Ratio & Proportion

(i) Given,

60 paise and ₹ 1.35

₹ 1 = 100 paise

₹ 1.35 = 135 paise

Required ratio = $\frac{60}{135} = \frac{4}{9}$

= 4 : 9

Hence, ratio = 4 : 9.

(ii) Given,

1.8 m and 75 cm

1 m = 100 cm

1.8 m = 180 cm

Required ratio = $\frac{180}{75} = \frac{12}{5}$

= 12 : 5.

Hence, ratio = 12 : 5.

(iii) Given,

1.5 kg and 600 g

1 kg = 1000 g

1.5 kg = 1500 g

Required ratio = $\frac{1500}{600} = \frac{5}{2}$

= 5 : 2.

Hence, ratio = 5 : 2.

(iv) Given,

35 min and $1\frac{3}{4}$ hrs

1 hour = 60 min

$\Rightarrow 1\frac{3}{4} = \frac{7}{4}$ hrs

= $\frac{7}{4} \times 60$ = 105 mins.

Required ratio = $\frac{35}{105} = \frac{1}{3}$.

Hence, ratio = 1 : 3.

Given,

A : B = 5 : 6

B : C = 9 : 11

Taking L.C.M of two values of B i.e. 6 and 9 = 18.

So, $\frac{A}{B} = \frac{5 \times 3}{6 \times 3} = \frac{15}{18}$ = 15 : 18

and $\frac{B}{C} = \frac{9 \times 2}{11 \times 2} = \frac{18}{22}$ = 18 : 22

∴ A : B : C = 15 : 18 : 22

⇒ A : C = 15 : 22.

(i)

Hence, A : C = 15 : 22.

(ii)

Hence, A : B : C = 15 : 18 : 22.

Given,

P : Q = 7 : 4

Q : R = 5 : 14

To find R : P and P : Q : R, we will make Q same in both cases.

Taking L.C.M of two values of Q i.e. 4 and 5 = 20

So, $\frac{P}{Q} = \frac{7 \times 5}{4 \times 5} = \frac{35}{20}$ = 35 : 20

and $\frac{Q}{R} = \frac{5 \times 4}{14 \times 4} = \frac{20}{56}$ = 20 : 56

$$\begin{aligned}\Rightarrow \frac{P}{Q} \times \frac{Q}{R} = \frac{35}{20} \times \frac{20}{56} \\[1em] \Rightarrow \frac{P}{R} = \frac{35}{56} \\[1em] \Rightarrow \frac{P}{R} = \frac{5}{8} \\[1em] \Rightarrow \frac{R}{P} = \frac{8}{5}.\end{aligned}$$

Hence, R : P = 8 : 5.

∴ P : Q : R = 35 : 20 : 56

Hence, ratio of P : Q : R = 35 : 20 : 56.

Let 3A = 5B = 6C = k.

$A = \frac{k}{3}, B = \frac{k}{5}, C = \frac{k}{6}$

$$\begin{aligned}\Rightarrow A : B : C = \frac{k}{3} : \frac{k}{5} : \frac{k}{6} \\[1em] \Rightarrow A : B : C = \frac{1}{3} : \frac{1}{5} : \frac{1}{6}\end{aligned}$$

Taking L.C.M of values of 3, 5, 6 = 30.

$$\begin{aligned}\Rightarrow A : B : C = \frac{1}{3} \times 30 : \frac{1}{5} \times 30 : \frac{1}{6} \times 30 \\[1em] \Rightarrow A : B : C = 10 : 6 : 5.\end{aligned}$$

Hence, A : B : C = 10 : 6 : 5.

Let, $\frac{A}{2} = \frac{B}{3} = \frac{C}{6}$ = k.

∴ A = 2k, B = 3k, C = 6k.

A : B : C = 2k : 3k : 6k

A : B : C = 2 : 3 : 6.

Hence, A : B : C = 2 : 3 : 6.

Given,

a : b = 8 : 5

Let a = 8x and b = 5x.

Substituting values in $\frac{7a + 5b}{8a − 9b}$ we get,

$$\begin{aligned}\Rightarrow \frac{7(8x) + 5(5x)}{8(8x) - 9(5x)} \\[1em] \Rightarrow \frac{56x + 25x}{64x - 45x} \\[1em] \Rightarrow \frac{81x}{19x} \\[1em] \Rightarrow \frac{81}{19}.\end{aligned}$$

Hence, 7a + 5b : 8a − 9b = 81 : 19.

Given,

x : y = 3 : 2

Let x = 3a and y = 2a.

Substituting values in $\frac{5x − 3y}{7x + 2y}$ we get,

$$\begin{aligned}\Rightarrow \frac{5(3a) - 3(2a)}{7(3a) + 2(2a)} \\[1em] \Rightarrow \frac{15a - 6a}{21a + 4a} \\[1em] \Rightarrow \frac{9a}{25a} \\[1em] \Rightarrow \frac{9}{25}.\end{aligned}$$

Hence, 5x − 3y : 7x + 2y = 9 : 25.

Given,

x : y = 10 : 3

Let x = 10a and y = 3a.

Substituting values in $\frac{3x^2 + 2y^2}{3x^2 − 2y^2}$ we get,

$$\begin{aligned}\Rightarrow \frac{3(10a)^2 + 2(3a)^2}{3(10a)^2 - 2(3a)^2} \\[1em] \Rightarrow \frac{3(100a^2) + 2(9a^2)}{3(100a^2) - 2(9a^2)} \\[1em] \Rightarrow \frac{300a^2 + 18a^2}{300a^2 - 18a^2} \\[1em] \Rightarrow \frac{318a^2}{282a^2} \\[1em] \Rightarrow \frac{53}{47}.\end{aligned}$$

Hence, (3x² + 2y²) : (3x² − 2y²) = 53 : 47.

Given,

a : b = 2 : 5

Let a = 2x, then b = 5x.

Substituting values in $\frac{3a^2 − 2ab + 5b^2}{a^2 + 7ab − 2b^2}$ we get,

$$\begin{aligned}\Rightarrow \frac{3 \times (2x)^2 − 2 \times (2x) \times (5x) + 5 \times (5x)^2}{(2x)^2 + 7 \times (2x) \times (5x) − 2 \times (5x)^2} \\[1em] \Rightarrow \frac{12x^2 − 20x^2 + 125x^2}{4x^2 + 70x^2 − 50x^2} \\[1em] \Rightarrow \frac{137x^2 − 20x^2}{74x^2 − 50x^2} \\[1em] \Rightarrow \frac{117x^2}{24x^2} \\[1em] \Rightarrow \frac{39}{8}.\end{aligned}$$

Hence, 3a² − 2ab + 5b² : a² + 7ab − 2b² = 39 : 8 .

Given,

(5x + 2y) : (7x + 4y) = 13 : 20,

$$\begin{aligned}\Rightarrow \frac{5x + 2y}{7x + 4y} = \frac{13}{20} \\[1em] \Rightarrow 20(5x + 2y) = 13(7x + 4y) \\[1em] \Rightarrow 100x + 40y = 91x + 52y \\[1em] \Rightarrow 100x - 91x = 52y - 40y \\[1em] \Rightarrow 9x = 12y \\[1em] \Rightarrow \frac{x}{y} = \frac{12}{9} \\[1em] \Rightarrow \frac{x}{y} =\frac{4}{3}.\end{aligned}$$

Hence, x : y = 4 : 3.

Given,

(3x + 5y) : (3x − 5y) = 7 : 3,

$$\begin{aligned}\Rightarrow \frac{3x + 5y}{3x - 5y} = \frac{7}{3} \\[1em] \Rightarrow 3(3x + 5y) = 7(3x - 5y) \\[1em] \Rightarrow 9x + 15y = 21x - 35y \\[1em] \Rightarrow 15y + 35y = 21x - 9x \\[1em] \Rightarrow 50y = 12x \\[1em] \Rightarrow \frac{x}{y} = \frac{50}{12} \\[1em] \Rightarrow \frac{x}{y} = \frac{25}{6}.\end{aligned}$$

Hence, x : y = 25 : 6.

Given,

(6x² − xy) : (2xy − y²) = 6 : 1,

Dividing numerator and denominator by y², we get :

$$\begin{aligned}\Rightarrow \frac{\frac{6x^2 − xy}{y^2}}{\frac{2xy − y^2}{y^2}} = \frac{6}{1} \\[1em] \Rightarrow \frac{6\Big(\frac{x}{y}\Big)^2 - \frac{x}{y}}{2\frac{x}{y} - 1} = 6\end{aligned}$$

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

$$\begin{aligned}\Rightarrow \frac{6t^2 - t}{2t - 1} = 6 \\[1em] \Rightarrow 6t^2 - t = (2t - 1) \times 6 \\[1em] \Rightarrow 6t^2 - t = 12t - 6 \\[1em] \Rightarrow 6t^2 - t - 12t + 6 = 0 \\[1em] \Rightarrow 6t^2 - 13t + 6 = 0 \\[1em] \Rightarrow 6t^2 - 4t - 9t + 6 = 0 \\[1em] \Rightarrow 2t(3t - 2) - 3(3t - 2) = 0 \\[1em] \Rightarrow (2t - 3)(3t - 2) = 0 \\[1em] \Rightarrow (2t - 3) = 0 \text{ or }(3t - 2) = 0 \text{ [Using zero - product rule] } \\[1em] \Rightarrow 2t = 3 \text{ or } 3t = 2 \\[1em] \Rightarrow t = \frac{3}{2} \text{ or } t = \frac{2}{3} .\end{aligned}$$

Thus,

$\frac{x}{y} = \frac{3}{2} \text{ or } \frac{x}{y} = \frac{2}{3}$.

Hence, x : y = 3 : 2 or 2 : 3.

Given,

(4x² − 3y²) : (2x² + 5y²) = 12 : 19,

Dividing numerator and denominator both by y² we get,

$$\begin{aligned}\Rightarrow \frac{\frac{4x^2 − 3y^2}{y^2}}{\frac{2x^2 + 5y^2}{y^2}} = \frac{12}{19} \\[1em] \Rightarrow \frac{4\Big(\frac{x}{y}\Big)^2 - 3\frac{y^2}{y^2}}{2\Big(\frac{x}{y}\Big)^2 + 5\frac{y^2}{y^2}} = \frac{12}{19} \\[1em] \Rightarrow \frac{4\Big(\frac{x}{y}\Big)^2 - 3}{2\Big(\frac{x}{y}\Big)^2 + 5} = \frac{12}{19}\end{aligned}$$

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

$$\begin{aligned}\Rightarrow \frac{4t^2 - 3}{2t^2 + 5} = \frac{12}{19} \\[1em] \Rightarrow (4t^2 - 3) \times 19 = (2t^2 + 5) \times 12 \\[1em] \Rightarrow 76t^2 - 57 = 24t^2 + 60 \\[1em] \Rightarrow 76t^2 - 24t^2 = 60 + 57 \\[1em] \Rightarrow 52t^2 = 117 \\[1em] \Rightarrow t^2 = \frac{117}{52} \\[1em] \Rightarrow t^2 = \frac{9}{4} \\[1em] \Rightarrow t = \sqrt{\frac{9}{4}} \\[1em] \Rightarrow t = \frac{3}{2}.\end{aligned}$$

Hence, x : y = 3 : 2.

Given,

x² + 4y² = 4xy

Dividing both sides by xy we get,

$$\begin{aligned}\Rightarrow \frac{x^2 + 4y^2}{xy} = \frac{4xy}{xy} \\[1em] \Rightarrow \frac{x}{y} + 4\frac{y}{x} = 4\end{aligned}$$

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

$$\begin{aligned}\Rightarrow t + 4\frac{1}{t} = 4 \\[1em] \Rightarrow \frac{t^2 + 4}{t} = 4 \\[1em] \Rightarrow t^2 + 4 = 4t \\[1em] \Rightarrow t^2 - 4t + 4 = 0 \\[1em] \Rightarrow t^2 - 2t - 2t + 4 = 0 \\[1em] \Rightarrow t(t - 2) - 2(t - 2) = 0 \\[1em] \Rightarrow (t - 2)(t - 2) = 0 \\[1em] \Rightarrow t - 2 = 0 \text{ or } t - 2 = 0 \\[1em] \Rightarrow t = 2 \\[1em] \Rightarrow \frac{x}{y} = 2.\end{aligned}$$

Hence, x : y = 2 : 1.

Given,

10x² − 23xy + 9y² = 0

Dividing both sides by xy we get,

$$\begin{aligned}\Rightarrow \frac{10x^2 − 23xy + 9y^2}{xy} = 0 \\[1em] \Rightarrow 10\frac{x}{y} - 23 + 9\frac{y}{x} = 0 \\[1em] \Rightarrow 10\frac{x}{y} + 9\frac{y}{x} = 23\end{aligned}$$

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

$$\begin{aligned}\Rightarrow 10t + 9\frac{1}{t} = 23 \\[1em] \Rightarrow \frac{10t^2 + 9}{t} = 23 \\[1em] \Rightarrow 10t^2 + 9 = 23t \\[1em] \Rightarrow 10t^2 - 23t + 9 = 0 \\[1em] \Rightarrow 10t^2 - 5t - 18t + 9 = 0 \\[1em] \Rightarrow 5t(2t - 1) - 9(2t - 1) = 0 \\[1em] \Rightarrow (5t - 9)(2t - 1) = 0 \\[1em] \Rightarrow 5t - 9 = 0 \text{ or } 2t - 1 = 0 \\[1em] \Rightarrow 5t = 9 \text{ or } 2t = 1 \\[1em] \Rightarrow t = \frac{9}{5} \text{ or } t = \frac{1}{2} \\[1em] \Rightarrow \frac{x}{y} = \frac{9}{5} \text{ or } \frac{x}{y} = \frac{1}{2}\end{aligned}$$

Hence, x : y = 9 : 5 or 1 : 2.

Let the antecedent be x. Then,

3 : 4 = x : 144

$$\begin{aligned}\Rightarrow \frac{3}{4} = \frac{x}{144} \\[1em] \Rightarrow 3 \times 144 = 4x \\[1em] \Rightarrow 432 = 4x \\[1em] \Rightarrow x = \frac{432}{4} \\[1em] \Rightarrow x = 108\end{aligned}$$

Hence, the antecedent = 108.

Let the numbers be 8x and 13x.

Given,

If 14 is added to each of the numbers, the ratio becomes 2 : 3.

$$\begin{aligned}\Rightarrow \frac{8x + 14}{13x + 14} = \frac{2}{3} \\[1em] \Rightarrow (8x + 14) \times 3 = (13x + 14) \times 2 \\[1em] \Rightarrow 24x + 42 = 26x + 28 \\[1em] \Rightarrow 26x - 24x = 42 - 28 \\[1em] \Rightarrow 2x = 14 \\[1em] \Rightarrow x = \frac{14}{2} \\[1em] \Rightarrow x = 7.\end{aligned}$$

The two numbers are 8x and 13x

= 8 × 7, 13 × 7

= 56, 91.

Hence, the numbers are 56 and 91.

Let the numbers be 5x and 7x.

Given,

If 8 is subtracted from each, the ratio becomes 3 : 5.

$$\begin{aligned}\Rightarrow \frac{5x - 8}{7x - 8} = \frac{3}{5} \\[1em] \Rightarrow (5x - 8) \times 5 = (7x - 8) \times 3 \\[1em] \Rightarrow 25x - 40 = 21x - 24\\[1em] \Rightarrow 25x - 21x = 40 - 24 \\[1em] \Rightarrow 4x = 16 \\[1em] \Rightarrow x = \frac{16}{4} \\[1em] \Rightarrow x = 4.\end{aligned}$$

The two numbers are 5x and 7x

= 5 × 4, 7 × 4

= 20, 28.

Hence, the numbers are 20 and 28.

Let x be the least number must be added to each term of the ratio 5 : 7 to make it 8 : 9.

$$\begin{aligned}\Rightarrow \frac{5 + x}{7 + x} = \frac{8}{9} \\[1em] \Rightarrow (5 + x) \times 9 = (7 + x) \times 8 \\[1em] \Rightarrow 45 + 9x = 56 + 8x \\[1em] \Rightarrow 9x - 8x = 56 - 45 \\[1em] \Rightarrow x = 11.\end{aligned}$$

Hence, the least number to be added is 11.

Given,

Income = ₹ 45,000

Expenditure = ₹ 31,500

Total savings = Income - Expenditure = ₹ 45,000 - 31,500

= ₹ 13,500.

(i) Required ratio,

Income : Expenditure = 45000 : 31500

$= \frac{45000}{31500} = \frac{10}{7}$

= 10 : 7.

Hence, required ratio = 10 : 7.

(ii) Required ratio,

Income : Savings = 45000 : 13500

$= \frac{45000}{13500} = \frac{10}{3}$

= 10 : 3.

Hence, required ratio = 10 : 3.

(iii) Required ratio is savings to expenditure.

Savings : expenditure = 13500 : 31500

$=\frac{13500}{31500} = \frac{3}{7}$

= 3 : 7.

Hence, required ratio = 3 : 7.

Given,

The ratio of material : labour : overheads = 6 : 4 : 1

Let the cost of material be 6x, labour be 4x and overheads be x.

Given,

Cost of material = ₹ 132

⇒ 6x = 132

⇒ x = $\frac{132}{6}$

⇒ x = 22.

The total cost of production of an umbrella = 6x + 4x + 1x = 11x

= 11 × 22

= ₹ 242.

Hence, cost of production of an umbrella = ₹ 242.

Given,

Let ₹ 6,720 be divided in two parts A and B.

A = 5a and B = 3a

To find A's part,

$$\begin{aligned}\Rightarrow \frac{5a}{5a + 3a} \times 6,720 \\[1em] \Rightarrow \frac{5a}{8a} \times 6,720 \\[1em] \Rightarrow \frac{5}{8} \times 6,720 \\[1em] \Rightarrow ₹ 4,200.\end{aligned}$$

To find B's part,

$$\begin{aligned}\Rightarrow \frac{3a}{5a + 3a} \times 6,720 \\[1em] \Rightarrow \frac{3a}{8a} \times 6,720 \\[1em] \Rightarrow \frac{3}{8} \times 6,720 \\[1em] \Rightarrow ₹ 2,520\end{aligned}$$

Hence, ₹ 6,720 can be divided into ₹ 4,200 and ₹ 2,520.

Given,

Let A = 35a and B = 28a and C = 20a,

To find A's part,

$$\begin{aligned}\Rightarrow \frac{35a}{35a + 28a + 20a} \times 11,620 \\[1em] \Rightarrow \frac{35a}{83a} \times 11,620 \\[1em] \Rightarrow \frac{35}{83} \times 11,620 \\[1em] \Rightarrow ₹ 4,900.\end{aligned}$$

To find B's part,

$$\begin{aligned}\Rightarrow \frac{28a}{35a + 28a + 20a} \times 11,620 \\[1em] \Rightarrow \frac{28a}{83a} \times 11,620 \\[1em] \Rightarrow \frac{28}{83} \times 11,620 \\[1em] \Rightarrow ₹ 3,920.\end{aligned}$$

To find C's part,

$$\begin{aligned}\Rightarrow \frac{20a}{35a + 28a + 20a} \times 11,620 \\[1em] \Rightarrow \frac{20a}{83a} \times 11,620 \\[1em] \Rightarrow \frac{20}{83} \times 11,620 \\[1em] \Rightarrow ₹ 2,800\end{aligned}$$

Hence, A = ₹ 4,900, B = ₹ 3,920 and C = ₹ 2,800.

Given,

P : Q : R = $\frac{1}{2} : \frac{2}{3} : \frac{3}{4}$.

Multiply each ratio by 12 (LCM of denominators) to clear fractions :

= $\frac{1}{2} \times 12 : \frac{2}{3} \times 12 : \frac{3}{4} \times 12$

= 6 : 4 : 9.

Let P = 6a and Q = 8a and R = 9a.

To find P's part,

$$\begin{aligned}\Rightarrow \frac{6a}{6a + 8a + 9a} \times 782 \\[1em] \Rightarrow \frac{6a}{23a} \times 782 \\[1em] \Rightarrow \frac{6}{23} \times 782 \\[1em] \Rightarrow 204\end{aligned}$$

To find Q's part,

$$\begin{aligned}\Rightarrow \frac{8a}{6a + 8a + 9a} \times 782 \\[1em] \Rightarrow \frac{8a}{23a} \times 782 \\[1em] \Rightarrow \frac{8}{23} \times 782 \\[1em] \Rightarrow 272\end{aligned}$$

To find Q's part,

$$\begin{aligned}\Rightarrow \frac{9a}{6a + 8a + 9a} \times 782 \\[1em] \Rightarrow \frac{9a}{23a} \times 782 \\[1em] \Rightarrow \frac{9}{23} \times 782 \\[1em] \Rightarrow 306\end{aligned}$$

Hence, ₹ 782 can be divided into P = ₹ 204, Q = ₹ 272 and R = ₹ 306.

Given,

A = $\frac{2}{3}$ of B and B = $\frac{1}{4}$ of C

Express all in terms of B:

A = $\frac{2}{3}B$, B = B and C = 4B

Total amount divided among A, B and C = ₹ 5,100

So,

$$\begin{aligned}\Rightarrow \frac{2}{3}B + B + 4B = 5100 \\[1em] \Rightarrow \frac{2B + 3B + 12B}{3} = 5100 \\[1em] \Rightarrow \frac{17B}{3} = 5100 \\[1em] \Rightarrow B = \frac{5100 \times 3}{17} \\[1em] \Rightarrow B = ₹ 900.\end{aligned}$$

B's share = ₹ 900

Therefore,

A's share = $\frac{2}{3} \times 900$ = ₹ 600

C's share = 4 × 900 = ₹ 3,600

Hence, A = ₹ 600, B = ₹ 900 and C = ₹ 3,600.

Given,

Let, 4A = 5B = 7C = k

Then, A = $\frac{k}{4}$, B = $\frac{k}{5}$, C = $\frac{k}{7}$

Total amount divided among A, B and C = ₹ 8,300

So,

$$\begin{aligned}\Rightarrow \frac{k}{4} + \frac{k}{5} + \frac{k}{7} = 8300 \\[1em] \Rightarrow k\Big(\frac{1}{4} + \frac{1}{5} + \frac{1}{7}\Big) = 8300 \\[1em] \Rightarrow k\Big(\frac{35 + 28 + 20}{140}\Big) = 8300 \\[1em] \Rightarrow k\Big(\frac{83}{140}\Big) = 8300 \\[1em] \Rightarrow k = \frac{8300 \times 140}{83} \\[1em] \Rightarrow k = 14000\end{aligned}$$

Therefore,

A' share = $\frac{14000}{4}$ = ₹ 3,500

B's share = $\frac{14000}{5}$ = ₹ 2,800

C"s share = $\frac{14000}{7}$ = ₹ 2,000

Hence, A = ₹ 3,500, B = ₹ 2,800 and C = ₹ 2,000.

(i) Let A's share be ₹ x.

Then,

$$\begin{aligned}\Rightarrow \frac{6}{11} = \frac{x}{7315} \\[1em] \Rightarrow x = \frac{6}{11} \times 7315 \\[1em] \Rightarrow x = ₹ 3,990.\end{aligned}$$

Hence, A' share of money = ₹ 3,990.

(ii) Total sum of money = A's share + B's share

= ₹ 3,990 + ₹ 7,315

= ₹ 11,305.

Hence, total sum of money = ₹ 11,305.

Given,

The ages of Tanvy and Divya are in the ratio 5 : 7.

Let the present age of Tanvy be 5x and the present age of Divya be 7x.

Five years from now :

Tanvy's age will be 5x + 5.

Divya's age will be 7x + 5.

Given,

The ratio of their ages in five years will be 3 : 4.

$\Rightarrow \frac{5x + 5}{7x + 5} = \frac{3}{4}$

⇒ 4(5x + 5) = 3(7x + 5)

⇒ 20x + 20 = 21x + 15

⇒ 21x - 20x = 20 - 15

⇒ x = 20 - 15

⇒ x = 5.

Tanvy's present age = 5x = 5(5) = 25 years.

Divya's present age = 7x = 7(5) = 35 years.

Hence, present age of Tnavy and Divya are 25 years and 35 years respectively.

Given,

The ratio of Amit’s and Arun’s ages was 6 : 7.

Let Amit's age one year ago be 6x and Arun's age one year ago be 7x.

Their present ages are :

Amit's present age = 6x + 1

Arun's present age = 7x + 1

After four years their ages will be:

Amit's age will be (6x + 1) + 4 = 6x + 5

Arun's age will be (7x + 1) + 4 = 7x + 5

Given,

The ratio of their ages four years hence will be 7 : 8.

$\therefore \frac{6x + 5}{7x + 5} = \frac{7}{8}$

⇒ 8(6x + 5) = 7(7x + 5)

⇒ 48x + 40 = 49x + 35

⇒ 40 - 35 = 49x - 48x

⇒ x = 5

Amit's present age is 6x + 1.

= 6(5) + 1 = 30 + 1

= 31 years.

Hence, Amit's present age = 31 years.

Given,

Reena reduces her weight in the ratio 5 : 4

Let her orignal weight be 5x and new weight be 4x.

Reena's orignal weight = 70 kg

⇒ 5x = 70

⇒ x = $\frac{70}{5}$

⇒ x = 14 kg

Reena's new weight = 4x

= 4 × 14

= 56 kg.

Hence, Reena's present weight = 56 kg.

Given,

Milk : Water = 27 : 7 and total = 68 kg

Milk = $\frac{27}{27 + 7} \times 68 = \frac{27}{34} \times 68 = 54\text{ kg}$

Water = $\frac{7}{27 + 7} \times 68 = \frac{7}{34} \times 68 = 14\text{ kg}$

Let x kg of water be added to make the ratio of milk to water as 3 : 1.

$$\begin{aligned}\Rightarrow \frac{54}{14 + x} = \frac{3}{1} \\[1em] \Rightarrow 54 = 3(14 + x) \\[1em] \Rightarrow 54 = 42 + 3x \\[1em] \Rightarrow 54 - 42 = 3x \\[1em] \Rightarrow 12 = 3x \\[1em] \Rightarrow x = 4.\end{aligned}$$

Hence, 4 kg of water must be added.

Given,

Milk : Water = 5 : 1

Let Milk = 5x and Water = x

After adding 5 litres of water, the ratio of milk to water becomes 5 : 2.

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

Milk in the original mixture = 5x = 5 × 5 = 25 litres.

Hence, the quantity of milk in the original mixture = 25 litres.

Given,

Ratio of Passes to Failures = 4 : 1

Let students who Pass = 4k and Fail = k

Total students appeared for examination = 4k + k = 5k

Given,

30 students didn't appear for examination.

Now total students appeared for exam = 5k − 30

If 20 less students passed the exam, then students that pass the exam now = 4k − 20

Number of students that fail in exam = (5k − 30) − (4k − 20) = k − 10.

The new ratio of passes to failures = 5 : 1

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

Total students who appeared exam = 5k = 5(30) = 150.

Hence, total number of students appeared for the examination = 150.

Given,

Let the angles of triangle be 5x, 4x and 3x.

We know that,

Sum of angles of a triangle = 180°

$$\begin{aligned}\Rightarrow 5x + 4x + 3x = 180 \\[1em] \Rightarrow 12x = 180 \\[1em] \Rightarrow x = \frac{180}{12} = 15.\end{aligned}$$

⇒ 5x = 5(15) = 75°

⇒ 4x = 4(15) = 60°

⇒ 3x = 3(15) = 45°

Hence, the angles of the triangle are 75°, 60° and 45°.

Given,

Side₁ : Side₂ : Side₃ = $\frac{1}{2} : \frac{1}{3} : \frac{1}{4}$

= $\frac{1}{2} \times 12 : \frac{1}{3} \times 12 : \frac{1}{4} \times 12$

= 6 : 4 : 3.

Let Side₁ = 6a and Side₂ = 4a and Side₃ = 3a

Sum of sides of triangle = 6a + 4a + 3a = 13a

Given,

Perimeter of triangle = 91 cm

⇒ 13a = 91

⇒ a = $\frac{91}{13}$

⇒ a = 7.

Therefore,

Length of Side₁ = 6a = 6(7) = 42 cm

Length of Side₂ = 4a = 4(7) = 28 cm

Length of Side₃ = 3a = 3(7) = 21 cm

Hence, the lengths of the sides are 42 cm, 28 cm and 21 cm.

Given,

Boys : Girls = 9 : 5

Let number of Boys = 9x and Girls = 5x.

Total number of girls in school = 425

So,

⇒ 5x = 425

⇒ x = $\frac{425}{5}$

⇒ x = 85.

Therefore,

Number of Boys in school = 9x = 9(85) = 765.

Total number of students in school = 765 + 425 = 1190.

Hence, the total number of students in the school = 1190.

(i) To compare 2 ratios, the consequent of the first ratio and 2nd ratio must be made equal.

Given,

A : B = 7 : 9 and C : D = 11: 16

L.C.M. of 9 and 16 is 144.

$$\begin{aligned}\Rightarrow \frac{A}{B} = \frac{7 \times 16}{9 \times 16} = \frac{112}{144} \\[1em] \Rightarrow \frac{C}{D} = \frac{11 \times 9}{16 \times 9} = \frac{99}{144} \\[1em] \Rightarrow \frac{112}{144} \gt \frac{99}{144} \\[1em] \Rightarrow \frac{7}{9} \gt \frac{11}{16}\end{aligned}$$

Hence, 7 : 9 > 11 : 16.

(ii) To compare 2 ratios, the consequent of the first ratio and 2nd ratio must be made equal.

Let A : B = 19 : 25 and C : D = 17 : 20.

L.C.M. of 25 and 20 is 100.

$$\begin{aligned}\Rightarrow \frac{A}{B} = \frac{19 \times 4}{25 \times 4} = \frac{76}{100} \\[1em] \Rightarrow \frac{C}{D} = \frac{17 \times 5}{20 \times 5} = \frac{85}{100} \\[1em] \Rightarrow \frac{76}{100} \lt \frac{85}{100} \\[1em] \Rightarrow \frac{19}{25} \lt \frac{17}{20}\end{aligned}$$

Hence, 19 : 25 < 17 : 20.

(ii) To compare 2 ratios, the consequent of the first ratio and 2nd ratio must be made equal.

Let A : B = $\Big(\frac{1}{2} : \frac{1}{5}\Big)$ and C : D = 5:2

First simplifying A : B,

L.C.M of 2 and 5 is 10.

$$\begin{aligned}\Rightarrow \Big(\frac{1}{2} \times 10 : \frac{1}{5} \times 10\Big) \\[1em] \Rightarrow 5:2\end{aligned}$$

Since both ratios are same.

Hence, $\Big(\frac{1}{2} : \frac{1}{5}\Big)$ = 5 : 2.

(i) Given,

(5 : 6), (8 : 9), (13 : 18) and (19 : 24)

We convert them into equivalent like fractions.

L.C.M of 6, 9, 18, 24 is 72.

$$\begin{aligned}\Rightarrow \frac{5 \times 12}{6 \times 12} = \frac{60}{72} \\[1em] \Rightarrow \frac{8 \times 8}{9 \times 8} = \frac{64}{72} \\[1em] \Rightarrow \frac{13 \times 4}{18 \times 4} = \frac{52}{72} \\[1em] \Rightarrow \frac{19 \times 3}{24 \times 3} = \frac{57}{72} \\[1em] \Rightarrow \frac{64}{72} \gt \frac{60}{72} \gt \frac{57}{72} \gt \frac{52}{72}\end{aligned}$$

8 : 9 > 5 : 6 > 19 : 24 > 13 : 18

Hence, the ratios in descending order are (8 : 9) > (5 : 6) > (19 : 24) > (13 : 18).

(ii) Given,

(6 : 7), (13 : 14), (19 : 21) and (23 : 28)

We convert them into equivalent like fractions.

L.C.M of 7, 14, 21, 28 is 84

$$\begin{aligned}\Rightarrow \frac{6 \times 12}{7 \times 12} = \frac{72}{84} \\[1em] \Rightarrow \frac{13 \times 6}{14 \times 6} = \frac{78}{84} \\[1em] \Rightarrow \frac{19 \times 4}{21 \times 4} = \frac{76}{84} \\[1em] \Rightarrow \frac{23 \times 3}{28 \times 3} = \frac{69}{84} \\[1em] \Rightarrow \frac{78}{84} \gt \frac{76}{84} \gt \frac{72}{84} \gt \frac{69}{84}\end{aligned}$$

(13 : 14) > (19 : 21) > (6 : 7) > (23 : 28).

Hence, the ratios in descending order are (13 : 14) > (19 : 21) > (6 : 7) > (23 : 28).

(iii) Given,

(7 : 12), (9 : 16), (13 : 20) and (5 : 8)

We convert them into equivalent like fractions.

L.C.M of 12, 16, 20, 8 is 240.

$$\begin{aligned}\Rightarrow \frac{7 \times 20}{12 \times 20} = \frac{140}{240} \\[1em] \Rightarrow \frac{9 \times 15}{16 \times 15} = \frac{135}{240} \\[1em] \Rightarrow \frac{13 \times 12}{20 \times 12} = \frac{156}{240} \\[1em] \Rightarrow \frac{5 \times 30}{8 \times 30} = \frac{150}{240} \\[1em] \Rightarrow \frac{156}{240} \gt \frac{150}{240} \gt \frac{140}{240} \gt \frac{135}{240}\end{aligned}$$

(13 : 20) > (5 : 8) > (7 : 12) > (9 : 16).

Hence, the ratios in descending order are (13 : 20) > (5 : 8) > (7 : 12) > (9 : 16).

(i) Given,

(4 : 9), (6 : 11), (7 : 13) and (27 : 50)

We convert them into decimals.

⇒ $\frac{4}{9}$ ≈ 0.444

⇒ $\frac{6}{11}$ ≈ 0.5455

⇒ $\frac{7}{13}$ ≈ 0.5385

⇒ $\frac{27}{50}$ ≈ 0.5400

⇒ 0.444 < 0.5385 < 0.5400 < 0.5455

⇒ (4 : 9) < (7 : 13) < (27 : 50) < (6 : 11).

Hence, the ratios in ascending order are (4 : 9), (7 : 13), (27 : 50), (6 : 11).

(ii) Given,

(2 : 3), (8 : 15), (11 : 12) and (7 : 16)

We convert them into decimals.

$\frac{2}{3}$ = 0.667

$\frac{8}{15}$ ≈ 0.533

$\frac{11}{12}$ ≈ 0.917

$\frac{7}{16}$ = 0.4375

0.4375 < 0.533 < 0.667 < 0.917

(7 : 16) < (8 : 15) < (2 : 3) < (11 : 12)

Hence, the ratios in ascending order are (7 : 16), (8 : 15), (2 : 3), (11 : 12).

(iii) Given,

(3 : 5), (4 : 9), (5 : 11) and (10 : 17)

We convert them into decimals.

$\frac{3}{5}$ = 0.600

$\frac{4}{9}$ ≈ 0.444

$\frac{5}{11}$ ≈ 0.455

$\frac{10}{17}$ = 0.588

⇒ 0.444 < 0.455 < 0.588 < 0.600

⇒ (4 : 9) < (5 : 11) < (10 : 17) < (3 : 5)

Hence, the ratios in ascending order are (4 : 9), (5 : 11), (10 : 17), (3 : 5).

Given,

(3a + 2b) : (5a + 3b) = 18 : 29

Solving,

$\frac{3a + 2b}{5a + 3b} = \frac{18}{29}$

⇒ 29(3a + 2b) = 18(5a + 3b)

⇒ 87a + 58b = 90a + 54b

⇒ 58b - 54b = 90a - 87a

⇒ 4b = 3a

⇒ $\frac{a}{b} = \frac{4}{3}$

⇒ a : b = 4 : 3

Hence, a : b = 4 : 3.

(i) Given,

3 : 4 :: 2.4 : x

Solving for x,

$$\begin{aligned}\Rightarrow \frac{3}{4} = \frac{2.4}{x} \\[1em] \Rightarrow 3 \times x = 4 \times 2.4 \\[1em] \Rightarrow 3x = 9.6 \\[1em] \Rightarrow x = \frac{9.6}{3} \\[1em] \Rightarrow x = 3.2\end{aligned}$$

Hence, x = 3.2

(ii) Given,

1 : 3 :: x : 7

Solving for x,

$$\begin{aligned}\Rightarrow \frac{1}{3} = \frac{x}{7} \\[1em] \Rightarrow x = \frac{7}{3} = 2\frac{1}{3}.\end{aligned}$$

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

(iii) Given,

x : 1.5 :: 3 : 5

Solving for x,

$$\begin{aligned}\Rightarrow \frac{x}{1.5} = \frac{3}{5} \\[1em] \Rightarrow 5 \times x = 1.5 \times 3 \\[1em] \Rightarrow x = \frac{4.5}{5} \\[1em] \Rightarrow x = 0.9\end{aligned}$$

Hence, x = 0.9

(i) Given,

3, 8 and 21

Let the fourth proportional to 3, 8 and 21 be x,

⇒ 3 : 8 = 21 : x

$$\begin{aligned}\Rightarrow \frac{3}{8} = \frac{21}{x} \\[1em] \Rightarrow x = \frac{21 \times 8}{3} \\[1em] \Rightarrow x = \frac{168}{3} \\[1em] \Rightarrow x = 56.\end{aligned}$$

Hence, the fourth proportional is 56.

(ii) Given,

1.4, 3.2 and 7

Let the fourth proportional to 1.4, 3.2 and 7 be x,

⇒ 1.4 : 3.2 = 7 : x

$$\begin{aligned}\Rightarrow \frac{1.4}{3.2} = \frac{7}{x} \\[1em] \Rightarrow x = \frac{7 \times 3.2}{1.4} \\[1em] \Rightarrow x = \frac{22.4}{1.4} \\[1em] \Rightarrow x = 16.\end{aligned}$$

Hence, the fourth proportional is 16.

(iii) Given,

1.5, 4.5 and 3.6

Let the fourth proportional to 1.5, 4.5 and 3.6 be x,

⇒ 1.5 : 4.5 = 3.6 : x

$$\begin{aligned}\Rightarrow \frac{1.5}{4.5} = \frac{3.6}{x} \\[1em] \Rightarrow x = \frac{4.5 \times 3.6}{1.5} \\[1em] \Rightarrow x = 3 \times 3.6 \\[1em] \Rightarrow x = 10.8.\end{aligned}$$

Hence, the fourth proportional is 10.8.

(iv) Given,

a², ab and b²

Let the fourth proportional to a², ab and b² be x,

⇒ a² : ab = b² : x

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

Hence, the fourth proportional is $\frac{b^3}{a}$.

(v) Given,

(a² − ab + b²), (a³ + b³) and (a − b)

Let the fourth proportional to (a² − ab + b²), (a³ + b³) and (a − b) be x,

⇒ a² : ab = b²:x

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

Hence, the fourth proportional is a² - b².

(i) Given,

9 and 6

Let third proportional to 9 and 6 be x.

⇒ 9 : 6 = 6 : x

⇒ $\frac{9}{6} = \frac{6}{x}$

⇒ x = $\frac{6^2}{9} = \frac{36}{4}$

⇒ x = 4.

Hence, the third proportional is 4.

(ii) Given,

$2\frac{2}{3}$ and 4

Let third proportional to $\frac{8}{3}$ and 4 be x

$\frac{8}{3}$ : 4 = 4 : x

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

Hence, the third proportional is 6.

(iii) Given,

1.6 and 2.4

Let third proportional to 1.6 and 2.4 be x.

1.6 : 2.4 = 2.4 : x

⇒ $\frac{1.6}{2.4} = \frac{2.4}{x}$

⇒ x = $\frac{(2.4)^2}{1.6} = \frac{5.76}{1.6}$

⇒ x = 3.6

Hence, the third proportional is 3.6.

(iv) Given,

(2 + $\sqrt{3}$) and (5 + 4 $\sqrt{3}$)

Let third proportional to (2 + $\sqrt{3}$) and (5 + 4 $\sqrt{3}$) be x.

$(2 + \sqrt{3}) : (5 + 4 \sqrt{3}) = (5 + 4\sqrt{3}) : x$

Thus,

$$\begin{aligned}\Rightarrow \frac{(2 + \sqrt{3})}{(5 + 4 \sqrt{3})} = \frac{(5 + 4 \sqrt{3})}{x} \\[1em] \Rightarrow x = \frac{(5 + 4 \sqrt{3})^2}{(2 + \sqrt{3})} \\[1em] = \frac{5^2 + 2(5)(4\sqrt3) + (4\sqrt3)^2}{(2 + \sqrt{3})} \\[1em] = \frac{25 + 40\sqrt3 + 16 \times 3}{(2 + \sqrt{3})} \\[1em] = \frac{25 + 40\sqrt3 + 48}{(2 + \sqrt{3})} \\[1em] = \frac{73 + 40\sqrt3}{(2 + \sqrt{3})}\end{aligned}$$

Multiplying numerator and denominator by $(2 - \sqrt{3})$, we get :

$$\begin{aligned}= \frac{(73 + 40\sqrt3)(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} \\[1em] = \frac{146 - 73\sqrt3 + 80\sqrt{3} - 40(\sqrt3)^2}{2^2 - (\sqrt{3})^2} \\[1em] = \frac{146 + 7\sqrt3 - 120}{4 - 3} \\[1em] = 26 + 7\sqrt3.\end{aligned}$$

Hence, the third proportional is $26 + 7\sqrt3$.

(v) Given,

$\Big(\frac{a}{b} + \frac{b}{a}\Big)$ and $\sqrt{a^{2} + b^{2}}$

Let third proportional to $\Big(\frac{a}{b} + \frac{b}{a}\Big) \text{ and } \sqrt{a^{2} + b^{2}}$ be x.

$\Big(\frac{a}{b} + \frac{b}{a}\Big): \sqrt{a^{2} + b^{2}} = \sqrt{a^{2} + b^{2}}:x$

$$\begin{aligned}\frac{\Big(\frac{a}{b} + \frac{b}{a}\Big)}{\sqrt{a^{2} + b^{2}}} = \frac{\sqrt{a^{2} + b^{2}}}{x} \\[1em] x = \frac{(\sqrt{a^2 + b^2})^2}{\Big(\frac{a}{b} + \frac{b}{a}\Big)} \\[1em] = \frac{a^2 + b^2}{\frac{a^2 + b^2}{ab}} \\[1em] = (a^2 + b^2) \times \frac{ab}{a^2 + b^2} \\[1em] = ab.\end{aligned}$$

Hence, the third proportional is ab.

(i) Given,

28 and 63

Let mean proportional between 28 and 63 be x.

28 : x :: x : 63

$$\begin{aligned}\Rightarrow \frac{28}{x} = \frac{x}{63} \\[1em] \Rightarrow x^2 = 28 \times 63 \\[1em] \Rightarrow x^2 = 1764 \\[1em] \Rightarrow x = \sqrt{1764} = 42\end{aligned}$$

Hence, the mean proportional is 42.

(ii) Given,

2.5 and 0.9

Let mean proportional between 2.5 and 0.9 be x.

2.5 : x :: x : 0.9

$$\begin{aligned}\Rightarrow \frac{2.5}{x} = \frac{x}{0.9} \\[1em] \Rightarrow x^2 = 2.5 \times 0.9 \\[1em] \Rightarrow x^2 = 2.25 \\[1em] \Rightarrow x = \sqrt{2.25} = 1.5\end{aligned}$$

Hence, the mean proportional is 1.5.

(iii) Given,

6.25 and 1.6

Let mean proportional between 6.25 and 1.6 be x.

6.25 : x :: x : 1.6

$$\begin{aligned}\Rightarrow \frac{6.25}{x} = \frac{x}{1.6} \\[1em] \Rightarrow x^2 = 6.25 \times 1.6 \\[1em] \Rightarrow x^2 = 10 \\[1em] \Rightarrow x = \sqrt{10}\end{aligned}$$

Hence, the mean proportional is $\sqrt{10}$.

(iv) Given,

$\Big(\sqrt{26} - \sqrt{17}\Big)$ and $\Big(\sqrt{26} + \sqrt{17}\Big)$

Let mean proportional between $\Big(\sqrt{26} - \sqrt{17}\Big)$ and $\Big(\sqrt{26} + \sqrt{17}\Big)$ be x

$\Big(\sqrt{26} - \sqrt{17}\Big):x::x:\Big(\sqrt{26} + \sqrt{17}\Big)$

$$\begin{aligned}\Rightarrow \frac{\sqrt{26} - \sqrt{17}}{x} = \frac{x}{\sqrt{26} + \sqrt{17}} \\[1em] \Rightarrow x^2 = (\sqrt{26} - \sqrt{17}) \times (\sqrt{26} + \sqrt{17}) \\[1em]\end{aligned}$$

Hence, the mean proportional is 3.

(v) Given,

6 + $3\sqrt{3}$ and 8 − $4\sqrt{3}$.

Let mean proportion between 6 + $3\sqrt{3}$ and 8 − $4\sqrt{3}$ be x.

6 + 3 $\sqrt{3} : x :: x : 8 − 4 \sqrt{3}$

$$\begin{aligned}\Rightarrow \frac{6 + 3\sqrt{3}}{x} = \frac{x}{8 − 4\sqrt{3}} \\[1em] \Rightarrow x^2 = (6 + 3\sqrt{3}) \times (8 − 4\sqrt{3}) \\[1em] \Rightarrow x^2 = 48 - 24\sqrt{3} + 24\sqrt{3} - 36 \\[1em] \Rightarrow x^2 = 12 \\[1em] \Rightarrow x = 2\sqrt3\end{aligned}$$

Hence, the mean proportional is $2\sqrt3$.

Let two numbers be x and y.

Given,

6 is mean proportion between x and y,

$$\begin{aligned}\therefore \frac{x}{6} = \frac{6}{y} \\[1em] \Rightarrow xy = 36 \space .....(1)\end{aligned}$$

Given,

48 is third proportional to x and y,

$$\begin{aligned}\therefore \frac{x}{y} = \frac{y}{48} \\[1em] \Rightarrow y^2 = 48x \\[1em] \Rightarrow x = \frac{y^2}{48} \space .....(2)\end{aligned}$$

Substituting value of x from equation (2) in (1) we get,

$$\begin{aligned}\Rightarrow \frac{y^2}{48}.y = 36 \\[1em] \Rightarrow y^3 = 36 \times 48 \\[1em] \Rightarrow y^3 = 1728 \\[1em] \Rightarrow y = \sqrt[3]{1728} \\[1em] \Rightarrow y = 12.\end{aligned}$$

Substituting value of y in equation (2), we get :

$x = \frac{12^2}{48} = \frac{144}{48}$ = 3.

Hence, numbers are 3 and 12.

Let least number to be added to numbers be x.

∴ 5 + x : 11 + x :: 19 + x : 37 + x

$$\begin{aligned}\Rightarrow \frac{5 + x}{11 + x} = \frac{19 + x}{37 + x} \\[1em] \Rightarrow (5 + x)(37 + x) = (19 + x)(11 + x) \\[1em] \Rightarrow 185 + 5x + 37x + x^2 = 209 + 19x + 11x + x^2 \\[1em] \Rightarrow x^2 + 42x + 185 = x^2 + 30x + 209 \\[1em] \Rightarrow x^2 - x^2 + 42x - 30x = 209 - 185 \\[1em] \Rightarrow 12x = 24 \\[1em] \Rightarrow x = 2.\end{aligned}$$

Hence, least number to be added to make numbers proportional is 2.

Let the number to be added to numbers be x.

∴ 4 + x : 6 + x :: 8 + x : 11 + x

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

Hence, the number to be added to make numbers proportional is 4.

Let least number to be subtracted to numbers be x.

∴ 23 - x : 30 - x :: 57 - x : 78 - x

$$\begin{aligned}\Rightarrow \frac{23 - x}{30 - x} = \frac{57 - x}{78 - x} \\[1em] \Rightarrow (23 - x)(78 - x) = (57 - x)(30 - x) \\[1em] \Rightarrow 1794 - 23x - 78x + x^2 = 1710 - 57x - 30x + x^2 \\[1em] \Rightarrow x^2 - 101x + 1794 = x^2 - 87x + 1710 \\[1em] \Rightarrow x^2 - x^2 - 101x + 87x = 1710 - 1794 \\[1em] \Rightarrow -14x = -84 \\[1em] \Rightarrow x = \frac{-84}{-14} \\[1em] \Rightarrow x = 6.\end{aligned}$$

Hence, least number to be subtracted to make numbers proportional is 6.

Let,

∴ x - 2 : x + 2 :: 2x + 1 : 2x + 19

$$\begin{aligned}\Rightarrow \frac{x - 2}{x + 2} = \frac{2x + 1}{2x + 19} \\[1em] \Rightarrow (x - 2)(2x + 19) = (2x + 1)(x + 2) \\[1em] \Rightarrow 2x^2 + 19x - 4x - 38 = 2x^2 + 4x + x + 2 \\[1em] \Rightarrow 2x^2 + 15x - 38 = 2x^2 + 5x + 2 \\[1em] \Rightarrow 2x^2 - 2x^2 + 15x - 5x = 2 + 38 \\[1em] \Rightarrow 10x = 40 \\[1em] \Rightarrow x = 4.\end{aligned}$$

Hence, the value of x = 4.

Let,

∴ K + 3 : K + 2 :: 3K − 7 : 2K − 3

$$\begin{aligned}\Rightarrow \frac{K + 3}{K + 2} = \frac{3K - 7}{2K - 3} \\[1em] \Rightarrow (K + 3)(2K - 3) = (3K - 7)(K + 2) \\[1em] \Rightarrow 2K^2 - 3K + 6k - 9 = 3K^2 + 6K - 7K - 14 \\[1em] \Rightarrow 2K^2 + 3K - 9 = 3K^2 - K - 14 \\[1em] \Rightarrow 0 = 3K^2 - K - 14 - 2K^2 - 3K + 9 \\[1em] \Rightarrow K^2 - 4K - 5 = 0 \\[1em] \Rightarrow K^2 + 1K - 5K - 5 = 0 \\[1em] \Rightarrow K(K + 1) - 5(K + 1) = 0 \\[1em] \Rightarrow (K - 5)(K + 1) = 0 \\[1em] \Rightarrow (K - 5) = 0 \text{ or }(K + 1) = 0 \text{[Using Zero - product rule]}\\[1em] \Rightarrow K = 5 \text{ or } K = - 1\end{aligned}$$

Hence, K = 5 or K = −1.

Given, x + 5 is G.M. between x + 2 and x + 9.

$\therefore \frac{x + 2}{x + 5} = \frac{x + 5}{x + 9}$

⇒ (x + 5)² = (x + 2)(x + 9)

⇒ x² + 10x + 25 = x² + 9x + 2x + 18

⇒ x² + 10x + 25 = x² + 11x + 18

⇒ x² - x² + 10x - 11x = 18 - 25

⇒ -x = -7

⇒ x = 7.

Hence, the value of x = 7.

Let the two numbers be x and y.

Thus, 36 is the mean proportion between x and y.

$$\begin{aligned}\Rightarrow \frac{x}{36} = \frac{36}{y} \\[1em] \Rightarrow xy = 36^2 \\[1em] \Rightarrow xy = 1296 \\[1em] \Rightarrow x = \frac{1296}{y}\text{.....(1)}\end{aligned}$$

The third proportional for x and y is 288.

$$\begin{aligned}\Rightarrow \frac{x}{y} = \frac{y}{288} \\[1em] \Rightarrow y^2 = 288x \text{ ..........(2)}\end{aligned}$$

Substituting value of x from equation (1) in (2), we get :

$$\begin{aligned}\Rightarrow y^2 = 288 \times \frac{1296}{y} \\[1em] \Rightarrow y^3 = 373248 \\[1em] \Rightarrow y = \sqrt[3]{373248} = 72.\end{aligned}$$

Substituting the value of y in equation (1), we get :

$$\begin{aligned}\Rightarrow x = \frac{1296}{y} \\[1em] \Rightarrow x = \frac{1296}{72} \\[1em] \Rightarrow x = 18\end{aligned}$$

Hence, the two numbers are 18 and 72.

(i) Given,

⇒ a : b :: c : d

∴ a : b = c : d

$$\begin{aligned}\therefore \frac{a}{b} = \frac{c}{d} \\[1em] \Rightarrow \frac{a}{c} = \frac{b}{d} = k \text{(let)}\end{aligned}$$

⇒ a = ck and b = dk.

Substituting value of a and b in L.H.S. of (a² + ab) : (c² + cd) = (b² − 2ab) : (d² − 2cd), we get :

$$\begin{aligned}\Rightarrow \frac{a^2 + ab}{c^2 + cd} \\[1em] \Rightarrow \frac{(ck)^2 + ck.dk}{c^2 + cd} \\[1em] \Rightarrow \frac{k^2 c^2 + k^2 cd}{c^2 + cd} \\[1em] \Rightarrow \frac{k^2(c^2 + cd)}{c^2 + cd} \\[1em] \Rightarrow k^2.\end{aligned}$$

Substituting value of a and b in R.H.S. of (a² + ab) : (c² + cd) = (b² − 2ab) : (d² − 2cd), we get :

$$\begin{aligned}\Rightarrow \frac{b^2 - 2ab}{d^2 - 2cd} \\[1em] \Rightarrow \frac{(kd)^2 - 2 \times ck \times dk}{d^2 - 2cd} \\[1em] \Rightarrow \frac{k^2 d^2 - 2k^2 cd}{d^2 - 2cd} \\[1em] \Rightarrow \frac{k^2(d^2 - 2cd)}{d^2 - 2cd} \\[1em] \Rightarrow k^2.\end{aligned}$$

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

Hence, proved that (a² + ab) : (c² + cd) = (b² − 2ab) : (d² − 2cd).

(ii) Given,

⇒ a : b :: c : d

∴ a : b = c : d

$$\begin{aligned}\therefore \frac{a}{b} = \frac{c}{d} \\[1em] \Rightarrow \frac{a}{c} = \frac{b}{d} = k \text{(let)}\end{aligned}$$

⇒ a = ck and b = dk.

Substituting value of a and b in L.H.S. of (a² + b²) : (c² + d²) = (ab + ad − bc) : (cd − ad + bc)

$$\begin{aligned}\Rightarrow \frac{a^2 + b^2}{c^2 + d^2} \\[1em] \Rightarrow \frac{(kc)^2 + (kd)^2}{c^2 + d^2} \\[1em] \Rightarrow \frac{k^2c^2 + k^2d^2}{c^2 + d^2} \\[1em] \Rightarrow \frac{k^2(c^2 + d^2)}{c^2 + d^2} \\[1em] \Rightarrow k^2.\end{aligned}$$

Substituting value of a and b in R.H.S. of (a² + b²) : (c² + d²) = (ab + ad − bc) : (cd − ad + bc)

$$\begin{aligned}\Rightarrow \frac{ab + ad - bc}{cd - ad + bc} \\[1em] \Rightarrow \frac{(kc)(kd) + (kc)d - (kd)c}{cd - (kc)d + (kd)c} \\[1em] \Rightarrow \frac{k^2 cd + kcd - kcd}{cd - kcd + kcd} \\[1em] \Rightarrow \frac{k^2 (cd)}{(cd)} \\[1em] \Rightarrow k^2.\end{aligned}$$

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

Hence, proved that (a² + b²) : (c² + d²) = (ab + ad − bc) : (cd − ad + bc).

(iii) Given,

⇒ a : b :: c : d

∴ a : b = c : d

$$\begin{aligned}\therefore \frac{a}{b} = \frac{c}{d} \\[1em] \Rightarrow \frac{a}{c} = \frac{b}{d} = k \text{(let)}\end{aligned}$$

⇒ a = ck and b = dk.

Substituting value of a and b in L.H.S. of (a² + ac + c²) : (a² − ac + c²) = (b² + bd + d²) : (b² − bd + d²),

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

Substituting value of a and b in R.H.S. of (a² + ac + c²) : (a² − ac + c²) = (b² + bd + d²) : (b² − bd + d²),

$$\begin{aligned}\Rightarrow \frac{b^2 + bd + d^2}{b^2 - bd + d^2} \\[1em] \Rightarrow \frac{(kd)^2 + (kd)d + d^2}{(kd)^2 - (kd)d + d^2} \\[1em] \Rightarrow \frac{k^2 d^2 + kd^2 + d^2}{k^2 d^2 - kd^2 + d^2} \\[1em] \Rightarrow \frac{d^2(k^2 + k + 1)}{d^2(k^2 - k + 1)} \\[1em] \Rightarrow \frac{k^2 + k + 1}{k^2 - k + 1}.\end{aligned}$$

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

Hence, proved that (a² + ac + c²) : (a² − ac + c²) = (b² + bd + d²) : (b² − bd + d²).

(i) Given,

⇒ a : b :: c : d

∴ a : b = c : d

$$\begin{aligned}\therefore \frac{a}{b} = \frac{c}{d} \\[1em] \Rightarrow \frac{a}{c} = \frac{b}{d} = k \text{(let)}\end{aligned}$$

⇒ a = ck and b = dk.

Substituting value of a and b in L.H.S. of $\frac{a + b}{c + d} = \sqrt{\frac{2a^{2} + 7b^{2}}{2c^{2} + 7d^{2}}}$, we get :

$$\begin{aligned}\Rightarrow \frac{a + b}{c + d} \\[1em] \Rightarrow \frac{kc + kd}{c + d} \\[1em] \Rightarrow \frac{k(c + d)}{(c + d)} \\[1em] \Rightarrow k.\end{aligned}$$

Substituting value of a and b in R.H.S. of $\frac{a + b}{c + d} = \sqrt{\frac{2a^{2} + 7b^{2}}{2c^{2} + 7d^{2}}}$, we get :

$$\begin{aligned}\Rightarrow \sqrt{\frac{2a^2 + 7b^2}{2c^2 + 7d^2}} \\[1em] \Rightarrow \sqrt{\frac{2(kc)^2 + 7(dk)^2}{2c^2 + 7d^2}} \\[1em] \Rightarrow \sqrt{\frac{k^2(2c^2 + 7d^2)}{2c^2 + 7d^2}} \\[1em] \Rightarrow \sqrt{k^2} \\[1em] \Rightarrow k.\end{aligned}$$

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

Hence, proved that $\frac{a + b}{c + d} = \sqrt{\frac{2a^{2} + 7b^{2}}{2c^{2} + 7d^{2}}}$.

(ii) Given,

⇒ a : b :: c : d

∴ a : b = c : d

$$\begin{aligned}\therefore \frac{a}{b} = \frac{c}{d} \\[1em] \Rightarrow \frac{a}{c} = \frac{b}{d} = k \text{(let)}\end{aligned}$$

⇒ a = ck and b = dk.

Substituting value of a and b in L.H.S. of $\frac{ma^{2} + nc^{2}}{mb^{2} + nd^{2}} = \sqrt{\frac{a^{4} + c^{4}}{b^{4} + d^{4}}}$, we get :

$$\begin{aligned}\Rightarrow \frac{ma^2 + nc^2}{mb^2 + nd^2} \\[1em] \Rightarrow \frac{m(kc)^2 + nc^2}{m(kd)^2 + nd^2} \\[1em] \Rightarrow \frac{m k^2 c^2 + n c^2}{m k^2 d^2 + n d^2} \\[1em] \Rightarrow \frac{c^2(m k^2 + n)}{d^2(m k^2 + n)} \\[1em] \Rightarrow \frac{c^2}{d^2}.\end{aligned}$$

Substituting value of a and b in R.H.S. of $\frac{ma^{2} + nc^{2}}{mb^{2} + nd^{2}} = \sqrt{\frac{a^{4} + c^{4}}{b^{4} + d^{4}}}$, we get :

$$\begin{aligned}\Rightarrow \sqrt{\frac{a^4 + c^4}{b^4 + d^4}} \\[1em] \Rightarrow \sqrt{\frac{(kc)^4 + c^4}{(kd)^4 + d^4}} \\[1em] \Rightarrow \sqrt{\frac{k^4 c^4 + c^4}{k^4 d^4 + d^4}} \\[1em] \Rightarrow \sqrt{\frac{c^4(k^4 + 1)}{d^4(k^4 + 1)}} \\[1em] \Rightarrow \sqrt{\frac{c^4}{d^4}} \\[1em] \Rightarrow \frac{c^2}{d^2}.\end{aligned}$$

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

Hence, proved that $\frac{ma^{2} + nc^{2}}{mb^{2} + nd^{2}} = \sqrt{\frac{a^{4} + c^{4}}{b^{4} + d^{4}}}$.

(iii) Given,

⇒ a : b :: c : d

∴ a : b = c : d

$$\begin{aligned}\therefore \frac{a}{b} = \frac{c}{d} \\[1em] \Rightarrow \frac{a}{c} = \frac{b}{d} = k \text{(let)}\end{aligned}$$

⇒ a = ck and b = dk.

Substituting values of a and b in L.H.S. of $\frac{a^{2} + ab + b^{2}}{a^{2} - ab + b^{2}} = \frac{c^{2} + cd + d^{2}}{c^{2} - cd + d^{2}}$, we get:

$$\begin{aligned}\Rightarrow \frac{a^2 + ab + b^2}{a^2 - ab + b^2} \\[1em] \Rightarrow \frac{(kc)^2 + (kc)(kd) + (kd)^2}{(kc)^2 - (kc)(kd) + (kd)^2} \\[1em] \Rightarrow \frac{k^2 c^2 + k^2 cd + k^2 d^2}{k^2 c^2 - k^2 cd + k^2 d^2} \\[1em] \Rightarrow \frac{k^2(c^2 + cd + d^2)}{k^2(c^2 - cd + d^2)} \\[1em] \Rightarrow \frac{c^2 + cd + d^2}{c^2 - cd + d^2}.\end{aligned}$$

Substituting values of a and b in R.H.S. of $\frac{a^{2} + ab + b^{2}}{a^{2} - ab + b^{2}} = \frac{c^{2} + cd + d^{2}}{c^{2} - cd + d^{2}}$ ,we get:

$\Rightarrow \frac{c^2 + cd + d^2}{c^2 - cd + d^2}.$

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

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

(iv) Given,

⇒ a : b :: c : d

∴ a : b = c : d

$$\begin{aligned}\therefore \frac{a}{b} = \frac{c}{d} \\[1em] \Rightarrow \frac{a}{c} = \frac{b}{d} = k \text{(let)}\end{aligned}$$

⇒ a = ck and b = dk.

Substituting values of a and b in L.H.S. of $\frac{(a + c)^{3}}{(b + d)^{3}} = \frac{a(a - c)^{2}}{b(b - d)^{2}}$, we get:

$$\begin{aligned}\Rightarrow \frac{(a + c)^3}{(b + d)^3} \\[1em] \Rightarrow \frac{(kc + c)^3}{(kd + d)^3} \\[1em] \Rightarrow \frac{c^3(k + 1)^3}{d^3(k + 1)^3} \\[1em] \Rightarrow \frac{c^3}{d^3}.\end{aligned}$$

Substituting values of a and b in R.H.S. of $\frac{(a + c)^{3}}{(b + d)^{3}} = \frac{a(a - c)^{2}}{b(b - d)^{2}}$, we get:

$$\begin{aligned}\Rightarrow \frac{a(a - c)^2}{b(b - d)^2} \\[1em] \Rightarrow \frac{kc(kc - c)^2}{kd(kd - d)^2} \\[1em] \Rightarrow \frac{kc \cdot c^2(k - 1)^2}{kd \cdot d^2(k - 1)^2} \\[1em] \Rightarrow \frac{kc^3}{kd^3} \\[1em] \Rightarrow \frac{c^3}{d^3}.\end{aligned}$$

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

Hence, proved that $\frac{(a + c)^{3}}{(b + d)^{3}} = \frac{a(a - c)^{2}}{b(b - d)^{2}}$.

(i) Given,

⇒ $\frac{x}{a} = \frac{y}{b} = \frac{z}{c}$ = k (let)

⇒ x = ak, y = bk, z = ck.

Substituting values of x, y and z in L.H.S. of the equation $\frac{x^{2} + y^{2} + z^{2}}{a^{2} + b^{2} + c^{2}} = \Big(\frac{px + qy + rz}{pa + qb + rc}\Big)^{2}$, we get :

$$\begin{aligned}\Rightarrow \frac{x^2 + y^2 + z^2}{a^2 + b^2 + c^2} \\[1em] \Rightarrow \frac{k^2 a^2 + k^2 b^2 + k^2 c^2}{a^2 + b^2 + c^2} \\[1em] \Rightarrow \frac{k^2 (a^2 + b^2 + c^2)}{a^2 + b^2 + c^2} \\[1em] \Rightarrow k^2.\end{aligned}$$

Substituting values of x, y and z in R.H.S. of the equation $\frac{x^{2} + y^{2} + z^{2}}{a^{2} + b^{2} + c^{2}} = \Big(\frac{px + qy + rz}{pa + qb + rc}\Big)^{2}$, we get :

$$\begin{aligned}\Rightarrow \Big(\frac{px + qy + rz}{pa + qb + rc}\Big)^2 \\[1em] \Rightarrow \Big(\frac{p(ka) + q(kb) + r(kc)}{pa + qb + rc}\Big)^2 \\[1em] \Rightarrow \Big(\frac{k(pa + qb + rc)}{pa + qb + rc}\Big)^2 \\[1em] \Rightarrow k^2.\end{aligned}$$

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

Hence, proved that $\frac{x^{2} + y^{2} + z^{2}}{a^{2} + b^{2} + c^{2}} = \Big(\frac{px + qy + rz}{pa + qb + rc}\Big)^{2}$.

(ii) Given,

⇒ $\frac{x}{a} = \frac{y}{b} = \frac{z}{c}$ = k (let)

⇒ x = ak, y = bk, z = ck.

Substituting values of x, y and z in L.H.S. of the equation $\frac{x^{3}}{a^{3}} + \frac{y^{3}}{b^{3}} + \frac{z^{3}}{c^{3}} = \frac{3xyz}{abc}$, we get:

$$\begin{aligned}\Rightarrow \frac{x^3}{a^3} + \frac{y^3}{b^3} + \frac{z^3}{c^3} \\[1em] \Rightarrow \frac{k^3 a^3}{a^3} + \frac{k^3 b^3}{b^3} + \frac{k^3 c^3}{c^3} \\[1em] \Rightarrow k^3 + k^3 + k^3 \\[1em] \Rightarrow 3k^3.\end{aligned}$$

Substituting values of x, y and z in R.H.S. of the equation $\frac{x^{3}}{a^{3}} + \frac{y^{3}}{b^{3}} + \frac{z^{3}}{c^{3}} = \frac{3xyz}{abc}$, we get:

$$\begin{aligned}\Rightarrow \frac{3xyz}{abc} \\[1em] \Rightarrow \frac{3(ka)(kb)(kc)}{abc} \\[1em] \Rightarrow \frac{3k^3 abc}{abc} \\[1em] \Rightarrow 3k^3.\end{aligned}$$

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

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

(iii) Given,

⇒ $\frac{x}{a} = \frac{y}{b} = \frac{z}{c}$ = k (let)

⇒ x = ak, y = bk, z = ck.

Substituting values of x, y and z in L.H.S. of the equation $\frac{x^{3}}{a^{2}} + \frac{y^{3}}{b^{2}} + \frac{z^{3}}{c^{2}} = \frac{(x + y + z)^{3}}{(a + b + c)^{2}}$, we get:

$$\begin{aligned}\Rightarrow \frac{x^3}{a^2} + \frac{y^3}{b^2} + \frac{z^3}{c^2} \\[1em] \Rightarrow \frac{k^3 a^3}{a^2} + \frac{k^3 b^3}{b^2} + \frac{k^3 c^3}{c^2} \\[1em] \Rightarrow k^3 a + k^3 b + k^3 c \\[1em] \Rightarrow k^3 (a + b + c).\end{aligned}$$

Substituting values of x, y and z in R.H.S. of the equation $\frac{x^{3}}{a^{2}} + \frac{y^{3}}{b^{2}} + \frac{z^{3}}{c^{2}} = \frac{(x + y + z)^{3}}{(a + b + c)^{2}}$, we get:

$$\begin{aligned}\Rightarrow \frac{(x + y + z)^3}{(a + b + c)^2} \\[1em] \Rightarrow \frac{(k(a + b + c))^3}{(a + b + c)^2} \\[1em] \Rightarrow \frac{k^3 (a + b + c)^3}{(a + b + c)^2} \\[1em] \Rightarrow k^3 (a + b + c).\end{aligned}$$

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

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

(iv) Given,

⇒ $\frac{x}{a} = \frac{y}{b} = \frac{z}{c}$ = k (let)

⇒ x = ak, y = bk, z = ck.

Substituting values of x, y and z in $\frac{ax - by}{(a + b)(x - y)}$, we get:

$$\begin{aligned}\Rightarrow \frac{ax - by}{(a + b)(x - y)} \\[1em] \Rightarrow \frac{a(ka) - b(kb)}{(a + b)(ka - kb)} \\[1em] \Rightarrow \frac{k(a^2 - b^2)}{k(a + b)(a - b)} \\[1em] \Rightarrow \frac{k(a - b)(a + b)}{k(a + b)(a - b)} \\[1em] \Rightarrow 1....(1)\end{aligned}$$

Substituting values of x, y and z in $\frac{by - cz}{(b + c)(y - z)}$,we get:

$$\begin{aligned}\Rightarrow \frac{by - cz}{(b + c)(y - z)} \\[1em] \Rightarrow \frac{b(kb) - c(kc)}{(b + c)(kb - kc)} \\[1em] \Rightarrow \frac{k(b^2 - c^2)}{k(b + c)(b - c)} \\[1em] \Rightarrow \frac{k(b - c)(b + c)}{k(b + c)(b - c)} \\[1em] \Rightarrow 1....(2)\end{aligned}$$

Substituting values of x, y and z in $\frac{cz - ax}{(c + a)(z - x)}$,we get:

$$\begin{aligned}\Rightarrow \frac{cz - ax}{(c + a)(z - x)} \\[1em] \Rightarrow \frac{c(kc) - a(ka)}{(c + a)(kc - ka)} \\[1em] \Rightarrow \frac{k(c^2 - a^2)}{k(c + a)(c - a)} \\[1em] \Rightarrow \frac{k(c - a)(c + a)}{k(c + a)(c - a)} \\[1em] \Rightarrow 1....(3)\end{aligned}$$

Adding 1,2 and 3 we get,

$$\begin{aligned}\Rightarrow \frac{ax - by}{(a + b)(x - y)} + \frac{by - cz}{(b + c)(y - z)} + \frac{cz - ax}{(c + a)(z - x)} \\[1em] \Rightarrow 1 + 1 + 1 \\[1em] \Rightarrow 3.\end{aligned}$$

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) Given,

⇒ $\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k$(let)

⇒ a = kb, c = kd, e = kf.

Substituting values of a,c and e in L.H.S. of (b² + d² + f²)(a² + c² + e²) = (ab + cd + ef)², we get:

$$\begin{aligned}\Rightarrow (b^2 + d^2 + f^2)(a^2 + c^2 + e^2) \\[1em] \Rightarrow (b^2 + d^2 + f^2)(k^2 b^2 + k^2 d^2 + k^2 f^2) \\[1em] \Rightarrow (b^2 + d^2 + f^2)[k^2(b^2 + d^2 + f^2)] \\[1em] \Rightarrow k^2(b^2 + d^2 + f^2)^2.\end{aligned}$$

Substituting values of a,c and e in R.H.S. of (b² + d² + f²)(a² + c² + e²) = (ab + cd + ef)², we get:

$$\begin{aligned}\Rightarrow ab + cd + ef \\[1em] \Rightarrow (k b)b + (k d)d + (k f)f \\[1em] \Rightarrow k(b^2 + d^2 + f^2) \\[1em] \therefore (ab + cd + ef)^2 = k^2(b^2 + d^2 + f^2)^2.\end{aligned}$$

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

Hence, proved that (b² + d² + f²)(a² + c² + e²) = (ab + cd + ef)² .

(ii) Given,

⇒ $\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k$(let)

⇒ a = kb, c = kd, e = kf.

Substituting values of a,c and e in L.H.S. of $\frac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}} = \frac{ace}{bdf}$, we get:

$$\begin{aligned}\Rightarrow \frac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}} \\[1em] \Rightarrow \frac{k^3 b^3 + k^3 d^3 + k^3 f^3}{b^{3} + d^{3} + f^{3}} \\[1em] \Rightarrow \frac{k^3(b^3 + d^3 + f^3)}{b^{3} + d^{3} + f^{3}} \\[1em] \Rightarrow k^3.\end{aligned}$$

SSubstituting values of a,c and e in R.H.S. of $\frac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}} = \frac{ace}{bdf}$, we get:

$$\begin{aligned}\Rightarrow \frac{ace}{bdf} \\[1em] \Rightarrow \frac{(k b)(k d)(k f)}{b d f} \\[1em] \Rightarrow k^3.\end{aligned}$$

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

Hence, proved that $\frac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}} = \frac{ace}{bdf}$.

(iii) Given,

⇒ $\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k$(let)

⇒ a = kb, c = kd, e = kf.

Substituting values of a,c and e in L.H.S. of $\Big(\frac{a^{2}}{b^{2}} + \frac{c^{2}}{d^{2}} + \frac{e^{2}}{f^{2}}\Big) = \Big(\frac{ac}{bd} + \frac{ce}{df} + \frac{ae}{bf}\Big)$, we get:

$$\begin{aligned}\Rightarrow \frac{a^2}{b^2} + \frac{c^2}{d^2} + \frac{e^2}{f^2} \\[1em] \Rightarrow \frac{k^2 b^2}{b^2} + \frac{k^2 d^2}{d^2} + \frac{k^2 f^2}{f^2} \\[1em] \Rightarrow k^2 + k^2 + k^2 \\[1em] \Rightarrow 3k^2.\end{aligned}$$

Substituting values of a,c and e in R.H.S. of $\Big(\frac{a^{2}}{b^{2}} + \frac{c^{2}}{d^{2}} + \frac{e^{2}}{f^{2}}\Big) = \Big(\frac{ac}{bd} + \frac{ce}{df} + \frac{ae}{bf}\Big)$, we get:

$$\begin{aligned}\Rightarrow \frac{ac}{bd} + \frac{ce}{df} + \frac{ae}{bf} \\[1em] \Rightarrow \frac{(k b)(k d)}{b d} + \frac{(k d)(k f)}{d f} + \frac{(k b)(k f)}{b f} \\[1em] \Rightarrow k^2 + k^2 + k^2 \\[1em] \Rightarrow 3k^2.\end{aligned}$$

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

Hence, proved that $\Big(\frac{a^{2}}{b^{2}} + \frac{c^{2}}{d^{2}} + \frac{e^{2}}{f^{2}}\Big) = \Big(\frac{ac}{bd} + \frac{ce}{df} + \frac{ae}{bf}\Big)$.

(iv) Given,

⇒ $\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k$(let)

⇒ a = kb, c = kd, e = kf.

Substituting values of a, c and d in L.H.S. of (bdf)·$\Big(\frac{a + b}{b} + \frac{c + d}{d} + \frac{e + f}{f}\Big)^{3} = 27(a + b)(c + d)(e + f)$, we get:

$$\begin{aligned}\Rightarrow bdf \cdot \Big(\frac{a + b}{b} + \frac{c + d}{d} + \frac{e + f}{f}\Big)^{3} \\[1em] \Rightarrow bdf \cdot \Big(\frac{kb + b}{b} + \frac{kd + d}{d} + \frac{kf + f}{f}\Big)^3 \\[1em] \Rightarrow bdf \cdot [(k + 1) + (k + 1) + (k + 1)] \\[1em] \Rightarrow bdf \cdot [3(k + 1)]^3 \\[1em] \Rightarrow bdf\cdot 27(k + 1)^3.\end{aligned}$$

Substituting values of a, c and d in R.H.S. of (bdf)·$\Big(\frac{a + b}{b} + \frac{c + d}{d} + \frac{e + f}{f}\Big)^{3} = 27(a + b)(c + d)(e + f)$, we get:

$$\begin{aligned}\Rightarrow 27(a + b)(c + d)(e + f) \\[1em] \Rightarrow 27(kb + b)(kd + d)(kf + f) \\[1em] \Rightarrow 27\big(b(k + 1)\big)\big(d(k + 1)\big)\big(f(k + 1)\big) \\[1em] \Rightarrow 27 bdf(k + 1)^3.\end{aligned}$$

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

Hence, proved that (bdf).$\Big(\frac{a + b}{b} + \frac{c + d}{d} + \frac{e + f}{f}\Big)^{3} = 27(a + b)(c + d)(e + f)$.

(i) Given,

⇒ a, b, c are in continued proportion

∴ a : b = b : c

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

⇒ b = ck, a = bk = (ck)k = ck².

Substituting values of a and b in L.H.S. of equation $\frac{a + b}{b + c} = \frac{a^{2}(b - c)}{b^{2}(a - b)}$, we get :

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

Substituting values of a and b in R.H.S. of equation $\frac{a + b}{b + c} = \frac{a^{2}(b - c)}{b^{2}(a - b)}$, we get :

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

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) Given,

⇒ a, b, c are in continued proportion

∴ a : b = b : c

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

⇒ b = ck, a = bk = (ck)k = ck².

Substituting values of a and b in L.H.S. of equation $\frac{a + b + c}{a - b + c} = \frac{(a + b + c)^{2}}{(a^{2} + b^{2} + c^{2})}$, we get :

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

Substituting values of a and b in R.H.S. of equation $\frac{a + b + c}{a - b + c} = \frac{(a + b + c)^{2}}{(a^{2} + b^{2} + c^{2})}$, we get :

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

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

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

(iii) Given,

⇒ a, b, c are in continued proportion

∴ a : b = b : c

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

⇒ b = ck, a = bk = (ck)k = ck².

Substituting values of a and b in L.H.S. of equation $\frac{a^{2} + ab + b^{2}}{b^{2} + bc + c^{2}} = \frac{a}{c}$, we get :

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

Substituting values of a and b in R.H.S. of equation $\frac{a^{2} + ab + b^{2}}{b^{2} + bc + c^{2}} = \frac{a}{c}$, we get :

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

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

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

(iv) Given,

⇒ a, b, c are in continued proportion

∴ a : b = b : c

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

⇒ b = ck, a = bk = (ck)k = ck².

Substituting values of a and b in L.H.S. of (a + b + c)(a − b + c) = (a² + b² + c²), we get:

⇒ (a + b + c)(a − b + c)

⇒ ((ck²) + (ck) + c)((ck²) − (ck) + c)

⇒ c(k² + k + 1). c(k² - k + 1)

⇒ c²(k⁴ + k² + 1)

Substituting values of a and b in R.H.S. of (a + b + c)(a − b + c) = (a² + b² + c²), we get:

⇒ (a² + b² + c²)

⇒ (ck²)² + (ck)² + c²

⇒ (c²k⁴ + c²k² + c²)

⇒ c²(k⁴ + k² + 1)

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

Hence, proved that (a + b + c)(a − b + c) = (a² + b² + c²).

(v) Given,

⇒ a, b, c are in continued proportion

∴ a : b = b : c

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

⇒ b = ck, a = bk = (ck)k = ck².

Substituting values of a and b in L.H.S. of a²b²c²(a⁻³ + b⁻³ + c⁻³) = (a³ + b³ + c³), we get:

$$\begin{aligned}\Rightarrow a^2 b^2 c^2\big(a^{ - 3} + b^{ - 3} + c^{ - 3}\big) \\[1em] \Rightarrow c^6 k^6\Big(\frac{1}{c^3 k^6} + \frac{1}{c^3 k^3} + \frac{1}{c^3}\Big) \\[1em] \Rightarrow c^3\big(1 + k^3 + k^6\big).\end{aligned}$$

Substituting values of a and b in R.H.S. of a²b²c²(a⁻³ + b⁻³ + c⁻³) = (a³ + b³ + c³), we get:

$$\begin{aligned}\Rightarrow a^3 + b^3 + c^3 \\[1em] \Rightarrow (ck^2)^3 + (ck)^3 + c^3 \\[1em] \Rightarrow c^3k^6 + c^3k^3 + c^3 \\[1em] \Rightarrow c^3(k^6 + k^3 + 1).\end{aligned}$$

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

Hence, proved that a²b²c²(a⁻³ + b⁻³ + c⁻³) = (a³ + b³ + c³).

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

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

c = dk, b = ck = (dk)k = dk², a = bk = (dk²)k = dk³.

Substituting values in L.H.S. of the equation ad(c² + d²) = c³(b + d), we get :

L.H.S = ad(c² + d²)

= dk³.(d).[(dk)² + d²]

= d²k³.[d²(k² + 1)]

= d⁴k³(k² + 1).

Substituting values in R.H.S. of the equation ad(c² + d²) = c³(b + d), we get :

R.H.S = c³(b + d)

= (dk)³.(dk² + d)

= d³k³[d(k² + 1)]

= d⁴k³(k² + 1).

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

Hence, proved that ad(c² + d²) = c³(b + d).

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

Given,

⇒ a, b, c, d are in continued proportion

∴ a : b = b : c = c : d

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

⇒ c = dk, b = ck = (dk)k = dk², a = bk = (dk²)k = dk³.

Substituting values of a, b and c in L.H.S. of equation (b + c)(b + d) = (c + a)(c + d), we get :

⇒ (b + c)(b + d)

⇒ (d k² + d k)(d k² + d)

⇒ d²(k² + k)(k² + 1)

⇒ d²k(k + 1)(k² + 1).

Substituting values of a, b and c in R.H.S. of equation (b + c)(b + d) = (c + a)(c + d), we get :

⇒ (c + a)(c + d)

⇒ (dk + dk³)(dk + d)

⇒ d²(k + k³)(k + 1)

⇒ d²k(1 + k²)(k + 1).

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

Hence, (b + c)(b + d) = (c + a)(c + d).

Given a, b, c, d are in continued proportion.

∴ a : b = b : c = c : d

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

⇒ c = dk, b = ck = (dk)k = dk², a = bk = (dk²)k = dk³.

Substituting values of a, b and c in L.H.S. of (a + b)(b + c) - (a + c)(b + d) = (b - c)², we get :

⇒ (dk³ + d)(dk² + dk) - (dk³ + dk)(dk² + d)

⇒ d(k³ + 1) dk(k + 1) - dk(k² + 1) d(k² + 1)

⇒ d²k(k³ + 1)(k + 1) - d²k(k² + 1)(k² + 1)

⇒ d²k[(k⁴ + k³ + k + 1) - (k⁴ + 2k² + 1)]

⇒ d²k[k⁴ + k³ + k + 1 - k⁴ - 2k² - 1]

⇒ d²k[k³ - 2k² + k]

⇒ d²k²[k² - 2k + 1]

⇒ d²k²(k - 1)²

Substituting values of a, b and c in R.H.S. of (a + b)(b + c) - (a + c)(b + d) = (b - c)², we get :

⇒ (b - c)²

⇒ (dk² - dk)²

⇒ (dk[k - 1])²

⇒ d²k²(k - 1)²

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

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

Given a, b, c, d are in continued proportion.

∴ a : b = b : c = c : d

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

⇒ c = dk, b = ck = (dk)k = dk², a = bk = (dk²)k = dk³.

Substituting values of a, b and c in L.H.S. of (a² − b²)(c² − d²) = (b² − c²)², we get :

⇒ (a² - b²)(c² - d²)

⇒ [(dk³)² - (dk²)²] [(dk)² - d²]

⇒ [d²k⁶ - d²k⁴] [d²k² - d²]

⇒ d⁴(k⁶ - k⁴)(k² - 1)

⇒ d⁴ k⁴(k² - 1)(k² - 1)

⇒ d⁴ k⁴ (k² - 1)².

Substituting values of a, b and c in R.H.S. of (a² − b²)(c² − d²) = (b² − c²)², we get :

⇒ (b² - c²)²

⇒ [(dk²)² - (dk)²]²

⇒ [d²k⁴ - d²k²]²

⇒ [d²k²(k² - 1)]²

⇒ d⁴ k⁴ (k² - 1)².

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

Hence, proved that (a² − b²)(c² − d²) = (b² − c²)² .

Given a, b, c, d are in continued proportion.

∴ a : b = b : c = c : d

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

⇒ c = dk, b = ck = (dk)k = dk², a = bk = (dk²)k = dk³.

Substituting values of a, b and c in L.H.S. of $\Big(\frac{a - b}{c} + \frac{a - c}{c}\Big)^2 - \Big(\frac{d - b}{c} + \frac{d - c}{b}\Big)^2 = (a - d)^2 \Big(\frac{1}{c^2} + \frac{1}{b^2}\Big)$ , we get :

$$\begin{aligned}\Rightarrow \Big(\frac{dk^3 - dk^2}{dk} + \frac{dk^3 - dk^2}{dk^2}\Big)^2 - \Big(\frac{d - dk^2}{dk} + \frac{d - dk}{dk^2}\Big)^2 \\[1em] \Rightarrow \Big(\frac{k(dk^3 - dk^2) + dk^3 - dk}{dk^2}\Big)^2 - \Big(\frac{k(d - dk^2) + d - dk}{dk^2}\Big)^2 \\[1em] \Rightarrow \Big(\frac{dk^4 - dk^3 + dk^3 - dk}{dk^2}\Big)^2 - \Big(\frac{kd - dk^3 + d - dk}{dk^2}\Big)^2 \\[1em] \Rightarrow \Big(\frac{dk^4 - dk}{dk^2}\Big)^2 - \Big(\frac{d - dk^3}{dk^2}\Big)^2 \\[1em] \Rightarrow \Big(\frac{dk(k^3 - 1)}{dk^2}\Big)^2 - \Big(\frac{d(1 - k^3)}{dk^2}\Big)^2 \\[1em] \Rightarrow \Big(\frac{d^2k^2(k^3 - 1)^2}{d^2k^4}\Big) - \Big(\frac{d^2(1 - k^3)^2}{d^2k^4}\Big) \\[1em] \Rightarrow \Big(\frac{(k^3 - 1)^2}{k^2}\Big) - \Big(\frac{(1 - k^3)^2}{k^4}\Big) \\[1em] \Rightarrow \Big(\frac{k^6 + 1 - 2k^3}{k^2}\Big) - \Big(\frac{1 + k^6 - 2k^3}{k^4}\Big) \\[1em] \Rightarrow \Big(\frac{k^2(k^6 + 1 - 2k^3) - (1 + k^6 - 2k^3)}{k^4}\Big) \\[1em] \Rightarrow \Big(\frac{k^8 + k^2 - 2k^5 - 1 - k^6 + 2k^3}{k^4}\Big).\end{aligned}$$

Substituting values of a, b and c in R.H.S. of $\Big(\frac{a - b}{c} + \frac{a - c}{c}\Big)^2 - \Big(\frac{d - b}{c} + \frac{d - c}{b}\Big)^2 = (a - d)^2 \Big(\frac{1}{c^2} + \frac{1}{b^2}\Big)$ , we get :

$$\begin{aligned}= (a - d)^2 \Big(\frac{1}{c^2} + \frac{1}{b^2}\Big) \\[1em] = (dk^3 - d)^2 \Big(\frac{1}{(dk)^2} + \frac{1}{(dk^2)^2}\Big) \\[1em] = (dk^3 - d)^2 \Big(\frac{1}{d^2k^2} + \frac{1}{d^2k^4}\Big) \\[1em] = \frac{d^2}{d^2k^2}(k^3 - 1)^2 \Big(1 - \frac{1}{k^2} \Big) \\[1em] = \frac{(k^3 - 1)^2(k^2 - 1)}{k^4} \\[1em] = \frac{(k^6 + 1 - 2k^3)(k^2 - 1)}{k^4} \\[1em] = \frac{k^8 - k^6 + k^2 - 1 + 2k^3 - 2k^5}{k^4}.\end{aligned}$$

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

Hence, proved that $\Big(\frac{a - b}{c} + \frac{a - c}{c}\Big)^2 - \Big(\frac{d - b}{c} + \frac{d - c}{b}\Big)^2 = (a - d)^2 \Big(\frac{1}{c^2} + \frac{1}{b^2}\Big)$ .

Given,

ax = by = cz

$\Rightarrow \frac{ax}{abc} = \frac{by}{abc} = \frac{cz}{abc}$

$\Rightarrow \frac{x}{bc} = \frac{y}{ca} = \frac{z}{ab} = k$ (let).

Thus,

x = kbc, y = kca, z = kab

Substitute values of x, y , z in L.H.S , we get :

$$\begin{aligned}\Rightarrow \frac{x^{2}}{yz} + \frac{y^{2}}{zx} + \frac{z^{2}}{xy} \\[1em] \Rightarrow \frac{(kbc)^{2}}{(kca)(kab)} + \frac{(kca)^{2}}{(kab)(kbc)} + \frac{(kab)^{2}}{(kbc)(kca)} \\[1em] \Rightarrow \frac{k^2b^2c^2}{k^2a^2bc} + \frac{k^2c^2a^2}{k^2b^2ca} + \frac{k^2a^2b^2}{k^2c^2ab} \\[1em] \Rightarrow \frac{b^2c^2}{a^2bc} + \frac{c^2a^2}{b^2ca} + \frac{a^2b^2}{c^2ab} \\[1em] \Rightarrow \frac{bc}{a^2} + \frac{ca}{b^2} + \frac{ab}{c^2}.\end{aligned}$$

Hence, proved that $\frac{x^{2}}{yz} + \frac{y^{2}}{zx} + \frac{z^{2}}{xy} = \frac{bc}{a^{2}} + \frac{ca}{b^{2}} + \frac{ab}{c^{2}}$.

Given,

$\frac{x}{b + c - a} = \frac{y}{c + a - b} = \frac{z}{a + b - c}$

Let the common value of the given ratios be k.

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

Therefore,

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

Adding x,y and z, we get:

⇒ x + y + z = k[(b + c − a) + (c + a − b) + (a + b − c)]

⇒ x + y + z = k(a + b + c)

⇒ k = $\frac{x + y + z}{a + b + c}$

Therefore,

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

Given,

(b − c)x + (c − a)y + (a − b)z = 0.

Substituting value of x, y, z in L.H.S of above equation, we get :

⇒ (b − c)[k(b + c - a)] + (c − a)[ k(c + a - b)] + (a − b)[k(a + b - c)]

⇒ k[(b − c)(b + c - a) + (c − a)(c + a - b) + (a − b)(a + b - c)]

⇒ k[(b² - c²) - a(b - c) + (c² - a²) - b(c - a) + (a² - b²) - c(a - b)]

⇒ k[b² - c² - ab + ac + c² - a² - bc + ab + a² - b² - ca + bc]

⇒ k[b² - b² + c² - c² + a² - a² - ab + ab + ac - ac - bc + bc]

⇒ k(0)

⇒ 0.

Hence, proved that $\frac{x}{b + c - a} = \frac{y}{c + a - b} = \frac{z}{a + b - c} = \frac{x + y + z}{a + b + c}$ and (b − c)x + (c − a)y + (a − b)z = 0.

Given,

Since b is the mean proportional between a and c, we have

⇒ a : b :: b : c

⇒ $\frac{a}{b} = \frac{b}{c}$

⇒ b² = ac

Substituting value of b² in $\frac{a^{4} + a^{2}b^{2} + b^{4}}{b^{4} + b^{2}c^{2} + c^{4}}$, we get :

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

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

Given,

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

Multiplying both sides by $\frac{9}{13}$,

$\frac{9a}{13b} = \frac{9c}{13d}$

Applying componendo and dividendo:

$\frac{9a + 13b}{9a - 13b} = \frac{9c + 13d}{9c - 13d}$

Hence, proved that (9a + 13b) : (9a − 13b) = (9c + 13d) : (9c − 13d).

Given,

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

Multiplying both sides by $\frac{3}{2}$,

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

Applying componendo and dividendo:

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

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

Given,

$\frac{3a + 5b}{3a - 5b} = \frac{3c + 5d}{3c - 5d}$

Applying componendo and dividendo, we get :

$$\begin{aligned}\Rightarrow \frac{(3a + 5b) + (3a - 5b)}{(3a + 5b) - (3a - 5b)} = \frac{(3c + 5d) + (3c - 5d)}{(3c + 5d) - (3c - 5d)} \\[1em] \Rightarrow \frac{3a + 5b + 3a - 5b}{3a + 5b - 3a + 5b} = \frac{3c + 5d + 3c - 5d}{3c +5d - 3c + 5d} \\[1em] \Rightarrow \frac{6a}{10b} = \frac{6c}{10d} \\[1em] \Rightarrow \frac{a}{b} = \frac{c}{d}\\[1em]\end{aligned}$$

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

Given,

(4a² + 7b²) : (4a² − 7b²) = (4c² + 7d²) : (4c² − 7d²)

$\Rightarrow \frac{4a^2 + 7b^2}{4a^2 - 7b^2} = \frac{4c^2 + 7d^2}{4c^2 - 7d^2}$

Applying componendo and dividendo:

$$\begin{aligned}\Rightarrow \frac{4a^2 + 7b^2 + 4a^2 - 7b^2}{4a^2 + 7b^2 - (4a^2 - 7b^2)} = \frac{4c^2 + 7d^2 + 4c^2 - 7d^2}{4c^2 + 7d^2 - (4c^2 - 7d^2)} \\[1em] \Rightarrow \frac{8a^2}{4a^2 + 7b^2 - 4a^2 + 7b^2} = \frac{8c^2}{4c^2 + 7d^2 - 4c^2 + 7d^2} \\[1em] \Rightarrow \frac{8a^2}{14b^2} = \frac{8c^2}{14d^2} \\[1em] \Rightarrow \frac{a^2}{b^2} = \frac{c^2}{d^2} \\[1em] \Rightarrow \sqrt{\frac{a^2}{b^2}} = \sqrt{\frac{c^2}{d^2}} \\[1em] \Rightarrow \frac{a}{b} = \frac{c}{d}.\end{aligned}$$

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

Given,

(ma + nb) : (mc + nd) = (ma − nb) : (mc − nd)

$\Rightarrow \frac{ma + nb}{mc + nd} = \frac{ma - nb}{mc - nd}$

Apply Alternendo,

$\frac{ma + nb}{ma - nb} = \frac{mc + nd}{mc - nd}$

Applying componendo and dividendo:

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

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

Apply Componendo & Dividendo :

$$\begin{aligned}\Rightarrow \frac{(5x + 6y) + (5x - 6y)}{(5x + 6y) - (5x - 6y)} = \frac{(5u + 6v) + (5u - 6v)}{(5u + 6v) - (5u - 6v)} \\[1em] \Rightarrow \frac{5x + 6y + 5x - 6y}{5x + 6y - 5x + 6y} = \frac{5u + 6v + 5u - 6v}{5u + 6v - 5u + 6v} \\[1em] \Rightarrow \frac{10x}{12y} = \frac{10u}{12v} \\[1em] \Rightarrow \frac{x}{y} = \frac{u}{v}.\end{aligned}$$

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

Given,

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

Applying componendo and dividendo rule,

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

Again solving x,

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

Apply the Componendo and Dividendo rule,

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

Adding equations (1) and (2), we get :

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

Hence, proved that $\Big(\frac{x + 3a}{x - 3a} + \frac{x + 3b}{x - 3b}\Big) = 2$.

Given,

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

Solving L.H.S:

Applying Componendo and Dividendo, we get :

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

Solving R.H.S:

Apply Componendo and Dividendo:

$$\begin{aligned}\Rightarrow \frac{341 + 91}{341 - 91} \\[1em] \Rightarrow \frac{432}{250}.\end{aligned}$$

Equating L.H.S. and R.H.S.,

$$\begin{aligned}\Rightarrow \Big(\frac{x + 1}{x - 1}\Big)^3 = \frac{432}{250} \\[1em] \Rightarrow \Big(\frac{x + 1}{x - 1}\Big)^3 = \frac{216}{125} \\[1em] \Rightarrow \Big(\frac{x + 1}{x - 1}\Big)^3 = \Big(\frac{6}{5}\Big)^3 \\[1em] \Rightarrow \Big(\frac{x + 1}{x - 1}\Big) = \Big(\frac{6}{5}\Big) \\[1em] \Rightarrow 5(x + 1) = 6(x - 1) \\[1em] \Rightarrow 5x + 5 = 6x - 6 \\[1em] \Rightarrow 6x - 5x = 6 + 5 \\[1em] \Rightarrow x = 11.\end{aligned}$$

Hence, proved that x = 11.

Given,

$\frac{\sqrt{x + 2} + \sqrt{x - 3}}{\sqrt{x + 2} - \sqrt{x - 3}} = 5$

Applying Componendo and Dividendo, we get :

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

Squaring both sides, we get :

$$\begin{aligned}\Rightarrow \frac{x + 2}{x - 3} = \Big(\frac{9}{4}\Big) \\[1em] \Rightarrow 4(x + 2) = 9(x - 3) \\[1em] \Rightarrow 4x + 8 = 9x - 27 \\[1em] \Rightarrow 9x - 4x = 27 + 8 \\[1em] \Rightarrow 5x = 35 \\[1em] \Rightarrow x = \frac{35}{5} = 7.\end{aligned}$$

Hence, proved that x = 7.

Given,

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

Applying Componendo and Dividendo, we get :

$$\begin{aligned}\Rightarrow \frac{(\sqrt{x + 5} + \sqrt{x - 16}) + (\sqrt{x + 5} - \sqrt{x - 16})}{(\sqrt{x + 5} + \sqrt{x - 16}) - (\sqrt{x + 5} - \sqrt{x - 16})} = \frac{7 + 3}{7 - 3} \\[1em] \Rightarrow \frac{(\sqrt{x + 5} + \sqrt{x - 16} + \sqrt{x + 5} - \sqrt{x - 16})}{(\sqrt{x + 5} + \sqrt{x - 16} - \sqrt{x + 5} + \sqrt{x - 16})} = \frac{10}{4} \\[1em] \Rightarrow \frac{2\sqrt{x + 5}}{2\sqrt{x - 16}} = \frac{5}{2} \\[1em] \Rightarrow \frac{\sqrt{x + 5}}{\sqrt{x - 16}} = \frac{5}{2} \\[1em]\end{aligned}$$

Squaring both sides, we get :

$$\begin{aligned}\Rightarrow \Big(\frac{\sqrt{x + 5}}{\sqrt{x - 16}}\Big)^2 = \Big(\frac{5}{2}\Big)^2 \\[1em] \Rightarrow \Big(\frac{x + 5}{x - 16}\Big) = \Big(\frac{25}{4}\Big) \\[1em] \Rightarrow 4(x + 5) = 25(x - 16) \\[1em] \Rightarrow 4x + 20 = 25x - 400 \\[1em] \Rightarrow 25x - 4x = 400 + 20 \\[1em] \Rightarrow 21x = 420 \\[1em] \Rightarrow x = \frac{420}{21} \\[1em] \Rightarrow x = 20.\end{aligned}$$

Hence, proved that x = 20.

(i) Given,

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

Applying Componendo and Dividendo, we get :

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

Squaring both sides, we get :

$$\begin{aligned}\Rightarrow \Big(\frac{\sqrt{3x}}{ \sqrt{2x - 1}}\Big)^2 = \Big(\frac{3}{2}\Big)^2 \\[1em] \Rightarrow \Big(\frac{3x}{2x - 1}\Big) = \Big(\frac{9}{4}\Big) \\[1em] \Rightarrow 4(3x) = 9(2x - 1) \\[1em] \Rightarrow 12x = 18x - 9 \\[1em] \Rightarrow 18x - 12x = 9 \\[1em] \Rightarrow 6x = 9 \\[1em] \Rightarrow x = \frac{9}{6} = \frac{3}{2}\end{aligned}$$

Hence, proved that x = $\frac{3}{2}$.

Given,

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

Applying Componendo and Dividendo, we get :

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

Squaring both sides, we get :

$$\begin{aligned}\Rightarrow \Big(\frac{2x}{\sqrt{4x^2 - 1}}\Big)^2 = \Big(\frac{5}{3}\Big)^2 \\[1em] \Rightarrow \Big(\frac{4x^2}{4x^2 - 1}\Big) = \Big(\frac{25}{9}\Big) \\[1em] \Rightarrow 9(4x^2) = 25(4x^2 - 1) \\[1em] \Rightarrow 36x^2 = 100x^2 - 25 \\[1em] \Rightarrow 100x^2 - 36x^2 = 25 \\[1em] \Rightarrow 64x^2 = 25 \\[1em] \Rightarrow x^2 = \frac{25}{64} \\[1em] \Rightarrow x = \sqrt{\frac{25}{64}} \\[1em] \Rightarrow x = \frac{5}{8}\end{aligned}$$

Hence, x = $\frac{5}{8}$.

Given,

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

Applying Componendo and Dividendo, we get :

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

Squaring both sides, we get :

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

Hence, x = 30.

Given,

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

Applying Componendo and Dividendo, we get :

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

Squaring both sides, we get :

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

Hence, x = 1.

Given,

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

Let, $r = \frac{a + x}{a - x} , \frac{1}{r} = \frac{a - x}{a + x}.$

Substituting value of r and $\frac{1}{r}$ in $16\Big(\frac{a - x}{a + x}\Big)^3 = \Big(\frac{a + x}{a - x}\Big)$, we get :

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

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

(i) 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, $\frac{a + b}{a - b} = \frac{4}{3}.$

(ii) 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,

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

Applying componendo and dividendo, we get :

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

Squaring both sides, we get :

$$\begin{aligned}\Rightarrow \frac{(x + 1)^2}{(x - 1)^2} = \frac{2a + 1}{2a - 1} \\[1em] \Rightarrow \frac{x^2 + 1 + 2x}{x^2 + 1 - 2x} = \frac{2a + 1}{2a - 1} \\[1em] \Rightarrow (x^2 + 1 + 2x)(2a - 1) = (x^2 + 1 - 2x)(2a + 1) \\[1em] \Rightarrow 2ax^2 - x^2 + 2a - 1 + 4ax - 2x = 2ax^2 + x^2 + 2a + 1 - 4ax - 2x \\[1em] \Rightarrow 2ax^2 - 2ax^2 + x^2 + x^2 + 2a - 2a + 1 + 1 - 4ax - 4ax - 2x + 2x = 0 \\[1em] \Rightarrow 2x^2 - 8ax + 2 = 0 \\[1em] \Rightarrow 2(x^2 - 4ax + 1) = 0 \\[1em] \Rightarrow x^2 - 4ax + 1 = 0.\end{aligned}$$

Hence, proved that x² - 4ax + 1 = 0.

Given,

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

Applying componendo and dividendo,

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

Squaring both sides:

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

Hence, proved that 3ax² - 2bx + 3a = 0.

Given,

$\Rightarrow x = \frac{\sqrt{2a + 3b} + \sqrt{2a - 3b}}{\sqrt{2a + 3b} - \sqrt{2a - 3b}}$

Applying componendo and dividendo, we get :

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

Squaring both sides, we get :

$$\begin{aligned}\Rightarrow \frac{(x + 1)^2}{(x - 1)^2} = \frac{2a + 3b}{2a - 3b} \\[1em] \Rightarrow \frac{x^2 + 1 + 2x}{x^2 + 1 - 2x} = \frac{2a + 3b}{2a - 3b} \\[1em] \Rightarrow (x^2 + 1 + 2x)(2a - 3b) = (x^2 + 1 - 2x)(2a + 3b) \\[1em] \Rightarrow 2ax^2 + 2a + 4ax - 3bx^2 - 3b - 6bx = 2ax^2 + 2a - 4ax + 3bx^2 + 3b - 6bx \\[1em] \Rightarrow 2ax^2 - 2ax^2 + 2a - 2a + 4ax + 4ax - 3bx^2 - 3bx^2 - 3b - 3b - 6bx + 6bx = 0 \\[1em] \Rightarrow 8ax - 6bx^2 - 6b = 0 \\[1em] \Rightarrow 6bx^2 - 8ax + 6b = 0 \\[1em] \Rightarrow 2(3bx^2 - 4ax + 3b) = 0 \\[1em] \Rightarrow 3bx^2 - 4ax + 3b = 0.\end{aligned}$$

Hence, proved that 3bx² - 4ax + 3b = 0.

Given,

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

Applying componendo and dividendo, we get :

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

Cubing both sides, we get :

$$\begin{aligned}\Rightarrow \Big(\frac{x + 1}{x - 1}\Big)^3 = \frac{m + 1}{m - 1} \\[1em] \Rightarrow (x + 1)^3 (m - 1) = (x - 1)^3 (m + 1) \\[1em] \Rightarrow (x^3 + 3x^2 + 3x + 1)(m - 1) = (x^3 - 3x^2 + 3x - 1)(m + 1) \\[1em] \Rightarrow m x^3 + 3m x^2 + 3m x + m - x^3 - 3x^2 - 3x - 1 = m x^3 - 3m x^2 + 3m x - m + x^3 - 3x^2 + 3x - 1 \\[1em] \Rightarrow m x^3 + 3m x^2 + 3m x + m - x^3 - 3x^2 - 3x - 1 -( m x^3 - 3m x^2 + 3m x - m + x^3 - 3x^2 + 3x - 1) = 0 \\[1em] \Rightarrow m x^3 - m x^3 + 3m x^2 + 3m x^2 + 3m x - 3m x + m + m - x^3 - x^3 - 3x^2 + 3x^2 - 3x - 3x - 1 + 1 = 0 \\[1em] \Rightarrow 6m x^2 + 2m - 2x^3 - 6x = 0 \\[1em] \Rightarrow 2(3mx^2 + m - x^3 - 3x) = 0 \\[1em] \Rightarrow 3mx^2 + m - x^3 - 3x = 0 \\[1em] \Rightarrow x^3 - 3mx^2 + 3x - m = 0.\end{aligned}$$

Hence, proved that x³ - 3x²m + 3x - m = 0.

Let x be added.

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

⇒ d(a + x) = c(b + x)

⇒ ad + dx = cb + cx

⇒ cx - dx = ad - cb

⇒ x(c - d) = ad - bc

⇒ x = $\frac{ad - bc}{c - d}$

Hence, quantity that must be added = $\frac{ad - bc}{c - d}$.

Given,

$\Rightarrow \frac{a + 3b + 2c + 6d}{a - 3b + 2c - 6d} = \frac{a + 3b - 2c - 6d}{a - 3b - 2c + 6d}$

Applying componendo and dividendo, we get :

$$\begin{aligned}\Rightarrow \frac{(a + 3b + 2c + 6d) + (a - 3b + 2c - 6d)}{(a + 3b + 2c + 6d) - (a - 3b + 2c - 6d)} = \frac{(a + 3b - 2c - 6d) + (a - 3b - 2c + 6d)}{(a + 3b - 2c - 6d) - (a - 3b - 2c + 6d)} \\[1em] \Rightarrow \frac{(a + 3b + 2c + 6d) + (a - 3b + 2c - 6d)}{(a + 3b + 2c + 6d) - a + 3b - 2c + 6d} = \frac{(a + 3b - 2c - 6d) + (a - 3b - 2c + 6d)}{(a + 3b - 2c - 6d) - a + 3b + 2c - 6d} \\[1em] \Rightarrow \frac{2a + 4c}{6b + 12d} = \frac{2a - 4c}{6b - 12d} \\[1em] \Rightarrow \frac{2(a + 2c)}{6(b + 2d)} = \frac{2(a - 2c)}{6(b - 2d)} \\[1em] \Rightarrow \frac{a + 2c}{b + 2d} = \frac{a - 2c}{b - 2d} \\[1em] \Rightarrow \frac{a + 2c}{a - 2c} = \frac{b + 2d}{b - 2d}\end{aligned}$$

Applying componendo and dividendo again,

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

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

Given,

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

Applying componendo and dividendo, we get :

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

Cross - multiplying and simplifying:

$$\begin{aligned}\Rightarrow (2a - 3c)(b - 4d) = (a - 4c)(2b - 3d) \\[1em] \Rightarrow 2ab - 8ad - 3bc + 12cd = 2ab - 3ad - 8bc + 12cd \\[1em] \Rightarrow 2ab - 2ab - 8ad + 3ad - 3bc + 8bc + 12cd - 12cd = 0 \\[1em] \Rightarrow 5bc - 5ad = 0 \\[1em] \Rightarrow 5bc = 5ad \\[1em] \Rightarrow bc = ad \\[1em] \therefore \frac{a}{b} = \frac{c}{d}.\end{aligned}$$

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

Given,

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

Applying componendo and dividendo, we get :

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

Applying componendo and dividendo again:

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

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

Given,

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

Applying componendo and dividendo we get,

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

Applying componendo and dividendo again we get,

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

Hence, x : y = 2 : 3.

Given,

$\Rightarrow \frac{x^{2} + 2x}{2x + 4} = \frac{y^{2} + 3y}{3y + 9}$

Applying componendo and dividendo, we get :

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

Applying componendo and dividendo again we get,

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

Hence, x : y = 2 : 3.

Given,

$\frac{6x^2 + 3x - 5}{3x - 5} = \frac{9x^2 + 2x + 5}{2x + 5}$

Applying componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{6x^2 + 3x - 5 + (3x - 5)}{6x^2 + 3x - 5 - (3x - 5)} = \frac{9x^2 + 2x + 5 + (2x + 5)}{9x^2 + 2x + 5 - (2x + 5)} \\[1em] \Rightarrow \frac{6x^2 + 3x - 5 + 3x - 5}{6x^2 + 3x - 5 - 3x + 5} = \frac{9x^2 + 2x + 5 + 2x + 5}{9x^2 + 2x + 5 - 2x - 5} \\[1em] \Rightarrow \frac{6x^2 + 6x - 10}{6x^2} = \frac{9x^2 + 4x + 10}{9x^2} \\[1em] \Rightarrow \frac{6x^2 + 6x - 10}{6} = \frac{9x^2 + 4x + 10}{9} \\[1em] \Rightarrow 9(6x^2 + 6x - 10) = 6(9x^2 + 4x + 10) \\[1em] \Rightarrow 54x^2 + 54x - 90 = 54x^2 + 24x + 60 \\[1em] \Rightarrow 54x^2 + 54x - 90 - 54x^2 - 24x - 60 = 0 \\[1em] \Rightarrow 30x - 150 = 0 \\[1em] \Rightarrow 30x = 150 \\[1em] \Rightarrow x = \frac{150}{30} \\[1em] \Rightarrow x = 5\end{aligned}$$

Hence, x = 5.

Let the number be x.

Given,

Number obtained by adding number and one-fifth of that number.

$$\begin{aligned}\Rightarrow x + \frac{x}{5} \\[1em] \Rightarrow x\Big(1 + \frac{1}{5}\Big) \\[1em] \Rightarrow x\Big(\frac{5 + 1}{5}\Big) \\[1em] \Rightarrow \Big(\frac{6x}{5}\Big).\end{aligned}$$

Ratio = x : $\frac{6}{5}$x

= $\frac{x}{\frac{6}{5}x}$

= $\frac{5x}{6x}$

= $\frac{5}{6}$

= 5 : 6.

Hence, option 3 is the correct option.

Let the number of jeans Gopal brought be x and shirts be y.

Given,

Cost of jeans = ₹ 1,200.

Cost of shirt = ₹ 300.

Total amount spent = ₹ 8,100.

⇒ 1200x + 300y = 8100

⇒ 1200x + 300y - 8100 = 0

⇒ 300(4x + y - 27) = 0

⇒ 4x + y - 27 = 0

To maximize x keep y ≥ 0

⇒ 4x = 27

Maximum value of x can be 6.

Substitute value of x to get y,

⇒ 4(6) + y - 27 = 0

⇒ 24 + y - 27 = 0

⇒ y - 3 = 0

⇒ y = 3.

Ratio of shirt(y) : jeans(x) = 3 : 6 = 1 : 2.

Hence, option 1 is the correct option.

If we consider,

$\Rightarrow \frac{32}{x} = \frac{y}{2}$

⇒ 32(2) = xy

⇒ xy = 64.

Hence, option 4 is the correct option.

Let ratio of broken mirrors to unbroken mirrors = p : q

Number of mirrors in dozen = 12

Number of broken mirrors = $\frac{p}{p + q} \times 12$

Number of broken mirrors = $\frac{q}{p + q} \times 12$

If we consider 3 : 2,

3 + 2 = 5

$\frac{3}{5} \times 12 = 7.2$

Since 7.2 is not integer. Thus, 3 : 2 is the ratio that cannot exist.

Hence, option 3 is the correct option.

Given,

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

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

Hence, option 4 is the correct option.

Given,

⇒ (x + 1) : 8 = 3.75 : 7

$$\begin{aligned}\Rightarrow \frac{x + 1}{8} = \frac{3.75}{7} \\[1em] \Rightarrow \frac{x + 1}{8} = \frac{3.75}{7} \\[1em] \Rightarrow \frac{x + 1}{8} = \frac{375}{700} \\[1em] \Rightarrow \frac{x + 1}{8} = \frac{15}{28} \\[1em] \Rightarrow x + 1 = \frac{15}{28} \times 8 \\[1em] \Rightarrow x + 1 = \frac{30}{7} \\[1em] \Rightarrow x = \frac{30}{7} - 1 \\[1em] \Rightarrow x = \frac{30 - 7}{7} \\[1em] \Rightarrow x = \frac{23}{7} \\[1em] \Rightarrow x = 3\frac{2}{7}.\end{aligned}$$

Hence, option 3 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \sqrt2:(1 + \sqrt3)::\sqrt6:x \\[1em] \Rightarrow \frac{\sqrt{2}}{1 + \sqrt3} = \frac{\sqrt6}{x} \\[1em] \Rightarrow x = \frac{\sqrt6(1 + \sqrt3)}{\sqrt2} \\[1em] \Rightarrow x = \sqrt3(1 + \sqrt3) \\[1em] \Rightarrow x = \sqrt3 + 3.\end{aligned}$$

Hence, option 1 is the correct option.

Let the numbers be x and y.

Given,

⇒ 0.5x = 0.07y

⇒ $\frac{x}{y} = \frac{0.07}{0.5} = \frac{7}{50}$

⇒ x : y = 7 : 50

Hence, option 3 is the correct option.

Given,

25% of x = 35% of y

$$\begin{aligned}\Rightarrow \frac{25}{100}x = \frac{35}{100}y \\[1em] \Rightarrow 25x = 35y \\[1em] \Rightarrow \frac{x}{y} = \frac{35}{25} \\[1em] \Rightarrow \frac{x}{y} = \frac{7}{5}.\end{aligned}$$

Thus,

x : y = 7 : 5

Hence, option 2 is the correct option.

Given,

A = $\frac{1}{3}$B

B = 3A

B = $\frac{1}{2}$ C

C = 2B

C = 2(3A)

C = 6A

A : B : C = A : 3A : 6A

= 1 : 3 : 6.

Hence, option 2 is the correct option.

Let, 2p = 3q = 4r = k

p = $\frac{k}{2}$

q = $\frac{k}{3}$

r = $\frac{k}{4}$

$$\begin{aligned}\Rightarrow p : q : r = \frac{k}{2}:\frac{k}{3}:\frac{k}{4} \\[1em] = \frac{k}{2} \times 12 : \frac{k}{3} \times 12 : \frac{k}{4} \times 12 \\[1em] = 6k : 4k : 3k \\[1em] = 6 : 4 : 3.\end{aligned}$$

Hence, option 4 is the correct option.

Let, $\frac{a}{3} = \frac{b}{4} = \frac{c}{7} = k$

We can express,

a = 3k, b = 4k, c = 7k

Substituting values in $\frac{a + b + c}{c}$, we get :

$$\begin{aligned}\Rightarrow \frac{3k + 4k + 7k}{7k} \\[1em] \Rightarrow \frac{14k}{7k} \\[1em] \Rightarrow 2.\end{aligned}$$

Hence, option 3 is the correct option.

Given,

x, 5.4, 5, 9 are in proportion.

⇒ $\frac{x}{5.4} = \frac{5}{9}$

⇒ $x = \frac{5}{9} \times 5.4$

⇒ x = 0.6 × 5

⇒ x = 3.

Hence, option 1 is the correct option.

Let forth proportional be x.

⇒ 5 : 8 = 15 : x

⇒ $\frac{5}{8} = \frac{15}{x}$

⇒ 5x = 8 × 15

⇒ x = $\frac{120}{5}$

⇒ x = 24.

Hence, option 4 is the correct option.

Let fourth proportional be x.

⇒ 0.12 : 0.21 = 8 : x

⇒ $\frac{0.12}{0.21} = \frac{8}{x}$

⇒ 0.12x = 0.21 × 8

⇒ x = $\frac{1.68}{0.12}$

⇒ x = 14.

Hence, option 2 is the correct option.

Let third proportional be x.

⇒ 38 : 15 = 15 : x

⇒ $\frac{38}{15} = \frac{15}{x}$

⇒ 38x = 15 × 15

⇒ x = $\frac{15 \times 15}{38}$

Hence, option 3 is the correct option.

Let third proportional be p.

⇒ (x² − y²) : (x − y) = (x − y) : p

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

Hence, option 1 is the correct option.

Let mean proportional be x.

⇒ 9 : x = x : 16

⇒ $\frac{9}{x} = \frac{x}{16}$

⇒ x² = 9(16)

⇒ x = $\sqrt{144}$

⇒ x = 12.

Hence, option 2 is the correct option.

Let mean proportion be x.

⇒ 0.02 : x = x : 0.32

⇒ $\frac{0.02}{x} = \frac{x}{0.32}$

⇒ x² = 0.02 × 0.32

⇒ x² = 0.0064

⇒ x = $\sqrt{0.0064}$

⇒ x = 0.08

Hence, option 1 is the correct option.

Let mean proportion be x.

(3 + $\sqrt{2}$) : x = x : (12 − $\sqrt{32}$)

$$\begin{aligned}\Rightarrow \frac{(3 + \sqrt{2})}{x} = \frac{x}{(12 − \sqrt{32})} \\[1em] \Rightarrow x^2 = (3 + \sqrt{2}) \times (12 − \sqrt{32}) \\[1em] \Rightarrow x^2 = (3 + \sqrt{2}) \times 4(3 − \sqrt{2}) \\[1em] \Rightarrow x^2 = 4[(3)^2 − (\sqrt{2})]^2 \\[1em] \Rightarrow x^2 = 4(9 - 2) \\[1em] \Rightarrow x^2 = 4 \times 7 \\[1em] \Rightarrow x^2 = 28 \\[1em] \Rightarrow x = \sqrt{28} \\[1em] \Rightarrow x = 2\sqrt7.\end{aligned}$$

Hence, option 2 is the correct option.

Given,

Mean proportion between x and y is 6.

⇒ x : 6 = 6 : y

⇒ $\frac{x}{6} = \frac{6}{y}$

⇒ y = $\frac{36}{x}$ ......(1)

Given,

The third proportional to x and y is 48.

⇒ x : y = y : 48

⇒ $\frac{x}{y} = \frac{y}{48}$

⇒ y² = 48x .......(2)

Substitute value of y from equation 1 in equation 2

$$\begin{aligned}\Rightarrow 48x = \Big(\frac{36}{x}\Big)^2 \\[1em] \Rightarrow 48x = \Big(\frac{1296}{x^2}\Big) \\[1em] \Rightarrow 48x \times x^2 = 1296 \\[1em] \Rightarrow 48x^3 = 1296 \\[1em] \Rightarrow x^3 = \frac{1296}{48} \\[1em] \Rightarrow x^3 = 27 \\[1em] \Rightarrow x = \sqrt[3]{27} \\[1em] \Rightarrow x = 3.\end{aligned}$$

Substituting value of x in equation 1, we get :

⇒ y = $\frac{36}{3}$

⇒ y = 12

⇒ x : y = 3 : 12

⇒ x : y = 1 : 4.

Hence, option 2 is the correct option.

Let, third proportional to 12 and 30 be t.

If 12 : 30 :: 30 : t, then,

$$\begin{aligned}\Rightarrow \frac{12}{30} = \frac{30}{t} \\[1em] \Rightarrow t = \frac{30 \times 30}{12} = \frac{30^2}{12} \\[1em] \Rightarrow t = \frac{900}{12} \\[1em] \Rightarrow t = 75.\end{aligned}$$

Let Mean proportion between 9 and 25 be m.

$$\begin{aligned}\Rightarrow \frac{9}{m} = \frac{m}{25} \\[1em] \Rightarrow m^2 = 9 \times 25 \\[1em] \Rightarrow m = \sqrt{9 \times 25} \\[1em] \Rightarrow m = \sqrt{225} \\[1em] \Rightarrow m = 15.\end{aligned}$$

Required ratio = t : m = 75 : 15 = 5 : 1.

Hence, option 2 is the correct option.

Let the first number be x and the second number be y.

According to question,

⇒ y - 30% of (x) = $\frac{4}{5}$y

$$\begin{aligned}\Rightarrow y - \frac{30}{100}x = \frac{4}{5}y \\[1em] \Rightarrow y - \frac{3}{10}x = \frac{4}{5}y \\[1em] \Rightarrow y - \frac{4}{5}y = \frac{3}{10}x \\[1em] \Rightarrow \frac{y}{5} = \frac{3}{10}x \\[1em] \Rightarrow \frac{x}{y} = \frac{1}{5} \times \frac{10}{3} \\[1em] \Rightarrow \frac{x}{y} = \frac{10}{15} \\[1em] \Rightarrow \frac{x}{y} = \frac{2}{3}.\end{aligned}$$

⇒ x : y = 2 : 3.

Hence, option 2 is the correct option.

Let the cost of a mangoe be M and the cost of a guava be G.

Given,

Five mangoes and four guavas cost as much as three mangoes and seven guavas.

$$\begin{aligned}\Rightarrow 5M + 4G = 3M + 7G \\[1em] \Rightarrow 5M - 3M = 7G - 4G \\[1em] \Rightarrow 2M = 3G \\[1em] \Rightarrow \frac{M}{G} = \frac{3}{2}.\end{aligned}$$

Hence, option 2 is the correct option.

Given,

Total examinees of a class = 132 and ratio of passed to failed = 9 : 2.

Let number of passed students be 9x and failed students be 2x.

Then,

⇒ 9x + 2x = 132

⇒ 11x = 132

⇒ x = $\frac{132}{11}$ = 12.

The number of passed students = 9 × 12 = 108

The number of failed students = 2 × 12 = 24

If 4 more students passed, then

New number of passed students = 108 + 4 = 112

New number of failed students = 24 - 4 = 20

Then the new ratio of passed to failed = 112 : 20 = 28 : 5.

Hence, option 2 is the correct option.

Given,

Initial ratio,

P : Q : R = 1 : 2 : 3

Let initial no. of marbles in P, Q and R be x, 2x and 3x respectively.

Total number of marbles = 60.

⇒ x + 2x + 3x = 60

⇒ 6x = 60

⇒ x = $\frac{60}{6}$

⇒ x = 10

P = 1 × 10 = 10, Q = 2 × 10 = 20, R = 3 × 10 = 30.

Given,

New ratio of marbles = 3 : 4 : 5.

Let now the no. of marbles in box P, Q and R be 3x, 4x and 5x respectively.

⇒ 3x + 4x + 5x = 60

⇒ 12x = 60

⇒ x = $\frac{60}{12}$ = 5.

P = 3 × 5 = 15, Q = 4 × 5 = 20, R = 5 × 5 = 25.

Thus, if initially 5 marbles are transferred from R to P then the ratio changes from 1 : 2 : 3 to 3 : 4 : 5.

Hence, option 4 is the correct option.

Given,

x² + 4y² = 4xy

Dividing the equation by y²,

$\therefore \Big(\frac{x}{y}\Big)^2 + 4 = 4\Big(\frac{x}{y}\Big)$

Let $\frac{x}{y} = k$. Then

⇒ k² + 4 = 4k

⇒ k² - 4k + 4 = 0

⇒ k² - 2k - 2k + 4 = 0

⇒ k(k - 2) - 2(k - 2) = 0

⇒ (k - 2)(k - 2) = 0

⇒ (k - 2)= 0      [Using Zero-product rule]

⇒ k = 2

Therefore, $\frac{x}{y} = k = \frac{2}{1}$

Hence, option 4 is the correct option.

Given,

3A = 5B

$\Rightarrow \frac{A}{B} = \frac{5}{3}$ ....(1)

Also, 4B = 6C

$\Rightarrow \frac{B}{C} = \frac{6}{4} = \frac{3}{2}$ ....(2)

From (1) and (2):

$$\begin{aligned}\Rightarrow \frac{A}{C} = \frac{A}{B} \times \frac{B}{C} \\[1em] \Rightarrow \frac{A}{C} = \frac{5}{3} \times \frac{3}{2} \\[1em] \Rightarrow \frac{A}{C} = \frac{5}{2}.\end{aligned}$$

Therefore, A : C = 5 : 2.

Hence, option 4 is the correct option.

Given,

A : B = 7 : 9 and B : C = 5 : 4

L.C.M of two values of B, that are 9 and 5 is 45.

$$\begin{aligned}\Rightarrow \frac{A}{B} = \frac{7 \times 5}{9\times 5} = \frac{35}{45} \\[1em] \Rightarrow \frac{B}{C} = \frac{5 \times 9}{4 \times 9} = \frac{45}{36}.\end{aligned}$$

A : B : C = 35 : 45 : 36.

Hence, option 3 is the correct option.

Given,

8a = 9b

$\therefore \frac{a}{b} = \frac{9}{8}$

Solving,

$$\begin{aligned}\Rightarrow \frac{\frac{a}{9}}{\frac{b}{8}} \\[1em] \Rightarrow \frac{a}{9} \times \frac{8}{b} \\[1em] \Rightarrow \frac{8a}{9b} \\[1em] \Rightarrow \frac{8}{9} \times \frac{9}{8} \\[1em] \Rightarrow \frac{9}{9} \\[1em] \Rightarrow \frac{1}{1} \\[1em] \Rightarrow 1 : 1.\end{aligned}$$

Hence, option 1 is the correct option.

Given,

a, b, c and d are proportional.

∴ a : b = c : d

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

Applying componendo and dividendo, we get :

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

Hence, option 4 is the correct option.

Given,

x : y = 3 : 2

Let x = 3k and y = 2k for some k.

For ratio (2x² + 3y²) : (3x² − 2y²),

Substituting value of x and y in the antecedent,

⇒ 2x² + 3y²

⇒ 2(3k)² + 3(2k)²

⇒ 2 × 9k² + 3 × 4k²

⇒ 18k² + 12k²

⇒ 30k².

Substituting value of x and y in the consequent,

⇒ 3x² - 2y²

⇒ 3(3k)² - 2(2k)²

⇒ 3 × 9k² - 2 × 4k²

⇒ 27k² - 8k²

⇒ 19k².

Therefore the required ratio is:

$$\begin{aligned}\Rightarrow \frac{2x^2 + 3y^2}{3x^2 - 2y^2} \\[1em] \Rightarrow \frac{30k^2}{19k^2} \\[1em] \Rightarrow \frac{30}{19} \\[1em] \Rightarrow 30 : 19.\end{aligned}$$

Hence, option 4 is the correct option.

Given,

$\therefore \frac{5a + 3b}{2a - 3b} = \frac{23}{5}$

Cross multiplying:

$$\begin{aligned}\Rightarrow 5(5a + 3b) = 23(2a - 3b) \\[1em] \Rightarrow 25a + 15b = 46a - 69b \\[1em] \Rightarrow 25a - 46a = -69b - 15b \\[1em] \Rightarrow -21a = -84b \\[1em] \Rightarrow 21a = 84b \\[1em] \Rightarrow \frac{a}{b} = \frac{84}{21} = 4.\end{aligned}$$

Thus,

a : b = 4 : 1.

Hence, option 4 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \frac{4x^2 - 3y^2}{2x^2 + 5y^2} = \frac{12}{19} \\[1em] \Rightarrow 19(4x^2 - 3y^2) = 12(2x^2 + 5y^2) \\[1em] \Rightarrow 76x^2 - 57y^2 = 24x^2 + 60y^2 \\[1em] \Rightarrow 76x^2 - 24x^2 = 60y^2 + 57y^2 \\[1em] \Rightarrow 52x^2 = 117y^2 \\[1em] \Rightarrow \frac{x^2}{y^2} = \frac{117}{52} \\[1em] \Rightarrow \frac{x^2}{y^2} = \frac{9}{4} \\[1em] \Rightarrow \frac{x}{y} = \sqrt{\frac{9}{4}} \\[1em] \Rightarrow \frac{x}{y} = \frac{3}{2}.\end{aligned}$$

Thus,

x : y = 3 : 2.

Hence, option 4 is the correct option.

Given,

a : b = b : c

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

⇒ b² = ac

⇒ b⁴ = (ac)²

⇒ b⁴ = a²c².

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

⇒ a⁴ : b⁴ = a² : c²

Hence, option 2 is the correct option.

Given,

a : b : c = 2 : 3 : 4

Let a = 2k, b = 3k, c = 4k.

Now,

$$\begin{aligned}\Rightarrow \frac{1}{a} : \frac{1}{b} : \frac{1}{c} = \frac{1}{2k} : \frac{1}{3k} : \frac{1}{4k} \\[1em] = \frac{1}{2k} \times 12k : \frac{1}{3k} \times 12k : \frac{1}{4k} \times 12k \\[1em] = 6 : 4 : 3.\end{aligned}$$

Hence, option 3 is the correct option.

Given,

$\frac{1}{x} : \frac{1}{y} : \frac{1}{z} = 2 : 3 : 5$

Thus,

x : y : z = $\frac{1}{2} : \frac{1}{3} : \frac{1}{5}$

Since, L.C.M of 2, 3, 5 is 30.

$x : y : z = \frac{1}{2} \times 30 : \frac{1}{3} \times 30 : \frac{1}{5} \times 30$

= 15 : 10 : 6.

Hence, option 3 is the correct option.

Given,

⇒ (x + y) : (x − y) = 4 : 1

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

Applying Componendo and Dividendo, we get :

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

Let x = 5k and y = 3k for some constant k.

Substituting value of x and y in $\frac{x^2 + y^2}{x^2 - y^2}$, we get:

$$\begin{aligned}\Rightarrow \frac{(5k)^2 + (3k)^2}{(5k)^2 - (3k)^2} \\[1em] \Rightarrow \frac{25k^2 + 9k^2}{25k^2 - 9k^2} \\[1em] \Rightarrow \frac{34k^2}{16k^2} \\[1em] \Rightarrow \frac{17}{8}.\end{aligned}$$

Thus, (x² + y²) : (x² − y²) = 17 : 8.

Hence, option 2 is the correct option.

Given,

a : b : c = 2 : 3 : 4

Let a = 2k, b = 3k and c = 4k for some constant k.

Substituting value of a, b and c in 2a − 3b + 4c = 33, we get :

⇒ 2(2k) − 3(3k) + 4(4k) = 33

⇒ 4k - 9k + 16k = 33

⇒ 11k = 33

⇒ k = $\frac{33}{11}$

⇒ k = 3.

⇒ 4k = 4(3) = 12.

Hence, option 4 is the correct option.

Given,

x, y, z are in continued proportion

Let $\frac{x}{y} = \frac{y}{z} = k$ for some constant ratio k.

⇒ $\frac{x}{y} = k, \frac{y}{z} = k$

⇒ y = zk and x = yk = (zk)k = zk²

Substitute value of x and y in $\frac{y^2 + z^2}{x^2 + y^2}$ we get:

$$\begin{aligned}\Rightarrow \frac{(zk)^2 + z^2}{(zk^2)^2 + (zk)^2} \\[1em] \Rightarrow \frac{z^2k^2 + z^2}{z^2k^4 + z^2k^2} \\[1em] \Rightarrow \frac{z^2(k^2 + 1)}{z^2k^2(k^2 + 1)} \\[1em] \Rightarrow \frac{1}{k^2} \\[1em] \Rightarrow \frac{1}{\frac{x}{y} \times \frac{y}{z}} \\[1em] \Rightarrow \frac{1}{\frac{x}{z}} \\[1em] \Rightarrow \frac{z}{x} \\[1em] \Rightarrow z : x.\end{aligned}$$

Hence, option 1 is the correct option.

Let $\frac{a}{b} = \frac{b}{c} = \frac{c}{d}$ = k where k is the constant ratio.

Therefore,

c = dk

b = ck = (dk)k = dk²

a = bk = (dk²)k = dk³

k³ = $\frac{a}{d}$

Substitute value of a,b and c in $\frac{b^3 + c^3 + d^3}{a^3 + b^3 + c^3}$, we get:

$$\begin{aligned}\Rightarrow \frac{(dk^2)^3 + (dk)^3 + d^3}{(dk^3)^3 + (dk^2)^3 + (dk)^3} \\[1em] \Rightarrow \frac{d^3k^6 + d^3k^3 + d^3}{d^3k^9 + d^3k^6 + d^3k^3} \\[1em] \Rightarrow \frac{d^3(k^6 + k^3 + 1)}{d^3k^3(k^6 + k^3 + 1)} \\[1em] \Rightarrow \frac{1}{k^3} \\[1em] \Rightarrow \frac{1}{\frac{a}{d}} \\[1em] \Rightarrow \frac{d}{a}.\end{aligned}$$

Hence, option 4 is the correct option.

Given,

⇒ a : b = c : d

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

Let, $\frac{a}{b} = \frac{c}{d}$ = k where k is the constant ratio.

Therefore, a = bk and c = dk

Substituting values of a and c in $\frac{ma + nc}{mb + nd}$, we get :

$$\begin{aligned}\Rightarrow \frac{m(bk) + n(dk)}{mb + nd} \\[1em] \Rightarrow \frac{k(mb + nd)}{mb + nd} \\[1em] \Rightarrow k \\[1em] \Rightarrow \frac{a}{b} \\[1em] \Rightarrow a : b.\end{aligned}$$

Hence, option 4 is the correct option.

Given,

Total weight of alloy = 17 kg 500 g = 17.5 kg

Given,

Ratio of copper and zinc is 5 : 2.

Let initial quantity of copper be 5x and zinc be 2x.

$$\begin{aligned}\text{Initial quantity of copper} = \frac{5x}{5x + 2x} \times 17.5 \\[1em] = \frac{5x}{7x} \times 17.5 \\[1em] = 5 \times 2.5 \\[1em] = 12.5\text{ kg} \\[1em] \text{Initial quantity of zinc} = \frac{2x}{5x + 2x} \times 17.5 \\[1em] = \frac{2x}{7x} \times 17.5 \\[1em] = 2 \times 2.5 \\[1em] = 5\text{ kg}.\end{aligned}$$

1.250 kg of zinc is mixed in 17 kg 500 g of alloy, so new quantity of zinc = 5 + 1.250 = 6.25 kg.

Ratio of copper to zinc = 12.5 : 6.25 = 2 : 1

Hence, option 2 is the correct option.

Wages ₹ 6,400 divided into 3 workers in ratio = $\frac{3}{5} : 2 : \frac{5}{3}$

L.C.M of 5 and 3 is 15

$\Rightarrow \frac{3}{5} \times 15:15 \times 2: \frac{5}{3} \times 15$

⇒ 9 : 30 : 25

Sum of the ratio = 9 + 30 + 25 = 64

The share of the second worker = $\frac{30}{64} \times 6400$

= 30 × 100

= ₹ 3,000.

Hence, option 2 is the correct option.

The sum ₹ x is divided between A and B in the ratio :

A : B = $\frac{a}{b}:\frac{c}{d}$

L.C.M of denominators bd:

A : B = $\frac{a}{b} \times bd :\frac{c}{d} \times bd$

A : B = ad : bc

Sum of the ratio = ad + bc.

The sum of money A get = $\frac{ad}{ad + bc} \times x = \frac{adx}{ad + bc}$.

Hence, option 3 is the correct option.

Given,

The total number of students is 720.

The initial ratio of boys (B) to girls (G) is 7 : 5.

The total number of ratio parts is 7 + 5 = 12.

Initial number of boys = $\frac{7}{12} \times 720$ = 420.

Initial number of girls = $\frac{5}{12} \times 720$ = 300.

In order to make the ratio 1 : 1, the number of girls and boys must be equal.

Thus, 420 - 300 = 120, more girls should be admitted.

Hence, option 2 is the correct option.

Let the two numbers be 7x and 11x.

Given,

When 7 is added to each number, the new ratio is 2 : 3.

$\therefore \frac{7x + 7}{11x + 7} = \frac{2}{3}$

⇒ 3(7x + 7) = 2(11x + 7)

⇒ 21x + 21 = 22x + 14

⇒ 22x - 21x = 21 - 14

⇒ x = 7.

The smaller number is 7x = 7(7) = 49.

Hence, option 2 is the correct option.

Let the number to be subtracted be x.

Thus, the numbers 7 - x, 9 - x, 11 - x and 15 - x are in proportion.

$\Rightarrow \frac{7 - x}{9 - x} = \frac{11 - x}{15 - x}$

⇒ (7 - x)(15 - x) = (11 - x)(9 - x)

⇒ 105 - 7x - 15x + x² = 99 - 11x - 9x + x²

⇒ 105 - 22x + x² = 99 - 20x + x²

⇒ 105 - 99 = 22x - 20x

⇒ 6 = 2x

⇒ x = $\frac{6}{2}$

⇒ x = 3.

Hence, option 3 is the correct option.

Since, p, q, and r are in continued proportion.

$\therefore \frac{p}{q} = \frac{q}{r}$

⇒ q² = pr.....(1)

Solving,

p:r = p²:q²

$$\begin{aligned}\Rightarrow \frac{p}{r} = \frac{p^2}{q^2} \\[1em] \Rightarrow q^2 = \frac{p^2 \times r}{p} \\[1em] \Rightarrow q^2 = pr...(2)\end{aligned}$$

Since, equation (1) and (2) are equal.

Hence, Option 4 is the correct option.

Let the initial quantities of milk and water be 7x and 3x respectively.

Given,

The total volume of the mixture is 80 litres.

⇒ 7x + 3x = 80

⇒ 10x = 80

⇒ x = $\frac{80}{10}$

⇒ x = 8.

Initial Quantity of Milk : 7x = 7 × 8 = 56 litres

Initial Quantity of Water : 3x = 3 × 8 = 24 litres

Let W be the quantity of water added to the mixture.

The quantity of milk remains unchanged.

Water = 24 + W

The new ratio of milk to water is required to be 2 : 1.

⇒ $\frac{56}{24 + W} = \frac{2}{1}$

⇒ 56 = 2(24 + W)

⇒ 56 = 48 + 2W

⇒ 2W = 56 - 48

⇒ 2W = 8

⇒ W = $\frac{8}{2}$

⇒ W = 4.

The quantity of water to be added is 4 litres.

Hence, option 1 is the correct option.

Let the number to be added be x.

Thus, numbers 7 + x, 11 + x and 19 + x will be in continued proportion.

$$\begin{aligned}\therefore \frac{7 + x}{11 + x} = \frac{11 + x}{19 + x} \\[1em] \Rightarrow (7 + x)(19 + x) = (11 + x)(11 + x) \\[1em] \Rightarrow 133 + 7x + 19x + x^2 = 121 + 11x + 11x + x^2 \\[1em] \Rightarrow x^2 + 26x + 133 = x^2 + 22x + 121 \\[1em] \Rightarrow x^2 - x^2 + 26x - 22x = 121 - 133 \\[1em] \Rightarrow 4x = -12 \\[1em] \Rightarrow x = \frac{-12}{4} \\[1em] \Rightarrow x = -3.\end{aligned}$$

Hence, option 2 is the correct option.

Given,

Let the two numbers be 3k and 5k.

After subtracting 9 from each number, the ratio changes to 12 : 23.

$$\begin{aligned}\therefore \frac{3k - 9}{5k - 9} = \frac{12}{23} \\[1em] \Rightarrow 23(3k - 9) = 12(5k - 9) \\[1em] \Rightarrow 69k - 207 = 60k - 108 \\[1em] \Rightarrow 69k - 60k = 207 - 108 \\[1em] \Rightarrow 9k = 99 \\[1em] \Rightarrow k = \frac{99}{9} \\[1em] \Rightarrow k = 11.\end{aligned}$$

The larger number = 5k = 5 × 11 = 55.

Hence, option 3 is the correct option.

Let the number to be added be x.

$$\begin{aligned}\therefore \frac{3 + x}{5 + x} = \frac{5}{6} \\[1em] \Rightarrow 6(3 + x) = 5(5 + x) \\[1em] \Rightarrow 18 + 6x = 25 + 5x \\[1em] \Rightarrow 6x - 5x = 25 - 18 \\[1em] \Rightarrow x = 7.\end{aligned}$$

Hence, option 2 is the correct option.

Given,

ab = 1, bc = $\frac{1}{2}$, cd = 6, de = 2, ef = $\frac{1}{2}$.

$$\begin{aligned}\Rightarrow ef = \frac{1}{2} \\[1em] \Rightarrow e = \frac{1}{2f} \\[1em] \Rightarrow de = 2 \\[1em] \Rightarrow d = \frac{2}{e} = \frac{2}{\frac{1}{2f}} = 4f \\[1em] \Rightarrow cd = 6 \\[1em] \Rightarrow c = \frac{6}{d} = \frac{6}{4f} = \frac{3}{2f} \\[1em] \Rightarrow bc = \frac{1}{2} \\[1em] \Rightarrow b = \frac{1}{2c} = \frac{1}{2 \times \frac{3}{2f}} = \frac{f}{3} \\[1em] \Rightarrow ab = 1 \\[1em] \Rightarrow a = \frac{1}{b} = \frac{1}{\frac{f}{3}} = \frac{3}{f}\end{aligned}$$

ad : be : cf = $\frac{3}{f} \times 4f : \frac{f}{3} \times \frac{1}{2f}: \frac{3}{2f} \times f$

= 12 : $\frac{1}{6} : \frac{3}{2}$

Since, L.C.M. of 2 and 6 is 6.

= 12 × 6 : $\frac{1}{6} \times 6 : \frac{3}{2} \times 6$

⇒ 72 : 1 : 9.

Hence, option 4 is the correct option.

Given,

a = $\frac{4xy}{x+y}$

Solving,

$$\begin{aligned}\Rightarrow \frac{a+2x}{a-2x} + \frac{a+2y}{a-2y} = \frac{\frac{4xy}{x+y} + 2x}{\frac{4xy}{x+y} - 2x} + \frac{\frac{4xy}{x+y} + 2y}{\frac{4xy}{x+y} - 2y} \\[1em] \Rightarrow \text{Multiply numerator and denominator by (x + y)} \\[1em] \Rightarrow \frac{\Big(\frac{4xy}{x+y} + 2x\Big)\times (x + y)}{\Big(\frac{4xy}{x+y} - 2x\Big) \times (x + y)} + \frac{\Big(\frac{4xy}{x+y} + 2y\Big)\times (x + y)}{\Big(\frac{4xy}{x+y} - 2y\Big) \times (x + y)}\\[1em] \Rightarrow \frac{4xy + 2x(x+y)}{4xy - 2x(x+y)} + \frac{4xy + 2y(x+y)}{4xy - 2y(x+y)} \\[1em] \Rightarrow \frac{4xy + 2x^2 + 2xy}{4xy - 2x^2 - 2xy} + \frac{4xy + 2xy + 2y^2}{4xy - 2xy - 2y^2} \\[1em] \Rightarrow \frac{2x^2 + 6xy}{-2x^2 + 2xy} + \frac{6xy + 2y^2}{2xy - 2y^2} \\[1em] \Rightarrow \frac{2x(x + 3y)}{2x(y - x)} + \frac{2y(3x+y)}{2y(x-y)} \\[1em] \Rightarrow \frac{x + 3y}{y - x} + \frac{3x + y}{x - y} \\[1em] \Rightarrow \frac{x + 3y}{-(x - y)} + \frac{3x + y}{x - y} \\[1em] \Rightarrow \frac{-(x + 3y)}{x - y} + \frac{3x + y}{x - y} \\[1em] \Rightarrow \frac{-x - 3y + 3x + y}{x - y} \\[1em] \Rightarrow \frac{-x + 3x + y - 3y}{x - y} \\[1em] \Rightarrow \frac{2x - 2y}{x - y} \\[1em] \Rightarrow \frac{2(x - y)}{x - y} \\[1em] \Rightarrow 2.\end{aligned}$$

Hence, option 2 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} - \sqrt{1-x}} = \frac{a}{b} \\[1em] \Rightarrow b\Big(\sqrt{1+x} + \sqrt{1-x}\Big) = a\Big(\sqrt{1+x} - \sqrt{1-x}\Big) \\[1em] \Rightarrow b\sqrt{1+x} + b\sqrt{1-x} = a\sqrt{1+x} - a\sqrt{1-x} \\[1em] \Rightarrow b\sqrt{1+x} - a\sqrt{1+x} = -a\sqrt{1-x} - b\sqrt{1-x} \\[1em] \Rightarrow (b-a)\sqrt{1+x} = -(a+b)\sqrt{1-x} \\[1em] \text{Squaring on Both Sides,} \\[1em] \Rightarrow (b-a)^2(1+x) = (a+b)^2(1-x) \\[1em] \Rightarrow (b^2 - 2ab + a^2)(1+x) = (a^2 + 2ab + b^2)(1-x) \\[1em] \Rightarrow (a^2 + b^2 - 2ab)(1+x) = (a^2 + b^2 + 2ab)(1-x) \\[1em] \Rightarrow (a^2 + b^2 - 2ab) + (a^2 + b^2 - 2ab)x = (a^2 + b^2 + 2ab) - (a^2 + b^2 + 2ab)x \\[1em] \Rightarrow (a^2 + b^2 - 2ab) - (a^2 + b^2 + 2ab) = -(a^2 + b^2 - 2ab)x - (a^2 + b^2 + 2ab)x \\[1em] \Rightarrow a^2 + b^2 - 2ab - a^2 - b^2 - 2ab = - a^2x - b^2x + 2abx - a^2x - b^2x - 2abx \\[1em] \Rightarrow a^2 - a^2 - b^2 + b^2 - 2ab - 2ab = - a^2x - a^2x - b^2x - b^2x + 2abx - 2abx \\[1em] \Rightarrow -4ab = - 2a^2x - 2b^2x \\[1em] \Rightarrow -4ab = -2(a^2 + b^2)x \\[1em] \Rightarrow x = \frac{-4ab}{-2(a^2 + b^2)} \\[1em] \Rightarrow x = \frac{2ab}{a^2 + b^2}.\end{aligned}$$

Hence, option 4 is the correct option.

Given,

$$\begin{aligned}x = \frac{\sqrt{2a+1} + \sqrt{2a-1}}{\sqrt{2a+1} - \sqrt{2a-1}} \\[1em] x\Big(\sqrt{2a+1} - \sqrt{2a-1}\Big) = \sqrt{2a+1} + \sqrt{2a-1} \\[1em] x\sqrt{2a+1} - x\sqrt{2a-1} = \sqrt{2a+1} + \sqrt{2a-1} \\[1em] (x-1)\sqrt{2a+1} = (x+1)\sqrt{2a-1} \\[1em] \text{Squaring on Both Sides,} \\[1em] (x-1)^2(2a+1) = (x+1)^2(2a-1) \\[1em] (2a+1)(x^2 - 2x + 1) = (2a-1)(x^2 + 2x + 1) \\[1em] 2ax^2 - 4ax + 2a + x^2 - 2x + 1 = 2ax^2 + 4ax + 2a - x^2 - 2x - 1 \\[1em] 2ax^2 - 2ax^2 + x^2 + x^2 - 4ax - 4ax + 2a - 2a - 2x + 2x + 1 + 1 = 0\\[1em] 2x^2 - 8ax + 2 = 0 \\[1em] 2(x^2 - 4ax + 1) = 0 \\[1em] x^2 - 4ax + 1 = 0\end{aligned}$$

Hence, option 4 is the correct option.

Given,

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

From the first relation: a = k(b + c) ....(1)

From the second relation: b = k(c + a) ....(2)

From the third relation: c = k(a + b) ....(3)

Adding (1), (2) and (3):

a + b + c = k(b + c) + k(c + a) + k(a + b)

a + b + c = k((b + c) + (c + a) + (a + b))

a + b + c = k(b + c + c + a + a + b)

a + b + c = k(a + a + b + b + c + c)

a + b + c = k(2a + 2b + 2c)

a + b + c = 2k(a + b + c)

If a + b + c ≠ 0, then,

$2k =\frac{(a + b + c)}{(a + b + c)}$

2k = 1

$k = \frac{1}{2}.$

If a + b + c = 0, then the equations will be,

a + b = -c ....(4)

b + c = -a ....(5)

c + a = -b ....(6)

Subtituting values in $\frac{a}{b+c} = \frac{b}{c+a} = \frac{c}{a+b}$

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

k = -1

Hence, Option 3 is the correct option.

Since b is the mean proportion between a and c,

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

$\Rightarrow b^2 = ac$

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

Hence, option 2 is the correct option.

Since b is the mean proportion between a and c,

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

Let the required mean proportion be x. Then,

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

Now substitute b² = ac:

⇒ x² = (a² + b²)(b² + c²)

⇒ x² = a²b² + a²c² + b⁴ + b²c²

⇒ x² = a²b² + (ac)² + b⁴ + b²c²

⇒ x² = a²b² + (b)² + b⁴ + b²c²

⇒ x² = a²b² + b⁴ + b⁴ + b²c²

⇒ x² = b²(a² + 2b² + c²)

⇒ x² = b²(a² + 2ac + c²)

⇒ x² = b²(a + c)².

Therefore,

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

Hence, option 2 is the correct option.

We know that,

Each ratio = $\frac{\text{sum of antecedents}}{\text{sum of consequents}}$.

Let $\frac{a}{b} = \frac{c}{d}$ = k for some constant k.

a = kb, c = kd

Substituting value of a and c in $\frac{a+c}{b+d}$, we get :

$$\begin{aligned}\Rightarrow \frac{kb + kd}{b + d} \\[1em] \Rightarrow \frac{k(b + d)}{(b + d)} \\[1em] \Rightarrow k \\[1em] \Rightarrow \frac{a}{b} \text{ or } \frac{c}{d}.\end{aligned}$$

∴ Both A and R are true, and R is the correct explanation of A.

Hence, option 1 is the correct option.

We know that,

Mean proportion between two numbers

= $\sqrt{\text{First number} \times \text{Second number}}$

Thus,

The mean proportion between x and y is given by $\sqrt{xy}$.

∴ Reason (R) is true.

Thus,

The mean proportion between a²b and $\frac{1}{b}$ = $\sqrt{a^2b \times \frac{1}{b}}$

$= \sqrt{a^2} = a$.

∴ Assertion (A) is false.

Hence, option 4 is the correct option.

Given,

⇒ 2x = 3y

⇒ $\frac{x}{y} = \frac{3}{2}$

⇒ x : y = 3 : 2

⇒ 4y = 5z

⇒ $\frac{y}{z} = \frac{5}{4}$

⇒ y : z = 5 : 4

Since, x : y = 3 : 2 and y : z = 5 : 4. L.C.M of 2 and 5 is 10.

⇒ x : y = 3 : 2 = (3 × 5) : (2 × 5) = 15 : 10

⇒ y : z = 5 : 4 = (5 × 2) : (4 × 2) = 10 : 8

⇒ x : y : z = 15 : 10 : 8.

∴ Assertion (A) is true.

If x = y and y = z, then by the transitive property of equality, it follows that x = y = z.

x : y : z = x : x : x = 1 : 1 : 1.

∴ Reason (R) is false.

Hence, option 3 is the correct option.

Let the numbers be a and b.

Given,

Mean proportion between the numbers is 6.

$\therefore \frac{a}{6} = \frac{6}{b}$

⇒ ab = 36

⇒ a = $\frac{36}{b}$ .....(1)

Third proportion of the numbers is 48.

$\therefore \frac{a}{b} = \frac{b}{45}$

⇒ b² = 48a

⇒ b² = 48 × $\frac{36}{b}$

⇒ b³ = 1728

⇒ b = $\sqrt[3]{1728}$

⇒ b = 12.

Substituting value of b in equation (1), we get :

⇒ a = $\frac{36}{12}$ = 3

Hence, the numbers are 3 and 12.

(a) Since,

$\frac{1400}{28000} = \frac{5600}{112000} = \frac{1}{20}$

Thus, number of customers and profit per week are in direct proportion not in continued proportion.

Hence, number of customers and profit per week are not in continued proportion.

(b) As,

Number of customers and profit per week are in proportion.

$\therefore \frac{x}{32140} = \frac{1}{20}$

$\Rightarrow x = 32140 \times \frac{1}{20} = 1607$

and

$\therefore \frac{3212}{y} = \frac{1}{20}$

⇒ y = 3212 × 20 = 64240.

Hence, x = 1607 and y = 64240.