Ratio & Proportion

Given,

⇒ A : B = 7 : 5

$\Rightarrow \frac{A}{B} = \frac{7}{5}$ ........(1)

⇒ B : C = 7 : 5

$\Rightarrow \frac{B}{C} = \frac{7}{5}$ ......(2)

Multiplying equation (1) and (2), we get :

$$\begin{aligned}\Rightarrow \frac{A}{B} \times \frac{B}{C} = \frac{7}{5} \times \frac{7}{5} \\[1em] \Rightarrow \frac{A}{C} = \frac{49}{25}.\end{aligned}$$

A : C = 49 : 25.

Hence, Option 2 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \frac{x^2 - 1}{x^2 + 1} =\frac{3}{5} \\[1em] \Rightarrow 5(x^2 - 1) = 3(x^2 + 1) \\[1em] \Rightarrow 5x^2 - 5 = 3x^2 + 3 \\[1em] \Rightarrow 5x^2 - 3x^2 = 3 + 5 \\[1em] \Rightarrow 2x^2 = 8 \\[1em] \Rightarrow x^2 = 4 \\[1em] \Rightarrow x = \sqrt{4} \\[1em] \Rightarrow x = \pm 2.\end{aligned}$$

Hence, Option 4 is the correct option.

Given,

$$\begin{aligned}\Rightarrow (2x + 3y) : (3x + 2y) = 4 : 3 \\[1em] \Rightarrow \frac{2x + 3y}{3x + 2y} = \frac{4}{3} \\[1em] \Rightarrow 3(2x + 3y) = 4(3x + 2y) \\[1em] \Rightarrow 6x + 9y = 12x + 8y \\[1em] \Rightarrow 12x - 6x = 9y - 8y \\[1em] \Rightarrow 6x = y \\[1em] \Rightarrow \frac{x}{y} = \frac{1}{6} \\[1em] \Rightarrow x : y = 1 : 6.\end{aligned}$$

Hence, Option 3 is the correct option.

Let two numbers be x and y.

Given,

Ratio of two numbers = 5 : 8.

$$\begin{aligned}\Rightarrow \frac{x}{y} = \frac{5}{8} \\[1em] \Rightarrow x = \frac{5}{8}y \space .....(1)\end{aligned}$$

Given,

If 10 is subtracted from each number, the ratio becomes 4 : 7.

$$\begin{aligned}\Rightarrow \frac{x - 10}{y - 10} = \frac{4}{7} \\[1em] \Rightarrow 7(x - 10) = 4(y - 10) \\[1em] \Rightarrow 7x - 70 = 4y - 40\end{aligned}$$

Substituting value of x from equation (1), in above equation we get :

$$\begin{aligned}\Rightarrow 7 \times \frac{5}{8}y - 70 = 4y - 40 \\[1em] \Rightarrow \frac{35y}{8} - 70 = 4y - 40 \\[1em] \Rightarrow \frac{35y}{8} - 4y = 70 - 40 \\[1em] \Rightarrow \frac{35y - 32y}{8} = 30 \\[1em] \Rightarrow \frac{3y}{8} = 30 \\[1em] \Rightarrow y = \frac{30 \times 8}{3} \\[1em] \Rightarrow y = 80. \\[1em] \Rightarrow x = \frac{5}{8}y \\[1em] \Rightarrow x = \frac{5}{8} \times 80 = 50.\end{aligned}$$

Numbers = 50 and 80.

Hence, Option 2 is the correct option.

Compounded ratio of x - y : x + y and (x + y)² : x² - y² :

$$\begin{aligned}\Rightarrow \frac{x - y}{x + y} \times \frac{(x + y)^2}{x^2 - y^2} \\[1em] \Rightarrow \frac{x - y}{x + y} \times \frac{(x + y)^2}{(x + y)(x - y)} \\[1em] \Rightarrow \frac{(x - y)(x + y)^2}{(x + y)^2(x - y)} \\[1em] \Rightarrow \frac{1}{1} \\[1em] \Rightarrow 1 : 1.\end{aligned}$$

Hence, Option 3 is the correct option.

Sub-duplicate ratio of 6x² : 24y² is :

$$\begin{aligned}\Rightarrow \frac{\sqrt{6x^2}}{\sqrt{24y^2}} \\[1em] \Rightarrow \frac{\sqrt{6}x}{2\sqrt{6}y} \\[1em] \Rightarrow \frac{x}{2y} \\[1em] \Rightarrow x : 2y.\end{aligned}$$

Hence, Option 1 is the correct option.

Given,

$$\begin{aligned}\phantom{\Rightarrow} \frac{m + n}{m + 3n} = \frac{2}{3} \\[1em] \Rightarrow 3(m + n) = 2(m + 3n) \\[1em] \Rightarrow 3m + 3n = 2m + 6n \\[1em] \Rightarrow 3m - 2m = 6n - 3n \\[1em] \Rightarrow m = 3n.\end{aligned}$$

Substituting value of m in $\frac{2n^2}{3m^2 + mn}$ we get,

$$\begin{aligned}\Rightarrow \frac{2n^2}{3(3n)^2 + (3n)n} \\[1em] \Rightarrow \frac{2n^2}{3(9n^2) + 3n^2} \\[1em] \Rightarrow \frac{2n^2}{27n^2 + 3n^2} \\[1em] \Rightarrow \frac{2n^2}{30n^2} \\[1em] \Rightarrow \frac{1}{15}.\end{aligned}$$

Hence, $\frac{2n^2}{3m^2 + mn} = \frac{1}{15}.$

According to question,

$$\begin{aligned}\phantom{\Rightarrow} \frac{2x - y}{x + 2y} = \frac{8}{11} \\[1em] \Rightarrow 11(2x - y) = 8(x + 2y) \\[1em] \Rightarrow 22x - 11y = 8x + 16y \\[1em] 22x - 8x = 16y + 11y \\[1em] 14x = 27y \\[1em] x = \frac{27y}{14}.\end{aligned}$$

Substituting value of x in $\frac{7x}{9y}$ we get,

$$\begin{aligned}\frac{7x}{9y} = \frac{7}{9y} \times \frac{27y}{14} \\[1em] = \frac{3}{2}.\end{aligned}$$

Hence, $\frac{7x}{9y} = \frac{3}{2}$.

Given, B : C = 4 : 3

If B = 4a, then C = 3a

Given, A is $\frac{2}{5}$ of B.

∴ A = $\frac{2}{5} \times 4a = \frac{8a}{5}$

Share of A,

$$\begin{aligned}= \frac{A}{A + B + C} \times 1290 \\[1em] = \frac{\frac{8a}{5}}{\frac{8a}{5} + 4a + 3a} \times 1290 \\[1em] = \frac{\frac{8a}{5}}{\frac{8a + 20a + 15a}{5}} \times 1290 \\[1em] = \frac{8a}{43a} \times 1290 \\[1em] = \frac{8}{43} \times 1290 \\[1em] = 240.\end{aligned}$$

Share of B,

$$\begin{aligned}= \frac{B}{A + B + C} \times 1290 \\[1em] = \frac{4a}{\frac{8a}{5} + 4a + 3a} \times 1290 \\[1em] = \frac{4a}{\frac{8a + 20a + 15a}{5}} \times 1290 \\[1em] = \frac{20a}{43a} \times 1290 \\[1em] = \frac{20}{43} \times 1290 \\[1em] = 600.\end{aligned}$$

Share of C,

$$\begin{aligned}= \frac{C}{A + B + C} \times 1290 \\[1em] = \frac{3a}{\frac{8a}{5} + 4a + 3a} \times 1290 \\[1em] = \frac{3a}{\frac{8a + 20a + 15a}{5}} \times 1290 \\[1em] = \frac{15a}{43a} \times 1290 \\[1em] = \frac{15}{43} \times 1290 \\[1em] = 450.\end{aligned}$$

Hence, the amount of money with A = ₹ 240, B = ₹ 600 and C = ₹ 450.

Ratio of boys to girls = 3 : 2.

No. of boys = $\frac{3}{3 + 2} \times 630 = \frac{3}{5} \times 630 = 378.$

No. of girls = $\frac{2}{3 + 2} \times 630 = \frac{2}{5} \times 630 = 252.$

Let no. of newly admitted boys be x and so girls = 90 - x.

According to question on admission of 90 new students ratio changes to 7 : 5.

$$\begin{aligned}\therefore \frac{378 + x}{252 + 90 - x} = \frac{7}{5} \\[1em] \Rightarrow \frac{378 + x}{342 - x} = \frac{7}{5} \\[1em] \Rightarrow 5(378 + x) = 7(342 - x) \\[1em] \Rightarrow 5(378 + x) = 7(342 - x) \\[1em] \Rightarrow 1890 + 5x = 2394 - 7x \\[1em] \Rightarrow 5x + 7x = 2394 - 1890 \\[1em] \Rightarrow 12x = 504 \\[1em] \Rightarrow x = \frac{504}{12} \\[1em] \Rightarrow x = 42.\end{aligned}$$

Hence, there are 42 newly admitted boys.

Let number to be subtracted be x,

$$\begin{aligned}\therefore \frac{9 - x}{17 - x} = \frac{1}{3} \\[1em] \Rightarrow 3(9 - x) = 1(17 - x) \\[1em] \Rightarrow 27 - 3x = 17 - x \\[1em] \Rightarrow 3x - x = 27 - 17 \\[1em] \Rightarrow 2x = 10 \\[1em] \Rightarrow x = 5.\end{aligned}$$

Hence, quantity to be subtracted = 5.

Amount of work done by (3x - 1) men in (2x + 3) days = (3x - 1)(2x + 3),

Amount of work done by (3x - 4) men in (2x + 1) days = (3x - 4)(2x + 1)

According to question,

$\Rightarrow \frac{(3x - 1)(2x + 3)}{(3x - 4)(2x + 1)} = \frac{4}{3}$

3(3x − 1)(2x + 3) = 4(3x - 4)(2x + 1)

3(6x² + 9x − 2x − 3) = 4(6x² + 3x − 8x − 4)

3(6x² + 7x − 3) = 4(6x² − 5x − 4)

18x² + 21x − 9 = 24x² − 20x − 16

18x² − 24x² + 21x + 20x − 9 + 16 = 0

−6x² + 41x + 7 = 0

6x² − 41x − 7 = 0

6x² - 42x + x - 7 = 0

6x(x − 7) + 1(x − 7) = 0

(6x + 1)(x − 7) = 0

6x + 1 = 0 or x - 7 = 0

x = −$\frac{1}{6}$ or x = 7

x ≠ −$\frac{1}{6}$ as that will make number of men negative which is not possible.

Hence, value of x = 7.

3A = 4B

$\frac{A}{B} = \frac{4}{3}$

⇒ A : B = 4 : 3

4B = 6C

$\frac{B}{C} = \frac{6}{4} = \frac{3}{2}$

⇒ B : C = 3 : 2

∴ A : B : C = 4 : 3 : 2.

Hence, A : B : C = 4 : 3 : 2.

2a = 3b

$\frac{a}{b} = \frac{3}{2}$

⇒ a : b = 3 : 2

4b = 5c

$\frac{b}{c} = \frac{5}{4}$

⇒ b : c = 5 : 4

Multiplying a : b by 5 and b : c by 2 we get,

a : b = 15 : 10

b : c = 10 : 8

⇒ a : b : c = 15 : 10 : 8

Hence, a : c = 15 : 8.

Compound ratio of 2 : 3, 9 : 14 and 14 : 27,

$$\begin{aligned}= \frac{2}{3} \times \frac{9}{14} \times \frac{14}{27} \\[1em] = \frac{252}{1134} \\[1em] = \frac{2}{9} = 2 : 9.\end{aligned}$$

Hence, compound ratio of 2 : 3, 9 : 14 and 14 : 27 is 2 : 9.

Compound ratio of 2a : 3b, mn : x² and x : n,

$$\begin{aligned}= \frac{2a}{3b} \times \frac{mn}{x^2} \times \frac{x}{n} \\[1em] = \frac{2amnx}{3bx^2n} \\[1em] = \frac{2am}{3bx} = 2am : 3bx.\end{aligned}$$

Hence, compound ratio of 2a : 3b, mn : x² and x : n is 2am : 3bx.

According to question,

$$\begin{aligned}\Rightarrow \frac{x + 3}{4x + 1} = \frac{3 \times 3}{5 \times 5} \\[1em] \Rightarrow \frac{x + 3}{4x + 1} = \frac{9}{25} \\[1em] \Rightarrow 25(x + 3) = 9(4x + 1) \\[1em] \Rightarrow 25x + 75 = 36x + 9 \\[1em] \Rightarrow 36x - 25x = 75 - 9 \\[1em] \Rightarrow 11x = 66 \\[1em] \Rightarrow x = 6.\end{aligned}$$

Hence, x = 6.

According to question,

$$\begin{aligned}\Rightarrow \frac{m}{n} = \frac{(m + x)^2}{(n + x)^2} \\[1em] \Rightarrow \frac{m}{n} = \frac{m^2 + x^2 + 2mx}{n^2 + x^2 + 2nx} \\[1em] \Rightarrow m(n^2 + x^2 + 2nx) = n(m^2 + x^2 + 2mx) \\[1em] \Rightarrow mn^2 + mx^2 + 2mnx = nm^2 + nx^2 + 2mnx \\[1em] \Rightarrow mx^2 - nx^2 = nm^2 - mn^2 + 2mnx - 2mnx \\[1em] \Rightarrow x^2(m - n) = mn(m - n) \\[1em] \Rightarrow x^2 = mn.\end{aligned}$$

Hence, proved that x² = mn.

According to question,

$$\begin{aligned}\Rightarrow \frac{3x - 9}{5x + 4} = \frac{3^3}{4^3} \\[1em] \Rightarrow \frac{3x - 9}{5x + 4} = \frac{27}{64} \\[1em] \Rightarrow 64(3x - 9) = 27(5x + 4) \\[1em] \Rightarrow 192x - 576 = 135x + 108 \\[1em] \Rightarrow 192x - 135x = 108 + 576 \\[1em] \Rightarrow 57x = 684 \\[1em] \Rightarrow x = 12.\end{aligned}$$

Hence, x = 12.

Given,

x, 2, 10 and y are in proportion.

$\therefore \frac{x}{2} = \frac{2}{10} = \frac{10}{y}$ .......(1)

Considering L.H.S. of the equation (1) :

$$\begin{aligned}\Rightarrow \frac{x}{2} = \frac{2}{10} \\[1em] \Rightarrow x = \frac{2 \times 2}{10} \\[1em] \Rightarrow x = \frac{4}{10} = 0.4\end{aligned}$$

Considering R.H.S. of the equation (1) :

$$\begin{aligned}\Rightarrow \frac{2}{10} = \frac{10}{y} \\[1em] \Rightarrow y = \frac{10 \times 10}{2} \\[1em] \Rightarrow y = \frac{100}{2} \\[1em] \Rightarrow y = 50.\end{aligned}$$

Hence, Option 3 is the correct option.

Given,

⇒ x : y = y : z

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

Substituting value of y² in x² : y², we get :

⇒ x² : xz

⇒ x : z.

Hence, Option 3 is the correct option.

Let mean proportion be x.

$$\begin{aligned}\Rightarrow \frac{3 + 2\sqrt{2}}{x} = \frac{x}{3 - 2\sqrt{2}} \\[1em] \Rightarrow x^2 = (3 + 2\sqrt{2})(3 - 2\sqrt{2}) \\[1em] \Rightarrow x^2 = 3^2 - (2\sqrt{2})^2 \\[1em] \Rightarrow x^2 = 9 - (4 \times 2) \\[1em] \Rightarrow x^2 = 9 - 8 \\[1em] \Rightarrow x^2 = 1 \\[1em] \Rightarrow x = \sqrt{1} = \pm 1.\end{aligned}$$

Since, geometrical mean cannot be negative,

∴ x = 1.

Hence, Option 1 is the correct option.

Given,

2x, 9 and 18 are in continued proportion.

$$\begin{aligned}\therefore \frac{2x}{9} = \frac{9}{18} \\[1em] \Rightarrow x = \frac{9 \times 9}{18 \times 2} \\[1em] \Rightarrow x = \frac{9}{4} = 2\frac{1}{4}.\end{aligned}$$

Hence, Option 1 is the correct option.

Let fourth proportional to 1.5, 4.5 and 3.5 be x

$$\begin{aligned}\Rightarrow 1.5 : 4.5 = 3.5 : x \\[1em] \Rightarrow \frac{1.5}{4.5} = \frac{3.5}{x} \\[1em] \Rightarrow x = \frac{3.5 \times 4.5}{1.5} \\[1em] \Rightarrow x = 3.5 \times 3 \\[1em] \Rightarrow x = 10.5\end{aligned}$$

Hence, fourth proportional to 1.5, 4.5 and 3.5 is 10.5

Let fourth proportional to 3a, 6a² and 2ab² be x

$$\begin{aligned}\Rightarrow 3a : 6a^2 = 2ab^2 : x \\[1em] \Rightarrow \frac{3a}{6a^2} = \frac{2ab^2}{x} \\[1em] \Rightarrow x = \frac{2ab^2 \times 6a^2}{3a} \\[1em] \Rightarrow x = 4a^2b^2\end{aligned}$$

Hence, fourth proportional to 3a, 6a² and 2ab² is 4a²b².

Let the third proportion to $2\frac{2}{3}$ and 4 be x,

$$\begin{aligned}\Rightarrow 2\frac{2}{3} : 4 = 4 : x \\[1em] \Rightarrow \frac{8}{3} : 4 = 4 : x \\[1em] \Rightarrow \frac{\frac{8}{3}}{4} = \frac{4}{x} \\[1em] \Rightarrow \frac{8}{12} = \frac{4}{x} \\[1em] \Rightarrow x = \frac{4 \times 12}{8} \\[1em] \Rightarrow x = 6.\end{aligned}$$

Hence, third proportional to $2\frac{2}{3}$ and 4 is 6.

Let the third proportion to a - b and a² - b² be x,

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

Hence, third proportional to a - b and a² - b² is (a + b)(a² - b²).

Let mean proportional between 6 + $3\sqrt{3} \text{ and } 8 - 4\sqrt{3}$ be x

$$\begin{aligned}\therefore 6 + 3\sqrt{3} : x = x : 8 - 4\sqrt{3} \\[1em] \Rightarrow \frac{6 + 3\sqrt{3}}{x} = \frac{x}{8 - 4\sqrt{3}} \\[1em] \Rightarrow x^2 = (6 + 3\sqrt{3})(8 - 4\sqrt{3}) \\[1em] \Rightarrow x^2 = 48 - 24\sqrt{3} + 24\sqrt{3} - 36 \\[1em] \Rightarrow x^2 = 48 - 36 \\[1em] \Rightarrow x^2 = 12 \\[1em] \Rightarrow x = 2\sqrt{3}.\end{aligned}$$

Hence, x = $2\sqrt{3}$.

Let mean proportional between a - b and a³ - a²b be x

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

Hence, mean proportional between a - b and a³ - a²b = a(a - b).

Given,

x + 5 is the mean proportion between x + 2 and x + 9

$$\begin{aligned}\therefore x + 2 : x + 5 = x + 5 : x + 9 \\[1em] \Rightarrow \frac{x + 2}{x + 5} = \frac{x + 5}{x + 9} \\[1em] \Rightarrow (x + 2)(x + 9) = (x + 5)(x + 5) \\[1em] \Rightarrow x^2 + 9x + 2x + 18 = x^2 + 5x + 5x + 25 \\[1em] \Rightarrow x^2 + 11x + 18 = x^2 + 10x + 25 \\[1em] \Rightarrow 11x - 10x = 25 - 18 \\[1em] \Rightarrow x = 7.\end{aligned}$$

Hence, x = 7.

Given,

x², 4 and 9 are in continued proportion.

$$\begin{aligned}\therefore \frac{x^2}{4} = \frac{4}{9} \\[1em] \Rightarrow x^2 = \frac{16}{9} \\[1em] \Rightarrow x = \sqrt{\frac{16}{9}} \\[1em] \Rightarrow x = \frac{4}{3}.\end{aligned}$$

Hence, x = $\frac{4}{3}.$

Given,

y is the mean proportional between x and z

$\therefore \frac{x}{y} = \frac{y}{z} \Rightarrow y^2 = xz.$

To prove,

xy + yz is the mean proportional between x² + y² and y² + z²

$$\begin{aligned}\therefore \frac{x^2 + y^2}{xy + yz} = \frac{xy + yz}{y^2 + z^2} \\[1em] \Rightarrow (xy + yz)(xy + yz) = (x^2 + y^2)(y^2 + z^2) \\[1em] \Rightarrow x^2y^2 + xy^2z + xy^2z + y^2z^2 = x^2y^2 + x^2z^2 + y^4 + y^2z^2 ....[\text{i}]\end{aligned}$$

Substituting y² = xz in L.H.S. of (i)

⇒ x²(xz) + x(xz)z + x(xz)z + (xz)z²

⇒ x³z + x²z² + x²z² + xz³

⇒ x³z + 2x²z² + xz³ .........(ii)

Substituting y² = xz in R.H.S. of (i)

⇒ x²(xz) + x²z² + (xz)² + (xz)z²

⇒ x³z + x²z² + x²z² + xz³

⇒ x³z + 2x²z² + xz³ .........(iii)

Since, (ii) = (iii)

Hence, proved that xy + yz is the mean proportional between x² + y² and y² + z².

Since, q is the mean proportional between p and r

$\therefore \frac{p}{q} = \frac{q}{r} \Rightarrow q^2 = pr.$

Substituting pr = q² in L.H.S. of pqr(p + q + r)³ = (pq + qr + pr)³

⇒ q.q²(p + q + r)³

⇒ q³(p + q + r)³

⇒ [q(p + q + r)]³

⇒ (pq + q² + qr)³

⇒ (pq + pr + qr)³ = R.H.S. (As q² = pr).

Hence, proved that pqr(p + q + r)³ = (pq + qr + pr)³.

Let x, y and z be in continued proportion.

∴ x : y = y : z

$\Rightarrow \frac{x}{y} = \frac{y}{z} \Rightarrow y^2 = xz$

To prove :

$\frac{x}{z} = \frac{x^2}{y^2}$

Substituting y² = xz in R.H.S. of above equation we get,

$\frac{x^2}{xz} = \frac{x}{z}$ = L.H.S.

Hence, proved that ratio of the first to third is duplicate ratio of first to the second.

Let third proportional be p.

$$\begin{aligned}\therefore \frac{\frac{x}{y} + \frac{y}{x}}{\sqrt{x^2 + y^2}} = \frac{\sqrt{x^2 + y^2}}{p} \\[1em] \Rightarrow \Big(\sqrt{x^2 + y^2}\Big)^2 = p\Big(\frac{x}{y} + \frac{y}{x}\Big) \\[1em] \Rightarrow x^2 + y^2 = p \times \frac{x^2 + y^2}{xy} \\[1em] \Rightarrow p = \frac{(x^2 + y^2)(xy)}{x^2 + y^2} \\[1em] \Rightarrow p = xy.\end{aligned}$$

Hence, third proportional to $\frac{x}{y} + \frac{y}{x} \text{ and } \sqrt{x^2 + y^2}$ is xy.

Given,

$\Rightarrow \frac{p}{q} = \frac{r}{s}$

Multiplying both sides by m:

$\Rightarrow \frac{mp}{q} = \frac{mr}{s}$

Adding n on both sides:

$$\begin{aligned}\Rightarrow \frac{mp}{q} + n = \frac{mr}{s} + n \\[1em] \Rightarrow \frac{mp + nq}{q}= \frac{mr + ns}{s}\end{aligned}$$

Hence, proved that mp + nq : q = mr + ns : s.

Given,

$$\begin{aligned}\Rightarrow \frac{1}{q} + \frac{1}{s} = \frac{m}{r} \\[1em] \Rightarrow \frac{s + q}{qs} = \frac{m}{r} \\[1em] \Rightarrow \frac{s + q}{s} = \frac{mq}{r} \\[1em] \Rightarrow \frac{s + q}{s} = \frac{p + r}{r} \text{ (As mq = p + r)} \\[1em] \Rightarrow 1 + \frac{q}{s} = \frac{p}{r} + 1 \\[1em] \Rightarrow \frac{q}{s} = \frac{p}{r} \\[1em] \Rightarrow \frac{p}{q} = \frac{r}{s}.\end{aligned}$$

Hence, proved that p : q = r : s.

Given,

⇒ x + y + z = 0

⇒ x = -(y + z).

Substituting value of x in $\frac{x}{y + z}$, we get :

$\Rightarrow \frac{x}{y + z} = \frac{-(y + z)}{y + z}$ = -1.

Hence, Option 4 is the correct option.

Given,

a, b, c and d are in proportion.

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

∴ a = bk and c = dk

Solving,

$$\begin{aligned}\Rightarrow \frac{8a^2 - 5b^2}{8c^2 - 5d^2} \\[1em] \Rightarrow \frac{8(bk)^2 - 5b^2}{8(dk)^2 - 5d^2} \\[1em] \Rightarrow \frac{8b^2k^2 - 5b^2}{8d^2k^2 - 5d^2} \\[1em] \Rightarrow \frac{b^2(8k^2 - 5)}{d^2(8k^2 - 5)} \\[1em] \Rightarrow \frac{b^2}{d^2} \space .....(1)\end{aligned}$$

As,

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

Substituting value of $\frac{b}{d}$ in (1), we get :

$\Rightarrow \frac{b^2}{d^2} = \frac{a^2}{c^2}$ = a² : c².

Hence, Option 2 is the correct option.

Given,

$\frac{x + y}{z} = \frac{y + z}{x} = \frac{z + x}{y}$

Applying componendo on each side we get :

$$\begin{aligned}\Rightarrow \frac{x + y}{z} + 1 = \frac{y + z}{x} + 1 = \frac{z + x}{y} + 1 \\[1em] \Rightarrow \frac{x + y + z}{z} = \frac{y + z + x}{x} = \frac{z + x + y}{y}.\end{aligned}$$

Since, above fractions are equal.

∴ We can conclude that,

x = y = z = a (let)

Substituting values of x, y and z in given equation we get :

$$\begin{aligned}\Rightarrow \frac{x + y}{z} \\[1em] \Rightarrow \frac{a + a}{a} \\[1em] \Rightarrow \frac{2a}{a} \\[1em] \Rightarrow 2.\end{aligned}$$

Hence, Option 3 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \frac{x^2 - 4}{x^2 + 4} = \frac{3}{5} \\[1em] \Rightarrow 5(x^2 - 4) = 3(x^2 + 4) \\[1em] \Rightarrow 5x^2 - 20 = 3x^2 + 12 \\[1em] \Rightarrow 5x^2 - 3x^2 = 12 + 20 \\[1em] \Rightarrow 2x^2 = 32 \\[1em] \Rightarrow x^2 = 16 \\[1em] \Rightarrow x= \sqrt{16} \\[1em] \Rightarrow x = \pm 4.\end{aligned}$$

Hence, Option 2 is the correct option.

Given,

⇒ a + b + c = 7

⇒ a + b = 7 - c

⇒ b + c = 7 - a

⇒ c + a = 7 - b

Given,

$$\begin{aligned}\phantom{\Rightarrow} \frac{x}{a + b - c} = 5 \\[1em] \Rightarrow \frac{x}{7 - c - c} = 5 \\[1em] \Rightarrow x = 5(7 - 2c). \\[2em] \phantom{\Rightarrow} \frac{y}{b + c - a} = 5 \\[1em] \Rightarrow \frac{y}{7 - a - a} = 5 \\[1em] \Rightarrow y = 5(7 - 2a). \\[2em] \phantom{\Rightarrow} \frac{z}{c + a - b} = 5 \\[1em] \Rightarrow \frac{z}{7 - b - b} = 5 \\[1em] \Rightarrow z = 5(7 - 2b). \\[1em]\end{aligned}$$

x + y + z = 5(7 - 2c) + 5(7 - 2a) + 5(7 - 2b)

= 35 - 10c + 35 - 10a + 35 - 10b

= 105 - 10(a + b + c)

= 105 - 10 × 7

= 105 - 70

= 35.

Hence, Option 1 is the correct option.

Given,

⇒ a : b = c : d

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

Multiplying both sides by $\frac{5}{7}$:

$\Rightarrow \frac{5a}{7b} = \frac{5c}{7d}$

Applying componendo and dividendo:

$\Rightarrow \frac{5a + 7b}{5a - 7b} = \frac{5c + 7d}{5c - 7d}$

Hence, proved that 5a + 7b : 5a - 7b = 5c + 7d : 5c - 7d.

Given,

a : b = c : d

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

Multiplying both sides by $\frac{x}{y}$:

$\Rightarrow \frac{xa}{yb} = \frac{xc}{yd}$

Applying componendo: $\Rightarrow \frac{xa + yb}{yb} = \frac{xc + yd}{yd}$

On cross-multiplication:

$$\begin{aligned}\Rightarrow \frac{xa + yb}{xc + yd} = \frac{yb}{yd} \\[1em] \Rightarrow \frac{xa + yb}{xc + yd} = \frac{b}{d}.\end{aligned}$$

Hence, proved that xa + yb : xc + yd = b : d.

Given,

(7a + 8b)(7c - 8d) = (7a - 8b)(7c + 8d)

$\therefore \frac{7a + 8b}{7a - 8b} = \frac{7c + 8d}{7c - 8d}$

Applying componendo and dividendo: $$\begin{aligned}\Rightarrow \frac{7a + 8b + 7a - 8b}{7a + 8b - (7a - 8b)} = \frac{7c + 8d + 7c - 8d}{7c + 8d - (7c - 8d)} \\[1em] \Rightarrow \frac{14a}{16b} = \frac{14c}{16d} \\[1em] \Rightarrow \frac{a}{b} = \frac{c}{d}.\end{aligned}$$

Hence, proved that a : b = c : d.

(i) Given,

$$\begin{aligned}\Rightarrow x = \frac{6ab}{a + b} \\[1em] \Rightarrow \frac{x}{3a} = \frac{2b}{a + b}\end{aligned}$$

Applying componendo and dividendo:

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

Again,

$$\begin{aligned}\Rightarrow x = \frac{6ab}{a + b} \\[1em] \Rightarrow \frac{x}{3b} = \frac{2a}{a + b}\end{aligned}$$

Applying componendo and dividendo:

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

Adding (i) and (ii) we get,

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

Hence, $\frac{x + 3a}{x - 3a} + \frac{x + 3b}{x - 3b}$ = 2.

Given,

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

Applying componendo and dividendo:

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

Again,

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

Applying componendo and dividendo:

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

Adding (i) and (ii) we get,

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

Hence, $\frac{a + 2\sqrt{2}}{a - 2\sqrt{2}} + \frac{a + 2\sqrt{3}}{a - 2\sqrt{3}}$ = 2.

Given,

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

Applying componendo and dividendo, we get :

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

Again applying componendo and dividendo, we get :

$$\begin{aligned}\Rightarrow \frac{(a + b) + (a - b)}{(a + b) - (a - b)} = \frac{(c + d) + (c - d)}{(c + d) - (c - d)} \\[1em] \Rightarrow \frac{2a}{2b} = \frac{2c}{2d} \\[1em] \Rightarrow \frac{a}{b} = \frac{c}{d} \\[1em] \Rightarrow a : b = c : d.\end{aligned}$$

Hence, proved that a : b = c : d.

Given,

$\Rightarrow \frac{a - 2b - 3c + 4d}{a + 2b - 3c - 4d} = \frac{a - 2b + 3c - 4d}{a + 2b + 3c + 4d}$

Applying componendo and dividendo:

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

Applying alternendo:

$$\begin{aligned}\Rightarrow \frac{(a - 3c)}{(a + 3c)} = \frac{(4d - 2b)}{(-4d - 2b)} \\[1em] \Rightarrow \frac{a - 3c + a + 3c}{a - 3c - (a + 3c)} = \frac{4d - 2b + (-4d - 2b)}{4d - 2b - (-4d - 2b)} \\[1em] \Rightarrow \frac{2a}{-6c} = \frac{-4b}{8d} \\[1em] \Rightarrow \frac{a}{-3c} = \frac{-b}{2d} \\[1em] \Rightarrow -2ad = -3bc \\[1em] \Rightarrow 2ad = 3bc.\end{aligned}$$

Hence, proved that 2ad = 3bc.

Given,

a, b and c are in continued proportion

$$\begin{aligned}\therefore \frac{a}{b} = \frac{b}{c} \\[1em] \text{Let } \frac{a}{b} = \frac{b}{c} = k \\[1em] \Rightarrow a = bk, b = ck \\[1em] \Rightarrow a = (ck)k = ck^2\end{aligned}$$

To prove :

$\frac{a^2 + ab + b^2}{b^2 + bc + c^2} = \frac{a}{c}$

Substituting value of a and b in L.H.S. of above equation,

$$\begin{aligned}\Rightarrow \frac{(ck^2)^2 + (ck^2)(ck) + (ck)^2}{(ck)^2 + (ck)c + c^2} \\[1em] \Rightarrow \frac{c^2k^4 + c^2k^3 + c^2k^2}{c^2k^2 + c^2k + c^2} \\[1em] \Rightarrow \frac{c^2k^2(k^2 + k + 1)}{c^2(k^2 + k + 1)} \\[1em] \Rightarrow k^2.\end{aligned}$$

Substituting value of a and b in R.H.S. of above equation,

$$\begin{aligned}\Rightarrow \frac{ck^2}{c} \\[1em] \Rightarrow k^2.\end{aligned}$$

Since, L.H.S. = R.H.S. = k².

Hence, proved that $\frac{a^2 + ab + b^2}{b^2 + bc + c^2} = \frac{a}{c}$.

Given,

$\Rightarrow \frac{\sqrt{x + 5} + \sqrt{x - 16}}{\sqrt{x + 5} - \sqrt{x - 16}} = \frac{7}{3}$

Applying componendo and dividendo:

$$\begin{aligned}\Rightarrow \frac{\sqrt{x + 5} + \sqrt{x - 16} + \sqrt{x + 5} - \sqrt{x - 16}}{\sqrt{x + 5} + \sqrt{x - 16} - (\sqrt{x + 5} - \sqrt{x - 16})} = \frac{7 + 3}{7 - 3} \\[1em] \Rightarrow \frac{2\sqrt{x + 5}}{2\sqrt{x - 16}} = \frac{10}{4} \\[1em] \Rightarrow \frac{\sqrt{x + 5}}{\sqrt{x - 16}} = \frac{10}{4} \\[1em] \Rightarrow \frac{x + 5}{x - 16} = \frac{100}{16} \\[1em] \Rightarrow 16(x + 5) = 100(x - 16) \\[1em] \Rightarrow 16x + 80 = 100x - 1600 \\[1em] \Rightarrow 100x - 16x = 1680 \\[1em] \Rightarrow 84x = 1680 \\[1em] \Rightarrow x = 20.\end{aligned}$$

Hence, x = 20.

Applying componendo and dividendo,

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

Squaring both sides of the equation,

$$\begin{aligned}\Rightarrow 9x^2 - 5 = 4x^2 \\[1em] \Rightarrow 9x^2 - 4x^2 = 5 \\[1em] \Rightarrow 5x^2 = 5 \\[1em] \Rightarrow x^2 = 1 \\[1em] \Rightarrow x = \pm1.\end{aligned}$$

Hence, x = 1.

Given,

$\Rightarrow x = \frac{\sqrt{a + 3b} + \sqrt{a - 3b}}{\sqrt{a + 3b} - \sqrt{a - 3b}}$

Applying componendo and dividendo,

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

Squaring both sides:

$$\begin{aligned}\Rightarrow \frac{(x + 1)^2}{(x - 1)^2} = \frac{a + 3b}{a - 3b} \\[1em] \Rightarrow \frac{x^2 + 1 + 2x}{x^2 + 1 - 2x} = \frac{a + 3b}{a - 3b} \\[1em] \Rightarrow (x^2 + 1 + 2x)(a - 3b) = (x^2 + 1 - 2x)(a + 3b) \\[1em] \Rightarrow ax^2 + a + 2ax - 3bx^2 - 3b - 6bx = ax^2 + a - 2ax + 3bx^2 + 3b - 6bx \\[1em] \Rightarrow ax^2 - ax^2 + a - a + 2ax + 2ax - 3bx^2 - 3bx^2 - 3b - 3b - 6bx + 6bx = 0 \\[1em] \Rightarrow 4ax - 6bx^2 - 6b = 0 \\[1em] \Rightarrow 6bx^2 - 4ax + 6b = 0 \\[1em] \Rightarrow 2(3bx^2 - 2ax + 3b) = 0 \\[1em] \Rightarrow 3bx^2 - 2ax + 3b = 0.\end{aligned}$$

Hence, proved that 3bx² - 2ax + 3b = 0.

Given,

$\Rightarrow \frac{x^4 + 1}{2x^2} = \frac{17}{8}$

Applying componendo and dividendo, we get

$$\begin{aligned}\Rightarrow \frac{x^4 + 1 + 2x^2}{x^4 + 1 - 2x^2} = \frac{17 + 8}{17 - 8} \\[1em] \Rightarrow \frac{(x^2 + 1)^2}{(x^2 - 1)^2} = \frac{25}{9} \\[1em] \Rightarrow \frac{x^2 + 1}{x^2 - 1} = \pm \frac{5}{3}\end{aligned}$$

Considering, $\frac{x^2 + 1}{x^2 - 1} = \frac{5}{3}$

Applying componendo and dividendo, we get

$$\begin{aligned}\Rightarrow \frac{x^2 + 1 + x^2 - 1}{x^2 + 1 - (x^2 - 1)} = \frac{5 + 3}{5 - 3} \\[1em] \Rightarrow \frac{2x^2}{2} = \frac{8}{2} \\[1em] \Rightarrow x^2 = 4 \\[1em] \Rightarrow x = \pm 2.\end{aligned}$$

Considering, $\frac{x^2 + 1}{x^2 - 1} = -\frac{5}{3}$

Applying componendo and dividendo, we get

$$\begin{aligned}\Rightarrow \frac{x^2 + 1 + x^2 - 1}{x^2 + 1 - (x^2 - 1)} = \frac{-5 + 3}{-5 - 3} \\[1em] \Rightarrow \frac{2x^2}{2} = \frac{-2}{-8} \\[1em] \Rightarrow x^2 = \frac{1}{4} \\[1em] \Rightarrow x = \sqrt{\frac{1}{4}} \\[1em] \Rightarrow x = \pm \frac{1}{2}.\end{aligned}$$

Hence, x = ±2 and $\pm \frac{1}{2}$.

Given,

$x = \frac{\sqrt{m + n} + \sqrt{m - n}}{\sqrt{m + n} - \sqrt{m - n}}$

Applying componendo and dividendo,

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

Applying componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{m + n + m - n}{m + n - (m - n)} = \frac{x^2 + 1 + 2x + x^2 + 1 - 2x}{x^2 + 1 + 2x - (x^2 + 1 - 2x)} \\[1em] \Rightarrow \frac{2m}{2n} = \frac{2(x^2 + 1)}{4x} \\[1em] \Rightarrow \frac{m}{n} = \frac{x^2 + 1}{2x} \\[1em] \Rightarrow n = \frac{2mx}{x^2 + 1}.\end{aligned}$$

Hence, n = $\frac{2mx}{x^2 + 1}.$

Given,

$\frac{x^3 + 3xy^2}{3x^2y + y^3} = \frac{m^3 + 3mn^2}{3m^2n + n^3}$

Applying componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{x^3 + 3xy^2 + 3x^2y + y^3}{x^3 + 3xy^2 - (3x^2y + y^3)} = \frac{m^3 + 3mn^2 + 3m^2n + n^3}{m^3 + 3mn^2 - (3m^2n + n^3)} \\[1em] \Rightarrow \frac{(x + y)^3}{(x - y)^3} = \frac{(m + n)^3}{(m - n)^3} \\[1em] \Rightarrow \frac{(x + y)}{(x - y)} = \frac{(m + n)}{(m - n)}\end{aligned}$$

Applying componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{x + y + x - y}{x + y - (x - y)} = \frac{m + n + m - n}{m + n - (m - n)} \\[1em] \Rightarrow \frac{2x}{2y} = \frac{2m}{2n} \\[1em] \Rightarrow \frac{x}{y} = \frac{m}{n} \\[1em] \Rightarrow nx = my.\end{aligned}$$

Hence, proved that nx = my.

It is a case of direct variation:

$\therefore \frac{10}{40} = \frac{a}{100} = \frac{40}{b} = \frac{c}{180}$

Taking,

$\frac{10}{40} = \frac{a}{100}$

⇒ 10 x 100 = 40 x a

⇒ a = $\frac{1000}{40}$

⇒ a = 25 m

Taking,

$\frac{10}{40} = \frac{40}{b}$

⇒ 10 x b = 40 x 40

⇒ b = $\frac{1600}{10}$

⇒ b = ₹ 160

Taking,

$\frac{10}{40} = \frac{c}{180}$

⇒ 10 x 180 = 40 x c

⇒ c = $\frac{1800}{40}$

⇒ c = 45 m

Hence, the value of a = 25 m, b = ₹ 160 and c = 45 m.

It is a case of inverse variation:

∴ 130 x a = 70 x 39 = b x 30 = 120 x c

Taking,

130 x a = 70 x 39

⇒ a = $\frac{70 \times 39}{130}$

⇒ a = 7 x 3

⇒ a = 21

Taking,

70 x 39 = b x 30

⇒ b = $\frac{70 \times 39}{30}$

⇒ b = 7 x 13

⇒ b = 91

Taking,

70 x 39 = 120 x c

⇒ c = $\frac{70 \times 39}{120}$

⇒ c = 22.75

Hence, the value of a = 21, b = 91 and c = 22.75

Let x be the mean proportion of $\sqrt{3} + \sqrt{2} \text{ and } \sqrt{3} - \sqrt{2}$ :

$$\begin{aligned}\therefore \frac{\sqrt{3} + \sqrt{2}}{x} = \frac{x}{\sqrt{3} - \sqrt{2}} \\[1em] \Rightarrow (\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = x^2 \\[1em] \Rightarrow x^2 = (\sqrt{3})^2 - (\sqrt{2})^2 \quad [\because (a + b)(a - b) = a^2 - b^2] \\[1em] \Rightarrow x^2 = 3 - 2 \\[1em] \Rightarrow x^2 = 1 \\[1em] \Rightarrow x = \sqrt{1} = \pm 1.\end{aligned}$$

Since, geometrical mean is always positive.

∴ x = 1.

Hence, Option 3 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \frac{a + b}{a - b} = \frac{13}{3} \\[1em] \Rightarrow 3(a + b) = 13(a - b) \\[1em] \Rightarrow 3a + 3b = 13a - 13b \\[1em] \Rightarrow 13a - 3a = 3b + 13b \\[1em] \Rightarrow 10a = 16b \\[1em] \Rightarrow \frac{a}{b} = \frac{16}{10} \\[1em] \Rightarrow \frac{a}{b} = \frac{8}{5}.\end{aligned}$$

Hence, Option 4 is the correct option.

Given,

x is proportional to y.

$\therefore \frac{A}{24} = \frac{12}{B}$ ........(1)

and

$\therefore \frac{24}{15} = \frac{B}{20}$ ........(2)

Solving equation (2), we get :

$$\begin{aligned}\Rightarrow B = \frac{24 \times 20}{15} \\[1em] \Rightarrow B = 32.\end{aligned}$$

Substituting value of B in equation 1 :

$$\begin{aligned}\Rightarrow \frac{A}{24} = \frac{12}{32} \\[1em] \Rightarrow A = \frac{12 \times 24}{32} \\[1em] \Rightarrow A = 9.\end{aligned}$$

Hence, Option 3 is the correct option.

Given,

(m + n) : (n - m) = 5 : 2

$$\begin{aligned}\therefore \frac{m + n}{n - m} = \frac{5}{2} \\[1em] \Rightarrow 2(m + n) = 5(n - m) \\[1em] \Rightarrow 2m + 2n = 5n - 5m \\[1em] \Rightarrow 2m + 5m = 5n - 2n \\[1em] \Rightarrow 7m = 3n \\[1em] \Rightarrow \frac{m}{n} = \frac{3}{7} \\[1em] \Rightarrow m : n = 3 : 7.\end{aligned}$$

Hence, Option 1 is the correct option.

Substituting x = y in (3x + y) : (5x - 3y), we get :

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

Hence, Option 2 is the correct option.

Both A and R are true and R is the correct reason for R.

Reason

Given,

x : y = 3 : 2

Let the value of x be 3k and y be 2k.

The value of

$$\begin{aligned}\Rightarrow\frac{x^2 + y^2}{x^2 - y^2}\\[1em] = \frac{(3k)^2 + (2k)^2}{(3k)^2 - (2k)^2}\\[1em] = \frac{9k^2 + 4k^2}{9k^2 - 4k^2}\\[1em] = \frac{13k^2}{5k^2}\\[1em] = \frac{13}{5}\end{aligned}$$

According to Assertion; the value of (x² + y²) : (x² - y²) = 13 : 5, which is true.

According to Reason; x : y = 3 : 2

⇒ $\frac{x^2 + y^2}{x^2 - y^2} = \frac{(3k)^2 + (2k)^2}{(3k)^2 - (2k)^2} = \frac{13}{5}$ ; k ≠ 0, which is true.

Hence, option 3 is the correct option.

Both A and R are true and R is the correct reason for R.

Reason

Given,

$$\begin{aligned}\Rightarrow \frac{(x + y)^4}{(x - y)^4} = \frac{16}{1}\\[1em] \Rightarrow \frac{(x + y)}{(x - y)} = \sqrt[4]{\frac{16}{1}}\\[1em] \Rightarrow \frac{(x + y)}{(x - y)} = \frac{2}{1}\\[1em] \Rightarrow \frac{(x + y) + (x - y)}{(x + y) - (x - y)} = \frac{2 + 1}{2 - 1}\\[1em] \Rightarrow \frac{x + y + x - y}{x + y - x + y} = \frac{3}{1}\\[1em] \Rightarrow \frac{2x}{2y} = \frac{3}{1}\\[1em] \Rightarrow \frac{x}{y} = \frac{3}{1}\end{aligned}$$

According to Assertion; x : y = 3 : 1 , which is true.

According to Reason; $\frac{x + y}{x - y} = \frac{2}{1}$ and $\frac{x + y + x - y}{x + y - x + y} = \frac{2 + 1}{2 - 1}$, which is true.

Hence, option 3 is the correct option.

Statement 1 is false, and statement 2 is true.

Reason

The mean proportion of two positive real numbers x and y is $\sqrt{x \times y}$

The mean proportion of $\sqrt{6}$ and $\sqrt{5}$ = $\sqrt{\sqrt{6} \times \sqrt{5}}$

= $\sqrt{\sqrt{30}} = \sqrt[4]{30}$

So, statement 1 is false but statement 2 is true.

Hence, option 4 is the correct option.

Both the statements are true.

Reason

Numbers a, b, c are in continued proportion.

Now, (a + b + c)(a - b + c)

= [(a + c) + b][(a + c) - b]

= (a + c)² - b²

= a² + c² + 2ac - b²

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

$$\begin{aligned}\therefore \frac{a}{b} = \frac{b}{c} \\[1em] \Rightarrow b^2 = ac\end{aligned}$$

Substituting the value b² = ac in above equation,

= a² + c² + 2b² - b²

= a² + c² + b²

So, Statement 1 correctly states that: (a + b + c)(a - b + c) = a² + b² + c².

and

Statement 2 correctly states that: b² = ac

Hence, option 1 is the correct option.

Given, a : b = 3 : 5,

If a = 3k, b = 5k.

Substituting value of a and b in (10a + 3b) : (5a + 2b),

$$\begin{aligned}\Rightarrow \frac{10(3k) + 3(5k)}{5(3k) + 2(5k)} \\[1em] \Rightarrow \frac{30k + 15k}{15k + 10k} \\[1em] \Rightarrow \frac{45k}{25k} \\[1em] \Rightarrow \frac{9}{5}.\end{aligned}$$

Hence, (10a + 3b) : (5a + 2b) = 9 : 5.

Given,

5x + 6y : 8x + 5y = 8 : 9

$$\begin{aligned}\Rightarrow \frac{5x + 6y}{8x + 5y} = \frac{8}{9} \\[1em] \Rightarrow 9(5x + 6y) = 8(8x + 5y) \\[1em] \Rightarrow 45x + 54y = 64x + 40y \\[1em] \Rightarrow 64x - 45x = 54y - 40y \\[1em] \Rightarrow 19x = 14y \\[1em] \Rightarrow \frac{x}{y} = \frac{14}{19}\end{aligned}$$

Hence, x : y = 14 : 19.

Duplicate ratio of $2\sqrt{2} : 3\sqrt{5}$,

$$\begin{aligned}= (2\sqrt{2})^2 : (3\sqrt{5})^2 \\[1em] = 8 : 45.\end{aligned}$$

Hence, duplicate ratio of $2\sqrt{2} : 3\sqrt{5}$ = 8 : 45.

Triplicate ratio of 2a : 3b,

= (2a)³ : (3b)³

= 8a³ : 27b³

Hence, triplicate ratio of 2a : 3b = 8a³ : 27b³.

Sub-duplicate ratio of 9x²a⁴ : 25y⁶b²

$$\begin{aligned}= \sqrt{9x^2a^4} : \sqrt{25y^6b^2} \\[1em] = 3xa^2 : 5y^3b\end{aligned}$$

Hence, sub-duplicate ratio of 9x²a⁴ : 25y⁶b² = 3xa² : 5y³b.

Sub-triplicate ratio of 216 : 343

= $\sqrt[3]{216} : \sqrt[3]{343}$

= 6 : 7.

Hence, sub-triplicate ratio of 216 : 343 = 6 : 7.

Reciprocal ratio of 3 : 5,

$$\begin{aligned}\Rightarrow \frac{1}{3} : \frac{1}{5} \\[1em] \Rightarrow 5 : 3.\end{aligned}$$

Hence, reciprocal ratio of 3 : 5 = 5 : 3.

Ratio compounded of the duplicate ratio of 5 : 6, the reciprocal ratio of 25 : 42 and the sub-duplicate ratio of 36 : 49,

$$\begin{aligned}\Rightarrow \frac{5^2}{6^2} \times \frac{\frac{1}{25}}{\frac{1}{42}} \times \frac{\sqrt{36}}{\sqrt{49}} \\[1em] \Rightarrow \frac{25}{36} \times \frac{42}{25} \times \frac{6}{7} \\[1em] \Rightarrow \frac{1}{1}.\end{aligned}$$

Hence, resultant ratio = 1 : 1.

According to question,

$$\begin{aligned}\Rightarrow \frac{2x + 3}{5x - 38} = \frac{(\sqrt{5})^2}{(\sqrt{6})^2} \\[1em] \Rightarrow \frac{2x + 3}{5x - 38} = \frac{5}{6} \\[1em] \Rightarrow 6(2x + 3) = 5(5x - 38) \\[1em] \Rightarrow 12x + 18 = 25x - 190 \\[1em] \Rightarrow 25x - 12x = 190 + 18 \\[1em] \Rightarrow 13x = 208 \\[1em] \Rightarrow x = 16.\end{aligned}$$

Hence, x = 16.

According to question,

$$\begin{aligned}\Rightarrow \frac{2x + 1}{3x + 13} = \frac{\sqrt{9}}{\sqrt{25}} \\[1em] \Rightarrow \frac{2x + 1}{3x + 13} = \frac{3}{5} \\[1em] \Rightarrow 5(2x + 1) = 3(3x + 13) \\[1em] \Rightarrow 10x + 5 = 9x + 39 \\[1em] \Rightarrow 10x - 9x = 39 - 5 \\[1em] \Rightarrow x = 34.\end{aligned}$$

Hence, x = 34.

According to question,

$$\begin{aligned}\Rightarrow \frac{3x - 7}{4x + 3} = \frac{\sqrt[3]{8}}{\sqrt[3]{27}} \\[1em] \Rightarrow \frac{3x - 7}{4x + 3} = \frac{2}{3} \\[1em] \Rightarrow 3(3x - 7) = 2(4x + 3) \\[1em] \Rightarrow 9x - 21 = 8x + 6 \\[1em] \Rightarrow 9x - 8x = 6 + 21 \\[1em] \Rightarrow x = 27.\end{aligned}$$

Hence, x = 27.

Let the number to be added be a

$$\begin{aligned}\Rightarrow \frac{x + a}{y + a} = \frac{c}{d} \\[1em] \Rightarrow d(x + a) = c(y + a) \\[1em] \Rightarrow dx + da = cy + ca \\[1em] \Rightarrow da - ca = cy - dx \\[1em] \Rightarrow a(d - c) = cy - dx \\[1em] \Rightarrow a = \frac{cy - dx}{d - c}.\end{aligned}$$

Hence, number to be added = $\frac{cy - dx}{d - c}.$

Let woman's reduced weight become x kg.

Since, weight is reduced in the ratio 7 : 5.

$$\begin{aligned}\therefore \frac{84}{x} = \frac{7}{5} \\[1em] \Rightarrow x = \frac{84 \times 5}{7} \\[1em] \Rightarrow x = 60.\end{aligned}$$

Hence, reduced weight = 60 kg.

Given,

$$\begin{aligned}\Rightarrow 15(2x^2 - y^2) = 7xy \\[1em] \Rightarrow 30x^2 - 15y^2 = 7xy \\[1em] \Rightarrow \frac{30x^2 - 15y^2}{xy} = \frac{7xy}{xy} \\[1em] \Rightarrow 30\frac{x}{y} - 15\frac{y}{x} = 7\end{aligned}$$

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

$$\begin{aligned}\Rightarrow 30t - 15\frac{1}{t} = 7 \\[1em] \Rightarrow \frac{30t^2 - 15}{t} = 7 \\[1em] \Rightarrow 30t^2 - 15 = 7t \\[1em] \Rightarrow 30t^2 - 7t - 15 = 0 \\[1em] \Rightarrow 30t^2 - 25t + 18t - 15 = 0 \\[1em] \Rightarrow 5t(6t - 5) + 3(6t - 5) = 0 \\[1em] \Rightarrow (5t + 3)(6t - 5) = 0 \\[1em] \Rightarrow 5t + 3 = 0 \text{ or } 6t - 5 = 0 \\[1em] \Rightarrow t = -\frac{3}{5} \text{ or } t = \frac{5}{6}.\end{aligned}$$

Since, x and y both are positive,

∴ t ≠ $-\frac{3}{5}$.

Hence, x : y = 5 : 6.

Let fourth proportional to 2xy, x² and y² be n.

$$\begin{aligned}\therefore \frac{2xy}{x^2} = \frac{y^2}{n} \\[1em] \Rightarrow n = \frac{x^2y^2}{2xy}\\[1em] \Rightarrow n = \frac{xy}{2}.\end{aligned}$$

Hence, fourth proportional = $\frac{xy}{2}$.

Let third proportional to a² - b² and a + b be x,

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

Hence, third proportional to a² - b² and a + b = $\frac{(a + b)}{(a - b)}.$

Let mean proportion to (x - y) and (x³ - x²y) be a.

$$\begin{aligned}\therefore \frac{x - y}{a} = \frac{a}{x^3 - x^2y} \\[1em] \Rightarrow a^2 = (x - y)(x^3 - x^2y) \\[1em] \Rightarrow a^2 = (x - y).x^2.(x - y) \\[1em] \Rightarrow a^2 = x^2.(x - y)^2 \\[1em] \Rightarrow a = x(x - y).\end{aligned}$$

Hence, mean proportion to (x - y) and (x³ - x²y) = x(x - y).

Let two numbers be x and y.

Given, 14 is mean proportional to x and y,

$$\begin{aligned}\therefore \frac{x}{14} = \frac{14}{y} \\[1em] \Rightarrow xy = 196 \space .....(i)\end{aligned}$$

Given, 112 is third proportional to x and y,

$$\begin{aligned}\therefore \frac{x}{y} = \frac{y}{112} \\[1em] \Rightarrow y^2 = 112x \\[1em] \Rightarrow x = \frac{y^2}{112} \space .....(ii)\end{aligned}$$

Substituting value of x from (ii) in (i) we get,

$$\begin{aligned}\Rightarrow \frac{y^2}{112}.y = 196 \\[1em] \Rightarrow y^3 = 196 \times 112 \\[1em] \Rightarrow y^3 = 21952 \\[1em] \Rightarrow y = \sqrt[3]{21952} \\[1em] \Rightarrow y = 28.\end{aligned}$$

$x = \frac{28^2}{112} = \frac{784}{112}$ = 7.

Hence, numbers are 7 and 28.

According to question,

$$\begin{aligned}\Rightarrow \frac{x}{y} = \frac{(x + z)^2}{(y + z)^2} \\[1em] \Rightarrow \frac{x}{y} = \frac{x^2 + z^2 + 2xz}{y^2 + z^2 + 2yz} \\[1em] \Rightarrow x(y^2 + z^2 + 2yz) = y(x^2 + z^2 + 2xz) \\[1em] \Rightarrow xy^2 + xz^2 + 2xyz = yx^2 + yz^2 + 2xyz \\[1em] \Rightarrow xz^2 - yz^2 = 2xyz - 2xyz + yx^2- xy^2 \\[1em] \Rightarrow z^2(x - y) = xy(x - y) \\[1em] \Rightarrow z^2 = xy.\end{aligned}$$

Since, z² = xy hence, proved that z is mean proportional between x and y.

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

a = bk, c = dk.

Substituting a = bk, c = dk in L.H.S. of (a + b) : (c + d) = $\sqrt{a^2 + b^2} : \sqrt{c^2 + d^2}$

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

Substituting a = bk, c = dk in R.H.S. of (a + b) : (c + d) = $\sqrt{a^2 + b^2} : \sqrt{c^2 + d^2}$

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

Since, L.H.S. = R.H.S. = $\frac{b}{d}$

Hence, proved that (a + b) : (c + d) = $\sqrt{a^2 + b^2} : \sqrt{c^2 + d^2}$.

Ratio of the number of boys to the number of girls is 3 : 1.

Total members = 36

No. of boys = $\frac{3}{3 + 1} \times 36 = \frac{3}{4} \times 36 = 27.$

No. of girls = 36 - 27 = 9.

Let no. of girls to be added be x.

$$\begin{aligned}\therefore \frac{27}{9 + x} = \frac{9}{5} \\[1em] \Rightarrow 135 = 9(9 + x) \\[1em] \Rightarrow 135 = 81 + 9x \\[1em] \Rightarrow 54 = 9x \\[1em] \Rightarrow x = 6.\end{aligned}$$

Hence, 6 girls must be added to council so ratio of number of boys to number of girls becomes 9 : 5.

7x - 15y = 4x + y

⇒ 7x - 4x = y + 15y

⇒ 3x = 16y

⇒ $\frac{x}{y} = \frac{16}{3}$

(i) $\frac{9x + 5y}{9x - 5y}$

$$\begin{aligned}\phantom{\Rightarrow} \frac{x}{y} = \frac{16}{3} \\[1em] \frac{9x}{5y} = \frac{9 \times 16}{5 \times 3} \\[1em] \frac{9x}{5y} = \frac{144}{15}\end{aligned}$$

Applying componendo and dividendo:

$$\begin{aligned}\frac{9x + 5y}{9x - 5y} = \frac{144 + 15}{144 - 15} \\[1em] \frac{9x + 5y}{9x - 5y} = \frac{159}{129} = \frac{53}{43}.\end{aligned}$$

Hence, $\frac{9x + 5y}{9x - 5y} = \frac{53}{43}$.

(ii) $\frac{3x^2 + 2y^2}{3x^2 - 2y^2}$

$$\begin{aligned}\Rightarrow \frac{x}{y} = \frac{16}{3} \\[1em] \Rightarrow \frac{x^2}{y^2} = \frac{256}{9} \\[1em] \Rightarrow \frac{3x^2}{2y^2} = \frac{3 \times 256}{2 \times 9} \\[1em] \Rightarrow \frac{3x^2}{2y^2} = \frac{768}{18}\end{aligned}$$

Applying componendo and dividendo:

$$\begin{aligned}\Rightarrow \frac{3x^2 + 2y^2}{3x^2 - 2y^2} = \frac{768 + 18}{768 - 18} \\[1em] \Rightarrow \frac{3x^2 + 2y^2}{3x^2 - 2y^2} = \frac{786}{750} \\[1em] \Rightarrow \frac{3x^2 + 2y^2}{3x^2 - 2y^2} = \frac{131}{125}.\end{aligned}$$

Hence, $\frac{3x^2 + 2y^2}{3x^2 - 2y^2} = \frac{131}{125}$.

(i) Given,

$\frac{4m + 3n}{4m - 3n} = \frac{7}{4}$

Applying componendo and dividendo:

$$\begin{aligned}\frac{4m + 3n + 4m - 3n}{4m + 3n - (4m - 3n)} = \frac{7 + 4}{7 - 4} \\[1em] \Rightarrow \frac{8m}{6n} = \frac{11}{3} \\[1em] \Rightarrow \frac{m}{n} = \frac{11 \times 6}{3 \times 8} \\[1em] \Rightarrow \frac{m}{n} = \frac{11}{4}.\end{aligned}$$

Hence, m : n = 11 : 4.

(ii) We know,

$$\begin{aligned}\phantom{\Rightarrow} \frac{m}{n} = \frac{11}{4} \\[1em] \Rightarrow \frac{m^2}{n^2} = \frac{121}{16} \\[1em] \Rightarrow \frac{2m^2}{11n^2} = \frac{2 \times 121}{11 \times 16} \\[1em] \Rightarrow \frac{2m^2}{11n^2} = \frac{242}{176}\end{aligned}$$

Applying componendo and dividendo:

$$\begin{aligned}\Rightarrow \frac{2m^2 + 11n^2}{2m^2 - 11n^2} = \frac{242 + 176}{242 - 176} \\[1em] \Rightarrow \frac{2m^2 + 11n^2}{2m^2 - 11n^2} = \frac{418}{66} = \frac{19}{3}\end{aligned}$$

Applying invertendo:

$\Rightarrow \frac{2m^2 - 11n^2}{2m^2 + 11n^2} = \frac{3}{19}$

Hence, $\frac{2m^2 - 11n^2}{2m^2 + 11n^2} = \frac{3}{19}.$

Given, x, y and z are in continued proportion

$\therefore \frac{x}{y} = \frac{y}{z}$

⇒ y² = xz

Taking LHS,

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

Substituting y² = xz in above we get,

$$\begin{aligned}\Rightarrow \frac{x^2 + xz + 2xy}{xz + z^2 + 2yz} \\[1em] \Rightarrow \frac{x(x + z + 2y)}{z(x + z + 2y)} \\[1em] \Rightarrow \frac{x}{z} = \text{ RHS }\end{aligned}$$

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

Given,

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

Applying componendo and dividendo:

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

Applying componendo and dividendo:

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

Hence, proved that b² = $\frac{2a^2x}{x^2 + 1}$.

Given,

$$\begin{aligned}\Rightarrow \frac{x^2 + y^2}{x^2 - y^2} = 2\frac{1}{8} \\[1em] \Rightarrow \frac{x^2 + y^2}{x^2 - y^2} = \frac{17}{8}\end{aligned}$$

Applying componendo and dividendo:

$$\begin{aligned}\Rightarrow \frac{x^2 + y^2 + x^2 - y^2}{x^2 + y^2 - (x^2 - y^2)} = \frac{17 + 8}{17 - 8} \\[1em] \Rightarrow \frac{2x^2}{2y^2} = \frac{25}{9} \\[1em] \Rightarrow \frac{x^2}{y^2} = \frac{25}{9} \\[1em] \Rightarrow \frac{x}{y} = \frac{5}{3}.\end{aligned}$$

Hence, $\frac{x}{y} = \frac{5}{3} = 1\frac{2}{3}$.

(ii) We know that,

$$\begin{aligned}\phantom{\Rightarrow} \frac{x}{y} = \frac{5}{3} \\[1em] \Rightarrow \frac{x^3}{y^3} = \frac{125}{27} \\[1em] \Rightarrow \frac{x^3 + y^3}{x^3 - y^3} = \frac{125 + 27}{125 - 27} \\[1em] \Rightarrow \frac{x^3 + y^3}{x^3 - y^3} = \frac{152}{98} \\[1em] \Rightarrow \frac{x^3 + y^3}{x^3 - y^3} = \frac{76}{49} \\[1em] \Rightarrow \frac{x^3 + y^3}{x^3 - y^3} = 1\frac{27}{49}.\end{aligned}$$

Hence, $\frac{x^3 + y^3}{x^3 - y^3} = 1\frac{27}{49}$.

Given,

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

Applying componendo and dividendo we get,

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

Applying componendo and dividendo again we get,

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

Hence, x : y = 2 : 3.

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

∴ x = ak, y = bk and z = ck.

Substituting x = ak, y = bk and z = ck in L.H.S. of $\frac{x^3}{a^3} + \frac{y^3}{b^3} + \frac{z^3}{c^3} = \frac{3xyz}{abc}$,

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

Substituting x = ak, y = bk and z = ck in R.H.S. of $\frac{x^3}{a^3} + \frac{y^3}{b^3} + \frac{z^3}{c^3} = \frac{3xyz}{abc}$,

$$\begin{aligned}\Rightarrow \frac{3(ak)(bk)(ck)}{abc} \\[1em] \Rightarrow \frac{3abck^3}{abc} \\[1em] \Rightarrow 3k^3.\end{aligned}$$

Since, L.H.S. = R.H.S. = 3k³

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

Given,

b is the mean proportion between a and c

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

⇒ b² = ac

Substituting b² = ac in L.H.S. of the equation $\frac{a^4 + a^2b^2 + b^4}{b^4 + b^2c^2 + c^4} = \frac{a^2}{c^2}$ we get,

$$\begin{aligned}\Rightarrow \frac{a^4 + a^2.(ac) + (ac)^2}{(ac)^2 + (ac).c^2 + c^4} \\[1em] \Rightarrow \frac{a^2(a^2 + ac + c^2)}{c^2(a^2 + ac + c^2)} \\[1em] \Rightarrow \frac{a^2}{c^2}.\end{aligned}$$

Hence, proved that $\frac{a^4 + a^2b^2 + b^4}{b^4 + b^2c^2 + c^4} = \frac{a^2}{c^2}$.

Given,

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

In first case, let $\frac{a + x}{a - x} = 2$

Applying componendo and dividendo we get,

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

In second case, let $\frac{a + x}{a - x} = -2$

Applying componendo and dividendo we get,

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

Hence, x = 3a or $\frac{a}{3}$.

Given,

a : b :: c : d

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

If the original ratios are equal, their reciprocals are also equal (this is known as Invertendo) :

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

Applying dividendo, we get:

$$\begin{aligned}\Rightarrow \frac{b - a}{a} = \frac{d - c}{c} \\[1em] \Rightarrow -\frac{a - b}{a} = -\frac{c - d}{c} \\[1em] \Rightarrow \frac{a - b}{a} = \frac{c - d}{c} \\[1em] \text{ Now, take the reciprocal} \\[1em] \therefore \frac{a}{a - b} = \frac{c}{c - d}\end{aligned}$$

Hence, proved that a : (a - b) :: c : (c - d).

Given,

$\frac{k^5 + x^5}{x^5 - k^5} = \frac{122}{121}$

Applying componendo and dividendo we get,

$$\begin{aligned}\Rightarrow \frac{(k^5 + x^5) + (x^5 - k^5)}{(k^5 + x^5) - (x^5 - k^5)} = \frac{122 + 121}{122 - 121} \\[1em] \Rightarrow \frac{k^5 + x^5 + x^5 - k^5}{k^5 + x^5 - x^5 + k^5} = \frac{243}{1} \\[1em] \Rightarrow \frac{2x^5}{2k^5} = \frac{243}{1} \\[1em] \Rightarrow \Big(\frac{x}{k}\Big)^5 = (3)^5 \\[1em] \Rightarrow \frac{x}{k} = 3 \\[1em] \Rightarrow x = 3k.\end{aligned}$$

Hence, x = 3k.

Given,

$$\begin{aligned}\Rightarrow \frac{x^2 - x + 1}{x^2 + x + 1} = \frac{112(1 - x)}{104(1 + x)} \\[1em] \Rightarrow \frac{x^2 - x + 1}{x^2 + x + 1} = \frac{14(1 - x)}{13(1 + x)}.\end{aligned}$$

Applying componendo and dividendo we get,

$$\begin{aligned}\Rightarrow \frac{(x^2 - x + 1) + (x^2 + x + 1)}{(x^2 - x + 1) - (x^2 + x + 1)} = \frac{14(1 - x) + 13(1 + x)}{14(1 - x) - 13(1 + x)} \\[1em] \Rightarrow \frac{x^2 - x + 1 + x^2 + x + 1}{x^2 - x + 1 - x^2 - x - 1} = \frac{14 − 14x + 13 + 13x}{14 − 14x − 13 − 13x} \\[1em] \Rightarrow \frac{2x^2 + 2}{-2x} = \frac{27 − x}{1 − 27x} \\[1em] \Rightarrow \frac{2(x^2 + 1)}{-2x} = \frac{27 - x}{1 - 27x} \\[1em] \Rightarrow \frac{x^2 + 1}{-x} = \frac{27 - x}{1 - 27x}\end{aligned}$$

⇒ (x² + 1)(1 − 27x) = −x(27 − x)

⇒ x² + 1 − 27x³ − 27x = −27x + x²

⇒ −27x³ + 1 = 0

⇒ 27x³ = 1

⇒ x³ = $\frac{1}{27}$

⇒ x = $\sqrt[3]{\frac{1}{27}} = \frac{1}{3}$

Hence, x = $\frac{1}{3}$.

Given,

$\frac{x^2 - x + 1}{y^2 - y + 1} = \frac{x^2 + x + 1}{y^2 + y + 1}$

Applying alternendo we get,

$\frac{x^2 - x + 1}{x^2 + x + 1} = \frac{y^2 - y + 1}{y^2 + y + 1}$

Applying componendo and dividendo we get,

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

⇒ xy² + x = x²y + y

⇒ xy(y - x) - (y - x) = 0

⇒ (y − x)(xy − 1) = 0

⇒ xy = 1

Hence, proved that xy = 1.

Let x be the quantity of ₹ 240/kg variety and y be the quantity of ₹ 280/kg variety.

Amit sells the mixture at ₹ 265 per kg.

∴ 240x + 280y = 265(x + y)

⇒ 240x + 280y = 265x + 265y

⇒ 240x - 265x + 280y - 265y = 0

⇒ -25x + 15y = 0

⇒ 15y = 25x

⇒ $\frac{y}{x} = \frac{25}{15}$

⇒ $\frac{y}{x} = \frac{5}{3}$

⇒ y : x = 5 : 3.

Hence, the ratio of ₹ 280/kg to ₹ 240/kg varieties is 5 : 3.

Let x be the quantity of 55% solution and y be the quantity of 80% solution.

Required concentration of solution by mixing two solutions = 60%.

∴ 55x + 80y = 60(x + y)

⇒ 55x + 80y = 60x + 60y

⇒ 55x - 60x + 80y - 60y = 0

⇒ -5x + 20y = 0

⇒ 20y = 5x

⇒ $\frac{x}{y} = \frac{20}{5}$

⇒ $\frac{x}{y} = \frac{4}{1}$

⇒ x : y = 4 : 1.

Hence, the ratio of 55% solution to 80% solution is 4 : 1.