Chapter 6: Simultaneous (Linear) Equations [Including Problems]

Exercise 6(A)

Question 1(a)

Solution of equations x - 2y = 7 and 2x - y = 8 is :

x = 3, y = 2
x = -3, y = 2
x = 3, y = -2
x = -3, y = -2

Answer

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 = $-\dfrac{6}{3}$ = -2.

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

⇒ x = 7 + 2 × -2 = 7 - 4 = 3.

Hence, Option 3 is the correct option.

Question 1(b)

Solution of equations x + y = 2.1 and x - y = 0.3 is :

x = 1.2, y = 0.9
x = 1.2, y = -0.9
x = -1.2, y = 0.9
x = -1.2, y = -0.9

Answer

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 = $\dfrac{1.8}{2}$ = 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.

Question 1(c)

Solution of equations $\dfrac{1}{x} + \dfrac{1}{y} = 21 \text{ and } \dfrac{1}{x} - \dfrac{1}{y} + 9 = 0$ is :

$x = \dfrac{1}{6}, y = -\dfrac{1}{15}$
$x = -\dfrac{1}{6}, y = \dfrac{1}{15}$
$x = -\dfrac{1}{6}, y = -\dfrac{1}{15}$
$x = \dfrac{1}{6}, y = \dfrac{1}{15}$

Answer

Given,

Equations :

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

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

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

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

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{y} = 21 \\[1em] \Rightarrow \dfrac{1}{\dfrac{1}{6}} + \dfrac{1}{y} = 21 \\[1em] \Rightarrow 6 + \dfrac{1}{y} = 21 \\[1em] \Rightarrow \dfrac{1}{y} = 21 - 6 \\[1em] \Rightarrow \dfrac{1}{y} = 15 \\[1em] \Rightarrow y = \dfrac{1}{15}.$

Hence, Option 4 is the correct option.

Question 1(d)

Solution of equations $\dfrac{x}{2} - \dfrac{y}{3} = 0 \text{ and } \dfrac{x}{2} + \dfrac{y}{3} = 6$ is :

x = 6, y = -9
x = 6, y = 9
x = -6, y = -9
x = -6, y = 9

Answer

Given,

Equations :

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

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

$\Rightarrow \dfrac{x}{2} - \dfrac{y}{3} + \dfrac{x}{2} + \dfrac{y}{3} = 0 + 6 \\[1em] \Rightarrow \dfrac{2x}{2} = 6 \\[1em] \Rightarrow x = 6.$

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

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

Hence, Option 2 is the correct option.

Question 1(e)

Solution of equations $\dfrac{x + 8}{5} - \dfrac{2y - 4}{3} = 0$ and y - x = 3 is :

x = 2, y = 5
x = -2, y = 5
x = 2, y = -5
x = -2, y = -5

Answer

Given,

Equations :

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

⇒ y - x = 3

⇒ y = x + 3 .........(2)

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

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

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:

Question 2

2x + 3y = 8
2x = 2 + 3y

Answer

Given,

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

⇒ 2x + 3y = 8

⇒ 2x = 8 - 3y

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

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

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

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

$\Rightarrow x = \dfrac{8 - 3 \times 1}{2} \\[1em] = \dfrac{8 - 3}{2} \\[1em] = \dfrac{5}{2} = 2.5$

Hence, x = 2.5 and y = 1.

Question 3

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

Answer

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 = $\dfrac{25 - 0.1y}{0.2}$ ..........(1)

$\Rightarrow 2(x - 2) - 1.6y = 116 \\[1em] \Rightarrow 2x - 4 - 1.6y = 116 \\[1em] \Rightarrow 2x - 1.6y = 116 + 4 \\[1em] \Rightarrow 2x - 1.6y = 120 \\[1em]$

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

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

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

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

Hence, x = 100 and y = 50.

Question 4

6x = 7y + 7
7y - x = 8

Answer

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 = $\dfrac{55}{35} = \dfrac{11}{7}$ .

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

$\Rightarrow x = 7 \times \dfrac{11}{7} - 8 \\[1em] \Rightarrow x = 11 - 8 = 3.$

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

Question 5

y = 4x - 7
16x - 5y = 25

Answer

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 = $\dfrac{10}{4} = \dfrac{5}{2}$ .

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

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

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

Question 6

1.5x + 0.1y = 6.2
3x - 0.4y = 11.2

Answer

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 = $\dfrac{6.2 - 0.1y}{1.5}$ ........(1)

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

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

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

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

Hence, x = 4 and y = 2.

Question 7

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

Answer

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 = $\dfrac{21 - 3y}{2}$ ..........(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 :

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

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

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

Hence, x = -3 and y = 9.

Question 8

$\dfrac{2x + 1}{7} + \dfrac{5y - 3}{3} = 12$

$\dfrac{3x + 2}{2} - \dfrac{4y + 3}{9} = 13$

Answer

Simplifying first equation :

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

Simplifying second equation :

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

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

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

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

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

Hence, x = 10 and y = 6.

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

Question 9

3x - y = 23

$\dfrac{x}{3} + \dfrac{y}{4}$ = 4

Answer

Given equations :

⇒ 3x - y = 23 .........(1)

⇒ $\dfrac{x}{3} + \dfrac{y}{4}$ = 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 :

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

⇒ 4x + 3y = 48 ...........(4)

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

⇒ 9x - 3y + 4x + 3y = 69 + 48

⇒ 13x = 117

⇒ x = $\dfrac{117}{13}$ = 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.

Question 10

$\dfrac{5y}{2} - \dfrac{x}{3} = 8$

$\dfrac{y}{2} + \dfrac{5x}{3} = 12$

Answer

Given equations :

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

Multiplying equation (1) by 30, we get :

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

Multiplying equation (2) by 6, we get :

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

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

⇒ 75y - 10x + 3y + 10x = 240 + 72

⇒ 78y = 312

⇒ y = $\dfrac{312}{78}$ = 4.

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

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

Hence, x = 6 and y = 4.

Question 11

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

26x + 3y + 4 = 0

Answer

Simplifying first equation :

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

⇒ 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 = $-\dfrac{59}{118} = -\dfrac{1}{2}$

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

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

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

Question 12

$\dfrac{x - y}{6} = 2(4 - x)$

2x + y = 3(x - 4)

Answer

Simplifying first equation :

⇒ $\dfrac{x - y}{6} = 2(4 - x)$

⇒ 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 = $\dfrac{36}{12}$ = 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.

Question 13

2x - 3y - 3 = 0

$\dfrac{2x}{3} + 4y + \dfrac{1}{2} = 0$

Answer

Given, equations :

⇒ 2x - 3y - 3 = 0 .............(1)

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

Simplifying second equation :

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

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 = $-\dfrac{9}{30} = -\dfrac{3}{10}$ .

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

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

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

Question 14

13x + 11y = 70

11x + 13y = 74

Answer

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 = $\dfrac{192}{48}$ = 4

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

⇒ 13x + 11(4) = 70

⇒ 13x + 44 = 70

⇒ 13x = 70 - 44

⇒ 13x = 26

⇒ x = $\dfrac{26}{13}$ = 2.

Hence, x = 2 and y = 4.

Question 15

41x + 53y = 135

53x + 41y = 147

Answer

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 = $\dfrac{1128}{1128}$ = 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 = $\dfrac{82}{41}$ = 2.

Hence, x = 2 and y = 1.

Question 16

If 2x + y = 23 and 4x - y = 19; find the values of x - 3y and 5y - 2x.

Answer

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 = $\dfrac{42}{6}$ = 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.

Question 17

If 10y = 7x - 4 and 12x + 18y = 1; find the values of 4x + 6y and 8y - x.

Answer

Given,

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

⇒ 10y = 7x - 4

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

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

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

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

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

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

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

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

Question 18(i)

Solve for x and y :

$\dfrac{y + 7}{5} = \dfrac{2y - x}{4} + 3x - 5$

$\dfrac{7 - 5x}{2} + \dfrac{3 - 4y}{6} = 5y - 18$

Answer

Simplifying first equation :

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

Simplifying second equation :

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

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

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

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

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

Hence, x = 2 and y = 3.

Question 18(ii)

Solve for x and y :

$4x = 17 - \dfrac{x - y}{8}$

$2y + x = 2 + \dfrac{5y + 2}{3}$

Answer

Simplifying first equation :

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

Simplifying second equation :

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

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

⇒ 33x - 136 = 8 - 3x

⇒ 33x + 3x = 136 + 8

⇒ 36x = 144

⇒ x = $\dfrac{144}{36}$ = 4.

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

⇒ y = 8 - 3(4) = 8 - 12 = -4.

Hence, x = 4 and y = -4.

Exercise 6(B)

Question 1(a)

Solution of equations x + y = 8 and 3x - 5y = 0 is :

x = 5, y = -3
x = 5, y = 3
x = -5, y = 3
x = -5, y = -3

Answer

Given, equations :

⇒ x + y = 8 and 3x - 5y = 0

⇒ x + y - 8 = 0 ...........(1)

⇒ 3x - 5y = 0 ..........(2)

By cross-multiplication method :

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

Hence, Option 2 is the correct option.

Question 1(b)

Solution of equations $\dfrac{4a - 3}{6} + \dfrac{5b - 7}{2}$ = 18 - 5a and a + b = 5 is :

a = -3, b = 2
a = -3, b = -2
a = 3, b = 2
a = 3, b = -2

Answer

Given,

1st equation :

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

2nd equation :

⇒ a + b = 5

⇒ a + b - 5 = 0

By cross-multiplication method :

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

Hence, Option 3 is the correct option.

Question 1(c)

Solution of equations $\dfrac{x}{7} - \dfrac{y}{2} = 2 \text{ and } \dfrac{x - y}{3} = 3$ is :

x = 7, y = 2
x = -7, y = 2
x = -7, y = -2
x = 7, y = -2

Answer

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

Simplifying first equation :

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

Simplifying second equation :

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

By cross-multiplication method :

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

Hence, Option 4 is the correct option.

Question 1(d)

Solution of equations 11x - 7y = 29 and 7x - 11y = 25 is :

x = 2, y = 1
x = 2, y = -1
x = -2, y = 1
x = -2, y = -1

Answer

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 :

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

Hence, Option 2 is the correct option.

Question 1(e)

Solution of equations 15x + 6y = 27 and 6x + 15y = 36 is :

x = 1, y = -2
x = -1, y = 2
x = 1, y = 2
x = -1, y = -2

Answer

Given, equations can be written as :

⇒ 15x + 6y - 27 = 0 .......(1)

⇒ 6x + 15y - 36 = 0 ........(2)

By cross-multiplication method :

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

Hence, Option 3 is the correct option.

Question 1(f)

Solution of equations $\dfrac{2}{x} - \dfrac{3}{y} = 13 \text{ and } \dfrac{3}{x} + \dfrac{2}{y}$ = 0 is :

$x = \dfrac{1}{2}, y = \dfrac{1}{3}$
$x = -\dfrac{1}{2}, y = \dfrac{1}{3}$
$x = \dfrac{1}{2}, y = -\dfrac{1}{3}$
$x = -\dfrac{1}{2}, y = -\dfrac{1}{3}$

Answer

Given, equations :

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

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

Multiplying equation (1) by 2, we get :

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

Multiplying equation (2) by 3, we get :

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

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

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

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

$\Rightarrow \dfrac{3}{\dfrac{1}{2}} + \dfrac{2}{y} = 0 \\[1em] \Rightarrow 6 + \dfrac{2}{y} = 0 \\[1em] \Rightarrow \dfrac{2}{y} = 0 - 6 \\[1em] \Rightarrow \dfrac{2}{y} = -6 \\[1em] \Rightarrow y = \dfrac{2}{-6} = -\dfrac{1}{3}.$

Hence, Option 3 is the correct option.

Question 2

Solve, using cross-multiplication :

4x + 3y = 17

3x - 4y + 6 = 0

Answer

Given, equations :

⇒ 4x + 3y = 17 and 3x - 4y + 6 = 0

⇒ 4x + 3y - 17 = 0 .......(1)

⇒ 3x - 4y + 6 = 0 ........(2)

By cross-multiplication method :

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

Hence, x = 2 and y = 3.

Question 3

Solve, using cross-multiplication :

3x + 4y = 11

2x + 3y = 8

Answer

Given, equations :

⇒ 3x + 4y = 11 and 2x + 3y = 8

⇒ 3x + 4y - 11 = 0 .......(1)

⇒ 2x + 3y - 8 = 0 ........(2)

By cross-multiplication method :

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

Hence, x = 1 and y = 2.

Question 4

Solve, using cross-multiplication :

6x + 7y - 11 = 0

5x + 2y = 13

Answer

Given, equations :

⇒ 6x + 7y - 11 = 0 and 5x + 2y = 13

⇒ 6x + 7y - 11 = 0 ........(1)

⇒ 5x + 2y - 13 = 0 ........(2)

By cross-multiplication method :

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

Hence, x = 3 and y = -1.

Question 5

Solve, using cross-multiplication :

5x + 4y + 14 = 0

3x = -10 - 4y

Answer

Given, equations :

⇒ 5x + 4y + 14 = 0 and 3x = -10 - 4y

⇒ 5x + 4y + 14 = 0 ..........(1)

⇒ 3x + 4y + 10 = 0 ..........(2)

By cross-multiplication method :

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

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

Question 6

Solve, using cross-multiplication :

x - y + 2 = 0

7x + 9y = 130

Answer

Given, equations :

⇒ x - y + 2 = 0 and 7x + 9y = 130

⇒ x - y + 2 = 0 ........(1)

⇒ 7x + 9y - 130 = 0 ........(2)

By cross-multiplication method :

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

Hence, x = 7 and y = 9.

Question 7

Solve, using cross-multiplication :

4x - y = 5

5y - 4x = 7

Answer

Given, equations :

⇒ 4x - y = 5 and 5y - 4x = 7

⇒ 4x - y - 5 = 0 ..........(1)

⇒ -4x + 5y - 7 = 0 .........(2)

By cross-multiplication method :

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

Hence, x = 2 and y = 3.

Question 8

Solve, using cross-multiplication :

4x - 3y = 0

2x + 3y = 18

Answer

Given, equations :

⇒ 4x - 3y = 0 and 2x + 3y = 18

⇒ 4x - 3y = 0 .........(1)

⇒ 2x + 3y - 18 = 0 .....(2)

By cross-multiplication method :

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

Hence, x = 3 and y = 4.

Question 9

Solve, using cross-multiplication :

8x + 5y = 9

3x + 2y = 4

Answer

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 :

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

Hence, x = -2 and y = 5.

Question 10

Solve, using cross-multiplication :

4x - 3y - 11 = 0

6x + 7y - 5 = 0

Answer

Given, equations :

⇒ 4x - 3y - 11 = 0 ............(1)

⇒ 6x + 7y - 5 = 0 ............(2)

By cross-multiplication method :

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

Hence, x = 2 and y = -1.

Question 11

Solve :

$\dfrac{9}{x} - \dfrac{4}{y} = 8$

$\dfrac{13}{x} + \dfrac{7}{y} = 101$

Answer

Given, equations :

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

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

Multiplying equation (1) by 7, we get :

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

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

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

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

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

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

$\Rightarrow \dfrac{9}{\dfrac{1}{4}} - \dfrac{4}{y} = 8 \\[1em] \Rightarrow 36 - \dfrac{4}{y} = 8 \\[1em] \Rightarrow \dfrac{4}{y} = 36 - 8 \\[1em] \Rightarrow \dfrac{4}{y} = 28 \\[1em] \Rightarrow y = \dfrac{4}{28} = \dfrac{1}{7}.$

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

Question 12

Solve :

$\dfrac{3}{x} + \dfrac{2}{y} = 10$

$\dfrac{9}{x} - \dfrac{7}{y} = 10.5$

Answer

Given, equations :

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

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

Multiplying equation (1) by 3, we get :

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

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

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

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

$\Rightarrow \dfrac{3}{x} + \dfrac{2}{\dfrac{2}{3}} = 10 \\[1em] \Rightarrow \dfrac{3}{x} + \dfrac{6}{2} = 10 \\[1em] \Rightarrow \dfrac{3}{x} + 3 = 10 \\[1em] \Rightarrow \dfrac{3}{x} = 7 \\[1em] \Rightarrow x = \dfrac{3}{7}.$

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

Question 13

Solve :

$5x + \dfrac{8}{y} = 19$

$3x - \dfrac{4}{y} = 7$

Answer

Given, equations :

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

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

Multiplying equation (2) by 2, we get :

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

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

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

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

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

Hence, x = 3 and y = 2.

Question 14

Solve : $4x + \dfrac{6}{y} = 15 \text{ and } 3x - \dfrac{4}{y} = 7$ .

Hence, find 'a' if y = ax - 2.

Answer

Given, equations :

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

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

Multiplying equation (1) by 2, we get :

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

Multiplying equation (2) by 3, we get :

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

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

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

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

$\Rightarrow 3 \times 3 - \dfrac{4}{y} = 7 \\[1em] \Rightarrow 9 - \dfrac{4}{y} = 7 \\[1em] \Rightarrow \dfrac{4}{y} = 9 - 7 \\[1em] \Rightarrow \dfrac{4}{y} = 2 \\[1em] \Rightarrow y = \dfrac{4}{2} = 2.$

Given,

⇒ y = ax - 2

⇒ 2 = 3a - 2

⇒ 3a = 2 + 2

⇒ 3a = 4

⇒ a = $\dfrac{4}{3} = 1\dfrac{1}{3}$ .

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

Question 15

Solve :

$\dfrac{3}{x} - \dfrac{2}{y} = 0 \text{ and } \dfrac{2}{x} + \dfrac{5}{y} = 19$ .

Hence, find 'a' if y = ax + 3,

Answer

Given, equations :

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

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

Multiplying equation (1) by 2, we get :

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

Multiplying equation (2) by 3, we get :

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

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

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

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

$\Rightarrow \dfrac{3}{x} - \dfrac{2}{\dfrac{1}{3}} = 0 \\[1em] \Rightarrow \dfrac{3}{x} - 6 = 0 \\[1em] \Rightarrow \dfrac{3}{x} = 6 \\[1em] \Rightarrow x = \dfrac{3}{6} = \dfrac{1}{2}.$

Given,

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

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

Question 16(i)

Solve :

$\dfrac{20}{x + y} + \dfrac{3}{x - y} = 7$

$\dfrac{8}{x - y} - \dfrac{15}{x + y} = 5$

Answer

Given, equations :

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

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

Multiplying equation (1) by 8, we get :

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

Multiplying equation (2) by 3, we get :

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

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

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

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

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

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

⇒ (x + y) + (x - y) = 5 + 1

⇒ 2x = 6

⇒ x = $\dfrac{6}{2}$

⇒ x = 3.

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

⇒ 3 - y = 1

⇒ y = 3 - 1 = 2.

Hence, x = 3 and y = 2.

Question 16(ii)

Solve :

$\dfrac{34}{3x + 4y} + \dfrac{15}{3x - 2y} = 5$

$\dfrac{25}{3x - 2y} - \dfrac{8.50}{3x + 4y}$ = 4.5

Answer

Given, equations :

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

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

Multiplying equation (2) by 4, we get :

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

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

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

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

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

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

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

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

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

Hence, x = 3 and y = 2.

Question 17(i)

Solve :

x + y = 2xy

x - y = 6xy

Answer

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

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

$\Rightarrow \dfrac{x + y}{xy} = \dfrac{2xy}{xy} \\[1em] \Rightarrow \dfrac{x}{xy} + \dfrac{y}{xy} = 2 \\[1em] \Rightarrow \dfrac{1}{y} + \dfrac{1}{x} = 2 \text{ .......(1)}$

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

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

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

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

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

$\Rightarrow \dfrac{1}{\dfrac{1}{4}} + \dfrac{1}{x} = 2 \\[1em] \Rightarrow 4 + \dfrac{1}{x} = 2 \\[1em] \Rightarrow \dfrac{1}{x} = 2 - 4 \\[1em] \Rightarrow \dfrac{1}{x} = -2 \\[1em] \Rightarrow x = -\dfrac{1}{2}.$

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

Question 17(ii)

Solve :

x + y = 7xy

2x - 3y = -xy

Answer

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

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

$\Rightarrow \dfrac{x + y}{xy} = \dfrac{7xy}{xy} \\[1em] \Rightarrow \dfrac{x}{xy} + \dfrac{y}{xy} = 7 \\[1em] \Rightarrow \dfrac{1}{y} + \dfrac{1}{x} = 7 \\[1em]$

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

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

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

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

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

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

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

$\Rightarrow \dfrac{3}{\dfrac{1}{4}} + \dfrac{3}{x} = 21 \\[1em] \Rightarrow 12 + \dfrac{3}{x} = 21 \\[1em] \Rightarrow \dfrac{3}{x} = 21 - 12 \\[1em] \Rightarrow \dfrac{3}{x} = 9 \\[1em] \Rightarrow x = \dfrac{3}{9} = \dfrac{1}{3}.$

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

Question 18

Solve :

$\dfrac{a}{x} - \dfrac{b}{y} = 0$

$\dfrac{ab^2}{x} + \dfrac{a^2b}{y} = a^2 + b^2$

Answer

Let $\dfrac{1}{x} = p \text{ and } \dfrac{1}{y} = q$ . 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 :

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

Hence, x = a and y = b.

Question 19

Solve :

$\dfrac{2xy}{x + y} = \dfrac{3}{2}$

$\dfrac{xy}{2x - y} = -\dfrac{3}{10}$

x + y ≠ 0 and 2x - y ≠ 0

Answer

Simplifying first equation :

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

Simplifying second equation :

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

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

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

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

$\Rightarrow \dfrac{1}{-\dfrac{3}{2}} + \dfrac{1}{x} = \dfrac{4}{3} \\[1em] \Rightarrow -\dfrac{2}{3} + \dfrac{1}{x} = \dfrac{4}{3} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{4}{3} + \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{6}{3} \\[1em] \Rightarrow \dfrac{1}{x} = 2 \\[1em] \Rightarrow x = \dfrac{1}{2}.$

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

Exercise 6(C)

Question 1(a)

The sum of two positive whole numbers is 15 and their difference is 9; the numbers are :

12 and 3
9 and 0
7 and 8
15 and 0

Answer

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 = $\dfrac{24}{2}$ = 12.

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

⇒ 12 + y = 15

⇒ y = 15 - 12 = 3.

Hence, Option 1 is the correct option.

Question 1(b)

The sum of numerator and denominator of a fraction is 22. If numerator is 8 less than its denominator; the fraction is :

$\dfrac{9}{13}$
$\dfrac{7}{15}$
$\dfrac{13}{9}$
$\dfrac{15}{7}$

Answer

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 = $\dfrac{30}{2}$ = 15.

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

⇒ x + 15 = 22

⇒ x = 22 - 15 = 7.

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

Hence, Option 2 is the correct option.

Question 1(c)

The sum of the digits of a two digit number is 8. The difference between them is 4. If the ten's digit is greater than the unit's digit then the number is :

Answer

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 = $\dfrac{12}{2}$ = 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.

Question 1(d)

The present ages of two persons are in the ratio 1 : 3. After 5 years, the ratio between their ages will be 2 : 5, their ages are :

10 years and 30 years
15 years and 45 years
20 years and 60 years
12 years and 36 years

Answer

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.

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

Hence, Option 2 is the correct option.

Question 2

The ratio of two numbers is $\dfrac{2}{3}$ . If 2 is subtracted from the first and 8 from the second, the ratio becomes the reciprocal of the original ratio. Find the numbers.

Answer

Let the numbers be x and y.

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

$\Rightarrow \dfrac{x}{y} = \dfrac{2}{3} \\[1em] \Rightarrow x = \dfrac{2y}{3} ........(1)$

Given,

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

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

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

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

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

$\Rightarrow x = \dfrac{2 \times 12}{3} \\[1em] = \dfrac{24}{3} \\[1em] = 8.$

Hence, numbers are 8 and 12.

Question 3

Two numbers are in the ratio 4 : 7. If thrice the larger be added to twice the smaller, the sum is 59. Find the numbers.

Answer

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.

$\therefore \dfrac{y}{x} = \dfrac{4}{7} \\[1em] \Rightarrow y = \dfrac{4x}{7} .......(1)$

Given,

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

∴ 3x + 2y = 59

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

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

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

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

Question 4

When the greater of the two numbers increased by 1 divides the sum of the numbers, the result is $\dfrac{3}{2}$ . When the difference of these numbers is divided by the smaller, the result is $\dfrac{1}{2}$ . Find the numbers.

Answer

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 $\dfrac{3}{2}$ .

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

Given,

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

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

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

$\Rightarrow 2y - 3 = \dfrac{3y}{2} \\[1em] \Rightarrow 2(2y - 3) = 3y \\[1em] \Rightarrow 4y - 6 = 3y \\[1em] \Rightarrow 4y - 3y = 6 \\[1em] \Rightarrow y = 6.$

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

$\Rightarrow x = \dfrac{3 \times 6}{2} \\[1em] = \dfrac{18}{2} \\[1em] = 9.$

Hence, numbers are 6 and 9.

Question 5

The sum of two positive numbers x and y (x > y) is 50 and the difference of their squares is 720. Find the numbers.

Answer

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 = $\dfrac{1780}{100}$ = 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

Question 6

The sum of two numbers is 8 and the difference of their squares is 32. Find the numbers.

Answer

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 = $\dfrac{32}{16}$ = 2.

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

⇒ x = 8 - 2 = 6.

Hence, numbers are 2 and 6.

Question 7

The difference between two positive numbers x and y (x > y) is 4 and the difference between their reciprocal is $\dfrac{4}{21}$ . Find the numbers.

Answer

Given,

⇒ x > y

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

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

∴ x - y = 4

⇒ x = y + 4 ........(1)

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

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

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.

Question 8

Two numbers are in the ratio 4 : 5. If 30 is subtracted from each of the numbers, the ratio becomes 1 : 2. Find the numbers.

Answer

Let two numbers be x and y.

Given,

Numbers are in the ratio 4 : 5.

$\Rightarrow \dfrac{x}{y} = \dfrac{4}{5} \\[1em] \Rightarrow x = \dfrac{4y}{5} .......(1)$

Given,

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

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

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

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

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

$\Rightarrow x = \dfrac{4 \times 50}{5} \\[1em] = \dfrac{200}{5} \\[1em] = 40.$

Hence, numbers are 40 and 50.

Question 9

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

Answer

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 $\dfrac{2}{3}$ .

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

Given,

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

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

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 = $\dfrac{6}{3}$ = 2.

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

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

Question 10

The sum of the numerator and the denominator of a fraction is equal to 7. Four times the numerator is 8 less than 5 times the denominator. Find the fraction.

Answer

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 = $\dfrac{36}{9}$ = 4

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

⇒ x = 7 - 4 = 3.

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

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

Question 11

If the numerator of a fraction is multiplied by 2 and its denominator is increased by 1, it becomes 1. However, if the numerator is increased by 4 and denominator is multiplied by 2, the fraction becomes $\dfrac{1}{2}$ . Find the fraction.

Answer

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.

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

Given,

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

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

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 = $\dfrac{10}{2}$ = 5.

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

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

Question 12

A fraction becomes $\dfrac{1}{2}$ if 5 is subtracted from its numerator and 3 is subtracted from its denominator. If the denominator of this fraction is 5 more than its numerator, find the fraction.

Answer

Let numerator be x and denominator be y.

Given,

Denominator is 5 more than numerator.

∴ y = x + 5 ......(1)

Given,

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

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

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 = $\dfrac{x}{y} = \dfrac{12}{17}$ .

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

Question 13

The sum of the digits of a two digit number is 5. If the digits are reversed, the number is reduced by 27. Find the number.

Answer

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 = $\dfrac{8}{2}$ = 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.

Question 14

The sum of the digits of a two digit number is 7. If the digits are reversed, the new number decreased by 2, equals twice the original number. Find the number.

Answer

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 = $\dfrac{54}{27}$ = 2.

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

⇒ x = 7 - 2 = 5.

Number = 10y + x = 10 × 2 + 5 = 25.

Hence, number = 25.

Question 15

The ten's digit of a two digit number is three times the unit digit. The sum of the number and the unit digit is 32. Find the number.

Answer

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.

Question 16

A two-digit number is such that the ten's digit exceeds twice the unit's digit by 2 and the number obtained by inter-changing the digits is 5 more than three times the sum of the digits. Find the two digit number.

Answer

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 = $\dfrac{9}{3}$ = 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.

Question 17

Five years ago, A's age was four times the age of B. Five years hence, A's age will be twice the age of B. Find their present ages.

Answer

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 = $\dfrac{20}{2}$ = 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.

Question 18

A is 20 years older than B. 5 years ago, A was 3 times as old as B. Find their present ages.

Answer

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 = $\dfrac{30}{2}$ = 15 years.

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

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

Question 19

Four years ago, a mother was four times as old as her daughter. Six years later, the mother will be two and a half times as old as her daughter at that time. Find the present ages of mother and her daughter.

Answer

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 = $\dfrac{21}{1.5}$ = 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.

Question 20

The age of a man is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children at that time. Find the present age of the man.

Answer

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.

Question 21

The annual incomes of A and B are in the ratio 3 : 4 and their annual expenditure are in the ratio 5 : 7. If each saves ₹ 5,000; find their annual incomes.

Answer

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 = $\dfrac{30000}{3}$ = 10000.

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

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

Question 22

In an examination, the ratio of passes to failures was 4 : 1. Had 30 less appeared and 20 less passed, the ratio of passes to failures would have been 5 : 1. Find the number of students who appeared for the examination.

Answer

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.

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

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

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

Question 23

A and B both have some pencils. If A gives 10 pencils to B, then B will have twice as many as A. And if B gives 10 pencils to A, then they will have the same number of pencils. How many pencils does each have ?

Answer

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.

Question 24

1250 persons went to see a circus-show. Each adult paid ₹ 75 and each child paid ₹ 25 for the admission ticket. Find the number of adults and number of children, if the total collection from them amounts to ₹ 61,250.

Answer

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 = $\dfrac{32500}{50}$ = 650.

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

⇒ x = 1250 - 650 = 600.

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

Question 25

Rohit says to Ajay, "Give me a hundred, I shall then become twice as rich as you." Ajay replies, "if you give me ten, I shall be six times as rich as you." How much does each have originally?

Answer

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 = $\dfrac{440}{11}$ = ₹ 40.

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

⇒ 2y - 40 = 300

⇒ 2y = 300 + 40

⇒ 2y = 340

⇒ y = $\dfrac{340}{2}$ = ₹ 170.

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

Question 26

The sum of a two digit number and the number obtained by reversing the order of the digits is 99. Find the number, if the digits differ by 3.

Answer

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.

Question 27

Seven times a two digit number is equal to four times the number obtained by reversing the digits. If the difference between the digits is 3, find the number.

Answer

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 = $\dfrac{66x}{33}$

⇒ 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.

Question 28

From Delhi station, if we buy 2 tickets for station A and 3 tickets for station B, the total cost is ₹ 77. But if we buy 3 tickets for station A and 5 tickets for station B, the total cost is ₹ 124. What are the fares from Delhi to station A and to station B ?

Answer

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 = $\dfrac{26}{2}$ = ₹ 13.

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

Question 29

The sum of digits of a two digit number is 11. 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. Find the original number.

Answer

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 = $\dfrac{16}{2}$ = 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.

Question 30

90% acid solution (90% pure acid and 10% water) and 97% acid solution are mixed to obtain 21 litres of 95% acid solution. How many litres of each solution are mixed.

Answer

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

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

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 = $\dfrac{105}{7}$ = 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.

Question 31

Class XI students of a school wanted to give a farewell party to the outgoing students of class XII. They decided to purchase two kinds of sweets, one costing ₹ 250 per kg and the other costing ₹ 350 per kg. They estimated that 40 kg of sweets were needed. If the total budget for the sweets was ₹ 11800; find how much sweets of each kind were brought ?

Answer

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 = $\dfrac{180}{10}$ = 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.

Test Yourself

Question 1(a)

If 3y - 2x = 1 and 3x + 4y = 24, the value of x and y are :

3 and 4
3 and 3
4 and 4
4 and 3

Answer

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 = $\dfrac{51}{17}$

⇒ 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 = $\dfrac{12}{3}$

⇒ x = 4

∴ x = 4 and y = 3

Hence, option 4 is the correct option.

Question 1(b)

On simplifying the equation 7(x - 1) - 6y = 5(x - y), we get:

2x - y = 7
2x + y = 7
x - 2y = 7
3x - 2y = 7

Answer

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.

Question 1(c)

Solution of equation $\dfrac{2}{x} + \dfrac{3}{y}$ + 1 = 0 and $\dfrac{3}{x} + \dfrac{5}{y}$ + 2 = 0 is:

-1 and -1
-1 and -2
1 and -1
none of these

Answer

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

Let $\dfrac{1}{x}$ be u and $\dfrac{1}{y}$ 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

⇒ $\dfrac{1}{y}$ = -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 = $\dfrac{3}{3}$

⇒ u = 1

⇒ $\dfrac{1}{x}$ = 1

⇒ x = 1

∴ x = 1 and y = -1

Hence, option 3 is the correct option.

Question 1(d)

In a two digit number, the sum of the digit is 11 and the tens digit minus unit digit is 5. The number is :

Answer

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 = $\dfrac{16}{2}$

⇒ 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.

Question 1(e)

Statement 1: x = 5 and y = 2 are the solution of equations x - y = 3 and 2x + y = 11.

Statement 2: On substituting x = 5 and y = 2 in each of the above equation the value of left hand side and right hand side for each equation must be same.

Both the statements are true.
Both the statements are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.

Answer

Given,

First equation :

⇒ x - y = 3

⇒ x = 3 + y ........(1)

Second equation :

⇒ 2x + y = 11 ......................(2)

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

⇒ 2(3 + y) + y = 11

⇒ 6 + 2y + y = 11

⇒ 6 + 3y = 11

⇒ 3y = 11 - 6

⇒ 3y = 5

⇒ y = $\dfrac{5}{3}$

Substitute the value of y in equation (1),

⇒ x = 3 + $\dfrac{5}{3}$

⇒ x = $\dfrac{9 + 5}{3}$

⇒ x = $\dfrac{14}{3}$

Thus, x = $\dfrac{5}{3}$ and y = $\dfrac{14}{3}$ are the solution of equations.

So, statement 1 is false.

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 2 is false.

∴ Both the statements are false.

Hence, option 2 is the correct option.

Question 1(f)

Statement 1: The sum of two numbers x and y is 11. Twice the first number plus three times the second number equals to 25.

⇒ x + y = 11 and 2x + 3y = 25

Statement 2: The numbers are 7 and 4.

Both the statements are true.
Both the statements are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.

Answer

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.

Question 1(g)

Assertion (A): Solutions of equations 3y - 2x = 1 and 3x + 4y = 24 is x = 4 and y = 3.

Reason (R): ∵ 3y - 2x = 3 x 3 - 2 x 4 = 1 and, 3x + 4y = 3 x 4 + 4 x 3 = 24

A is true, but R is false.
A is false, but R is true.
Both A and R are correct, and R is the correct explanation for A.
Both A and R are correct, and R is correct explanation for A.

Answer

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.

Question 1(h)

Assertion (A): In a two digit number, three times the digit at ten's place is equals to two times the digit at the unit place and the sum of the digit is 5 then for number 10x + y, 3x = 2y and x + y = 5.

Reason (R): The number is 32.

A is true, but R is false.
A is false, but R is true.
Both A and R are correct, and R is the correct explanation for A.
Both A and R are correct, and R is not the correct explanation for A.

Answer

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 = $\dfrac{2}{3}$ 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 :

$\Rightarrow \dfrac{2}{3}y + y = 5\\[1em] \Rightarrow \dfrac{2y + 3y}{3} = 5\\[1em] \Rightarrow \dfrac{5y}{3} = 5\\[1em] \Rightarrow y = \dfrac{5 \times 3}{5}\\[1em] \Rightarrow y = 3$

Substituting the value of y in equation (1),

⇒ x = $\dfrac{2}{3} \times 3$

⇒ 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.

Question 2

Solve the following pair of (simultaneous) equations using method of elimination by substitution :

3x + 2y = 11

2x - 3y + 10 = 0

Answer

Given,

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

⇒ 3x + 2y = 11

⇒ 3x = 11 - 2y

⇒ x = $\dfrac{11 - 2y}{3}$ ........(1)

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

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

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

$\Rightarrow x = \dfrac{11 - 2 \times 4}{3} \\[1em] = \dfrac{11 - 8}{3} \\[1em] = \dfrac{3}{3} \\[1em] = 1.$

Hence, x = 1 and y = 4.

Question 3

Solve the following pair of (simultaneous) equations using method of elimination by substitution :

2x - 3y + 6 = 0

2x + 3y - 18 = 0

Answer

Given,

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

⇒ 2x - 3y + 6 = 0

⇒ 2x = 3y - 6

⇒ x = $\dfrac{3y - 6}{2}$ .......(1)

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

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

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

$\Rightarrow x = \dfrac{3 \times 4 - 6}{2} \\[1em] = \dfrac{12 - 6}{2} \\[1em] = \dfrac{6}{2} \\[1em] = 3.$

Hence, x = 3 and y = 4.

Question 4

Solve the following pair of (simultaneous) equations using method of elimination by substitution :

$\dfrac{3x}{2} - \dfrac{5y}{3} + 2 = 0$

$\dfrac{x}{3} + \dfrac{y}{2} = 2\dfrac{1}{6}$

Answer

Given,

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

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

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

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

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

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

Hence, x = 2 and y = 3.

Question 5

Solve the following pair of (simultaneous) equations using method of elimination by substitution :

$\dfrac{x}{6} + \dfrac{y}{15} = 4$

$\dfrac{x}{3} - \dfrac{y}{12} = 4\dfrac{3}{4}$

Answer

Given,

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

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

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

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

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

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

Hence, x = 18 and y = 15.

Question 6

Find the value of m, if x = 2, y = 1 is a solution of the equation 2x + 3y = m.

Answer

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.

Question 7

Solve :

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

Answer

Given,

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

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

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 = $\dfrac{700}{7}$ = 100.

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

⇒ x = 240 - 2(100) = 240 - 200 = 40.

Hence, x = 40 and y = 100.

Question 8

The value of expression mx - ny is 3 when x = 5 and y = 6 and its value is 8 when x = 6 and y = 5. Find the values of m and n.

Answer

Given,

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

∴ 5m - 6n = 3

⇒ 5m = 3 + 6n

⇒ m = $\dfrac{3 + 6n}{5}$ ............(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.

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

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

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

Hence, m = 3 and n = 2.

Question 9

Solve :

11(x - 5) + 10(y - 2) + 54 = 0

7(2x - 1) + 9(3y - 1) = 25

Answer

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 = $\dfrac{21 - 10y}{11}$ .........(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 :

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

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

$\Rightarrow x = \dfrac{21 - 10 \times 1}{11} \\[1em] = \dfrac{21 - 10}{11} \\[1em] = \dfrac{11}{11} \\[1em] = 1.$

Hence, x = 1 and y = 1.

Question 10

Solve :

$\dfrac{7 + x}{5} - \dfrac{2x - y}{4} = 3y - 5$

$\dfrac{5y - 7}{2} + \dfrac{4x - 3}{6} = 18 - 5x$

Answer

Given,

1st equation :

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

2nd equation :

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

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

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

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

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

Hence, x = 3 and y = 2.

Question 11

Solve :

$4x + \dfrac{x - y}{8}$ = 17

$2y + x - \dfrac{5y + 2}{3}$ = 2

Answer

Given,

1st equation :

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

2nd equation :

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

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

⇒ 33x - 136 = 8 - 3x

⇒ 33x + 3x = 8 + 136

⇒ 36x = 144

⇒ x = $\dfrac{144}{36}$ = 4.

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

⇒ y = 33(4) - 136 = 132 - 136 = -4.

Hence, x = 4 and y = -4.

Question 12

Solve using cross-multiplication :

0.4x - 1.5y = 6.5

0.3x + 0.2y = 0.9

Answer

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 :

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

Hence, x = 5 and y = -3.

Question 13

Solve using cross-multiplication :

$\sqrt{2}x - \sqrt{3}y = 0$

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

Answer

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 :

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

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

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

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

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

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

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

Hence, x = 0 and y = 0.

Question 14

Solve :

$\dfrac{3}{2x} + \dfrac{2}{3y} = -\dfrac{1}{3}$

$\dfrac{3}{4x} + \dfrac{1}{2y} = -\dfrac{1}{8}$

Answer

Given, equations :

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

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

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

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

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

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

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

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

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

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

Hence, x = -3 and y = 4.

Question 15

Four times a certain two digit number is seven times the number obtained on interchanging its digits. If the difference between the digits is 4; find the number.

Answer

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 = $\dfrac{66\text{y}}{33}$

⇒ 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.

Question 16

The sum of a two digit number and the number obtained by interchanging the digits of the number is 121. If the digits of the number differ by 3, find the number.

Answer

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)

⇒ 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 = $\dfrac{14}{2}$

⇒ 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.

Question 17

A two digit number is obtained by multiplying the sum of the digits by 8. Also, it is obtained by multiplying the difference of the digits by 14 and adding 2. Find the number.

Answer

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 = $\dfrac{7}{2}$ 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 :

$\Rightarrow 4 \times \dfrac{7}{2}y = 15y - 2 \\[1em] \Rightarrow 14y = 15y - 2 \\[1em] \Rightarrow 15y - 14y = 2 \\[1em] \Rightarrow y = 2.$

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

$\Rightarrow x = \dfrac{7}{2}y = \dfrac{7}{2} \times 2$ = 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 :

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

This is not possible as y cannot be a fraction.

Hence, number = 72.

Question 18

Two articles A and B are sold for ₹ 1,167 making 5% profit on A and 7% profit on B. If the two articles are sold for ₹ 1,165 a profit of 7% is made on A and a profit of 5% is made on B. Find the cost price of each article.

Answer

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.

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

Given,

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

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

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

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

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

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

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

Question 19

Pooja and Ritu can do a piece of work in $17\dfrac{1}{7}$ days. If one day work of Pooja be three fourth of one day work of Ritu; find in how many days each will do the work alone.

Answer

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

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

Given,

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

$\therefore \dfrac{1}{x} = \dfrac{3}{4} \times \dfrac{1}{y} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{3}{4y} \\[1em] \Rightarrow x = \dfrac{4y}{3} .......(1)$

Given,

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

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

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

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

$\Rightarrow \dfrac{1}{\dfrac{4y}{3}} + \dfrac{1}{y} = \dfrac{7}{120} \\[1em] \Rightarrow \dfrac{3}{4y} + \dfrac{1}{y} = \dfrac{7}{120} \\[1em] \Rightarrow \dfrac{3 + 4}{4y} = \dfrac{7}{120} \\[1em] \Rightarrow \dfrac{7}{4y} = \dfrac{7}{120} \\[1em] \Rightarrow y = \dfrac{120 \times 7}{7 \times 4} \\[1em] \Rightarrow y = 30.$

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

$\Rightarrow x = \dfrac{4 \times 30}{3} \\[1em] = 40.$

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

Question 20

Mr. and Mrs. Ahuja weigh x kg and y kg respectively. They both take a dieting course, at the end of which Mr. Ahuja loses 5 kg and weighs as much as his wife weighed before the course. Mrs. Ahuja loses 4 kg and weighs $\dfrac{7}{8}$ th of what her husband weighed before the course. Form two equations in x and y to find their weights before taking the dieting course.

Answer

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 $\dfrac{7}{8}$ th of what her husband weighed before the course.

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

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

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

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.

Question 21

A part of monthly expenses of a family is constant and the remaining vary with the number of members in the family. For a family of 4 persons, the total monthly expenses are ₹ 10,400; whereas for a family of 7 persons, the total monthly expenses are ₹ 15,800. Find the constant expenses per month and the monthly expenses on each member of a family.

Answer

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 = $\dfrac{5400}{3}$ = ₹ 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.

Question 22

The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹ 315 and for a journey of 15 km, the charge paid is ₹ 465. What are the fixed charges and the charge per kilometer ? How much does a person have to pay for travelling a distance of 32 km ?

Answer

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 = $\dfrac{150}{5}$ = ₹ 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 = ₹ 15m, charge per kilometer = ₹ 30 and for 32 km charge = ₹ 975.

Question 23

A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Geeta paid ₹ 27 for a book kept for seven days, while Mohit paid ₹ 21 for the book he kept for five days. Find the fixed charges and the charge for each extra day.

Answer

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 = $\dfrac{6}{2}$

⇒ 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.

Question 24

The area of rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. However, if the length of this rectangle increases by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Answer

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 = $\dfrac{171}{19}$ = 9.

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

⇒ 3x - 5(9) = 6

⇒ 3x - 45 = 6

⇒ 3x = 45 + 6

⇒ 3x = 51

⇒ x = $\dfrac{51}{3}$ = 17.

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

Question 25

It takes 12 hours to fill a swimming pool using two pipes. 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. How long would each pipe take to fill the swimming pool ?

Answer

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 $\dfrac{1}{x}$ th of the pool

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

Given,

It takes 12 hours to fill the pool.

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

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

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

Multiplying equation (1) by 4, we get :

$\Rightarrow \dfrac{4}{x} + \dfrac{4}{y} = \dfrac{4}{12} \\[1em] \Rightarrow \dfrac{4}{x} + \dfrac{4}{y} = \dfrac{1}{3} .......(3)$

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

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

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

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

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

Chapter 6

Simultaneous (Linear) Equations [Including Problems]

Class - 9 Concise Mathematics Selina

Exercise 6(A)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18(i)

Question 18(ii)

Exercise 6(B)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 1(f)

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16(i)

Question 16(ii)

Question 17(i)

Question 17(ii)

Question 18

Question 19

Exercise 6(C)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22