Ratio & Proportion — Exercise 7(C)

If x + y + z = 0, the value of $\dfrac{x}{y + z}$ is :

1. 2

2. $\dfrac{1}{2}$

3. 1

4. -1

Answer

Given,

⇒ x + y + z = 0

⇒ x = -(y + z).

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

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

Hence, Option 4 is the correct option.

If a, b, c and d are in proportion, the value of $\dfrac{8a^2 - 5b^2}{8c^2 - 5d^2}$ is equal to :

1. a² : b²

2. a² : c²

3. a² : d²

4. c² : d²

Answer

Given,

a, b, c and d are in proportion.

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

∴ a = bk and c = dk

Solving,

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

As,

$\Rightarrow \dfrac{a}{b} = \dfrac{c}{d} \\[1em] \Rightarrow \dfrac{b}{d} = \dfrac{a}{c}$

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

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

Hence, Option 2 is the correct option.

$\dfrac{x + y}{z} = \dfrac{y + z}{x} = \dfrac{z + x}{y}$ is equal to :

1. 0

2. 1

3. 2

4. -2

Answer

Given,

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

Applying componendo on each side we get :

$\Rightarrow \dfrac{x + y}{z} + 1 = \dfrac{y + z}{x} + 1 = \dfrac{z + x}{y} + 1 \\[1em] \Rightarrow \dfrac{x + y + z}{z} = \dfrac{y + z + x}{x} = \dfrac{z + x + y}{y}.$

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 :

$\Rightarrow \dfrac{x + y}{z} \\[1em] \Rightarrow \dfrac{a + a}{a} \\[1em] \Rightarrow \dfrac{2a}{a} \\[1em] \Rightarrow 2.$

Hence, Option 3 is the correct option.

If $\dfrac{x^2 - 4}{x^2 + 4} = \dfrac{3}{5}$, the value of x is :

1. 4

2. $\pm 4$

3. $\dfrac{1}{4}$

4. $\pm \dfrac{1}{4}$

Answer

Given,

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

Hence, Option 2 is the correct option.

If $\dfrac{x}{a + b - c} = \dfrac{y}{b + c - a} = \dfrac{z}{c + a - b}$ = 5 and a + b + c = 7; the value of x + y + z is :

1. 35

2. $\dfrac{7}{5}$

3. $\dfrac{5}{7}$

4. 42

Answer

Given,

⇒ a + b + c = 7

⇒ a + b = 7 - c

⇒ b + c = 7 - a

⇒ c + a = 7 - b

Given,

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

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.

If a : b = c : d, prove that :

5a + 7b : 5a - 7b = 5c + 7d : 5c - 7d

Answer

Given,

⇒ a : b = c : d

$\therefore \dfrac{a}{b} = \dfrac{c}{d} \\[1em]$

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

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

Applying componendo and dividendo:

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

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

If a : b = c : d, prove that :

xa + yb : xc + yd = b : d

Answer

Given,

a : b = c : d

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

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

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

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

On cross-multiplication:

$\Rightarrow \dfrac{xa + yb}{xc + yd} = \dfrac{yb}{yd} \\[1em] \Rightarrow \dfrac{xa + yb}{xc + yd} = \dfrac{b}{d}.$

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

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

prove that a : b = c : d.

Answer

Given,

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

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

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

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

If x = $\dfrac{6ab}{a + b}$, find the value of :

$\dfrac{x + 3a}{x - 3a} + \dfrac{x + 3b}{x - 3b}$.

Answer

(i) Given,

$\Rightarrow x = \dfrac{6ab}{a + b} \\[1em] \Rightarrow \dfrac{x}{3a} = \dfrac{2b}{a + b}$

Applying componendo and dividendo:

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

Again,

$\Rightarrow x = \dfrac{6ab}{a + b} \\[1em] \Rightarrow \dfrac{x}{3b} = \dfrac{2a}{a + b}$

Applying componendo and dividendo:

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

Adding (i) and (ii) we get,

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

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

If a = $\dfrac{4\sqrt{6}}{\sqrt{2} + \sqrt{3}}$, find the value of :

$\dfrac{a + 2\sqrt{2}}{a - 2\sqrt{2}} + \dfrac{a + 2\sqrt{3}}{a - 2\sqrt{3}}$.

Answer

Given,

$\Rightarrow a = \dfrac{4\sqrt{6}}{\sqrt{2} + \sqrt{3}} \\[1em] \Rightarrow \dfrac{a}{2\sqrt{2}} = \dfrac{2\sqrt{3}}{\sqrt{2} + \sqrt{3}}$

Applying componendo and dividendo:

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

Again,

$\Rightarrow a = \dfrac{4\sqrt{6}}{\sqrt{2} + \sqrt{3}} \\[1em] \Rightarrow \dfrac{a}{2\sqrt{3}} = \dfrac{2\sqrt{2}}{\sqrt{2} + \sqrt{3}}$

Applying componendo and dividendo:

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

Adding (i) and (ii) we get,

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

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

If (a + b + c + d)(a - b - c + d) = (a + b - c - d)(a - b + c - d); prove that a : b = c : d.

Answer

Given,

$\Rightarrow (a + b + c + d)(a - b - c + d) = (a + b - c - d)(a - b + c - d) \\[1em] \Rightarrow \dfrac{a + b + c + d}{a + b - c - d} = \dfrac{a - b + c - d}{a - b - c + d}$

Applying componendo and dividendo, we get :

$\Rightarrow \dfrac{(a + b + c + d) + (a + b - c - d)}{(a + b + c + d) - (a + b - c - d)} = \dfrac{(a - b + c - d) + (a - b - c + d)}{(a - b + c - d) - (a - b - c + d)} \\[1em] \Rightarrow \dfrac{a + a + b + b + c - c + d - d}{a - a + b - b + c + c + d + d} = \dfrac{a + a - b - b + c - c - d + d}{a - a - b + b + c + c - d - d} \\[1em] \Rightarrow \dfrac{2(a + b)}{2(c + d)} = \dfrac{2(a - b)}{2(c - d)} \\[1em] \Rightarrow \dfrac{(a + b)}{(c + d)} = \dfrac{(a - b)}{(c - d)} \\[1em] \Rightarrow \dfrac{a + b}{a - b} = \dfrac{c + d}{c - d}$

Again applying componendo and dividendo, we get :

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

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

If $\dfrac{a - 2b - 3c + 4d}{a + 2b - 3c - 4d} = \dfrac{a - 2b + 3c - 4d}{a + 2b + 3c + 4d}$

show that : 2ad = 3bc.

Answer

Given,

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

Applying componendo and dividendo:

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

Applying alternendo:

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

Hence, proved that 2ad = 3bc.

If a, b and c are in continued proportion, prove that :

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

Answer

Given,

a, b and c are in continued proportion

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

To prove :

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

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

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

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

$\Rightarrow \dfrac{ck^2}{c} \\[1em] \Rightarrow k^2.$

Since, L.H.S. = R.H.S. = k².

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

Using properties of proportion, solve for x :

$\dfrac{\sqrt{x + 5} + \sqrt{x - 16}}{\sqrt{x + 5} - \sqrt{x - 16}} = \dfrac{7}{3}.$

Answer

Given,

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

Applying componendo and dividendo:

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

Hence, x = 20.

Using properties of proportion, solve for x :

$\dfrac{3x + \sqrt{9x^2 - 5}}{3x - \sqrt{9x^2 - 5}} = 5$

Answer

Applying componendo and dividendo,

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

Squaring both sides of the equation,

$\Rightarrow 9x^2 - 5 = 4x^2 \\[1em] \Rightarrow 9x^2 - 4x^2 = 5 \\[1em] \Rightarrow 5x^2 = 5 \\[1em] \Rightarrow x^2 = 1 \\[1em] \Rightarrow x = \pm1.$

Hence, x = 1.

If x = $\dfrac{\sqrt{a + 3b} + \sqrt{a - 3b}}{\sqrt{a + 3b} - \sqrt{a - 3b}}$, prove that :

3bx² - 2ax + 3b = 0

Answer

Given,

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

Applying componendo and dividendo,

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

Squaring both sides:

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

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

Using the properties of proportion, solve for x,

given $\dfrac{x^4 + 1}{2x^2} = \dfrac{17}{8}$

Answer

Given,

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

Applying componendo and dividendo, we get

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

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

Applying componendo and dividendo, we get

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

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

Applying componendo and dividendo, we get

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

Hence, x = ±2 and $\pm \dfrac{1}{2}$.

If $x = \dfrac{\sqrt{m + n} + \sqrt{m - n}}{\sqrt{m + n} - \sqrt{m - n}}$, express n in terms of x and m.

Answer

Given,

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

Applying componendo and dividendo,

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

Applying componendo and dividendo,

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

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

If $\dfrac{x^3 + 3xy^2}{3x^2y + y^3} = \dfrac{m^3 + 3mn^2}{3m^2n + n^3}$,

show that : nx = my.

Answer

Given,

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

Applying componendo and dividendo,

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

Applying componendo and dividendo,

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

Hence, proved that nx = my.

From the following table, find the values of a, b and c:

Answer

It is a case of direct variation:

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

Taking,

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

⇒ 10 x 100 = 40 x a

⇒ a = $\dfrac{1000}{40}$

⇒ a = 25 m

Taking,

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

⇒ 10 x b = 40 x 40

⇒ b = $\dfrac{1600}{10}$

⇒ b = ₹ 160

Taking,

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

⇒ 10 x 180 = 40 x c

⇒ c = $\dfrac{1800}{40}$

⇒ c = 45 m

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

From the following table, find the values of a, b and c:

Answer

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 = $\dfrac{70 \times 39}{130}$

⇒ a = 7 x 3

⇒ a = 21

Taking,

70 x 39 = b x 30

⇒ b = $\dfrac{70 \times 39}{30}$

⇒ b = 7 x 13

⇒ b = 91

Taking,

70 x 39 = 120 x c

⇒ c = $\dfrac{70 \times 39}{120}$

⇒ c = 22.75

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