Simultaneous (Linear) Equations

Given,

Equations : x - 2y = 7 and 2x - y = 8.

⇒ x - 2y = 7

⇒ x = 7 + 2y ........(1)

Substituting value of x from equation (1) in 2x - y = 8, we get :

⇒ 2(7 + 2y) - y = 8

⇒ 14 + 4y - y = 8

⇒ 3y = 8 - 14

⇒ 3y = -6

⇒ y = $-\frac{6}{3}$ = -2.

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

⇒ x = 7 + 2 × -2 = 7 - 4 = 3.

Hence, Option 3 is the correct option.

Given,

Equations :

⇒ x + y = 2.1 ..........(1)

⇒ x - y = 0.3 ............(2)

Subtracting equation (2) from (1), we get :

⇒ x + y - (x - y) = 2.1 - 0.3

⇒ x - x + y - (-y) = 1.8

⇒ 2y = 1.8

⇒ y = $\frac{1.8}{2}$ = 0.9

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

⇒ x + 0.9 = 2.1

⇒ x = 2.1 - 0.9 = 1.2

Hence, Option 1 is the correct option.

Given,

Equations :

$$\begin{aligned}\Rightarrow \frac{1}{x} + \frac{1}{y} = 21 .......(1) \\[1em] \Rightarrow \frac{1}{x} - \frac{1}{y} + 9 = 0 \\[1em] \Rightarrow \frac{1}{x} - \frac{1}{y} = -9 .......(2)\end{aligned}$$

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

$$\begin{aligned}\Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{x} - \frac{1}{y} = 21 + (-9) \\[1em] \Rightarrow \frac{2}{x} = 12 \\[1em] \Rightarrow x = \frac{2}{12} = \frac{1}{6}.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow \frac{1}{x} + \frac{1}{y} = 21 \\[1em] \Rightarrow \frac{1}{\frac{1}{6}} + \frac{1}{y} = 21 \\[1em] \Rightarrow 6 + \frac{1}{y} = 21 \\[1em] \Rightarrow \frac{1}{y} = 21 - 6 \\[1em] \Rightarrow \frac{1}{y} = 15 \\[1em] \Rightarrow y = \frac{1}{15}.\end{aligned}$$

Hence, Option 4 is the correct option.

Given,

Equations :

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

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

$$\begin{aligned}\Rightarrow \frac{x}{2} - \frac{y}{3} + \frac{x}{2} + \frac{y}{3} = 0 + 6 \\[1em] \Rightarrow \frac{2x}{2} = 6 \\[1em] \Rightarrow x = 6.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow \frac{6}{2} - \frac{y}{3} = 0 ..........(1) \\[1em] \Rightarrow \frac{6}{2} = \frac{y}{3} \\[1em] \Rightarrow \frac{y}{3} = 3 \\[1em] \Rightarrow y = 9.\end{aligned}$$

Hence, Option 2 is the correct option.

Given,

Equations :

$\Rightarrow \frac{x + 8}{5} - \frac{2y - 4}{3} = 0$ ......(1)

⇒ y - x = 3

⇒ y = x + 3 .........(2)

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

$$\begin{aligned}\Rightarrow \frac{x + 8}{5} - \frac{2(x + 3) - 4}{3} = 0 \\[1em] \Rightarrow \frac{x + 8}{5} - \frac{2x + 6 - 4}{3} = 0 \\[1em] \Rightarrow \frac{x + 8}{5} - \frac{2x + 2}{3} = 0 \\[1em] \Rightarrow \frac{3(x + 8) - 5(2x + 2)}{15} = 0 \\[1em] \Rightarrow 3x + 24 - 10x - 10 = 0 \\[1em] \Rightarrow -7x + 14 = 0 \\[1em] \Rightarrow 7x = 14 \\[1em] \Rightarrow x = \frac{14}{7} = 2.\end{aligned}$$

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

⇒ y = x + 3 = 2 + 3 = 5.

Hence, Option 1 is the correct option.

Solve the following pairs of linear (simultaneously) equations using method of elimination by substitution:

Given,

Equations : 2x + 3y = 8 and 2x = 2 + 3y

⇒ 2x + 3y = 8

⇒ 2x = 8 - 3y

⇒ x = $\frac{8 - 3y}{2}$ ............(1)

Substituting value of x from equation (1) in 2x = 2 + 3y, we get :

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

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

$$\begin{aligned}\Rightarrow x = \frac{8 - 3 \times 1}{2} \\[1em] = \frac{8 - 3}{2} \\[1em] = \frac{5}{2} = 2.5\end{aligned}$$

Hence, x = 2.5 and y = 1.

Given,

Equations : 0.2x + 0.1y = 25 and 2(x - 2) - 1.6y = 116

⇒ 0.2x + 0.1y = 25

⇒ 0.2x = 25 - 0.1y

⇒ x = $\frac{25 - 0.1y}{0.2}$ ..........(1)

$$\begin{aligned}\Rightarrow 2(x - 2) - 1.6y = 116 \\[1em] \Rightarrow 2x - 4 - 1.6y = 116 \\[1em] \Rightarrow 2x - 1.6y = 116 + 4 \\[1em] \Rightarrow 2x - 1.6y = 120 \\[1em]\end{aligned}$$

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

$$\begin{aligned}\Rightarrow 2 \times \Big(\frac{25 - 0.1y}{0.2}\Big) - 1.6y = 120 \\[1em] \Rightarrow \frac{25 - 0.1y}{0.1} - 1.6y = 120 \\[1em] \Rightarrow \frac{25 - 0.1y - 0.16y}{0.1} = 120 \\[1em] \Rightarrow 25 - 0.26y = 120 \times 0.1 \\[1em] \Rightarrow 25 - 0.26y = 12 \\[1em] \Rightarrow 0.26y = 25 - 12 \\[1em] \Rightarrow 0.26y = 13 \\[1em] \Rightarrow y = \frac{13}{0.26} = \frac{1300}{26} = 50.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow x = \frac{25 - 0.1 \times 50}{0.2} \\[1em] \Rightarrow x = \frac{25 - 5}{0.2} \\[1em] \Rightarrow x = \frac{20}{0.2} = 100.\end{aligned}$$

Hence, x = 100 and y = 50.

Given,

Equations : 6x = 7y + 7 and 7y - x = 8

⇒ 7y - x = 8

⇒ x = 7y - 8 .........(1)

Substituting value of x from equation (1) in 6x = 7y + 7, we get :

⇒ 6(7y - 8) = 7y + 7

⇒ 42y - 48 = 7y + 7

⇒ 42y - 7y = 7 + 48

⇒ 35y = 55

⇒ y = $\frac{55}{35} = \frac{11}{7}$.

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

$$\begin{aligned}\Rightarrow x = 7 \times \frac{11}{7} - 8 \\[1em] \Rightarrow x = 11 - 8 = 3.\end{aligned}$$

Hence, x = 3 and y = $\frac{11}{7}$.

Given,

Equations :

⇒ y = 4x - 7 .........(1)

⇒ 16x - 5y = 25 .......(2)

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

⇒ 16x - 5(4x - 7) = 25

⇒ 16x - 20x + 35 = 25

⇒ -4x = 25 - 35

⇒ -4x = -10

⇒ 4x = 10

⇒ x = $\frac{10}{4} = \frac{5}{2}$.

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

$$\begin{aligned}\Rightarrow y = 4 \times \frac{5}{2} - 7 \\[1em] \Rightarrow y = 10 - 7 = 3.\end{aligned}$$

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

Given,

Equations : 1.5x + 0.1y = 6.2 and 3x - 0.4y = 11.2

⇒ 1.5x + 0.1y = 6.2

⇒ 1.5x = 6.2 - 0.1y

⇒ x = $\frac{6.2 - 0.1y}{1.5}$ ........(1)

Substituting value of x from equation (1) in 3x - 0.4y = 11.2, we get :

$$\begin{aligned}\Rightarrow 3 \times \Big(\frac{6.2 - 0.1y}{1.5}\Big) - 0.4y = 11.2 \\[1em] \Rightarrow \frac{6.2 - 0.1y}{0.5} - 0.4y = 11.2 \\[1em] \Rightarrow \frac{6.2 - 0.1y - 0.2y}{0.5} = 11.2 \\[1em] \Rightarrow 6.2 - 0.3y = 11.2 \times 0.5 \\[1em] \Rightarrow 6.2 - 0.3y = 5.6 \\[1em] \Rightarrow 0.3y = 6.2 - 5.6 \\[1em] \Rightarrow 0.3y = 0.6 \\[1em] \Rightarrow y = \frac{0.6}{0.3} = 2.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow x = \frac{6.2 - 0.1 \times 2}{1.5} \\[1em] = \frac{6.2 - 0.2}{1.5} \\[1em] = \frac{6}{1.5} \\[1em] = 4.\end{aligned}$$

Hence, x = 4 and y = 2.

Given,

Equations : 2(x - 3) + 3(y - 5) = 0 and 5(x - 1) + 4(y - 4) = 0

⇒ 2(x - 3) + 3(y - 5) = 0

⇒ 2x - 6 + 3y - 15 = 0

⇒ 2x + 3y - 21 = 0

⇒ 2x = 21 - 3y

⇒ x = $\frac{21 - 3y}{2}$ ..........(1)

⇒ 5(x - 1) + 4(y - 4) = 0

⇒ 5x - 5 + 4y - 16 = 0

⇒ 5x + 4y - 21 = 0

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

$$\begin{aligned}\Rightarrow 5 \times \Big(\frac{21 - 3y}{2}\Big) + 4y - 21 = 0 \\[1em] \Rightarrow \frac{105 - 15y}{2} + 4y - 21 = 0 \\[1em] \Rightarrow \frac{105 - 15y + 8y - 42}{2} = 0 \\[1em] \Rightarrow 63 - 7y = 0 \\[1em] \Rightarrow 7y = 63 \\[1em] \Rightarrow y = \frac{63}{7} = 9.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow x = \frac{21 - 3 \times 9}{2} \\[1em] = \frac{21 - 27}{2} \\[1em] = -\frac{6}{2} \\[1em] = -3.\end{aligned}$$

Hence, x = -3 and y = 9.

Simplifying first equation :

$$\begin{aligned}\Rightarrow \frac{2x + 1}{7} + \frac{5y - 3}{3} = 12 \\[1em] \Rightarrow \frac{3(2x + 1) + 7(5y - 3)}{21} = 12 \\[1em] \Rightarrow 6x + 3 + 35y - 21 = 12 \times 21 \\[1em] \Rightarrow 6x + 35y - 18 = 252 \\[1em] \Rightarrow 6x + 35y = 252 + 18 \\[1em] \Rightarrow 6x + 35y = 270 \\[1em] \Rightarrow 6x = 270 - 35y \\[1em] \Rightarrow x = \frac{270 - 35y}{6} \text{ ........(1)}\end{aligned}$$

Simplifying second equation :

$$\begin{aligned}\Rightarrow \frac{3x + 2}{2} - \frac{4y + 3}{9} = 13 \\[1em] \Rightarrow \frac{9(3x + 2) - 2(4y + 3)}{18} = 13 \\[1em] \Rightarrow 27x + 18 - 8y - 6 = 18 \times 13 \\[1em] \Rightarrow 27x - 8y + 12 = 234 \\[1em] \Rightarrow 27x - 8y = 234 -12 \\[1em] \Rightarrow 27x - 8y = 222 \text{ .......(2)}\end{aligned}$$

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

$$\begin{aligned}\Rightarrow 27 \times \Big(\frac{270 - 35y}{6}\Big) - 8y = 222 \\[1em] \Rightarrow \frac{9(270 - 35y)}{2} - 8y = 222 \\[1em] \Rightarrow \frac{2430 - 315y - 16y}{2} = 222 \\[1em] \Rightarrow 2430 - 331y = 444 \\[1em] \Rightarrow 331y = 2430 - 444 \\[1em] \Rightarrow 331y = 1986 \\[1em] \Rightarrow y = \frac{1986}{331} = 6.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow x = \frac{270 - 35 \times 6}{6} \\[1em] = \frac{270 - 210}{6} \\[1em] = \frac{60}{6} \\[1em] = 10.\end{aligned}$$

Hence, x = 10 and y = 6.

For solving each pair of equations, use the method of elimination by equating coefficients :

Given equations :

⇒ 3x - y = 23 .........(1)

⇒ $\frac{x}{3} + \frac{y}{4}$ = 4 ......(2)

Multiplying equation (1) by 3, we get :

⇒ 3(3x - y) = 3 × 23

⇒ 9x - 3y = 69 .........(3)

Multiplying equation (2) by 12, we get :

⇒ $12\Big(\frac{x}{3} + \frac{y}{4}\Big) = 4 \times 12$

⇒ 4x + 3y = 48 ...........(4)

Adding equation (3) and (4), we get :

⇒ 9x - 3y + 4x + 3y = 69 + 48

⇒ 13x = 117

⇒ x = $\frac{117}{13}$ = 9.

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

⇒ 3 × 9 - y = 23

⇒ 27 - y = 23

⇒ y = 27 - 23 = 4.

Hence, x = 9 and y = 4.

Given equations :

$$\begin{aligned}\Rightarrow \frac{5y}{2} - \frac{x}{3} = 8 ..........(1) \\[1em] \Rightarrow \frac{y}{2} + \frac{5x}{3} = 12 ......(2)\end{aligned}$$

Multiplying equation (1) by 30, we get :

$$\begin{aligned}\Rightarrow 30\Big(\frac{5y}{2} - \frac{x}{3}\Big) = 30 \times 8 \\[1em] \Rightarrow 75y - 10x = 240 ......(3)\end{aligned}$$

Multiplying equation (2) by 6, we get :

$$\begin{aligned}\Rightarrow 6\Big(\frac{y}{2} + \frac{5x}{3}\Big) = 12 \times 6 \\[1em] \Rightarrow 3y + 10x = 72 ...........(4)\end{aligned}$$

Adding equations (3) and (4), we get :

⇒ 75y - 10x + 3y + 10x = 240 + 72

⇒ 78y = 312

⇒ y = $\frac{312}{78}$ = 4.

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

$$\begin{aligned}\Rightarrow \frac{5 \times 4}{2} - \frac{x}{3} = 8 \\[1em] \Rightarrow 10 - \frac{x}{3} = 8 \\[1em] \Rightarrow \frac{x}{3} = 10 - 8 \\[1em] \Rightarrow x = 3 \times 2 = 6.\end{aligned}$$

Hence, x = 6 and y = 4.

Simplifying first equation :

⇒ $\frac{1}{5}(x - 2) = \frac{1}{4}(1 - y)$

⇒ 4(x - 2) = 5(1 - y)

⇒ 4x - 8 = 5 - 5y

⇒ 4x + 5y - 8 - 5 = 0

⇒ 4x + 5y - 13 = 0 .......(1)

⇒ 26x + 3y + 4 = 0 .......(2)

Multiplying equation (1) by 3, we get :

⇒ 3(4x + 5y - 13) = 0

⇒ 12x + 15y - 39 = 0 .......(3)

Multiplying equation (2) by 5, we get :

⇒ 5(26x + 3y + 4) = 0

⇒ 130x + 15y + 20 = 0 .......(4)

Subtracting equation (3) from (4), we get :

⇒ 130x + 15y + 20 - (12x + 15y - 39) = 0

⇒ 130x - 12x + 15y - 15y + 20 - (-39) = 0

⇒ 118x + 59 = 0

⇒ 118x = -59

⇒ x = $-\frac{59}{118} = -\frac{1}{2}$

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

$$\begin{aligned}\Rightarrow 4x + 5y - 13 = 0 \\[1em] \Rightarrow 4 \times -\frac{1}{2} + 5y - 13 = 0 \\[1em] \Rightarrow -2 + 5y - 13 = 0 \\[1em] \Rightarrow 5y - 15 = 0 \\[1em] \Rightarrow 5y = 15 \\[1em] \Rightarrow y = \frac{15}{5} = 3.\end{aligned}$$

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

Simplifying first equation :

⇒ $\frac{x - y}{6} = 2(4 - x)$

⇒ x - y = 12(4 - x)

⇒ x - y = 48 - 12x

⇒ x + 12x - y = 48

⇒ 13x - y = 48

⇒ 13x - y - 48 = 0 .......(1)

Simplifying second equation :

⇒ 2x + y = 3(x - 4)

⇒ 2x + y = 3x - 12

⇒ y + 2x - 3x + 12 = 0

⇒ y - x + 12 = 0 .......(2)

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

⇒ (13x - y - 48) + (y - x + 12) = 0

⇒ 13x - x - y + y - 48 + 12 = 0

⇒ 12x - 36 = 0

⇒ 12x = 36

⇒ x = $\frac{36}{12}$ = 3.

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

⇒ y - 3 + 12 = 0

⇒ y + 9 = 0

⇒ y = -9.

Hence, x = 3 and y = -9.

Given, equations :

⇒ 2x - 3y - 3 = 0 .............(1)

⇒ $\frac{2x}{3} + 4y + \frac{1}{2} = 0$ .......(2)

Simplifying second equation :

$$\begin{aligned}\Rightarrow \frac{2x}{3} + 4y + \frac{1}{2} = 0 \\[1em] \Rightarrow \frac{4x + 24y + 3}{6} = 0 \\[1em] \Rightarrow 4x + 24y + 3 = 0 \text{ .......(3)}\end{aligned}$$

Multiplying equation (1) by 2, we get :

⇒ 2(2x - 3y - 3) = 2 × 0

⇒ 4x - 6y - 6 = 0 .........(4)

Subtracting equation (4) from (3), we get :

⇒ 4x + 24y + 3 - (4x - 6y - 6) = 0

⇒ 4x - 4x + 24y + 6y + 3 + 6 = 0

⇒ 30y + 9 = 0

⇒ 30y = -9

⇒ y = $-\frac{9}{30} = -\frac{3}{10}$.

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

$$\begin{aligned}\Rightarrow 2x - 3y - 3 = 0 \\[1em] \Rightarrow 2x - 3 \times -\frac{3}{10} - 3 = 0 \\[1em] \Rightarrow 2x + \frac{9}{10} - 3 = 0\\[1em] \Rightarrow 2x = 3 - \frac{9}{10} \\[1em] \Rightarrow 2x = \frac{30 - 9}{10} \\[1em] \Rightarrow x = \frac{21}{20}.\end{aligned}$$

Hence, x = $\frac{21}{20} \text{ and } y = -\frac{3}{10}$.

Given, equations :

⇒ 13x + 11y = 70 ..........(1)

⇒ 11x + 13y = 74 ..........(2)

Multiplying equation (1) by 11, we get :

⇒ 11(13x + 11y) = 11 × 70

⇒ 143x + 121y = 770 ........(3)

Multiplying equation (2) by 13, we get :

⇒ 13(11x + 13y) = 13 × 74

⇒ 143x + 169y = 962 ........(4)

Subtracting equation (3) from (4), we get :

⇒ 143x + 169y - (143x + 121y) = 962 - 770

⇒ 143x - 143x + 169y - 121y = 192

⇒ 48y = 192

⇒ y = $\frac{192}{48}$ = 4

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

⇒ 13x + 11(4) = 70

⇒ 13x + 44 = 70

⇒ 13x = 70 - 44

⇒ 13x = 26

⇒ x = $\frac{26}{13}$ = 2.

Hence, x = 2 and y = 4.

Given, equations :

⇒ 41x + 53y = 135 .........(1)

⇒ 53x + 41y = 147 .........(2)

Multiplying equation (1) by 53, we get :

⇒ 53(41x + 53y) = 53 × 135

⇒ 2173x + 2809y = 7155 .......(3)

Multiplying equation (2) by 41, we get :

⇒ 41(53x + 41y) = 41 × 147

⇒ 2173x + 1681y = 6027 .......(4)

Subtracting equation (4) from (3), we get :

⇒ 2173x + 2809y - (2173x + 1681y) = 7155 - 6027

⇒ 2173x - 2173x + 2809y - 1681y = 1128

⇒ 1128y = 1128

⇒ y = $\frac{1128}{1128}$ = 1.

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

⇒ 41x + 53y = 135

⇒ 41x + 53(1) = 135

⇒ 41x + 53 = 135

⇒ 41x = 135 - 53

⇒ 41x = 82

⇒ x = $\frac{82}{41}$ = 2.

Hence, x = 2 and y = 1.

Given,

Equations : 2x + y = 23 and 4x - y = 19

⇒ 2x + y = 23

⇒ y = 23 - 2x .......(1)

Substituting value of y from equation (1) in 4x - y = 19, we get :

⇒ 4x - (23 - 2x) = 19

⇒ 4x - 23 + 2x = 19

⇒ 6x = 19 + 23

⇒ 6x = 42

⇒ x = $\frac{42}{6}$ = 7.

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

⇒ y = 23 - 2(7) = 23 - 14 = 9.

⇒ x - 3y = 7 - 3 × 9 = 7 - 27 = -20

⇒ 5y - 2x = 5 × 9 - 2 × 7 = 45 - 14 = 31.

Hence, x - 3y = -20 and 5y - 2x = 31.

Given,

Equations : 10y = 7x - 4 and 12x + 18y = 1

⇒ 10y = 7x - 4

⇒ y = $\frac{7x - 4}{10}$ .........(1)

Substituting value of y from equation (1) in 12x + 18y = 1, we get :

$$\begin{aligned}\Rightarrow 12x + 18 \times \Big(\frac{7x - 4}{10}\Big) = 1 \\[1em] \Rightarrow 12x + \frac{126x - 72}{10} = 1 \\[1em] \Rightarrow \frac{120x + 126x - 72}{10} = 1\\[1em] \Rightarrow 246x - 72 = 10 \\[1em] \Rightarrow 246x = 10 + 72 \\[1em] \Rightarrow 246x = 82 \\[1em] \Rightarrow x = \frac{82}{246} = \frac{1}{3}.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow y = \frac{7 \times \frac{1}{3} - 4}{10}\\[1em] = \frac{\frac{7}{3} - 4}{10} \\[1em] = \frac{\frac{7 - 12}{3}}{10} \\[1em] = \frac{-5}{3 \times 10} \\[1em] = \frac{-5}{30} \\[1em] = -\frac{1}{6}.\end{aligned}$$

Substituting value of x and y in 4x + 6y and 8y - x, we get :

$$\begin{aligned}\Rightarrow 4x + 6y = 4 \times \frac{1}{3} + 6 \times -\frac{1}{6} \\[1em] = \frac{4}{3} + (-1) \\[1em] = \frac{4 - 3}{3} \\[1em] = \frac{1}{3}. \\[1em] \Rightarrow 8y - x = 8 \times -\frac{1}{6} - \frac{1}{3} \\[1em] = -\frac{4}{3} - \frac{1}{3} \\[1em] = \frac{-4 - 1}{3} \\[1em] = -\frac{5}{3}.\end{aligned}$$

Hence, $4x + 6y = \frac{1}{3} \text{ and } 8y - x = -\frac{5}{3}$.

Simplifying first equation :

$$\begin{aligned}\Rightarrow \frac{y + 7}{5} = \frac{2y - x}{4} + 3x - 5 \\[1em] \Rightarrow \frac{y + 7}{5} = \frac{2y - x + 12x - 20}{4} \\[1em] \Rightarrow 4(y + 7) = 5(2y + 11x - 20) \\[1em] \Rightarrow 4y + 28 = 10y + 55x - 100 \\[1em] \Rightarrow 55x + 10y - 4y = 100 + 28 \\[1em] \Rightarrow 55x + 6y = 128 \\[1em] \Rightarrow 55x = 128 - 6y \\[1em] \Rightarrow x = \frac{128 - 6y}{55}\text{ .......(1)}\end{aligned}$$

Simplifying second equation :

$$\begin{aligned}\Rightarrow \frac{7 - 5x}{2} + \frac{3 - 4y}{6} = 5y - 18 \\[1em] \Rightarrow \frac{3(7 - 5x) + 3 - 4y}{6} = 5y - 18 \\[1em] \Rightarrow \frac{21 - 15x + 3 - 4y}{6} = 5y - 18 \\[1em] \Rightarrow 24 - 15x - 4y = 6(5y - 18) \\[1em] \Rightarrow 24 - 15x - 4y = 30y - 108 \\[1em] \Rightarrow 15x + 30y + 4y = 108 + 24 \\[1em] \Rightarrow 15x + 34y = 132 \text{ ......(2)}.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow 15 \times \frac{128 - 6y}{55} + 34y = 132 \\[1em] \Rightarrow \frac{3}{11} \times (128 - 6y) + 34y = 132 \\[1em] \Rightarrow \frac{384 - 18y + 374y}{11} = 132 \\[1em] \Rightarrow 384 - 18y + 374y = 1452 \\[1em] \Rightarrow 384 + 356y = 1452 \\[1em] \Rightarrow 356y = 1452 - 384 \\[1em] \Rightarrow 356y = 1068 \\[1em] \Rightarrow y = \frac{1068}{356} = 3.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow x = \frac{128 - 6 \times 3}{55} \\[1em] = \frac{128 - 18}{55}\\[1em] = \frac{110}{55} \\[1em] = 2.\end{aligned}$$

Hence, x = 2 and y = 3.

Simplifying first equation :

$$\begin{aligned}\Rightarrow 4x = 17 - \frac{x - y}{8} \\[1em] \Rightarrow 4x = \frac{136 - (x - y)}{8} \\[1em] \Rightarrow 4x = \frac{136 - x + y}{8} \\[1em] \Rightarrow 32x = 136 - x + y \\[1em] \Rightarrow 32x + x - 136 = y \\[1em] \Rightarrow y = 33x - 136 \text{ .......(1)}\end{aligned}$$

Simplifying second equation :

$$\begin{aligned}\Rightarrow 2y + x = 2 + \frac{5y + 2}{3} \\[1em] \Rightarrow 2y + x = \frac{6 + 5y + 2}{3} \\[1em] \Rightarrow 3(2y + x) = 5y + 8 \\[1em] \Rightarrow 6y + 3x = 5y + 8 \\[1em] \Rightarrow 6y - 5y = 8 - 3x \\[1em] \Rightarrow y = 8 - 3x \text{ .......(2)}\end{aligned}$$

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

⇒ 33x - 136 = 8 - 3x

⇒ 33x + 3x = 136 + 8

⇒ 36x = 144

⇒ x = $\frac{144}{36}$ = 4.

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

⇒ y = 8 - 3(4) = 8 - 12 = -4.

Hence, x = 4 and y = -4.

Given, equations :

⇒ x + y = 8 and 3x - 5y = 0

⇒ x + y - 8 = 0 ...........(1)

⇒ 3x - 5y = 0 ..........(2)

By cross-multiplication method :

$$\begin{aligned}\Rightarrow \frac{x}{1 \times 0 - (-5) \times (-8)} = \frac{y}{(-8) \times 3 - 0 \times 1} = \frac{1}{1 \times (-5) - 3 \times 1} \\[1em] \Rightarrow \frac{x}{0 - 40} = \frac{y}{-24 - 0} = \frac{1}{-5 - 3} \\[1em] \Rightarrow \frac{x}{-40} = \frac{y}{-24} = \frac{1}{-8} \\[1em] \Rightarrow \frac{x}{-40} = \frac{1}{-8} \text{ and } \frac{y}{-24} = \frac{1}{-8} \\[1em] \Rightarrow x = \frac{-40}{-8} \text{ and } y = \frac{-24}{-8} \\[1em] \Rightarrow x = 5 \text{ and } y = 3.\end{aligned}$$

Hence, Option 2 is the correct option.

Given,

1st equation :

$$\begin{aligned}\Rightarrow \frac{4a - 3}{6} + \frac{5b - 7}{2} = 18 - 5a \\[1em] \Rightarrow \frac{4a - 3 + 3(5b - 7)}{6} = 18 - 5a \\[1em] \Rightarrow \frac{4a - 3 + 15b - 21}{6} = 18 - 5a \\[1em] \Rightarrow 4a + 15b - 24 = 6(18 - 5a) \\[1em] \Rightarrow 4a + 15b - 24 = 108 - 30a \\[1em] \Rightarrow 4a + 30a + 15b = 108 + 24 \\[1em] \Rightarrow 34a + 15b = 132 \\[1em] \Rightarrow 34a + 15b - 132 = 0\end{aligned}$$

2nd equation :

⇒ a + b = 5

⇒ a + b - 5 = 0

By cross-multiplication method :

$$\begin{aligned}\Rightarrow \frac{a}{15 \times (-5) - 1\times (-132)} = \frac{b}{(-132) \times 1 - (-5) \times 34} = \frac{1}{34 \times 1 - 1 \times 15} \\[1em] \Rightarrow \frac{a}{-75 + 132} = \frac{b}{-132 + 170} = \frac{1}{34 - 15} \\[1em] \Rightarrow \frac{a}{57} = \frac{b}{38} = \frac{1}{19} \\[1em] \Rightarrow \frac{a}{57} = \frac{1}{19} \text{ and } \frac{b}{38} = \frac{1}{19}\\[1em] \Rightarrow a = \frac{57}{19} \text{ and } b = \frac{38}{19} \\[1em] \Rightarrow a = 3 \text{ and } b = 2.\end{aligned}$$

Hence, Option 3 is the correct option.

Given, equations : $\frac{x}{7} - \frac{y}{2} = 2 \text{ and } \frac{x - y}{3} = 3$

Simplifying first equation :

$$\begin{aligned}\Rightarrow \frac{x}{7} - \frac{y}{2} = 2 \\[1em] \Rightarrow \frac{2x - 7y}{14} = 2 \\[1em] \Rightarrow 2x - 7y = 28 \\[1em] \Rightarrow 2x - 7y - 28 = 0 \text{ ........(1)}\end{aligned}$$

Simplifying second equation :

$$\begin{aligned}\Rightarrow \frac{x - y}{3} = 3 \\[1em] \Rightarrow x - y = 9 \\[1em] \Rightarrow x - y - 9 = 0 \text{ .......(2)}\end{aligned}$$

By cross-multiplication method :

$$\begin{aligned}\Rightarrow \frac{x}{(-7) \times (-9) - (-1) \times (-28)} = \frac{y}{(-28) \times 1 - (-9) \times 2} = \frac{1}{2 \times (-1) - 1 \times (-7)} \\[1em] \Rightarrow \frac{x}{63 - 28} = \frac{y}{-28 - (-18)} = \frac{1}{-2 + 7} \\[1em] \Rightarrow \frac{x}{35} = \frac{y}{-10} = \frac{1}{5} \\[1em] \Rightarrow \frac{x}{35} = \frac{1}{5} \text{ and } \frac{y}{-10} = \frac{1}{5} \\[1em] \Rightarrow x = \frac{35}{5} \text{ and } y = \frac{-10}{5} \\[1em] \Rightarrow x = 7 \text{ and } y = -2.\end{aligned}$$

Hence, Option 4 is the correct option.

Given, equations : 11x - 7y = 29 and 7x - 11y = 25

⇒ 11x - 7y - 29 = 0 .......(1)

⇒ 7x - 11y - 25 = 0 .......(2)

By cross-multiplication method we have :

$$\begin{aligned}\Rightarrow \frac{x}{(-7) \times (-25) - (-11) \times (-29)} = \frac{y}{(-29) \times 7 - (-25) \times 11} = \frac{1}{11 \times (-11) - 7 \times (-7)} \\[1em] \Rightarrow \frac{x}{175 - 319} = \frac{y}{-203 + 275} = \frac{1}{-121 + 49} \\[1em] \Rightarrow \frac{x}{-144} = \frac{y}{72} = \frac{1}{-72} \\[1em] \Rightarrow \frac{x}{-144} = \frac{1}{-72} \text{ and } \frac{y}{72} = \frac{1}{-72} \\[1em] \Rightarrow x = \frac{-144}{-72} \text{ and } y = \frac{72}{-72} \\[1em] \Rightarrow x = 2 \text{ and } y = -1.\end{aligned}$$

Hence, Option 2 is the correct option.

Given, equations can be written as :

⇒ 15x + 6y - 27 = 0 .......(1)

⇒ 6x + 15y - 36 = 0 ........(2)

By cross-multiplication method :

$$\begin{aligned}\Rightarrow \frac{x}{6 \times (-36) - 15 \times (-27)} = \frac{y}{(-27) \times 6 - (-36) \times 15} = \frac{1}{15 \times 15 - 6 \times 6} \\[1em] \Rightarrow \frac{x}{-216 + 405} = \frac{y}{-162 - (-540)} = \frac{1}{225 - 36} \\[1em] \Rightarrow \frac{x}{189} = \frac{y}{-162 + 540} = \frac{1}{189}\\[1em] \Rightarrow \frac{x}{189} = \frac{y}{378} = \frac{1}{189}\\[1em] \Rightarrow \frac{x}{189} = \frac{1}{189} \text{ and } \frac{y}{378} = \frac{1}{189} \\[1em] \Rightarrow x = \frac{189}{189} \text{ and } y = \frac{378}{189} \\[1em] \Rightarrow x = 1 \text{ and } y = 2.\end{aligned}$$

Hence, Option 3 is the correct option.

Given, equations :

$\frac{2}{x} - \frac{3}{y} = 13$ .......(1)

$\frac{3}{x} + \frac{2}{y} = 0$ ........(2)

Multiplying equation (1) by 2, we get :

$$\begin{aligned}\Rightarrow 2\Big(\frac{2}{x} - \frac{3}{y}\Big) = 2 \times 13 \\[1em] \Rightarrow \frac{4}{x} - \frac{6}{y} = 26 \text{ .......(3)}\end{aligned}$$

Multiplying equation (2) by 3, we get :

$$\begin{aligned}\Rightarrow 3\Big(\frac{3}{x} + \frac{2}{y}\Big) = 3 \times 0 \\[1em] \Rightarrow \frac{9}{x} + \frac{6}{y} = 0 \text{ .......(4)}\end{aligned}$$

Adding equations (3) and (4), we get :

$$\begin{aligned}\Rightarrow \Big(\frac{4}{x} - \frac{6}{y}\Big) + \Big(\frac{9}{x} + \frac{6}{y}\Big) = 26 + 0 \\[1em] \Rightarrow \frac{4}{x} + \frac{9}{x} - \frac{6}{y} + \frac{6}{y} = 26 \\[1em] \Rightarrow \frac{13}{x} = 26 \\[1em] \Rightarrow x = \frac{13}{26} = \frac{1}{2}.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow \frac{3}{\frac{1}{2}} + \frac{2}{y} = 0 \\[1em] \Rightarrow 6 + \frac{2}{y} = 0 \\[1em] \Rightarrow \frac{2}{y} = 0 - 6 \\[1em] \Rightarrow \frac{2}{y} = -6 \\[1em] \Rightarrow y = \frac{2}{-6} = -\frac{1}{3}.\end{aligned}$$

Hence, Option 3 is the correct option.

Given, equations :

⇒ 4x + 3y = 17 and 3x - 4y + 6 = 0

⇒ 4x + 3y - 17 = 0 .......(1)

⇒ 3x - 4y + 6 = 0 ........(2)

By cross-multiplication method :

$$\begin{aligned}\Rightarrow \frac{x}{3 \times 6 - (-4) \times (-17)} = \frac{y}{(-17) \times 3 - 6 \times 4} = \frac{1}{4 \times (-4) - 3 \times 3} \\[1em] \Rightarrow \frac{x}{18 - 68} = \frac{y}{-51 - 24} = \frac{1}{-16 - 9} \\[1em] \Rightarrow \frac{x}{-50} = \frac{y}{-75} = \frac{1}{-25} \\[1em] \Rightarrow \frac{x}{-50} = \frac{1}{-25} \text{ and } \frac{y}{-75} = \frac{1}{-25} \\[1em] \Rightarrow x = \frac{-50}{-25} \text{ and } y = \frac{-75}{-25} \\[1em] \Rightarrow x = 2 \text{ and } y = 3.\end{aligned}$$

Hence, x = 2 and y = 3.

Given, equations :

⇒ 3x + 4y = 11 and 2x + 3y = 8

⇒ 3x + 4y - 11 = 0 .......(1)

⇒ 2x + 3y - 8 = 0 ........(2)

By cross-multiplication method :

$$\begin{aligned}\Rightarrow \frac{x}{4 \times (-8) - 3 \times (-11)} = \frac{y}{(-11) \times 2 - (-8) \times 3} = \frac{1}{3 \times 3 - 2 \times 4} \\[1em] \Rightarrow \frac{x}{-32 + 33} = \frac{y}{-22 + 24} = \frac{1}{9 - 8} \\[1em] \Rightarrow \frac{x}{1} = \frac{y}{2} = \frac{1}{1} \\[1em] \Rightarrow x = \frac{y}{2} = 1 \\[1em] \Rightarrow x = 1 \text{ and } \frac{y}{2} = 1 \\[1em] \Rightarrow x = 1 \text{ and } y = 2.\end{aligned}$$

Hence, x = 1 and y = 2.

Given, equations :

⇒ 6x + 7y - 11 = 0 and 5x + 2y = 13

⇒ 6x + 7y - 11 = 0 ........(1)

⇒ 5x + 2y - 13 = 0 ........(2)

By cross-multiplication method :

$$\begin{aligned}\Rightarrow \frac{x}{7 \times (-13) - 2 \times (-11)} = \frac{y}{(-11) \times 5 - (-13) \times 6} = \frac{1}{6 \times 2 - 5 \times 7} \\[1em] \Rightarrow \frac{x}{-91 + 22} = \frac{y}{-55 + 78} = \frac{1}{12 - 35} \\[1em] \Rightarrow \frac{x}{-69} = \frac{y}{23} = \frac{1}{-23} \\[1em] \Rightarrow \frac{x}{-69} = \frac{1}{-23} \text{ and } \frac{y}{23} = \frac{1}{-23} \\[1em] \Rightarrow x = \frac{-69}{-23} \text{ and } y = \frac{23}{-23}\\[1em] \Rightarrow x = 3 \text{ and } y = -1.\end{aligned}$$

Hence, x = 3 and y = -1.

Given, equations :

⇒ 5x + 4y + 14 = 0 and 3x = -10 - 4y

⇒ 5x + 4y + 14 = 0 ..........(1)

⇒ 3x + 4y + 10 = 0 ..........(2)

By cross-multiplication method :

$$\begin{aligned}\Rightarrow \frac{x}{4 \times 10 - 4 \times 14} = \frac{y}{14 \times 3 - 10 \times 5} = \frac{1}{5\times 4 - 3 \times 4} \\[1em] \Rightarrow \frac{x}{40 - 56} = \frac{y}{42 - 50} = \frac{1}{20 - 12} \\[1em] \Rightarrow \frac{x}{-16} = \frac{y}{-8} = \frac{1}{8} \\[1em] \Rightarrow \frac{x}{-16} = \frac{1}{8} \text{ and } \frac{y}{-8} = \frac{1}{8} \\[1em] \Rightarrow x = \frac{-16}{8} \text{ and } y = \frac{-8}{8} \\[1em] \Rightarrow x = -2 \text{ and } y = -1.\end{aligned}$$

Hence, x = -2 and y = -1.

Given, equations :

⇒ x - y + 2 = 0 and 7x + 9y = 130

⇒ x - y + 2 = 0 ........(1)

⇒ 7x + 9y - 130 = 0 ........(2)

By cross-multiplication method :

$$\begin{aligned}\Rightarrow \frac{x}{(-1) \times (-130) - 9 \times 2} = \frac{y}{2 \times 7 - 1 \times (-130)} = \frac{1}{1 \times 9 - 7 \times (-1)} \\[1em] \Rightarrow \frac{x}{130 - 18} = \frac{y}{14 + 130} = \frac{1}{9 + 7} \\[1em] \Rightarrow \frac{x}{112} = \frac{y}{144} = \frac{1}{16} \\[1em] \Rightarrow \frac{x}{112} = \frac{1}{16} \text{ and } \frac{y}{144} = \frac{1}{16} \\[1em] \Rightarrow x = \frac{112}{16} \text{ and } y = \frac{144}{16} \\[1em] \Rightarrow x = 7 \text{ and } y = 9.\end{aligned}$$

Hence, x = 7 and y = 9.

Given, equations :

⇒ 4x - y = 5 and 5y - 4x = 7

⇒ 4x - y - 5 = 0 ..........(1)

⇒ -4x + 5y - 7 = 0 .........(2)

By cross-multiplication method :

$$\begin{aligned}\Rightarrow \frac{x}{(-1) \times (-7) - 5 \times (-5)} = \frac{y}{(-5) \times (-4) - (-7) \times 4} = \frac{1}{4 \times 5 - (-4) \times (-1)} \\[1em] \Rightarrow \frac{x}{7 + 25} = \frac{y}{20 + 28} = \frac{1}{20 - 4} \\[1em] \Rightarrow \frac{x}{32} = \frac{y}{48} = \frac{1}{16} \\[1em] \Rightarrow \frac{x}{32} = \frac{1}{16} \text{ and } \frac{y}{48} = \frac{1}{16} \\[1em] \Rightarrow x = \frac{32}{16} \text{ and } y = \frac{48}{16} \\[1em] \Rightarrow x = 2 \text{ and } y = 3.\end{aligned}$$

Hence, x = 2 and y = 3.

Given, equations :

⇒ 4x - 3y = 0 and 2x + 3y = 18

⇒ 4x - 3y = 0 .........(1)

⇒ 2x + 3y - 18 = 0 .....(2)

By cross-multiplication method :

$$\begin{aligned}\Rightarrow \frac{x}{(-3) \times (-18) - 3 \times 0} = \frac{y}{0 \times 2 - (-18) \times 4} = \frac{1}{4 \times 3 - 2 \times (-3)} \\[1em] \Rightarrow \frac{x}{54 - 0} = \frac{y}{0 + 72} = \frac{1}{12 + 6} \\[1em] \Rightarrow \frac{x}{54} = \frac{y}{72} = \frac{1}{18} \\[1em] \Rightarrow \frac{x}{54} = \frac{1}{18} \text{ and } \frac{y}{72} = \frac{1}{18} \\[1em] \Rightarrow x = \frac{54}{18} \text{ and } y = \frac{72}{18}\\[1em] \Rightarrow x = 3 \text{ and } y = 4.\end{aligned}$$

Hence, x = 3 and y = 4.

Given, equations :

⇒ 8x + 5y = 9 and 3x + 2y = 4

⇒ 8x + 5y - 9 = 0 ..........(1)

⇒ 3x + 2y - 4 = 0 .........(2)

By cross-multiplication method we have :

$$\begin{aligned}\Rightarrow \frac{x}{5 \times (-4) - 2 \times (-9)} = \frac{y}{(-9) \times 3 - (-4) \times 8} = \frac{1}{8 \times 2 - 3 \times 5} \\[1em] \Rightarrow \frac{x}{-20 + 18} = \frac{y}{-27 + 32} = \frac{1}{16 - 15} \\[1em] \Rightarrow \frac{x}{-2} = \frac{y}{5} = \frac{1}{1} \\[1em] \Rightarrow \frac{x}{-2} = \frac{1}{1} \text{ and } \frac{y}{5} = \frac{1}{1} \\[1em] \Rightarrow x = \frac{-2}{1} \text{ and } y = \frac{5}{1} \\[1em] \Rightarrow x = -2 \text{ and } y = 5.\end{aligned}$$

Hence, x = -2 and y = 5.

Given, equations :

⇒ 4x - 3y - 11 = 0 ............(1)

⇒ 6x + 7y - 5 = 0 ............(2)

By cross-multiplication method :

$$\begin{aligned}\Rightarrow \frac{x}{(-3) \times (-5) - 7 \times (-11)} = \frac{y}{(-11) \times 6 - (-5) \times 4} = \frac{1}{4 \times 7 - 6 \times (-3)} \\[1em] \Rightarrow \frac{x}{15 + 77} = \frac{y}{-66 + 20} = \frac{1}{28+ 18} \\[1em] \Rightarrow \frac{x}{92} = \frac{y}{-46} = \frac{1}{46} \\[1em] \Rightarrow \frac{x}{92} = \frac{1}{46} \text{ and } \frac{y}{-46} = \frac{1}{46} \\[1em] \Rightarrow x = \frac{92}{46} \text{ and } y = \frac{-46}{46} \\[1em] \Rightarrow x = 2 \text{ and } y = -1.\end{aligned}$$

Hence, x = 2 and y = -1.

Given, equations :

$\Rightarrow \frac{9}{x} - \frac{4}{y} = 8$ .......(1)

$\Rightarrow \frac{13}{x} + \frac{7}{y} = 101$ .....(2)

Multiplying equation (1) by 7, we get :

$$\begin{aligned}\Rightarrow 7\Big(\frac{9}{x} - \frac{4}{y}\Big) = 7 \times 8 \\[1em] \Rightarrow \frac{63}{x} - \frac{28}{y} = 56 \text{ ......(3)}\end{aligned}$$

Multiplying equation by (2) by 4, we get :

$$\begin{aligned}\Rightarrow 4\Big(\frac{13}{x} + \frac{7}{y}\Big) = 4 \times 101 \\[1em] \Rightarrow \frac{52}{x} + \frac{28}{y} = 404 \text{ .......(4)}\end{aligned}$$

Adding equation (3) and (4), we get :

$$\begin{aligned}\Rightarrow \Big(\frac{63}{x} - \frac{28}{y}\Big) + \Big(\frac{52}{x} + \frac{28}{y} \Big) = 56 + 404 \\[1em] \Rightarrow \frac{63 + 52}{x} = 460 \\[1em] \Rightarrow \frac{115}{x} = 460 \\[1em] \Rightarrow x = \frac{115}{460} = \frac{1}{4}.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow \frac{9}{\frac{1}{4}} - \frac{4}{y} = 8 \\[1em] \Rightarrow 36 - \frac{4}{y} = 8 \\[1em] \Rightarrow \frac{4}{y} = 36 - 8 \\[1em] \Rightarrow \frac{4}{y} = 28 \\[1em] \Rightarrow y = \frac{4}{28} = \frac{1}{7}.\end{aligned}$$

Hence, x = $\frac{1}{4} \text{ and } y = \frac{1}{7}$.

Given, equations :

$\Rightarrow \frac{3}{x} + \frac{2}{y} = 10$ ............(1)

$\Rightarrow \frac{9}{x} - \frac{7}{y} = 10.5$ ..........(2)

Multiplying equation (1) by 3, we get :

$$\begin{aligned}\Rightarrow 3\Big(\frac{3}{x} + \frac{2}{y}\Big) = 3 \times 10 \\[1em] \Rightarrow \frac{9}{x} + \frac{6}{y} = 30 ..........(3)\end{aligned}$$

Subtracting equation (2) from (3), we get :

$$\begin{aligned}\Rightarrow \Big(\frac{9}{x} + \frac{6}{y}\Big) - \Big(\frac{9}{x} - \frac{7}{y}\Big) = 30 - 10.5 \\[1em] \Rightarrow \frac{9}{x} - \frac{9}{x} + \frac{6}{y} + \frac{7}{y} = 19.5 \\[1em] \Rightarrow \frac{13}{y} = 19.5 \\[1em] \Rightarrow y = \frac{13}{19.5} \\[1em] \Rightarrow y = \frac{1}{1.5} = \frac{2}{3}.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow \frac{3}{x} + \frac{2}{\frac{2}{3}} = 10 \\[1em] \Rightarrow \frac{3}{x} + \frac{6}{2} = 10 \\[1em] \Rightarrow \frac{3}{x} + 3 = 10 \\[1em] \Rightarrow \frac{3}{x} = 7 \\[1em] \Rightarrow x = \frac{3}{7}.\end{aligned}$$

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

Given, equations :

$\Rightarrow 5x + \frac{8}{y} = 19$ .........(1)

$\Rightarrow 3x - \frac{4}{y} = 7$ ........(2)

Multiplying equation (2) by 2, we get :

$$\begin{aligned}\Rightarrow 2\Big(3x - \frac{4}{y}\Big) = 2 \times 7 \\[1em] \Rightarrow 6x - \frac{8}{y} = 14 \text{ ........(3)}\end{aligned}$$

Adding equation (1) and (3), we get :

$$\begin{aligned}\Rightarrow \Big(5x + \frac{8}{y}\Big) + \Big(6x - \frac{8}{y}\Big) = 19 + 14 \\[1em] \Rightarrow 5x + 6x + \frac{8}{y} - \frac{8}{y} = 33 \\[1em] \Rightarrow 11x = 33 \\[1em] \Rightarrow x = \frac{33}{11} = 3.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow 5 \times 3 + \frac{8}{y} = 19 \\[1em] \Rightarrow 15 + \frac{8}{y} = 19 \\[1em] \Rightarrow \frac{8}{y} = 19 - 15 \\[1em] \Rightarrow \frac{8}{y} = 4 \\[1em] \Rightarrow y = \frac{8}{4} = 2.\end{aligned}$$

Hence, x = 3 and y = 2.

Given, equations :

$\Rightarrow 4x + \frac{6}{y} = 15$ ..........(1)

$\Rightarrow 3x - \frac{4}{y} = 7$ ..........(2)

Multiplying equation (1) by 2, we get :

$$\begin{aligned}\Rightarrow 2\Big(4x + \frac{6}{y}\Big) = 2 \times 15 \\[1em] \Rightarrow 8x + \frac{12}{y} = 30 \text{ .........(3)}\end{aligned}$$

Multiplying equation (2) by 3, we get :

$$\begin{aligned}\Rightarrow 3\Big(3x - \frac{4}{y}\Big) = 3 \times 7 \\[1em] \Rightarrow 9x - \frac{12}{y} = 21 \text{ .........(4)}\end{aligned}$$

Adding equation (3) and (4), we get :

$$\begin{aligned}\Rightarrow 8x + \frac{12}{y} + 9x - \frac{12}{y} = 30 + 21 \\[1em] \Rightarrow 17x = 51 \\[1em] \Rightarrow x = \frac{51}{17} = 3.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow 3 \times 3 - \frac{4}{y} = 7 \\[1em] \Rightarrow 9 - \frac{4}{y} = 7 \\[1em] \Rightarrow \frac{4}{y} = 9 - 7 \\[1em] \Rightarrow \frac{4}{y} = 2 \\[1em] \Rightarrow y = \frac{4}{2} = 2.\end{aligned}$$

Given,

⇒ y = ax - 2

⇒ 2 = 3a - 2

⇒ 3a = 2 + 2

⇒ 3a = 4

⇒ a = $\frac{4}{3} = 1\frac{1}{3}$.

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

Given, equations :

$\Rightarrow \frac{3}{x} - \frac{2}{y} = 0$ .......(1)

$\Rightarrow \frac{2}{x} + \frac{5}{y} = 19$ .......(2)

Multiplying equation (1) by 2, we get :

$$\begin{aligned}\Rightarrow 2\Big(\frac{3}{x} - \frac{2}{y}\Big) = 2 \times 0 \\[1em] \Rightarrow \frac{6}{x} - \frac{4}{y} = 0 \text{ .........(3)}\end{aligned}$$

Multiplying equation (2) by 3, we get :

$$\begin{aligned}\Rightarrow 3\Big(\frac{2}{x} + \frac{5}{y}\Big) = 3 \times 19 \\[1em] \Rightarrow \frac{6}{x} + \frac{15}{y} = 57 \text{ .........(4)}\end{aligned}$$

Subtracting equation (3) from (4), we get :

$$\begin{aligned}\Rightarrow \Big(\frac{6}{x} + \frac{15}{y}\Big) - \Big(\frac{6}{x} - \frac{4}{y}\Big) = 57 - 0 \\[1em] \Rightarrow \frac{6}{x} - \frac{6}{x} + \frac{15}{y} + \frac{4}{y} = 57 \\[1em] \Rightarrow \frac{19}{y} = 57 \\[1em] \Rightarrow y = \frac{19}{57} = \frac{1}{3}.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow \frac{3}{x} - \frac{2}{\frac{1}{3}} = 0 \\[1em] \Rightarrow \frac{3}{x} - 6 = 0 \\[1em] \Rightarrow \frac{3}{x} = 6 \\[1em] \Rightarrow x = \frac{3}{6} = \frac{1}{2}.\end{aligned}$$

Given,

$$\begin{aligned}\Rightarrow y = ax + 3 \\[1em] \Rightarrow \frac{1}{3} = a \times \frac{1}{2} + 3 \\[1em] \Rightarrow \frac{1}{3} = \frac{a}{2} + 3 \\[1em] \Rightarrow \frac{a}{2} = \frac{1}{3} - 3 \\[1em] \Rightarrow \frac{a}{2} = \frac{1 - 9}{3} \\[1em] \Rightarrow \frac{a}{2} = \frac{-8}{3} \\[1em] \Rightarrow a = -\frac{16}{3} = -5\frac{1}{3}.\end{aligned}$$

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

Given, equations :

$\frac{20}{x + y} + \frac{3}{x - y} = 7$ .......(1)

$\frac{8}{x - y} - \frac{15}{x + y} = 5$ .............(2)

Multiplying equation (1) by 8, we get :

$$\begin{aligned}\Rightarrow 8\Big(\frac{20}{x + y} + \frac{3}{x - y}\Big) = 8 \times 7 \\[1em] \Rightarrow \frac{160}{x + y} + \frac{24}{x - y} = 56 \text{ ..........(3)}\end{aligned}$$

Multiplying equation (2) by 3, we get :

$$\begin{aligned}\Rightarrow 3\Big(\frac{8}{x - y} - \frac{15}{x + y}\Big) = 3 \times 5 \\[1em] \Rightarrow \frac{24}{x - y} - \frac{45}{x + y} = 15 \text{ .........(4)}\end{aligned}$$

Subtracting equation (4) from (3), we get :

$$\begin{aligned}\Rightarrow \frac{160}{x + y} + \frac{24}{x - y} - \Big(\frac{24}{x - y} - \frac{45}{x + y}\Big) = 56 - 15 \\[1em] \Rightarrow \frac{160}{x + y} + \frac{45}{x + y} + \frac{24}{x - y} - \frac{24}{x - y} = 41 \\[1em] \Rightarrow \frac{205}{x + y} = 41 \\[1em] \Rightarrow x + y = \frac{205}{41} \\[1em] \Rightarrow x + y = 5 \text{ ........(5)}\end{aligned}$$

Substituting value of x + y from equation 5 in equation 1, we get :

$$\begin{aligned}\Rightarrow \frac{20}{x + y} + \frac{3}{x - y} = 7 \\[1em] \Rightarrow \frac{20}{5} + \frac{3}{x - y} = 7 \\[1em] \Rightarrow 4 + \frac{3}{x - y} = 7 \\[1em] \Rightarrow \frac{3}{x - y} = 7 - 4 \\[1em] \Rightarrow \frac{3}{x - y} = 3 \\[1em] \Rightarrow x - y = \frac{3}{3} \\[1em] \Rightarrow x - y = 1 \text{ ........(6)}\end{aligned}$$

Adding equation (5) and (6), we get :

⇒ (x + y) + (x - y) = 5 + 1

⇒ 2x = 6

⇒ x = $\frac{6}{2}$

⇒ x = 3.

Substituting value of x in equation (6), we get :

⇒ 3 - y = 1

⇒ y = 3 - 1 = 2.

Hence, x = 3 and y = 2.

Given, equations :

$\frac{34}{3x + 4y} + \frac{15}{3x - 2y} = 5$ .......(1)

$\frac{25}{3x - 2y} - \frac{8.50}{3x + 4y}$ = 4.5 .........(2)

Multiplying equation (2) by 4, we get :

$$\begin{aligned}\Rightarrow 4\Big(\frac{25}{3x - 2y} - \frac{8.50}{3x + 4y}\Big) = 4 \times 4.5 \\[1em] \Rightarrow \frac{100}{3x - 2y} - \frac{34}{3x + 4y} = 18 \text{ .........(3)}\end{aligned}$$

Adding equation (1) and (3), we get :

$$\begin{aligned}\Rightarrow \frac{34}{3x + 4y} + \frac{15}{3x - 2y} + \Big(\frac{100}{3x - 2y} - \frac{34}{3x + 4y}\Big) = 5 + 18 \\[1em] \Rightarrow \frac{115}{3x - 2y} = 23 \\[1em] \Rightarrow 3x- 2y = \frac{115}{23} \\[1em] \Rightarrow 3x - 2y = 5 \text{ .........(4)} \\[1em] \Rightarrow 3x = 5 + 2y \\[1em] \Rightarrow x = \frac{5 + 2y}{3} \text{ ......(5)}\end{aligned}$$

Substituting value of 3x - 2y from equation (4) in (2), we get :

$$\begin{aligned}\Rightarrow \frac{25}{5} - \frac{8.50}{3x + 4y} = 4.5 \\[1em] \Rightarrow \frac{25(3x + 4y) - 42.50}{5(3x + 4y)} = \frac{45}{10} \\[1em] \Rightarrow \frac{75x + 100y - 42.50}{15x + 20y} = \frac{9}{2} \\[1em] \Rightarrow 2(75x + 100y - 42.50) = 9(15x + 20y) \\[1em] \Rightarrow 150x + 200y - 85 = 135x + 180y \\[1em] \Rightarrow 150x - 135x + 200y - 180y = 85 \\[1em] \Rightarrow 15x + 20y = 85 \text{ ........(6)}\end{aligned}$$

Substituting value of x from equation (5) in (6), we get :

$$\begin{aligned}\Rightarrow 15 \times \frac{5 + 2y}{3} + 20y = 85 \\[1em] \Rightarrow 5(5 + 2y) + 20y = 85 \\[1em] \Rightarrow 25 + 10y + 20y = 85 \\[1em] \Rightarrow 30y = 85 - 25 \\[1em] \Rightarrow 30y = 60 \\[1em] \Rightarrow y = \frac{60}{30} = 2.\end{aligned}$$

Substituting value of y in equation (5), we get :

$$\begin{aligned}\Rightarrow x = \frac{5 + 2y}{3} \\[1em] = \frac{5 + 2 \times 2}{3} \\[1em] = \frac{5 + 4}{3} \\[1em] = \frac{9}{3} \\[1em] = 3.\end{aligned}$$

Hence, x = 3 and y = 2.

Given, equations : x + y = 2xy and x - y = 6xy

Dividing both the sides of first equation by xy, we get :

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

Dividing both the sides of second equation by xy, we get :

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

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

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

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

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

Hence, $x = -\frac{1}{2} \text{ and } y = \frac{1}{4}$.

Given, equations : x + y = 7xy and 2x - 3y = -xy

Dividing both the sides of first equation by xy, we get :

$$\begin{aligned}\Rightarrow \frac{x + y}{xy} = \frac{7xy}{xy} \\[1em] \Rightarrow \frac{x}{xy} + \frac{y}{xy} = 7 \\[1em] \Rightarrow \frac{1}{y} + \frac{1}{x} = 7 \\[1em]\end{aligned}$$

Multiplying both sides of the above equation by 3, we get :

$$\begin{aligned}\Rightarrow 3\Big(\frac{1}{y} + \frac{1}{x}\Big) = 3 \times 7 \\[1em] \Rightarrow \frac{3}{y} + \frac{3}{x} = 21\text{ .......(1)}\end{aligned}$$

Dividing both the sides of second equation by xy, we get :

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

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

$$\begin{aligned}\Rightarrow \Big(\frac{3}{y} + \frac{3}{x}\Big) + \Big(\frac{2}{y} - \frac{3}{x}\Big) = 21 + (-1) \\[1em] \Rightarrow \frac{5}{y} = 20 \\[1em] \Rightarrow y = \frac{5}{20} = \frac{1}{4}.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow \frac{3}{\frac{1}{4}} + \frac{3}{x} = 21 \\[1em] \Rightarrow 12 + \frac{3}{x} = 21 \\[1em] \Rightarrow \frac{3}{x} = 21 - 12 \\[1em] \Rightarrow \frac{3}{x} = 9 \\[1em] \Rightarrow x = \frac{3}{9} = \frac{1}{3}.\end{aligned}$$

Hence, $x = \frac{1}{3} \text{ and } y = \frac{1}{4}$.

Let $\frac{1}{x} = p \text{ and } \frac{1}{y} = q$. Substituting in equations, we get :

⇒ ap - bq = 0 ..........(1)

⇒ ab²p + a²bq = a² + b²

⇒ ab²p + a²bq - (a² + b²) = 0 .........(2)

By cross-multiplication method we have :

$$\begin{aligned}\Rightarrow \frac{p}{-b \times -(a^2 + b^2) - a^2b \times 0} = \frac{q}{0 \times ab^2 - [-(a^2 + b^2)] \times a} = \frac{1}{a \times a^2b - ab^2 \times (-b)} \\[1em] \Rightarrow \frac{p}{b(a^2 + b^2)} = \frac{q}{a(a^2 + b^2)} = \frac{1}{a^3b + ab^3} \\[1em] \Rightarrow \frac{p}{b(a^2 + b^2)} = \frac{1}{a^3b + ab^3} \text{ and } \frac{q}{a(a^2 + b^2)} = \frac{1}{a^3b + ab^3} \\[1em] \Rightarrow \frac{p}{b(a^2 + b^2)} = \frac{1}{ab(a^2 + b^2)} \text{ and } \frac{q}{a(a^2 + b^2)} = \frac{1}{ab(a^2 + b^2)} \\[1em] \Rightarrow p = \frac{b(a^2 + b^2)}{ab(a^2 + b^2)} \text{ and } q = \frac{a(a^2 + b^2)}{ab(a^2 + b^2)} \\[1em] \Rightarrow p = \frac{1}{a} \text{ and } q = \frac{1}{b} \\[1em] \Rightarrow \frac{1}{x} = p \text{ and } \frac{1}{y} = q \\[1em] \Rightarrow \frac{1}{x} = \frac{1}{a} \text{ and } \frac{1}{y} = \frac{1}{b} \\[1em] \Rightarrow x = a \text{ and } y = b.\end{aligned}$$

Hence, x = a and y = b.

Simplifying first equation :

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

Simplifying second equation :

$$\begin{aligned}\Rightarrow \frac{xy}{2x - y} = -\frac{3}{10} \\[1em] \Rightarrow \frac{2x - y}{xy} = -\frac{10}{3} \\[1em] \Rightarrow \frac{2x}{xy} - \frac{y}{xy} = -\frac{10}{3} \\[1em] \Rightarrow \frac{2}{y} - \frac{1}{x} = -\frac{10}{3} \text{ ..........(2)}\end{aligned}$$

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

$$\begin{aligned}\Rightarrow \frac{1}{y} + \frac{1}{x} + \Big(\frac{2}{y} - \frac{1}{x}\Big) = \frac{4}{3} + \Big(-\frac{10}{3}\Big) \\[1em] \Rightarrow \frac{1}{y} + \frac{2}{y} + \frac{1}{x} - \frac{1}{x} = \frac{4 - 10}{3} \\[1em] \Rightarrow \frac{3}{y} = \frac{-6}{3} \\[1em] \Rightarrow y = \frac{3 \times 3}{-6} \\[1em] \Rightarrow y = -\frac{9}{6} = -\frac{3}{2}.\end{aligned}$$

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

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

Hence, $x = \frac{1}{2} \text{ and } y = -\frac{3}{2}$.

Let two whole numbers be x and y, where x > y.

Given,

Sum of numbers = 15

⇒ x + y = 15 .........(1)

Difference of numbers = 9

⇒ x - y = 9 .........(2)

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

⇒ (x + y) + (x - y) = 15 + 9

⇒ x + x + y - y = 24

⇒ 2x = 24

⇒ x = $\frac{24}{2}$ = 12.

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

⇒ 12 + y = 15

⇒ y = 15 - 12 = 3.

Hence, Option 1 is the correct option.

Let numerator be x and denominator be y.

Given,

Sum of numerator and denominator of a fraction is 22.

∴ x + y = 22 ........(1)

Given,

Numerator is 8 less than its denominator.

∴ y - x = 8 ........(2)

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

⇒ (x + y) + y - x = 8 + 22

⇒ x - x + y + y = 30

⇒ 2y = 30

⇒ y = $\frac{30}{2}$ = 15.

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

⇒ x + 15 = 22

⇒ x = 22 - 15 = 7.

Fraction = $\frac{x}{y} = \frac{7}{15}$.

Hence, Option 2 is the correct option.

Let ten's digit be x and unit's digit be y.

Given,

Sum of digits = 8

∴ x + y = 8 .........(1)

Difference between digits = 4

∴ x - y = 4 .........(2)

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

⇒ (x + y) + (x - y) = 8 + 4

⇒ x + x + y - y = 12

⇒ 2x = 12

⇒ x = $\frac{12}{2}$ = 6.

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

⇒ 6 + y = 8

⇒ y = 8 - 6 = 2.

Number = 10x + y = 10(6) + 2 = 60 + 2 = 62.

Hence, Option 4 is the correct option.

Given,

The present ages of two persons are in the ratio 1 : 3.

Let present age be x and 3x.

Given,

After 5 years, the ratio between their ages will be 2 : 5.

$$\begin{aligned}\Rightarrow \frac{x + 5}{3x + 5} = \frac{2}{5} \\[1em] \Rightarrow 5(x + 5) = 2(3x + 5) \\[1em] \Rightarrow 5x + 25 = 6x + 10 \\[1em] \Rightarrow 6x - 5x = 25 - 10 \\[1em] \Rightarrow x = 15 \\[1em] \Rightarrow 3x = 3 \times 15 = 45.\end{aligned}$$

Hence, Option 2 is the correct option.

Let the numbers be x and y.

According to question ratio of numbers = $\frac{2}{3}$,

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

Given,

If 2 is subtracted from the first and 8 from the second, the ratio becomes the reciprocal of the original ratio.

$$\begin{aligned}\Rightarrow \frac{x - 2}{y - 8} = \frac{3}{2} \\[1em] \Rightarrow 2(x - 2) = 3(y - 8) \\[1em] \Rightarrow 2x - 4 = 3y - 24 \\[1em]\end{aligned}$$

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

$$\begin{aligned}\Rightarrow 2 \times \frac{2y}{3} - 4 = 3y - 24 \\[1em] \Rightarrow \frac{4y}{3} - 4 = 3y - 24 \\[1em] \Rightarrow -4 + 24 = 3y - \frac{4y}{3} \\[1em] \Rightarrow 20 = \frac{9y - 4y}{3} \\[1em] \Rightarrow 20 = \frac{5y}{3} \\[1em] \Rightarrow y = \frac{20 \times 3}{5} \\[1em] \Rightarrow y = 12.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow x = \frac{2 \times 12}{3} \\[1em] = \frac{24}{3} \\[1em] = 8.\end{aligned}$$

Hence, numbers are 8 and 12.

Let two numbers be x and y, where x is the larger and y is the smaller number.

Given,

Two numbers are in the ratio 4 : 7.

$$\begin{aligned}\therefore \frac{y}{x} = \frac{4}{7} \\[1em] \Rightarrow y = \frac{4x}{7} .......(1)\end{aligned}$$

Given,

If thrice the larger be added to twice the smaller, the sum is 59.

∴ 3x + 2y = 59

$$\begin{aligned}\Rightarrow 3x + 2 \times \frac{4x}{7} = 59 \\[1em] \Rightarrow 3x + \frac{8x}{7} = 59 \\[1em] \Rightarrow \frac{21x + 8x}{7} = 59 \\[1em] \Rightarrow \frac{29x}{7} = 59 \\[1em] \Rightarrow x = \frac{59 \times 7}{29} \\[1em] \Rightarrow x = 14\frac{7}{29}.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow y = \frac{4}{7} \times 14\frac{7}{29} \\[1em] = \frac{4}{7} \times \frac{413}{29} \\[1em] = \frac{4 \times 59}{29} \\[1em] = 8\frac{4}{29}.\end{aligned}$$

Hence, numbers are $8\frac{4}{29} \text{ and } 14\frac{7}{29}$.

Let two numbers be x and y, where x is the larger and y is the smaller number.

Given,

When the greater of the two numbers increased by 1 divides the sum of the numbers, the result is $\frac{3}{2}$.

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

Given,

When the difference of these numbers is divided by the smaller, the result is $\frac{1}{2}$.

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

From (1) and (2), we get :

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

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

$$\begin{aligned}\Rightarrow x = \frac{3 \times 6}{2} \\[1em] = \frac{18}{2} \\[1em] = 9.\end{aligned}$$

Hence, numbers are 6 and 9.

Given,

Sum of numbers = 50

∴ x + y = 50

⇒ x = 50 - y .........(1)

Difference of squares = 720

∴ x² - y² = 720

⇒ (50 - y)² - y² = 720

⇒ 50² + y² - 2 × 50 × y - y² = 720

⇒ 2500 - 100y = 720

⇒ 100y = 2500 - 720

⇒ 100y = 1780

⇒ y = $\frac{1780}{100}$ = 17.80

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

⇒ x = 50 - y = 50 - 17.80 = 32.20

Hence, numbers are 32.20 and 17.80

Let two numbers be x and y, such that x > y.

Given,

Sum of two numbers = 8

⇒ x + y = 8

⇒ x = 8 - y ........(1)

Given,

Difference of squares = 32

⇒ x² - y² = 32

⇒ (8 - y)² - y² = 32

⇒ 8² + y² - 2 × 8 × y - y² = 32

⇒ 64 - 16y = 32

⇒ 16y = 64 - 32

⇒ 16y = 32

⇒ y = $\frac{32}{16}$ = 2.

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

⇒ x = 8 - 2 = 6.

Hence, numbers are 2 and 6.

Given,

⇒ x > y

$\Rightarrow \frac{1}{x} \lt \frac{1}{y}$

Difference between two positive numbers x and y (x > y) is 4.

∴ x - y = 4

⇒ x = y + 4 ........(1)

Difference between their reciprocal is $\frac{4}{21}$.

$$\begin{aligned}\Rightarrow \frac{1}{y} - \frac{1}{x} = \frac{4}{21} \\[1em] \Rightarrow \frac{x - y}{xy} = \frac{4}{21}\\[1em] \Rightarrow 21(x - y) = 4xy \\[1em] \Rightarrow 21x - 21y - 4xy = 0 \text{ ..........(2)}\end{aligned}$$

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

⇒ 21(y + 4) - 21y - 4y(y + 4) = 0

⇒ 21y + 84 - 21y - 4y² - 16y = 0

⇒ 4y² + 16y - 84 = 0

⇒ 4(y² + 4y - 21) = 0

⇒ y² + 7y - 3y - 21 = 0

⇒ y(y + 7) - 3(y + 7) = 0

⇒ (y - 3)(y + 7) = 0

⇒ y - 3 = 0 or y + 7 = 0

⇒ y = 3 or y = -7.

Since, numbers are positive.

∴ y = 3.

⇒ x = y + 4 = 3 + 4 = 7.

Hence, numbers are 3 and 7.

Let two numbers be x and y.

Given,

Numbers are in the ratio 4 : 5.

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

Given,

If 30 is subtracted from each of the numbers, the ratio becomes 1 : 2.

$$\begin{aligned}\Rightarrow \frac{x - 30}{y - 30} = \frac{1}{2} \\[1em] \Rightarrow 2(x - 30) = y - 30 \\[1em] \Rightarrow 2x - 60 = y - 30 \\[1em] \Rightarrow 2x = y - 30 + 60 \\[1em] \Rightarrow 2x = y + 30 \\[1em] \Rightarrow x = \frac{y + 30}{2} .......(2)\end{aligned}$$

From (1) and (2), we get :

$$\begin{aligned}\Rightarrow \frac{4y}{5} = \frac{y + 30}{2} \\[1em] \Rightarrow 4y \times 2 = 5(y + 30) \\[1em] \Rightarrow 8y = 5y + 150 \\[1em] \Rightarrow 8y - 5y = 150 \\[1em] \Rightarrow 3y = 150 \\[1em] \Rightarrow y = \frac{150}{3} \\[1em] \Rightarrow y = 50.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow x = \frac{4 \times 50}{5} \\[1em] = \frac{200}{5} \\[1em] = 40.\end{aligned}$$

Hence, numbers are 40 and 50.

Let numerator be x and denominator be y.

Given,

If the numerator of a fraction is increased by 2 and denominator is decreased by 1, it becomes $\frac{2}{3}$.

$$\begin{aligned}\therefore \frac{x + 2}{y - 1} = \frac{2}{3} \\[1em] \Rightarrow 3(x + 2) = 2(y - 1) \\[1em] \Rightarrow 3x + 6 = 2y - 2 \\[1em] \Rightarrow 3x - 2y + 6 + 2 = 0 \\[1em] \Rightarrow 3x - 2y + 8 = 0 \text{ .........(1)}\end{aligned}$$

Given,

If the numerator is increased by 1 and denominator is increased by 2, it becomes $\frac{1}{3}$.

$$\begin{aligned}\Rightarrow \frac{x + 1}{y + 2} = \frac{1}{3} \\[1em] \Rightarrow 3(x + 1) = 1(y + 2) \\[1em] \Rightarrow 3x + 3 = y + 2 \\[1em] \Rightarrow 3x - y + 3 - 2 = 0 \\[1em] \Rightarrow 3x - y + 1 = 0 \text{ .........(2)}\end{aligned}$$

Subtracting equation (1) from (2), we get :

⇒ 3x - y + 1 - (3x - 2y + 8) = 0 - 0

⇒ 3x - 3x - y + 2y + 1 - 8 = 0

⇒ y - 7 = 0

⇒ y = 7.

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

⇒ 3x - 2(7) + 8 = 0

⇒ 3x - 14 + 8 = 0

⇒ 3x - 6 = 0

⇒ 3x = 6

⇒ x = $\frac{6}{3}$ = 2.

Fraction = $\frac{x}{y} = \frac{2}{7}$.

Hence, fraction = $\frac{2}{7}$.

Let numerator be x and denominator be y.

Given,

Sum of the numerator and the denominator of a fraction is equal to 7.

∴ x + y = 7

⇒ x = 7 - y ..........(1)

Given,

Four times the numerator is 8 less than 5 times the denominator.

⇒ 5y - 4x = 8 .........(2)

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

⇒ 5y - 4(7 - y) = 8

⇒ 5y - 28 + 4y = 8

⇒ 9y = 8 + 28

⇒ 9y = 36

⇒ y = $\frac{36}{9}$ = 4

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

⇒ x = 7 - 4 = 3.

Fraction = $\frac{x}{y} = \frac{3}{4}$.

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

Let numerator be x and denominator be y.

Given,

If the numerator of a fraction is multiplied by 2 and its denominator is increased by 1, it becomes 1.

$$\begin{aligned}\therefore \frac{2 \times x}{y + 1} = 1 \\[1em] \Rightarrow 2x = 1(y + 1) \\[1em] \Rightarrow 2x = y + 1 \\[1em] \Rightarrow 2x - y - 1 = 0 .........(1)\end{aligned}$$

Given,

If the numerator is increased by 4 and denominator is multiplied by 2, the fraction becomes $\frac{1}{2}$.

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

Subtracting equation (2) from (1), we get :

⇒ 2x - y - 1 - (2x - 2y + 8) = 0

⇒ 2x - 2x - y + 2y - 1 - 8 = 0

⇒ y - 9 = 0

⇒ y = 9.

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

⇒ 2x - 9 - 1 = 0

⇒ 2x - 10 = 0

⇒ 2x = 10

⇒ x = $\frac{10}{2}$ = 5.

Fraction = $\frac{x}{y} = \frac{5}{9}$.

Hence, fraction = $\frac{5}{9}$.

Let numerator be x and denominator be y.

Given,

Denominator is 5 more than numerator.

∴ y = x + 5 ......(1)

Given,

A fraction becomes $\frac{1}{2}$ if 5 is subtracted from its numerator and 3 is subtracted from its denominator.

$$\begin{aligned}\therefore \frac{x - 5}{y - 3} = \frac{1}{2} \\[1em] \Rightarrow 2(x - 5) = y - 3 \\[1em] \Rightarrow 2x - 10 = y - 3 \\[1em] \Rightarrow 2x - y = -3 + 10 \\[1em] \Rightarrow 2x - y = 7 \text{ .......(2)}\end{aligned}$$

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

⇒ 2x - (x + 5) = 7

⇒ 2x - x - 5 = 7

⇒ x = 7 + 5

⇒ x = 12.

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

⇒ y = 12 + 5 = 17.

Fraction = $\frac{x}{y} = \frac{12}{17}$.

Hence, fraction = $\frac{12}{17}$.

Let digit at unit's place be x and ten's place be y.

Number = 10 × y + x = 10y + x

Given,

Sum of the digits of a two digit number is 5.

∴ x + y = 5

⇒ x = 5 - y ........(1)

If the digits are reversed, then number = 10x + y.

Given,

If the digits are reversed, the number is reduced by 27.

∴ 10x + y = 10y + x - 27

⇒ 10x - x = 10y - y - 27

⇒ 9x = 9y - 27

⇒ 9x = 9(y - 3)

⇒ x = y - 3 ........(2)

From (1) and (2), we get :

⇒ y - 3 = 5 - y

⇒ y + y = 5 + 3

⇒ 2y = 8

⇒ y = $\frac{8}{2}$ = 4.

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

⇒ x = 4 - 3 = 1.

Number = 10y + x = 10(4) + 1 = 40 + 1 = 41.

Hence, the number = 41.

Let digit at unit's place be x and ten's place be y.

Number = 10 × y + x = 10y + x

Given,

Sum of digits = 7

∴ x + y = 7

⇒ x = 7 - y ........(1)

On reversing digits,

New number = 10 × x + y = 10x + y

Given,

The new number decreased by 2, equals twice the original number.

⇒ (10x + y) - 2 = 2(10y + x)

⇒ 10x + y - 2 = 20y + 2x

⇒ 10x - 2x = 20y - y + 2

⇒ 8x = 19y + 2

⇒ 8(7 - y) = 19y + 2 ........[From (1)]

⇒ 56 - 8y = 19y + 2

⇒ 56 - 2 = 19y + 8y

⇒ 54 = 27y

⇒ y = $\frac{54}{27}$ = 2.

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

⇒ x = 7 - 2 = 5.

Number = 10y + x = 10 × 2 + 5 = 25.

Hence, number = 25.

Let ten's digit be x and unit's digit be y.

Given,

The ten's digit of a two digit number is three times the unit digit.

∴ x = 3y ...........(1)

Given,

The sum of the number and the unit digit is 32.

Number = 10x + y, unit's digit = y

∴ 10x + y + y = 32

⇒ 10x + 2y = 32

⇒ 2(5x + y) = 32

⇒ 5x + y = 16

⇒ 5(3y) + y = 16 ........[From (1)]

⇒ 15y + y = 16

⇒ 16y = 16

⇒ y = 1.

⇒ x = 3y = 3(1) = 3.

Number = 10x + y = 10(3) + 1 = 31.

Hence, number = 31.

Let x be the digit at ten's place and y be the digit at unit's place.

Given,

Ten's digit exceeds twice the unit's digit by 2.

∴ x - 2y = 2

⇒ x = 2y + 2 ........(1)

Given,

Number obtained by inter-changing the digits is 5 more than three times the sum of the digits.

∴ 10y + x = 3(x + y) + 5

⇒ 10y + x = 3x + 3y + 5

⇒ 10y - 3y + x - 3x = 5

⇒ 7y - 2x = 5 ........(2)

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

⇒ 7y - 2(2y + 2) = 5

⇒ 7y - 4y - 4 = 5

⇒ 3y - 4 = 5

⇒ 3y = 9

⇒ y = $\frac{9}{3}$ = 3.

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

⇒ x = 2y + 2 = 2(3) + 2 = 6 + 2 = 8.

Original number = 10x + y = 10(8) + 3 = 80 + 3 = 83.

Hence, original number = 83.

Let present age of A be x years and B be y years.

Five years's ago their age will be :

A = (x - 5) years

B = (y - 5) years

Given,

Five years ago, A's age was four times the age of B.

⇒ (x - 5) = 4(y - 5)

⇒ x - 5 = 4y - 20

⇒ x = 4y - 20 + 5

⇒ x = 4y - 15 .........(1)

Five years's later their age will be :

A = (x + 5) years

B = (y + 5) years

Given,

Five years hence, A's age will be twice the age of B.

⇒ (x + 5) = 2(y + 5)

⇒ x + 5 = 2y + 10

⇒ x = 2y + 10 - 5

⇒ x = 2y + 5 .........(2)

From (1) and (2), we get :

⇒ 4y - 15 = 2y + 5

⇒ 4y - 2y = 5 + 15

⇒ 2y = 20

⇒ y = $\frac{20}{2}$ = 10 years.

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

⇒ x = 2(10) + 5 = 20 + 5 = 25 years.

Hence, present age of A = 25 years and B = 10 years.

Let present age of B be x years and so, age of A = (x + 20) years.

Given,

5 years ago, A was 3 times as old as B.

∴ (x + 20) - 5 = 3(x - 5)

⇒ x + 15 = 3x - 15

⇒ 3x - x = 15 + 15

⇒ 2x = 30

⇒ x = $\frac{30}{2}$ = 15 years.

Age of A = (x + 20) = 15 + 20 = 35 years.

Hence, present age of A = 35 years and B = 15 years.

Let present ages of mother and her daughter be x and y years respectively.

Given,

Four years ago, a mother was four times as old as her daughter.

⇒ (x - 4) = 4(y - 4)

⇒ x - 4 = 4y - 16

⇒ x = 4y - 16 + 4

⇒ x = 4y - 12 .......(1)

Given,

Six years later, the mother will be two and a half times as old as her daughter at that time.

⇒ (x + 6) = 2.5(y + 6)

⇒ x + 6 = 2.5y + 15

⇒ x = 2.5y + 15 - 6

⇒ x = 2.5y + 9 ........(2)

From (1) and (2), we get :

⇒ 4y - 12 = 2.5y + 9

⇒ 4y - 2.5y = 9 + 12

⇒ 1.5y = 21

⇒ y = $\frac{21}{1.5}$ = 14.

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

⇒ x = 4 × 14 - 12 = 56 - 12 = 44.

Hence, age of mother is 44 years and age of daughter is 14 years.

Let present age of man be x years and sum of the ages of his two children be y years.

Given,

The age of a man is twice the sum of the ages of his two children.

∴ x = 2y ........(1)

Given,

After 20 years, his age will be equal to the sum of the ages of his children at that time.

After 20 years,

Age of man = (x + 20) years

Sum of ages of his children = (y + 40), as each child's age will increase by 20 years.

⇒ x + 20 = y + 40

⇒ 2y + 20 = y + 40 .........[From (1)]

⇒ 2y - y = 40 - 20

⇒ y = 20.

⇒ x = 2y = 2(20) = 40.

Hence, present age of man = 40 years.

Given,

Annual incomes of A and B are in the ratio 3 : 4.

Let annual income of A be 3x and B be 4x.

Annual expenditure are in the ratio 5 : 7.

Let annual expenditure of A be 5y and B be 7y.

Given,

Each save ₹ 5,000.

For A,

⇒ 3x - 5y = 5000 ...........(1)

For B,

⇒ 4x - 7y = 5000 ..........(2)

Multiplying equation (1) by 4, we get :

⇒ 4(3x - 5y) = 4 × 5000

⇒ 12x - 20y = 20000 .........(3)

Multiplying equation (2) by 3, we get :

⇒ 3(4x - 7y) = 3 × 5000

⇒ 12x - 21y = 15000 .........(4)

Subtracting equation (4) from (3), we get :

⇒ 12x - 20y - (12x - 21y) = 20000 - 15000

⇒ 12x - 12x - 20y + 21y = 5000

⇒ y = 5000.

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

⇒ 3x - 5(5000) = 5000

⇒ 3x - 25000 = 5000

⇒ 3x = 25000 + 5000

⇒ 3x = 30000

⇒ x = $\frac{30000}{3}$ = 10000.

⇒ 3x = 3(10000) = 30000 and 4x = 4(10000) = 40000.

Hence, annual income of A = ₹ 30,000 and B = ₹ 40,000.

Given,

Ratio of passes to failures = 4 : 1

Let no. of students passed be 4x and failed be x.

So, total students who appeared for examination (originally) = 5x (4x + x)

Given,

Had 30 less appeared and 20 less passed, the ratio of passes to failures would have been 5 : 1.

So, now students appeared = 5x - 30 and no. of students passed = 4x - 20

No. of students failed = (5x - 30) - (4x - 20) = 5x - 4x - 30 + 20 = x - 10.

Now ratio = 5 : 1.

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

No. of students who appeared for examination originally = 5x = 5 x 30 = 150.

Hence, no. of students who appeared for examination are 150.

Let A have x pencils and B have y pencils.

Given,

If A gives 10 pencils to B, then B will have twice as many as A.

∴ 2(x - 10) = y + 10

⇒ 2x - 20 = y + 10

⇒ 2x - y = 10 + 20

⇒ 2x - y = 30 .........(1)

Given,

If B gives 10 pencils to A, then they will have the same number of pencils.

∴ x + 10 = y - 10

⇒ x - y = -10 - 10

⇒ x - y = -20 ..........(2)

Subtracting equation (2) from (1), we get :

⇒ 2x - y - (x - y) = 30 - (-20)

⇒ 2x - x - y + y = 30 + 20

⇒ x = 50

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

⇒ 2(50) - y = 30

⇒ 100 - y = 30

⇒ y = 100 - 30 = 70.

Hence, A has 50 pencils and B has 70 pencils.

Let there be x adults and y children.

Given,

There are total 1250 persons.

∴ x + y = 1250

⇒ x = 1250 - y .......(1)

Given,

Total collection is of ₹ 61,250.

∴ 75x + 25y = 61250 .......(2)

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

⇒ 75(1250 - y) + 25y = 61250

⇒ 93750 - 75y + 25y = 61250

⇒ 93750 - 50y = 61250

⇒ 50y = 93750 - 61250

⇒ 50y = 32500

⇒ y = $\frac{32500}{50}$ = 650.

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

⇒ x = 1250 - 650 = 600.

Hence, number of adults = 600 and number of children = 650.

Let Rohit have ₹ x and Ajay have ₹ y.

According to first part of question :

⇒ x + 100 = 2(y - 100)

⇒ x + 100 = 2y - 200

⇒ 2y - x = 100 + 200

⇒ 2y - x = 300 ...........(1)

According to second part of question :

⇒ y + 10 = 6(x - 10)

⇒ y + 10 = 6x - 60

⇒ y - 6x = -60 - 10

⇒ y - 6x = -70

Multiplying both sides of the above equation by 2, we get :

⇒ 2(y - 6x) = 2 × -70

⇒ 2y - 12x = -140 ...........(2)

Subtracting equation (2) from (1), we get :

⇒ 2y - x - (2y - 12x) = 300 - (-140)

⇒ 2y - x - 2y + 12x = 300 + 140

⇒ 11x = 440

⇒ x = $\frac{440}{11}$ = ₹ 40.

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

⇒ 2y - 40 = 300

⇒ 2y = 300 + 40

⇒ 2y = 340

⇒ y = $\frac{340}{2}$ = ₹ 170.

Hence, originally Rohit has ₹ 40 and Ajay has ₹ 170.

Let digit at ten's place be x and unit's place be y.

Number = 10(x) + y = 10x + y

On reversing the digits,

Reversed number = 10(y) + x = 10y + x

Given,

Sum of a two digit number and the number obtained by reversing the order of the digits is 99.

∴ (10x + y) + (10y + x) = 99

⇒ 11x + 11y = 99

⇒ 11(x + y) = 99

⇒ x + y = 9

⇒ x = 9 - y ............(1)

Given,

Digits differ by 3.

∴ x - y = 3 or y - x = 3

Considering x - y = 3

⇒ x = 3 + y ..........(2)

From (1) and (2), we get :

⇒ 9 - y = 3 + y

⇒ 9 - 3 = 2y

⇒ 2y = 6

⇒ y = 3.

Substituting value of y = 3 in equation (1), we get :

⇒ x = 9 - 3 = 6.

Number = 10x + y = 10(6) + 3 = 60 + 3 = 63.

Considering y - x = 3

⇒ x = y - 3 ..........(3)

From (1) and (3), we get :

⇒ 9 - y = y - 3

⇒ 9 + 3 = 2y

⇒ 2y = 12

⇒ y = 6.

Substituting value of y = 6 in equation (1), we get :

⇒ x = 9 - 6 = 3.

Number = 10x + y = 10(3) + 6 = 30 + 6 = 36.

Hence, number = 36 or 63.

Let digit at ten's place be x and digit at unit's place be y.

Number = 10(x) + y = 10x + y

On reversing the digits,

Reversed number = 10(y) + x = 10y + x

Given,

Seven times a two digit number is equal to four times the number obtained by reversing the digits.

⇒ 7(10x + y) = 4(10y + x)

⇒ 70x + 7y = 40y + 4x

⇒ 40y - 7y = 70x - 4x

⇒ 33y = 66x

⇒ y = $\frac{66x}{33}$

⇒ y = 2x ..........(1)

Given,

Difference between the digits is 3.

Let x > y

⇒ x - y = 3 ..........(2)

Let y > x

⇒ y - x = 3 ..........(3)

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

⇒ x - 2x = 3

⇒ -x = 3

⇒ x = -3

This is not possible as digits cannot be negative.

Substituting value of y from equation (1) in equation (3), we get :

⇒ 2x - x = 3

⇒ x = 3

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

⇒ y = 2x = 2(3) = 6.

Number = 10x + y = 10(3) + 6 = 30 + 6 = 36.

Hence, number = 36.

Let cost of ticket from Delhi to station A be ₹ x and cost of ticket from Delhi to station B be ₹ y.

Given,

From Delhi station, if we buy 2 tickets for station A and 3 tickets for station B, the total cost is ₹ 77.

⇒ 2x + 3y = 77 .......(1)

Given,

From Delhi station, if we buy 3 tickets for station A and 5 tickets for station B, the total cost is ₹ 124.

⇒ 3x + 5y = 124 .......(2)

Multiplying equation (1) by 3, we get :

⇒ 3(2x + 3y) = 3 × 77

⇒ 6x + 9y = 231 .......(3)

Multiplying equation (2) by 2, we get :

⇒ 2(3x + 5y) = 2 × 124

⇒ 6x + 10y = 248 .......(4)

Subtracting equation (3) from (4), we get :

⇒ (6x + 10y) - (6x + 9y) = 248 - 231

⇒ 6x - 6x + 10y - 9y = 17

⇒ y = ₹ 17.

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

⇒ 2x + 3(17) = 77

⇒ 2x + 51 = 77

⇒ 2x = 77 - 51

⇒ 2x = 26

⇒ x = $\frac{26}{2}$ = ₹ 13.

Hence, fares from Delhi to station A = ₹ 13 and to station B = ₹ 17.

Let digit at ten's place be x and digit at unit's place be y.

Given,

The sum of digits of a two digit number is 11.

∴ x + y = 11

⇒ x = 11 - y ..........(1)

Given,

If the digit at ten's place is increased by 5 and the digit at unit's place is decreased by 5, the digits of the number are found to be reversed.

∴ 10(x + 5) + (y - 5) = 10y + x

⇒ 10x + 50 + y - 5 = 10y + x

⇒ 10x + y - 10y - x + 45 = 0

⇒ 9x - 9y + 45 = 0

⇒ 9y = 9x + 45

⇒ y = x + 5

⇒ x = y - 5 ...........(2)

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

⇒ 11 - y = y - 5

⇒ y + y = 11 + 5

⇒ 2y = 16

⇒ y = $\frac{16}{2}$ = 8.

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

⇒ x = 8 - 5 = 3.

Number = 10x + y = 10(3) + 8 = 30 + 8 = 38.

Hence, number = 38.

Let x litres of 90% and y litres of 97% be mixed; then

⇒ x + y = 21

⇒ x = 21 - y .......(1)

and

⇒ 90% of x + 97% of y = 95% of 21

$$\begin{aligned}\Rightarrow \frac{90}{100}x + \frac{97}{100}y = \frac{95}{100} \times 21 \\[1em] \Rightarrow 90x + 97y = 95 \times 21 \\[1em] \Rightarrow 90x + 97y = 1995 \text{ ........(2)}\end{aligned}$$

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

⇒ 90(21 - y) + 97y = 1995

⇒ 1890 - 90y + 97y = 1995

⇒ 7y = 1995 - 1890

⇒ 7y = 105

⇒ y = $\frac{105}{7}$ = 15.

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

⇒ x = 21 - 15 = 6.

Hence, 6 litres of 90% acid solution and 15 litres of 97% of acid solution are mixed.

Let x kg of ₹ 250 per kg and y kg of ₹ 350 per kg sweets were purchased.

Given,

40 kg of sweets were needed.

∴ x + y = 40

⇒ x = 40 - y .........(1)

Given,

Total budget for sweets was ₹ 11800.

∴ 250x + 350y = 11800

⇒ 25x + 35y = 1180 ........(2)

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

⇒ 25(40 - y) + 35y = 1180

⇒ 1000 - 25y + 35y = 1180

⇒ 10y = 1180 - 1000

⇒ 10y = 180

⇒ y = $\frac{180}{10}$ = 18.

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

⇒ x = 40 - 18 = 22.

Hence, 22 kg of ₹ 250 per kg and 18 kg of ₹ 350 per kg sweets were purchased.

Given,

First equation :

⇒ 3y - 2x = 1

⇒ 2x - 3y + 1 = 0 ..............(1)

Second equation :

⇒ 3x + 4y = 24

⇒ 3x + 4y - 24 = 0 ................(2)

Multiplying eq. (1) by 3, we get :

⇒ 6x - 9y + 3 = 0 ................(3)

Multiplying eq. (2) by 2, we get :

⇒ 6x + 8y - 48 = 0 ................(4)

Subtracting eq. (3) from (4) we get,

⇒ (6x + 8y - 48) - (6x - 9y + 3) = 0 - 0

⇒ 6x + 8y - 48 - 6x + 9y - 3 = 0

⇒ 17y - 51 = 0

⇒ 17y = 51

⇒ y = $\frac{51}{17}$

⇒ y = 3.

Substituting value of y in eq. (2) we get

⇒ 3x + 4(3) - 24 = 0

⇒ 3x + 12 - 24 = 0

⇒ 3x - 12 = 0

⇒ 3x = 12

⇒ x = $\frac{12}{3}$

⇒ x = 4

∴ x = 4 and y = 3

Hence, option 4 is the correct option.

Given,

⇒ 7(x - 1) - 6y = 5(x - y)

⇒ 7x - 7 - 6y = 5x - 5y

⇒ 7x - 7 - 6y - 5x + 5y = 0

⇒ 2x - 7 - y = 0

⇒ 2x - y = 7

Hence, option 1 is the correct option.

Given, $\frac{2}{x} + \frac{3}{y}$ + 1 = 0 and $\frac{3}{x} + \frac{5}{y}$ + 2 = 0

Let $\frac{1}{x}$ be u and $\frac{1}{y}$ be v.

⇒ 2u + 3v + 1 = 0 ...................(1)

⇒ 3u + 5v + 2 = 0 ...................(2)

Multiplying eq. (1) by 3, we get,

⇒ 6u + 9v + 3 = 0 ................(3)

Multiplying eq. (2) by 2, we get,

⇒ 6u + 10v + 4 = 0 ................(4)

Subtracting eq. (3) from (4) we get,

⇒ 6u + 10v + 4 - (6u + 9v + 3) = 0 - 0

⇒ 6u + 10v + 4 - 6u - 9v - 3 = 0

⇒ v + 1 = 0

⇒ v = -1

⇒ $\frac{1}{y}$ = -1

⇒ y = -1

Substituting value of v in eq. (2) we get

⇒ 3u + 5(-1) + 2 = 0

⇒ 3u - 5 + 2 = 0

⇒ 3u - 3 = 0

⇒ 3u = 3

⇒ u = $\frac{3}{3}$

⇒ u = 1

⇒ $\frac{1}{x}$ = 1

⇒ x = 1

∴ x = 1 and y = -1

Hence, option 3 is the correct option.

Let digit at ten's place be x.

It is given that the sum of the digit is 11.

So, digit at one's place be 11 - x.

And, the tens digit minus unit digit is 5.

⇒ x - (11 - x) = 5

⇒ x - 11 + x = 5

⇒ 2x - 11 = 5

⇒ 2x = 5 + 11

⇒ 2x = 16

⇒ x = $\frac{16}{2}$

⇒ x = 8.

The digit at one's place = 11 - x = 11 - 8 = 3.

∴ Number = 10 × 8 + 3 = 80 + 3 = 83.

Hence, option 2 is the correct option.

Total stack of 5 glasses = 34 cm

Bottom 2 glasses = 19 cm

∴ Top 3 uniform glasses = 34 - 19 = 15

3 uniform glasses = 15

∴ 1 uniform glass = $\frac{15}{3}$ = 5 cm

Bottom 2 glasses = 19 cm

∴ Bottom glass = 19 - 5 = 14 cm.

Hence, option 2 is the correct option.

Given,

x = 5

y = 2

First equation :

⇒ x - y = 3

Substituting x = 5 and y = 2 in L.H.S. of first equation

⇒ 5 - 2

⇒ 3.

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

Second equation :

⇒ 2x + y = 11

Substituting x = 5 and y = 2 in L.H.S. of second equation

⇒ 2(5) + 2

⇒ 10 + 2

⇒ 12

L.H.S. ≠ R.H.S.

So, statement 1 is false.

A pair of values (x, y) is called a solution only if it makes both equations true, i.e., LHS = RHS for each equation.

So, statement 2 is true.

∴ Statement 1 is false , and statement 2 is true.

Hence, option 4 is the correct option.

Given, The sum of two numbers x and y is 11.

⇒ x + y = 11

⇒ x = 11 - y ....................(1)

Twice the first number plus three times the second number equals to 25.

⇒ 2x + 3y = 25 .........(2)

So, statement 1 is true.

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

⇒ 2(11 - y) + 3y = 25

⇒ 22 - 2y + 3y = 25

⇒ 22 + y = 25

⇒ y = 25 - 22

⇒ y = 3.

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

⇒ x = 11 - 3 = 8

The numbers are 8 and 3.

So, statement 2 is false.

∴ Statement 1 is true, and statement 2 is false.

Hence, option 3 is the correct option.

If x = 4 and y = 3 are the solution of equation 3y - 2x = 1, then substitute x = 4 and y = 3 in L.H.S. and R.H.S., both values should be same.

Given,

3y - 2x = 1

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

⇒ 3y - 2x

⇒ 3 x 3 - 2 x 4

⇒ 9 - 8

⇒ 1

As, L.H.S. = R.H.S.

So, x = 4 and y = 3 are the solution of equation 3y - 2x = 1.

3x + 4y = 24

Substituting value of x and y in L.H.S. of the equation :

⇒ 3x + 4y

⇒ 3 x 4 + 4 x 3

⇒ 12 + 12

⇒ 24

As, L.H.S. = R.H.S.

So, x = 4 and y = 3 are the solution of equation 3x + 4y = 24.

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

Hence, option 3 is the correct option.

Let the two digit number be 10x + y.

It is given that three times the digit at ten's place is equals to two times the digit at the unit place.

⇒ 3x = 2y

⇒ x = $\frac{2}{3}$ y ....................(1)

And, the sum of the digit is 5.

⇒ x + y = 5 ........................(2)

So, assertion (A) is true.

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

$$\begin{aligned}\Rightarrow \frac{2}{3}y + y = 5\\[1em] \Rightarrow \frac{2y + 3y}{3} = 5\\[1em] \Rightarrow \frac{5y}{3} = 5\\[1em] \Rightarrow y = \frac{5 \times 3}{5}\\[1em] \Rightarrow y = 3\end{aligned}$$

Substituting the value of y in equation (1),

⇒ x = $\frac{2}{3} \times 3$

⇒ x = 2

Thus, the number is 10x + y = 10(2) + 3 = 23.

So, reason (R) is false.

∴ A is true, but R is false.

Hence, option 1 is the correct option.

Given,

Equations : 3x + 2y = 11 and 2x - 3y + 10 = 0

⇒ 3x + 2y = 11

⇒ 3x = 11 - 2y

⇒ x = $\frac{11 - 2y}{3}$ ........(1)

Substituting value of x from equation (1) in 2x - 3y + 10 = 0,

$$\begin{aligned}\Rightarrow 2 \times \Big(\frac{11 - 2y}{3}\Big) - 3y + 10 = 0 \\[1em] \Rightarrow \frac{22 - 4y}{3} - 3y + 10 = 0 \\[1em] \Rightarrow \frac{22 - 4y - 9y + 30}{3} = 0 \\[1em] \Rightarrow \frac{52 - 13y}{3} = 0 \\[1em] \Rightarrow 52 - 13y = 0 \\[1em] \Rightarrow 13y = 52 \\[1em] \Rightarrow y = \frac{52}{13} = 4.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow x = \frac{11 - 2 \times 4}{3} \\[1em] = \frac{11 - 8}{3} \\[1em] = \frac{3}{3} \\[1em] = 1.\end{aligned}$$

Hence, x = 1 and y = 4.

Given,

Equations : 2x - 3y + 6 = 0 and 2x + 3y - 18 = 0

⇒ 2x - 3y + 6 = 0

⇒ 2x = 3y - 6

⇒ x = $\frac{3y - 6}{2}$ .......(1)

Substituting value of x from equation (1) in 2x + 3y - 18 = 0, we get :

$$\begin{aligned}\Rightarrow 2 \times \Big(\frac{3y - 6}{2}\Big) + 3y - 18 = 0 \\[1em] \Rightarrow 3y - 6 + 3y - 18 = 0 \\[1em] \Rightarrow 6y - 24 = 0 \\[1em] \Rightarrow 6y = 24 \\[1em] \Rightarrow y = \frac{24}{6} = 4.\end{aligned}$$

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

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

Hence, x = 3 and y = 4.

Given,

Equations : $\frac{3x}{2} - \frac{5y}{3} + 2 = 0, \frac{x}{3} + \frac{y}{2} = 2\frac{1}{6}$

$$\begin{aligned}\Rightarrow \frac{3x}{2} - \frac{5y}{3} + 2 = 0 \\[1em] \Rightarrow \frac{9x - 10y + 12}{6} = 0 \\[1em] \Rightarrow 9x - 10y + 12 = 0 \\[1em] \Rightarrow 10y = 9x + 12 \\[1em] \Rightarrow y = \frac{9x + 12}{10} .......(1)\end{aligned}$$

Substituting value of y from equation (1) in $\frac{x}{3} + \frac{y}{2} = 2\frac{1}{6}$, we get :

$$\begin{aligned}\Rightarrow \frac{x}{3} + \frac{\frac{9x + 12}{10}}{2} = 2\frac{1}{6} \\[1em] \Rightarrow \frac{x}{3} + \frac{9x + 12}{20} = \frac{13}{6} \\[1em] \Rightarrow \frac{20x + 3(9x + 12)}{60} = \frac{13}{6} \\[1em] \Rightarrow \frac{20x + 27x + 36}{60} = \frac{13}{6} \\[1em] \Rightarrow \frac{47x + 36}{60} = \frac{13}{6} \\[1em] \Rightarrow 47x + 36 = \frac{13}{6} \times 60 \\[1em] \Rightarrow 47x + 36 = 130 \\[1em] \Rightarrow 47x = 94 \\[1em] \Rightarrow x = \frac{94}{47} = 2.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow y = \frac{9x + 12}{10} \\[1em] \Rightarrow y = \frac{9 \times 2 + 12}{10} \\[1em] \Rightarrow y = \frac{18 + 12}{10} \\[1em] \Rightarrow y = \frac{30}{10} = 3.\end{aligned}$$

Hence, x = 2 and y = 3.

Given,

Equations : $\frac{x}{6} + \frac{y}{15} = 4, \frac{x}{3} - \frac{y}{12} = 4\frac{3}{4}$

$$\begin{aligned}\Rightarrow \frac{x}{6} + \frac{y}{15} = 4 \\[1em] \Rightarrow \frac{x}{6} = 4 - \frac{y}{15} \\[1em] \Rightarrow \frac{x}{6} = \frac{60 - y}{15} \\[1em] \Rightarrow x = \frac{6(60 - y)}{15} \\[1em] \Rightarrow x = \frac{2(60 - y)}{5} \\[1em] \Rightarrow x = \frac{120 - 2y}{5} ......(1)\end{aligned}$$

Substituting value of x from equation (1) in $\frac{x}{3} - \frac{y}{12} = 4\frac{3}{4}$, we get :

$$\begin{aligned}\Rightarrow \frac{\frac{120 - 2y}{5}}{3} - \frac{y}{12} = 4\frac{3}{4} \\[1em] \Rightarrow \frac{120 - 2y}{15} - \frac{y}{12} = \frac{19}{4} \\[1em] \Rightarrow \frac{4(120 - 2y) - 5y}{60} = \frac{19}{4} \\[1em] \Rightarrow 480 - 8y - 5y = \frac{19}{4} \times 60 \\[1em] \Rightarrow 480 - 13y = 19 \times 15 \\[1em] \Rightarrow 480 - 13y = 285 \\[1em] \Rightarrow 13y = 480 - 285 \\[1em] \Rightarrow 13y = 195 \\[1em] \Rightarrow y = \frac{195}{13} = 15.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow x = \frac{120 - 2 \times 15}{5} \\[1em] = \frac{120 - 30}{5} \\[1em] = \frac{90}{5} \\[1em] = 18.\end{aligned}$$

Hence, x = 18 and y = 15.

Substituting value of x and y in 2x + 3y = m, we get :

⇒ 2(2) + 3(1) = m

⇒ 4 + 3 = m

⇒ m = 7.

Hence, m = 7.

Given,

Equations : 10% of x + 20% of y = 24, 3x - y = 20

$$\begin{aligned}\Rightarrow 10% \text{ of } x + 20% \text{ of } y = 24 \\[1em] \Rightarrow \frac{10}{100} \times x + \frac{20}{100} \times y = 24 \\[1em] \Rightarrow \frac{x}{10} + \frac{y}{5} = 24 \\[1em] \Rightarrow \frac{x + 2y}{10} = 24 \\[1em] \Rightarrow x + 2y = 240 \\[1em] \Rightarrow x = 240 - 2y ........(1)\end{aligned}$$

Substituting value of x from equation (1) in 3x - y = 20, we get :

⇒ 3(240 - 2y) - y = 20

⇒ 720 - 6y - y = 20

⇒ 720 - 7y = 20

⇒ 7y = 720 - 20

⇒ 7y = 700

⇒ y = $\frac{700}{7}$ = 100.

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

⇒ x = 240 - 2(100) = 240 - 200 = 40.

Hence, x = 40 and y = 100.

Given,

Value of expression mx - ny is 3 when x = 5 and y = 6.

∴ 5m - 6n = 3

⇒ 5m = 3 + 6n

⇒ m = $\frac{3 + 6n}{5}$ ............(1)

Value of expression mx - ny is 8 when x = 6 and y = 5.

∴ 6m - 5n = 8

Substituting value of m from equation (1) in above equation.

$$\begin{aligned}\Rightarrow 6\Big(\frac{3 + 6n}{5}\Big) - 5n = 8 \\[1em] \Rightarrow \frac{18 + 36n}{5} - 5n = 8 \\[1em] \Rightarrow \frac{18 + 36n - 25n}{5} = 8 \\[1em] \Rightarrow 18 + 11n = 5 \times 8 \\[1em] \Rightarrow 18 + 11n = 40 \\[1em] \Rightarrow 11n = 40 - 18 \\[1em] \Rightarrow 11n = 22 \\[1em] \Rightarrow n = \frac{22}{11} \\[1em] \Rightarrow n = 2.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow m = \frac{3 + 6 \times 2}{5} \\[1em] = \frac{3 + 12}{5} \\[1em] = \frac{15}{5} \\[1em] = 3.\end{aligned}$$

Hence, m = 3 and n = 2.

Given,

Equations : 11(x - 5) + 10(y - 2) + 54 = 0, 7(2x - 1) + 9(3y - 1) = 25

1st equation :

⇒ 11(x - 5) + 10(y - 2) + 54 = 0

⇒ 11x - 55 + 10y - 20 + 54 = 0

⇒ 11x + 10y - 21 = 0

⇒ 11x = 21 - 10y

⇒ x = $\frac{21 - 10y}{11}$ .........(1)

2nd equation :

⇒ 7(2x - 1) + 9(3y - 1) = 25

⇒ 14x - 7 + 27y - 9 = 25

⇒ 14x + 27y - 16 = 25

⇒ 14x + 27y = 25 + 16

⇒ 14x + 27y = 41

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

$$\begin{aligned}\Rightarrow 14\Big(\frac{21 - 10y}{11}\Big) + 27y = 41 \\[1em] \Rightarrow \frac{294 - 140y}{11} + 27y = 41 \\[1em] \Rightarrow \frac{294 - 140y + 297y}{11} = 41 \\[1em] \Rightarrow 294 + 157y = 41 \times 11 \\[1em] \Rightarrow 294 + 157y = 451 \\[1em] \Rightarrow 157y = 157 \\[1em] \Rightarrow y = 1.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow x = \frac{21 - 10 \times 1}{11} \\[1em] = \frac{21 - 10}{11} \\[1em] = \frac{11}{11} \\[1em] = 1.\end{aligned}$$

Hence, x = 1 and y = 1.

Given,

1st equation :

$$\begin{aligned}\Rightarrow \frac{7 + x}{5} - \frac{2x - y}{4} = 3y - 5 \\[1em] \Rightarrow \frac{4(7 + x) - 5(2x - y)}{20} = 3y - 5 \\[1em] \Rightarrow 28 + 4x - 10x + 5y = 20(3y - 5) \\[1em] \Rightarrow 28 - 6x + 5y = 60y - 100 \\[1em] \Rightarrow 6x = 28 + 100 + 5y - 60y \\[1em] \Rightarrow 6x = 128 - 55y \\[1em] \Rightarrow x = \frac{128 - 55y}{6} ......(1)\end{aligned}$$

2nd equation :

$$\begin{aligned}\Rightarrow \frac{5y - 7}{2} + \frac{4x - 3}{6} = 18 - 5x \\[1em] \Rightarrow \frac{3(5y - 7) + 4x - 3}{6} = 18 - 5x \\[1em] \Rightarrow 15y - 21 + 4x - 3 = 6(18 - 5x) \\[1em] \Rightarrow 15y + 4x - 24 = 108 - 30x \\[1em] \Rightarrow 4x + 30x + 15y = 108 + 24 \\[1em] \Rightarrow 34x + 15y = 132.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow 34\Big(\frac{128 - 55y}{6}\Big) + 15y = 132 \\[1em] \Rightarrow \frac{4352 - 1870y}{6} + 15y = 132 \\[1em] \Rightarrow \frac{4352 - 1870y + 90y}{6} = 132 \\[1em] \Rightarrow 4352 - 1780y = 792 \\[1em] \Rightarrow 1780y = 4352 - 792 \\[1em] \Rightarrow 1780y = 3560 \\[1em] \Rightarrow y = \frac{3560}{1780} = 2.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow \frac{128 - 55 \times 2}{6}\\[1em] = \frac{128 - 110}{6} \\[1em] = \frac{18}{6} \\[1em] = 3.\end{aligned}$$

Hence, x = 3 and y = 2.

Given,

1st equation :

$$\begin{aligned}\Rightarrow 4x + \frac{x - y}{8} = 17 \\[1em] \Rightarrow \frac{32x + x - y}{8} = 17 \\[1em] \Rightarrow 33x - y = 136 \\[1em] \Rightarrow y = 33x - 136 .......(1)\end{aligned}$$

2nd equation :

$$\begin{aligned}\Rightarrow 2y + x - \frac{5y + 2}{3} = 2 \\[1em] \Rightarrow \frac{6y + 3x - 5y - 2}{3} = 2 \\[1em] \Rightarrow y + 3x - 2 = 6 \\[1em] \Rightarrow y = 8 - 3x ........(2)\end{aligned}$$

From (1) and (2), we get :

⇒ 33x - 136 = 8 - 3x

⇒ 33x + 3x = 8 + 136

⇒ 36x = 144

⇒ x = $\frac{144}{36}$ = 4.

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

⇒ y = 33(4) - 136 = 132 - 136 = -4.

Hence, x = 4 and y = -4.

Simplifying first equation, we get :

⇒ 0.4x - 1.5y = 6.5

Multiplying both sides of the above equation by 10, we get :

⇒ 10(0.4x - 1.5y) = 10 × 6.5

⇒ 4x - 15y = 65

⇒ 4x - 15y - 65 = 0

Simplifying second equation, we get :

⇒ 0.3x + 0.2y = 0.9

⇒ 10(0.3x + 0.2y) = 0.9 × 10

⇒ 3x + 2y = 9

⇒ 3x + 2y - 9 = 0

So, the equations are

⇒ 4x - 15y - 65 = 0 ........(1)

⇒ 3x + 2y - 9 = 0 ........(2)

By cross-multiplication method, we get :

$$\begin{aligned}\Rightarrow \frac{x}{(-15) \times (-9) - 2 \times (-65)} = \frac{y}{(-65) \times 3 - (-9) \times 4} = \frac{1}{4 \times 2 - 3 \times (-15)} \\[1em] \Rightarrow \frac{x}{135 + 130} = \frac{y}{-195 + 36} = \frac{1}{8 + 45} \\[1em] \Rightarrow \frac{x}{265} = \frac{y}{-159} = \frac{1}{53} \\[1em] \Rightarrow \frac{x}{265} = \frac{1}{53} \text{ and } \frac{y}{-159} = \frac{1}{53} \\[1em] \Rightarrow x = \frac{265}{53} \text{ and } y = \frac{-159}{53} \\[1em] \Rightarrow x = 5 \text{ and } y = -3.\end{aligned}$$

Hence, x = 5 and y = -3.

Given, equations :

$\sqrt{2}x - \sqrt{3}y = 0$ ........(1)

$\sqrt{5}x + \sqrt{2}y = 0$ ........(2)

Multiplying equation (1) by $\sqrt{2}$, we get :

$$\begin{aligned}\Rightarrow \sqrt{2}(\sqrt{2}x - \sqrt{3}y) = \sqrt{2} \times 0 \\[1em] \Rightarrow 2x - \sqrt{6}y = 0 \text{ .....(1)}\end{aligned}$$

Multiplying equation (2) by $\sqrt{3}$, we get :

$$\begin{aligned}\Rightarrow \sqrt{3}(\sqrt{5}x + \sqrt{2}y) = \sqrt{3} \times 0 \\[1em] \Rightarrow \sqrt{15}x + \sqrt{6}y = 0 \text{ .....(2)}\end{aligned}$$

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

$$\begin{aligned}\Rightarrow 2x - \sqrt{6}y + \sqrt{15}x + \sqrt{6}y = 0 \\[1em] \Rightarrow 2x + \sqrt{15}x = 0 \\[1em] \Rightarrow x(2 + \sqrt{15}) = 0 \\[1em] \Rightarrow x = 0.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow 2 \times 0 - \sqrt{6}y = 0 \\[1em] \Rightarrow 0 - \sqrt{6}y = 0 \\[1em] \Rightarrow \sqrt{6}y = 0 \\[1em] \Rightarrow y = 0.\end{aligned}$$

Hence, x = 0 and y = 0.

Given, equations :

$\frac{3}{2x} + \frac{2}{3y} = -\frac{1}{3}$ .......(1)

$\frac{3}{4x} + \frac{1}{2y} = -\frac{1}{8}$ .......(2)

Multiplying equation (1) by $\frac{1}{2}$, we get :

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

Subtracting equation (3) from (2), we get :

$$\begin{aligned}\Rightarrow \Big(\frac{3}{4x} + \frac{1}{2y}\Big) - \Big(\frac{3}{4x} + \frac{1}{3y}\Big) = -\frac{1}{8} - \Big(-\frac{1}{6}\Big) \\[1em] \Rightarrow \frac{3}{4x} - \frac{3}{4x} + \frac{1}{2y} - \frac{1}{3y} = -\frac{1}{8} + \frac{1}{6} \\[1em] \Rightarrow \frac{3 - 2}{6y} = \frac{-3 + 4}{24} \\[1em] \Rightarrow \frac{1}{6y} = \frac{1}{24} \\[1em] \Rightarrow y = \frac{24}{6} = 4.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow \frac{3}{2x} + \frac{2}{3 \times 4} = -\frac{1}{3} \\[1em] \Rightarrow \frac{3}{2x} + \frac{1}{6} = -\frac{1}{3} \\[1em] \Rightarrow \frac{3}{2x} = -\frac{1}{3} - \frac{1}{6} \\[1em] \Rightarrow \frac{3}{2x} = \frac{-2 - 1}{6} \\[1em] \Rightarrow \frac{3}{2x} = \frac{-3}{6} \\[1em] \Rightarrow x = \frac{3 \times 6}{2 \times -3}\\[1em] \Rightarrow x = \frac{18}{-6} = -3.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow \frac{3}{2 \times -3} + \frac{2}{3y} = -\frac{1}{3} \\[1em] \Rightarrow -\frac{1}{2} + \frac{2}{3y} = -\frac{1}{3} \\[1em] \Rightarrow \frac{2}{3y} = -\frac{1}{3} + \frac{1}{2} \\[1em] \Rightarrow \frac{2}{3y} = \frac{-2 + 3}{6} \\[1em] \Rightarrow \frac{2}{3y} = \frac{1}{6} \\[1em] \Rightarrow y = \frac{2 \times 6}{3} \\[1em] \Rightarrow y = \frac{12}{3} = 4.\end{aligned}$$

Hence, x = -3 and y = 4.

Let x be the digit at ten's place and y be the digit at unit's place.

Number = 10(x) + y = 10x + y

On reversing digits, the number becomes = 10(y) + x = 10y + x

Given,

Difference between digits = 4

⇒ x - y = 4

⇒ x = y + 4 .....(1)

Given,

Four times a certain two digit number is seven times the number obtained on interchanging its digits.

⇒ 4(10x + y) = 7(10y + x)

⇒ 40x + 4y = 70y + 7x

⇒ 40x - 7x = 70y - 4y

⇒ 33x = 66y

⇒ x = $\frac{66\text{y}}{33}$

⇒ x = 2y .........(2)

From (1) and (2), we get :

⇒ 2y = y + 4

⇒ 2y - y = 4

⇒ y = 4

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

⇒ x = y + 4 = 4 + 4 = 8.

Number = 10x + y = 10(8) + 4 = 80 + 4 = 84.

Hence, number = 84.

Let x be the digit at ten's place and y be the digit at unit's place.

Number = 10(x) + y = 10x + y

On reversing digits, the number becomes = 10(y) + x = 10y + x.

Given,

The sum of a two digit number and the number obtained by interchanging the digits of the number is 121.

∴ 10x + y + 10y + x = 121

⇒ 11x + 11y = 121

⇒ 11(x + y) = 121

⇒ x + y = 11 ..........(1)

Let digits of the number differ by 3.

⇒ x - y = 3 ........(2)

or

⇒ y - x = 3..........(3)

Considering x - y = 3,

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

⇒ x + y + x - y = 11 + 3

⇒ 2x = 14

⇒ x = 7.

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

⇒ 7 + y = 11

⇒ y = 11 - 7

⇒ y = 4.

Number = 10x + y = 10(7) + 4 = 70 + 4 = 74.

Considering y - x = 3,

Adding equation (1) and (3), we get :

⇒ x + y + y - x = 11 + 3

⇒ 2y = 14

⇒ y = $\frac{14}{2}$

⇒ y = 7.

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

⇒ x + 7 = 11

⇒ x = 11 - 7

⇒ x = 4.

Number = 10x + y = 10(4) + 7 = 40 + 7 = 47.

Hence, number = 47 or 74.

Let x be the digit at ten's place and y be the digit at unit's place.

Number = 10(x) + y = 10x + y

Given,

Number is obtained by multiplying the sum of the digits by 8.

∴ 8(x + y) = 10x + y

⇒ 8x + 8y = 10x + y

⇒ 10x - 8x = 8y - y

⇒ 2x = 7y

⇒ x = $\frac{7}{2}$y ..........(1)

Number is obtained by multiplying the difference of the digits by 14 and adding 2.

⇒ 14(x - y) + 2 = 10x + y .............(2)

or,

⇒ 14(y - x) + 2 = 10x + y .............(3)

Solving equation (2),

⇒ 14x - 14y + 2 = 10x + y

⇒ 14x - 10x = y + 14y - 2

⇒ 4x = 15y - 2 ...........(4)

Substituting value of x from equation (1) in equation (4), we get :

$$\begin{aligned}\Rightarrow 4 \times \frac{7}{2}y = 15y - 2 \\[1em] \Rightarrow 14y = 15y - 2 \\[1em] \Rightarrow 15y - 14y = 2 \\[1em] \Rightarrow y = 2.\end{aligned}$$

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

$\Rightarrow x = \frac{7}{2}y = \frac{7}{2} \times 2$ = 7.

Number = 10x + y = 10(7) + 2 = 70 + 2 = 72.

Solving equation (3),

⇒ 14(y - x) + 2 = 10x + y

⇒ 14y - 14x + 2 = 10x + y

⇒ 14y - y = 10x + 14x - 2

⇒ 13y = 24x + 2 ............(5)

Substituting value of x from equation (1) in equation (5), we get :

$$\begin{aligned}\Rightarrow 13y = 24 \times \frac{7}{2}y + 2 \\[1em] \Rightarrow 13y = 84y + 2 \\[1em] \Rightarrow 84y - 13y = -2 \\[1em] \Rightarrow 71y = -2 \\[1em] \Rightarrow y = -\frac{2}{71}.\end{aligned}$$

This is not possible as y cannot be a fraction.

Hence, number = 72.

Let cost price of article A be ₹ x and article B be ₹ y.

Given,

Two articles A and B are sold for ₹ 1,167 making 5% profit on A and 7% profit on B.

$$\begin{aligned}\Rightarrow \frac{105}{100}x + \frac{107}{100}y = 1167 \\[1em] \Rightarrow \frac{105x + 107y}{100} = 1167 \\[1em] \Rightarrow 105x + 107y = 116700 \\[1em] \Rightarrow 105x = 116700 - 107y \\[1em] \Rightarrow x = \frac{116700 - 107y}{105} .....(1)\end{aligned}$$

Given,

Two articles A and B are sold for ₹ 1,165 making 7% profit on A and 5% profit on B.

$$\begin{aligned}\Rightarrow \frac{107}{100}x + \frac{105}{100}y = 1165 \\[1em] \Rightarrow \frac{107x + 105y}{100} = 1165 \\[1em] \Rightarrow 107x + 105y = 116500 .....(2)\end{aligned}$$

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

$$\begin{aligned}\Rightarrow 107 \times \frac{116700 - 107y}{105} + 105y = 116500 \\[1em] \Rightarrow \frac{12486900 - 11449y}{105} + 105y = 116500 \\[1em] \Rightarrow \frac{12486900 - 11449y + 11025y}{105} = 116500 \\[1em] \Rightarrow 12486900 - 424y = 116500 \times 105 \\[1em] \Rightarrow 12486900 - 424y = 12232500 \\[1em] \Rightarrow 424y = 254400 \\[1em] \Rightarrow y = \frac{254400}{424} \\[1em] \Rightarrow y = 600\end{aligned}$$

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

$$\begin{aligned}\Rightarrow x = \frac{116700 - 107 \times 600}{105} \\[1em] = \frac{116700 - 64200}{105} \\[1em] = \frac{52500}{105} \\[1em] = 500.\end{aligned}$$

Hence, cost of article A = ₹ 500 and cost of article B = ₹ 600.

Let Pooja alone can do work in x days and Ritu alone can do in y days.

So, in one day Pooja can do $\frac{1}{x}$ th part of work and Ritu can do $\frac{1}{y}$ th part of work.

Given,

One day work of Pooja equals three fourth of one day work of Ritu.

$$\begin{aligned}\therefore \frac{1}{x} = \frac{3}{4} \times \frac{1}{y} \\[1em] \Rightarrow \frac{1}{x} = \frac{3}{4y} \\[1em] \Rightarrow x = \frac{4y}{3} .......(1)\end{aligned}$$

Given,

Pooja and Ritu can do a piece of work in $17\frac{1}{7} \text{ or } \frac{120}{7}$ days.

So, in one day both of them can do $\frac{7}{120}$ th part of work.

$\therefore \frac{1}{x} + \frac{1}{y} = \frac{7}{120}$

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

$$\begin{aligned}\Rightarrow \frac{1}{\frac{4y}{3}} + \frac{1}{y} = \frac{7}{120} \\[1em] \Rightarrow \frac{3}{4y} + \frac{1}{y} = \frac{7}{120} \\[1em] \Rightarrow \frac{3 + 4}{4y} = \frac{7}{120} \\[1em] \Rightarrow \frac{7}{4y} = \frac{7}{120} \\[1em] \Rightarrow y = \frac{120 \times 7}{7 \times 4} \\[1em] \Rightarrow y = 30.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow x = \frac{4 \times 30}{3} \\[1em] = 40.\end{aligned}$$

Hence, Pooja can do the work alone in 40 days and Ritu in 30 days.

Let Mr. Ahuja and Mrs. Ahuja weigh x kg and y kg respectively.

After one month,

Mr. Ahuja weighs (x - 5) kg and Mrs. Ahuja weighs (y - 4) kg.

Given,

After one month Mr. Ahuja weighs as much as his wife weighed before the course.

∴ x - 5 = y

⇒ x = y + 5 .........(1)

Given,

Mrs. Ahuja after one month weighs $\frac{7}{8}$ th of what her husband weighed before the course.

$$\begin{aligned}\therefore y - 4 = \frac{7}{8} \times x \\[1em] \Rightarrow 8(y - 4) = 7x \\[1em] \Rightarrow 8y - 32 = 7x \\[1em] \Rightarrow 8y = 7x + 32 \\[1em] \Rightarrow y = \frac{7x + 32}{8} .......(2)\end{aligned}$$

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

$$\begin{aligned}\Rightarrow y = \frac{7(y + 5) + 32}{8} \\[1em] \Rightarrow y = \frac{7y + 35 + 32}{8} \\[1em] \Rightarrow 8y = 7y + 67 \\[1em] \Rightarrow 8y - 7y = 67 \\[1em] \Rightarrow y = 67.\end{aligned}$$

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

⇒ x = 67 + 5 = 72.

Hence, weight of Mr. Ahuja and Mrs. Ahuja are 72 kg and 67 kg respectively.

Let constant monthly expense be ₹ x and for each member of the family monthly expense be ₹ y.

Given,

For a family of 4 persons, the total monthly expenses are ₹ 10,400.

⇒ x + 4y = 10400 .......(1)

For a family of 7 persons, the total monthly expenses are ₹ 15,800.

⇒ x + 7y = 15800 .......(2)

Subtracting equation (1) from (2), we get :

⇒ (x + 7y) - (x + 4y) = 15800 - 10400

⇒ x - x + 7y - 4y = 5400

⇒ 3y = 5400

⇒ y = $\frac{5400}{3}$ = ₹ 1800.

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

⇒ x + 4(1800) = 10400

⇒ x + 7200 = 10400

⇒ x = 10400 - 7200 = ₹ 3200.

Hence, constant expense = ₹ 3200 and monthly expense = ₹ 1800.

Let fixed charge be ₹ x and charge per kilometer be ₹ y.

Given,

For a distance of 10 km, the charge paid is ₹ 315.

⇒ x + 10y = 315 ........(1)

For a distance of 15 km, the charge paid is ₹ 465.

⇒ x + 15y = 465 ........(2)

Subtracting equation (1) from (2), we get :

⇒ (x + 15y) - (x + 10y) = 465 - 315

⇒ x - x + 15y - 10y = 150

⇒ 5y = 150

⇒ y = $\frac{150}{5}$ = ₹ 30.

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

⇒ x + 10 × 30 = 315

⇒ x + 300 = 315

⇒ x = 315 - 300 = ₹ 15.

For 32 km,

Charge = x + 32y = 15 + 32 × 30

= 15 + 960 = ₹ 975.

Hence, fixed charge = ₹ 15, charge per kilometer = ₹ 30 and for 32 km charge = ₹ 975.

Let for first three days charge be ₹ x and for the days after that charge be ₹ y per day.

Given,

Geeta paid ₹ 27 for a book kept for seven days.

∴ x + (7 - 3)y = 27

⇒ x + 4y = 27 ........(1)

Mohit paid ₹ 21 for a book kept for five days.

∴ x + (5 - 3)y = 21

⇒ x + 2y = 21 ........(2)

Subtracting equation (2) from (1), we get :

⇒ (x + 4y) - (x + 2y) = 27 - 21

⇒ x - x + 4y - 2y = 6

⇒ 2y = 6

⇒ y = $\frac{6}{2}$

⇒ y = ₹ 3.

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

⇒ x + 4(3) = 27

⇒ x + 12 = 27

⇒ x = 27 - 12

⇒ x = ₹ 15.

Hence, fixed charge = ₹ 15 and charge for each extra day = ₹ 3.

Let the length of rectangle be x units and breadth be y units.

Given,

Area of rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units.

⇒ (x - 5)(y + 3) = xy - 9

⇒ xy + 3x - 5y - 15 = xy - 9

⇒ xy - xy + 3x - 5y = -9 + 15

⇒ 3x - 5y = 6 ........(1)

Given,

If the length of this rectangle increases by 3 units and the breadth by 2 units, the area increases by 67 square units.

⇒ (x + 3)(y + 2) = xy + 67

⇒ xy + 2x + 3y + 6 = xy + 67

⇒ xy - xy + 2x + 3y = 67 - 6

⇒ 2x + 3y = 61 ........(2)

Multiplying equation (1) by 2, we get :

⇒ 2(3x - 5y) = 2 × 6

⇒ 6x - 10y = 12 .........(3)

Multiplying equation (2) by 3, we get :

⇒ 3(2x + 3y) = 3 × 61

⇒ 6x + 9y = 183 .........(4)

Subtracting equation (3) from (4), we get :

⇒ (6x + 9y) - (6x - 10y) = 183 - 12

⇒ 6x - 6x + 9y + 10y = 171

⇒ 19y = 171

⇒ y = $\frac{171}{19}$ = 9.

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

⇒ 3x - 5(9) = 6

⇒ 3x - 45 = 6

⇒ 3x = 45 + 6

⇒ 3x = 51

⇒ x = $\frac{51}{3}$ = 17.

Hence, length = 17 units and breadth = 9 units.

Let the time taken by the first pipe be x hours and the time taken by the second pipe be y hours.

In 1 hour,

The first pipe can fill $\frac{1}{x}$ th of the pool

The second pipe can fill $\frac{1}{y}$ th part of pool.

Given,

It takes 12 hours to fill the pool.

So, in 1 hour both of them will fill $\frac{1}{12}$ th part of the pool.

$\therefore \frac{1}{x} + \frac{1}{y} = \frac{1}{12}$ ...........(1)

Given,

If the pipe of larger diameter is used for 4 hours and the pipe of smaller diameter is used for 9 hours, only half of the pool is filled.

$\therefore \frac{4}{x} + \frac{9}{y} = \frac{1}{2}$ .............(2)

Multiplying equation (1) by 4, we get :

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

Subtracting equation (3) from (2), we get :

$$\begin{aligned}\Rightarrow \frac{4}{x} + \frac{9}{y} - \Big(\frac{4}{x} + \frac{4}{y}\Big)= \frac{1}{2} - \frac{1}{3} \\[1em] \Rightarrow \frac{4}{x} - \frac{4}{x} + \frac{9}{y} - \frac{4}{y} = \frac{3 - 2}{6} \\[1em] \Rightarrow \frac{5}{y} = \frac{1}{6} \\[1em] \Rightarrow y = 5 \times 6 \\[1em] \Rightarrow y = 30.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow \frac{1}{x} + \frac{1}{y} = \frac{1}{12} \\[1em] \Rightarrow \frac{1}{x} + \frac{1}{30} = \frac{1}{12} \\[1em] \Rightarrow \frac{1}{x} = \frac{1}{12} - \frac{1}{30} \\[1em] \Rightarrow \frac{1}{x} = \frac{5 - 2}{60} \\[1em] \Rightarrow \frac{1}{x} = \frac{3}{60} \\[1em] \Rightarrow \frac{1}{x} = \frac{1}{20} \\[1em] \Rightarrow x = 20.\end{aligned}$$

Hence, pipe will larger diameter will fill pool in 20 hours and with smaller diameter in 30 hours.

Given,

Distance (Connaught Place to Qutub Minar while going) = 13 km

Distance (Connaught Place to Qutub Minar while returning) = 16 km

Case 1:

Charges applied:

Fixed charge = ₹ x

Per km charge = ₹ y

Total fare = Fixed charge + (Distance × per km charge)

For trip while going (13 km):

Total fare = x + (13 × y) = x + 13y

For trip while returning (16 km):

Total fare = x + (16 × y) = x + 16y

Total fare of whole journey = (x + 13y) + (x + 16y)

= 2x + 29y.

Hence, total fare = ₹ (2x + 29y).

Case 2:

Charges :

Fixed charge = ₹ x for first kilometer

Per km = ₹ y for remaining distance

Going (13 km):

Charge for first km = ₹ x

Remaining distance = 12 km

Charge for remaining distance = 12y

Total fare = x + 12y

Returning (16 km):

First km = ₹ x

Remaining distance = 15 km

Charge for remaining distance = 15y

Total fare = x + 15y

Total fare of whole journey = (x + 12y) + (x + 15y)

= 2x + 27y.

Hence, total fare = ₹ (2x + 27y).

Given,

Volume of the common room = 144 cubic units.

Total surface area of common room = 192 sq. units

Height of the room = 3 units.

(a) Let, length = x units and breadth = y units.

Volume of cuboid = l × b × h

144 = x × y × 3

3xy = 144

xy = $\frac{144}{3}$

xy = 48 .........(1)

By formula,

Total surface area of cuboid = 2(lb + bh + hl)

192 = 2(x × y + y × 3 + 3 × x)

192 = 2(xy + 3y + 3x)

xy + 3y + 3x = $\frac{192}{2}$

xy + 3y + 3x = 96........(2)

Hence, equation for volume : 3xy = 144 and for TSA = 2(xy + 3y + 3x) = 192.

(b) Substituting the value of equation (1) in equation (2), we get :

48 + 3y + 3x = 96

3x + 3y = 96 - 48

3x + 3y = 48

3(x + y) = 48

x + y = $\frac{48}{3}$

x + y = 16

x = 16 - y .........(3)

Substituting the value of x from equation (3) in equation (1) :

(16 - y) × y = 48

16y - y² = 48

y² - 16y + 48 = 0

y² - 12y - 4y + 48 =0

y(y - 12) - 4(y - 12) = 0

(y - 12) (y - 4) = 0

y = 12 or y = 4

If y = 12, then

x = 16 - y = 16 - 12 = 4.

If y = 4, then

x = 16 - y = 16 - 4 = 12.

Since length is longer than breadth,

∴ x = 12 and y = 4

Hence, x = 12 and y = 4.

(c) Length of the biggest rod that can be put in a room = Diagonal of room

Diagonal of room (cuboid) = $\sqrt{l^2 + b^2 + h^2}$

$$\begin{aligned}\Rightarrow d = \sqrt{12^2 + 4^2 + 3^2} \\[1em] \Rightarrow d = \sqrt{144 + 16 + 9} \\[1em] \Rightarrow d = \sqrt{169} \\[1em] \Rightarrow d = 13 \text{ units}.\end{aligned}$$

Hence, length of the biggest iron rod can be put in the room = 13 units.