - 1} - \sqrt[3]{a + 1} + \sqrt[3]{a - 1}} \\[1em] \Rightarrow \frac{x + 1}{x - 1} = \frac{2\sqrt[3]{a + 1}}{2\sqrt[3]{a - 1}}\end{aligned}$$

Cubing both the sides,

$\Rightarrow \frac{(x + 1)^3}{(x - 1)^3} = \frac{a + 1}{a - 1}$

Again applying componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{(x + 1)^3 + (x - 1)^3}{(x + 1)^3 - (x - 1)^3} = \frac{a + \cancel{1} + a - \cancel{1}}{\cancel{a} + 1 - \cancel{a} + 1} \\[1em] \Rightarrow \frac{x^3 + 1 + 3x(x + 1) + x^3 - 1 - 3x(x - 1)}{x^3 + 1 + 3x(x + 1) - (x^3 - 1 - 3x(x - 1))} = \frac{2a}{2} \\[1em] \Rightarrow \frac{x^3 + x^3 + 1 - 1 + 3x^2 - 3x^2 + 3x + 3x}{x^3 - x^3 + 1 + 1 + 3x^2 + 3x^2 + 3x - 3x} = \frac{2a}{2} \\[1em] \Rightarrow \frac{2x^3 + 6x}{2 + 6x^2} = a \\[1em] \Rightarrow \frac{2(x^3 + 3x)}{2(1 + 3x^2)} = a \\[1em] \Rightarrow \frac{x^3 + 3x}{1 + 3x^2} = a\end{aligned}$$

On cross-multiplication,

$$\begin{aligned}\Rightarrow x^3 + 3x = a(1 + 3x^2) \\[1em] \Rightarrow x^3 + 3x = a + 3ax^2 \\[1em] \Rightarrow x^3 - 3ax^2 + 3x - a = 0.\end{aligned}$$

Hence, proved that x³ - 3ax² + 3x - a = 0.

(a) Solving,

$$\begin{aligned}\Rightarrow \frac{(a + b)^3}{(a - b)^3} = \frac{64}{27} \\[1em] \Rightarrow \frac{(a + b)^3}{(a - b)^3} = \frac{4^3}{3^3} \\[1em] \Rightarrow \Big(\frac{a + b}{a - b}\Big)^3 = \Big(\frac{4}{3}\Big)^3 \\[1em] \Rightarrow \frac{a + b}{a - b} = \frac{4}{3} \\[1em]\end{aligned}$$

Hence, $\frac{a + b}{a - b} = \frac{4}{3}.$

(b) Solving further,

$$\begin{aligned}\Rightarrow 3(a + b) = 4(a - b) \\[1em] \Rightarrow 3a + 3b = 4a - 4b \\[1em] \Rightarrow 4a - 3a = 3b + 4b \\[1em] \Rightarrow a = 7b \\[1em] \Rightarrow \frac{a}{b} = \frac{7}{1} \\[1em] \Rightarrow a : b = 7 : 1.\end{aligned}$$

Hence, a : b = 7 : 1.

Given,

x, y and z are in continued proportion.

$$\begin{aligned}\therefore \frac{x}{y} = \frac{y}{z} \\[1em] \Rightarrow y^2 = xz\end{aligned}$$

To prove :

$\frac{x}{y^2.z^2} + \frac{y}{z^2.x^2} + \frac{z}{x^2.y^2} = \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3}$

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{x}{y^2.z^2} + \frac{y}{z^2.x^2} + \frac{z}{x^2.y^2} \\[1em] \Rightarrow \frac{x^3 + y^3 + z^3}{x^2.y^2.z^2} \\[1em] \Rightarrow \frac{x^3 + y^3 + z^3}{x^2.xz.z^2} \\[1em] \Rightarrow \frac{x^3 + y^3 + z^3}{x^3.z^3} \\[1em] \Rightarrow \frac{x^3}{x^3.z^3} + \frac{y^3}{x^3z^3} + \frac{z^3}{x^3.z^3} \\[1em] \Rightarrow \frac{1}{z^3} + \frac{y^3}{(xz)^3} + \frac{1}{x^3} \\[1em] \Rightarrow \frac{1}{z^3} + \frac{y^3}{(y^2)^3} + \frac{1}{x^3} \\[1em] \Rightarrow \frac{1}{z^3} + \frac{y^3}{y^6} + \frac{1}{x^3} \\[1em] \Rightarrow \frac{1}{z^3} + \frac{1}{y^3} + \frac{1}{x^3}.\end{aligned}$$

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

Hence, proved that $\frac{x}{y^2.z^2} + \frac{y}{z^2.x^2} + \frac{z}{x^2.y^2} = \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3}$.