Chapter 5: Simultaneous Linear Equations

Exercise 5A

Question 1

Solve the following simultaneous equations:

x + 2y = 1, 3x - y = 17

Answer

Given,

Equations : x + 2y = 1, 3x - y = 17

⇒ x + 2y = 1

⇒ x = 1 - 2y ....(1)

Substituting value of x from equation (1) in 3x - y = 17, we get :

⇒ 3(1 - 2y) - y = 17

⇒ 3 - 6y - y = 17

⇒ -7y = 17 - 3

⇒ -7y = 14

⇒ y = $-\dfrac{14}{7}$ = -2.

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

⇒ x = 1 - 2y

⇒ x = 1 - 2(-2)

⇒ x = 1 + 4

⇒ x = 5.

Hence, x = 5, y = -2.

Question 2

Solve the following simultaneous equations:

5x + 4y = 4, x - 12y = 20

Answer

Given,

Equations : 5x + 4y = 4, x - 12y = 20

⇒ x - 12y = 20

⇒ x = 20 + 12y ....(1)

Substituting value of x from equation (1) in 5x + 4y = 4, we get :

⇒ 5(20 + 12y) + 4y = 4

⇒ 100 + 60y + 4y = 4

⇒ 100 + 64y = 4

⇒ 64y = 4 - 100

⇒ 64y = -96

⇒ y = $-\dfrac{96}{64} = -\dfrac{3}{2}$ .

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

⇒ x = 20 + $12 \times -\Big(\dfrac{3}{2}\Big)$

⇒ x = 20 + 6(-3)

⇒ x = 20 - 18

⇒ x = 2.

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

Question 3

Solve the following simultaneous equations:

x + 2y + 9 = 0, 3x + 4y + 17 = 0

Answer

Given,

Equations : x + 2y + 9 = 0, 3x + 4y + 17 = 0

⇒ x + 2y + 9 = 0

⇒ x = -9 - 2y ....(1)

Substituting value of x from equation (1) in 3x + 4y + 17 = 0, we get :

⇒ 3(-9 - 2y) + 4y + 17 = 0

⇒ -27 - 6y + 4y + 17 = 0

⇒ -10 - 2y = 0

⇒ -2y = 10

⇒ y = $-\dfrac{10}{2} = -5$ .

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

⇒ x = -9 - 2y

⇒ x = -9 - 2(-5)

⇒ x = -9 + 10

⇒ x = 1.

Hence, x = 1, y = -5.

Question 4

Solve the following simultaneous equations:

10x + 3y = 75, 6x - 5y = 11

Answer

Given,

Equations : 10x + 3y = 75, 6x - 5y = 11

⇒ 10x + 3y = 75

⇒ 10x = 75 - 3y

⇒ x = $\dfrac{75 - 3y}{10}$ ....(1)

Substituting value of x from equation (1) in 6x - 5y = 11, we get :

$\Rightarrow 6\Big(\dfrac{75 - 3y}{10}\Big) - 5y = 11 \\[1em] \Rightarrow 3\Big(\dfrac{75 - 3y}{5}\Big) - 5y = 11 \\[1em] \Rightarrow \Big(\dfrac{225 - 9y}{5}\Big) - 5y = 11 \\[1em] \Rightarrow \Big(\dfrac{225 - 9y - 25y}{5}\Big) = 11 \\[1em] \Rightarrow \Big(\dfrac{225 - 34y}{5}\Big) = 11 \\[1em] \Rightarrow 225 - 34y = 11 \times 5\\[1em] \Rightarrow -34y = 55 - 225 \\[1em] \Rightarrow -34y = -170 \\[1em] \Rightarrow y = \dfrac{170}{34} \\[1em] \Rightarrow y = 5.$

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

$\Rightarrow x = \dfrac{75 - 3y}{10} \\[1em] \Rightarrow x = \dfrac{75 - 3(5)}{10} \\[1em] \Rightarrow x = \dfrac{75 - 15}{10} \\[1em] \Rightarrow x = \dfrac{60}{10} \\[1em] \Rightarrow x = 6.$

Hence, x = 6, y = 5.

Question 5

Solve the following simultaneous equations:

7x - 2y = 20, 11x + 15y + 23 = 0

Answer

Given,

Equations : 7x - 2y = 20, 11x + 15y + 23 = 0

⇒ 7x - 2y = 20

⇒ 7x = 20 + 2y

⇒ x = $\dfrac{20 + 2y}{7}$ ....(1)

Substituting value of x from equation (1) in 11x + 15y + 23 = 0, we get :

$\Rightarrow 11\Big(\dfrac{20 + 2y}{7}\Big) + 15y + 23 = 0 \\[1em] \Rightarrow \Big(\dfrac{220 + 22y}{7}\Big) + 15y + 23 = 0 \\[1em] \Rightarrow \Big(\dfrac{220 + 22y + 105y + 161}{7}\Big) = 0 \\[1em] \Rightarrow 381 + 127y = 0 \\[1em] \Rightarrow 127y = - 381 \\[1em] \Rightarrow y = \dfrac{-381}{127}\\[1em] \Rightarrow y = -3.$

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

$\Rightarrow x = \dfrac{20 + 2y}{7} \\[1em] \Rightarrow x = \dfrac{20 + 2(-3)}{7} \\[1em] \Rightarrow x = \dfrac{20 - 6}{7} \\[1em] \Rightarrow x = \dfrac{14}{7} \\[1em] \Rightarrow x = 2.$

Hence, x = 2, y = -3.

Question 6

Solve the following simultaneous equations:

$\dfrac{7 - 4x}{3} = y$ , 2x + 3y + 1 = 0

Answer

Given,

Equations : $\dfrac{7 - 4x}{3} = y$ , 2x + 3y + 1 = 0

⇒ $y = \dfrac{7 - 4x}{3}$ ....(1)

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

$\Rightarrow 2x + 3\Big(\dfrac{7 - 4x}{3}\Big) + 1 = 0$

⇒ 2x + (7 - 4x) + 1 = 0

⇒ 8 - 2x = 0

⇒ 2x = 8

⇒ x = $\dfrac{8}{2}$

⇒ x = 4.

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

$\Rightarrow y = \dfrac{7 - 4x}{3} \\[1em] \Rightarrow y = \dfrac{7 - 4(4)}{3} \\[1em] \Rightarrow y = \dfrac{7 - 16}{3} \\[1em] \Rightarrow y = -\dfrac{9}{3} \\[1em] \Rightarrow y = -3.$ Hence, x = 4, y = -3.

Question 7

Solve the following simultaneous equations:

4x - 3y = 8, 18x - 3y = 29

Answer

Given,

Equations : 4x - 3y = 8, 18x - 3y = 29

⇒ 4x - 3y = 8

⇒ 4x = 3y + 8

⇒ x = $\dfrac{3y + 8}{4}$ ....(1)

Substituting value of x from equation (1) in 18x - 3y = 29, we get :

$\Rightarrow 18\Big(\dfrac{3y + 8}{4}\Big) - 3y = 29 \\[1em] \Rightarrow 9\Big(\dfrac{3y + 8}{2}\Big) - 3y = 29 \\[1em] \Rightarrow \Big(\dfrac{27y + 72}{2}\Big) - 3y = 29 \\[1em] \Rightarrow \Big(\dfrac{27y + 72 - 6y}{2}\Big) = 29 \\[1em] \Rightarrow 21y + 72 = 29 \times 2 \\[1em] \Rightarrow 21y + 72 = 58 \\[1em] \Rightarrow 21y = 58 - 72 \\[1em] \Rightarrow 21y = -14 \\[1em] \Rightarrow y = \dfrac{-14}{21} \\[1em] \Rightarrow y = -\dfrac{2}{3}.$

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

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

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

Question 8

Solve the following simultaneous equations:

$\dfrac{x + y - 8}{2} = \dfrac{x + 2y - 14}{3} = \dfrac{3x + y - 12}{11}$

Answer

Given,

$\dfrac{x + y - 8}{2} = \dfrac{x + 2y - 14}{3} = \dfrac{3x + y - 12}{11}$

Solving L.H.S. of the given equation,

$\Rightarrow \dfrac{x + y - 8}{2} = \dfrac{x + 2y - 14}{3} \\[1em] \Rightarrow 3(x + y - 8) = 2(x + 2y - 14) \\[1em] \Rightarrow 3x + 3y - 24 = 2x + 4y - 28 \\[1em] \Rightarrow 3x + 3y - 2x - 4y = -28 + 24 \\[1em] \Rightarrow x - y = -4 \\[1em] \Rightarrow x = -4 + y \text{ ....(1)}$

Solving R.H.S. of the given equation,

$\Rightarrow \dfrac{x + 2y - 14}{3} = \dfrac{3x + y - 12}{11} \\[1em] \Rightarrow 11(x + 2y - 14) = 3(3x + y - 12) \\[1em] \Rightarrow 11x + 22y - 154 = 9x + 3y - 36 \\[1em] \Rightarrow 11x + 22y - 9x - 3y = -36 + 154 \\[1em] \Rightarrow 2x + 19y = 118 \text{ ........(2)}$

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

⇒ 2(-4 + y) + 19y = 118

⇒ -8 + 2y + 19y = 118

⇒ 21y = 118 + 8

⇒ 21y = 126

⇒ y = $\dfrac{126}{21} = 6$ .

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

⇒ x = -4 + y

⇒ x = -4 + 6

⇒ x = 2.

Hence, x = 2, y = 6.

Question 9

Solve the following simultaneous equations:

$\dfrac{x}{7} + \dfrac{y}{3} = 5, \dfrac{x}{2} - \dfrac{y}{9} = 6$

Answer

Given,

$\dfrac{x}{7} + \dfrac{y}{3} = 5, \dfrac{x}{2} - \dfrac{y}{9} = 6$

Simplifying,

$\Rightarrow \dfrac{x}{7} + \dfrac{y}{3} = 5 \\[1em] \Rightarrow \dfrac{3x + 7y}{21} = 5 \\[1em] \Rightarrow 3x + 7y = 5 \times 21 \\[1em] \Rightarrow 3x + 7y = 105 \\[1em] \Rightarrow 3x = 105 - 7y \\[1em] \Rightarrow x = \dfrac{105 - 7y}{3} \text{ ....(1)}$

Substituting value of x from equation (1) in $\dfrac{x}{2} - \dfrac{y}{9} = 6$ , we get :

$\Rightarrow \dfrac{\dfrac{105 - 7y}{3}}{2} - \dfrac{y}{9} = 6 \\[1em] \Rightarrow \dfrac{(105 - 7y)}{6} - \dfrac{y}{9} = 6 \\[1em] \Rightarrow \dfrac{3(105 - 7y) - 2y}{18} = 6 \\[1em] \Rightarrow \dfrac{315 - 21y - 2y}{18} = 6 \\[1em] \Rightarrow 315 - 23y = 6 \times 18 \\[1em] \Rightarrow 315 - 23y = 108 \\[1em] \Rightarrow 23y = 315 - 108 \\[1em] \Rightarrow 23y = 207 \\[1em] \Rightarrow y = \dfrac{207}{23} = 9.$

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

$\Rightarrow x = \dfrac{105 - 7y}{3} \\[1em] \Rightarrow x = \dfrac{105 - 7(9)}{3} \\[1em] \Rightarrow x = \dfrac{105 - 63}{3} \\[1em] \Rightarrow x = \dfrac{42}{3} \\[1em] \Rightarrow x = 14.$

Hence, x = 14, y = 9.

Question 10

Solve the following simultaneous equations:

$\dfrac{x}{6} + 6 = y, \dfrac{3x}{4} = 1 + y$

Answer

Simplifying, equation : $\dfrac{x}{6} + 6 = y$

$\Rightarrow \dfrac{x}{6} + 6 = y \\[1em] \Rightarrow \dfrac{x + 36}{6} = y \\[1em] \Rightarrow x + 36 = 6y \\[1em] \Rightarrow x = 6y - 36 \text{ ....(1)}$

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

$\Rightarrow \dfrac{3(6y - 36)}{4} = 1 + y \\[1em] \Rightarrow \dfrac{18y - 108}{4} = 1 + y \\[1em] \Rightarrow 18y - 108 = 4(1 + y) \\[1em] \Rightarrow 18y - 108 = (4 + 4y) \\[1em] \Rightarrow 18y - 4y = 4 + 108 \\[1em] \Rightarrow 14y = 112 \\[1em] \Rightarrow y = \dfrac{112}{14} \\[1em] \Rightarrow y = 8.$

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

⇒ x = 6y - 36

⇒ x = 6(8) - 36

⇒ x = 48 - 36

⇒ x = 12.

Hence, x = 12, y = 8.

Question 11

Solve the following simultaneous equations:

$4x + \dfrac{x - y}{8} = 17, x + 2y = \dfrac{y - 2}{3} - 2$

Answer

Simplifying, equation : $4x + \dfrac{x - y}{8} = 17$

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

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

$\Rightarrow x + 2y = \dfrac{y - 2}{3} - 2 \\[1em] \Rightarrow x + 2(33x - 136) = \dfrac{33x - 136 - 2}{3} - 2 \\[1em] \Rightarrow x + 66x - 272 = \dfrac{33x -138}{3} - 2 \\[1em] \Rightarrow 67x - 272 = \dfrac{3(11x - 46)}{3} - 2 \\[1em] \Rightarrow 67x - 272 = (11x - 46) - 2 \\[1em] \Rightarrow 67x - 272 = (11x - 48) \\[1em] \Rightarrow 67x - 11x = -48 + 272 \\[1em] \Rightarrow 56x = 224 \\[1em] \Rightarrow x = \dfrac{224}{56} = 4.$

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

⇒ y = 33(4) - 136

⇒ y = 132 - 136

⇒ y = -4.

Hence, x = 4, y = -4.

Question 12

Solve the following simultaneous equations:

$\dfrac{x}{2} + y = \dfrac{4}{5}, x + \dfrac{y}{2} = \dfrac{7}{10}$

Answer

Simplifying, equation : $\dfrac{x}{2} + y = \dfrac{4}{5}$

$\Rightarrow \dfrac{x}{2} + y = \dfrac{4}{5} \\[1em] \Rightarrow \dfrac{x + 2y}{2} = \dfrac{4}{5} \\[1em] \Rightarrow 5(x + 2y) = 4 \times 2 \\[1em] \Rightarrow 5x + 10y = 8 \\[1em] \Rightarrow 10y = 8 - 5x \\[1em] \Rightarrow y = \dfrac{8 - 5x}{10} \text{ ....(1)}$

Substituting value of y from equation (1) in $x + \dfrac{y}{2} = \dfrac{7}{10}$ , we get :

$\Rightarrow x + \dfrac{\dfrac{8 - 5x}{10}}{2} = \dfrac{7}{10} \\[1em] \Rightarrow x + \dfrac{8 - 5x}{20} = \dfrac{7}{10} \\[1em] \Rightarrow \dfrac{20x + 8 - 5x}{20} = \dfrac{7}{10} \\[1em] \Rightarrow 15x + 8 = \dfrac{7}{10} \times 20 \\[1em] \Rightarrow 15x + 8 = 7 \times 2 \\[1em] \Rightarrow 15x = 14 - 8 \\[1em] \Rightarrow 15x = 6 \\[1em] \Rightarrow x = \dfrac{6}{15} = \dfrac{2}{5}.$

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

$\Rightarrow y = \dfrac{8 - 5x}{10} \\[1em] \Rightarrow y = \dfrac{8 - 5 \Big(\dfrac{2}{5}\Big)}{10} \\[1em] \Rightarrow y = \dfrac{8 - 2}{10} \\[1em] \Rightarrow y = \dfrac{6}{10} = \dfrac{3}{5}.$

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

Question 13

Solve the following simultaneous equations:

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

Answer

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

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

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

$\Rightarrow \dfrac{4x - 3}{6} + \dfrac{5y - 7}{2} = 18 - 5x \\[1em] \Rightarrow \dfrac{2(4x - 3) + 6(5y - 7)}{12} = 18 - 5x \\[1em] \Rightarrow \dfrac{8x - 6 + 30y - 42}{12} = 18 - 5x \\[1em] \Rightarrow 8x - 6 + 30y - 42 = 12(18 - 5x) \\[1em] \Rightarrow 8x + 30y - 48 = 216 - 60x \\[1em] \Rightarrow 8x + 30y + 60x = 216 + 48 \\[1em] \Rightarrow 68x + 30y = 264 \text{ ....(2) }$

Substituting value of x from equation (1) in 68x + 30y = 264, we get :

$\Rightarrow 68\Big(\dfrac{128 - 55y}{6}\Big) + 30y = 264 \\[1em] \Rightarrow 34\Big(\dfrac{128 - 55y}{3}\Big) + 30y = 264 \\[1em] \Rightarrow \Big(\dfrac{4352 - 1870y}{3}\Big) + 30y = 264 \\[1em] \Rightarrow \Big(\dfrac{4352 - 1870y + 90y}{3}\Big) = 264 \\[1em] \Rightarrow (4352 - 1780y) = 264 \times 3 \\[1em] \Rightarrow (4352 - 1780y) = 792 \\[1em] \Rightarrow 1780y = 4352 - 792 \\[1em] \Rightarrow 1780y = 3560 \\[1em] \Rightarrow y = \dfrac{3560}{1780} = 2.$

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

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

Hence, x = 3, y = 2.

Question 14

Solve the following simultaneous equations:

$4x + \dfrac{6}{y} = 15, 3x - \dfrac{4}{y} = 7$

Answer

Given,

Equations:

$\Rightarrow 4x + \dfrac{6}{y} = 15 \text{ ....(1)} \\[1em] \Rightarrow 3x - \dfrac{4}{y} = 7 \text{ ....(2)}$

Multiplying equation (1) by 4, we get :

$\Rightarrow 4\Big(4x + \dfrac{6}{y}\Big) = 15 \times 4 \\[1em] \Rightarrow 4\Big(4x + \dfrac{6}{y}\Big) = 15 \times 4 \\[1em] \Rightarrow 16x + \dfrac{24}{y} = 60\text{ ....(3)}$

Multiplying equation (2) by 6, we get :

$\Rightarrow 6\Big(3x - \dfrac{4}{y}\Big) = 7 \times 6 \\[1em] \Rightarrow 18x - \dfrac{24}{y} = 42\text{ ....(4)}$

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

$\Rightarrow \Big(16x + \dfrac{24}{y}\Big) + \Big(18x - \dfrac{24}{y}\Big) = 60 + 42 \\[1em] \Rightarrow 18x + \dfrac{24}{y} + 16x - \dfrac{24}{y} = 102 \\[1em] \Rightarrow 34x = 102 \\[1em] \Rightarrow x = \dfrac{102}{34} \\[1em] \Rightarrow x = 3.$

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

$\Rightarrow 4(3) + \dfrac{6}{y} = 15 \\[1em] \Rightarrow 12 + \dfrac{6}{y} = 15 \\[1em] \Rightarrow \dfrac{6}{y} = 15 - 12 \\[1em] \Rightarrow \dfrac{6}{y} = 3 \\[1em] \Rightarrow y = \dfrac{6}{3} \\[1em] \Rightarrow y = 2.$

Hence, x = 3, y = 2.

Question 15

Solve the following simultaneous equations:

$5x - 9 = \dfrac{1}{y}, x + \dfrac{1}{y} = 3$

Answer

Equations:

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

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

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

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

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

Hence, x = 2, y = 1.

Question 16

Solve the following simultaneous equations:

$\dfrac{2}{x} + \dfrac{2}{3y} = \dfrac{1}{6},;\dfrac{3}{x} + \dfrac{2}{y} = 0$

Answer

Given,

Equations:

$\dfrac{2}{x} + \dfrac{2}{3y} = \dfrac{1}{6} \text{ ....(1) }$ ,

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

Multiplying equation (1) by 3, we get:

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

Multiplying equation (2) by 2, we get:

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

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

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

Substituting value of y in equation (1),

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

Hence, x = 6, y = -4.

Question 17

Solve the following simultaneous equations:

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

Answer

Given,

Equations:

$\dfrac{3}{2x} + \dfrac{2}{3y} = 5 \text{ ....(1) },$

$\dfrac{5}{x} - \dfrac{3}{y} = 1 \text{ ....(2) }$

Multiplying equation (1) by 9, we get:

$\Rightarrow 9\Big(\dfrac{3}{2x} + \dfrac{2}{3y}\Big) = 5 \times 9 \\[1em] \Rightarrow \Big(\dfrac{27}{2x} + \dfrac{6}{y}\Big) = 45 \text{ ..........(3) }$

Multiplying equation (2) by 2, we get:

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

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

$\Rightarrow \Big(\dfrac{27}{2x} + \dfrac{6}{y}\Big) + \Big(\dfrac{10}{x} - \dfrac{6}{y}\Big) = 45 + 2 \\[1em] \Rightarrow \Big(\dfrac{27}{2x} + \dfrac{10}{x}\Big) = 47 \\[1em] \Rightarrow \Big(\dfrac{27 + 20}{2x}\Big) = 47 \\[1em] \Rightarrow \Big(\dfrac{47}{2x}\Big) = 47 \\[1em] \Rightarrow 2x = \dfrac{47}{47} \\[1em] \Rightarrow 2x = 1 \\[1em] \Rightarrow x = \dfrac{1}{2}.$

Substituting value of x in equation (1),

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

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

Question 18

Solve the following simultaneous equations:

x + y = 2xy, x − y = 6xy

Answer

Given,

Equations :

⇒ x + y = 2xy ....(1)

⇒ x - y = 6xy ....(2)

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

⇒ x + y + (x - y) = 2xy + 6xy

⇒ x + y + x - y = 8xy

⇒ 2x = 8xy

⇒ 2 = 8y

⇒ y = $\dfrac{2}{8} = \dfrac{1}{4}$ .

Substituting $y = \dfrac{1}{4}$ in equation (1), we get :

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

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

Question 19

Solve the following simultaneous equations:

$\dfrac{3}{x + y} + \dfrac{2}{x - y} = 3,;\dfrac{2}{x + y} + \dfrac{3}{x - y} = \dfrac{11}{3}$

Answer

Given,

Equations:

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

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

Multiplying equation (1) by 2, we get:

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

Multiplying equation (2) by 3, we get:

$\Rightarrow 3\Big(\dfrac{2}{x + y} + \dfrac{3}{x - y}\Big) = \dfrac{11}{3} \times 3 \\[1em] \Rightarrow \Big(\dfrac{6}{x + y} + \dfrac{9}{x - y}\Big) = 11 \text{ .........(4) }$

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

$\Rightarrow \Big(\dfrac{6}{x + y} + \dfrac{9}{x - y}\Big) - \Big(\dfrac{6}{x + y} + \dfrac{4}{x - y}\Big) = 11 - 6 \\[1em] \Rightarrow \Big(\dfrac{6}{x + y} + \dfrac{9}{x - y} - \dfrac{6}{x + y} - \dfrac{4}{x - y}\Big) = 5 \\[1em] \Rightarrow \Big(\dfrac{9}{x - y} - \dfrac{4}{x - y}\Big) = 5 \\[1em] \Rightarrow \Big(\dfrac{5}{x - y}\Big) = 5 \\[1em] \Rightarrow 5 = 5(x - y) \\[1em] \Rightarrow x - y = 1 \text{ .........(5) }$

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

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

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

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

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

⇒ x + y = 3

⇒ 2 + y = 3

⇒ y = 3 - 2

⇒ y = 1.

Hence, x = 2, y = 1.

Question 20

Solve the following simultaneous equations:

$\dfrac{22}{x + y} + \dfrac{15}{x - y} = 5,\dfrac{55}{x + y} + \dfrac{40}{x - y} = 13$

Answer

Given,

Equations:

$\dfrac{22}{x + y} + \dfrac{15}{x - y} = 5$ ..........(1)

$\dfrac{55}{x + y} + \dfrac{40}{x - y} = 13$ ..........(2)

Multiplying equation (1) by 5, we get:

$\Rightarrow 5\Big(\dfrac{22}{x + y} + \dfrac{15}{x - y}\Big) = 5 \times 5 \\[1em] \Rightarrow \Big(\dfrac{110}{x + y} + \dfrac{75}{x - y}\Big) = 25 \text{ ..........(3) }$

Multiplying equation (2) by 2, we get:

$\Rightarrow 2\Big(\dfrac{55}{x + y} + \dfrac{40}{x - y}\Big) = 13 \times 2 \\[1em] \Rightarrow \Big(\dfrac{110}{x + y} + \dfrac{80}{x - y}\Big) = 26 \text{ ..........(4) }$

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

$\Rightarrow \Big(\dfrac{110}{x + y} + \dfrac{80}{x - y}\Big) - \Big(\dfrac{110}{x + y} + \dfrac{75}{x - y}\Big) = 26 - 25 \\[1em] \Rightarrow \Big(\dfrac{110}{x + y} + \dfrac{80}{x - y} - \dfrac{110}{x + y} - \dfrac{75}{x - y}\Big) = 1 \\[1em] \Rightarrow \Big(\dfrac{80 - 75}{x - y}\Big) = 1 \\[1em] \Rightarrow \Big(\dfrac{5}{x - y}\Big) = 1 \\[1em] \Rightarrow 5 = (x - y) \\[1em] \Rightarrow x - y = 5 \text{ .........(5) }$

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

$\Rightarrow \dfrac{22}{x + y} + \dfrac{15}{x - y} = 5 \\[1em] \Rightarrow \dfrac{22}{x + y} + \dfrac{15}{5} = 5 \\[1em] \Rightarrow \dfrac{22}{x + y} + 3 = 5 \\[1em] \Rightarrow \dfrac{22}{x + y} = 5 - 3 \\[1em] \Rightarrow \dfrac{22}{x + y} = 2 \\[1em] \Rightarrow x + y = \dfrac{22}{2} \\[1em] \Rightarrow x + y = 11 \text{ .......(6) }$

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

$\Rightarrow x + y + x - y = 11 + 5 \\[1em] \Rightarrow 2x = 16 \\[1em] \Rightarrow x = \dfrac{16}{2} = 8.$

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

⇒ x + y = 11

⇒ 8 + y = 11

⇒ y = 11 - 8

⇒ y = 3

Hence, x = 8, y = 3.

Question 21

Solve the following simultaneous equations:

103x + 51y = 617, 97x + 49y = 583

Answer

Given,

Equations :

103x + 51y = 617 .......(1),

97x + 49y = 583 .......(2)

Adding equations (1) and (2),

⇒ 103x + 51y + (97x + 49y) = 617 + 583

⇒ 103x + 51y + 97x + 49y = 617 + 583

⇒ 200x + 100y = 1200

⇒ 100(2x + y) = 1200

⇒ 2x + y = $\dfrac{1200}{100}$

⇒ 2x + y = 12 ........(3)

Subtracting equation (2) from (1),

⇒ 103x + 51y - (97x + 49y) = 617 - 583

⇒ 103x + 51y - 97x - 49y = 34

⇒ 6x + 2y = 34

⇒ 2(3x + y) = 34

⇒ 3x + y = $\dfrac{34}{2}$

⇒ 3x + y = 17 ........(4)

Subtracting equation 4 from 3,

⇒ 2x + y - (3x + y) = 12 - 17

⇒ 2x + y - 3x - y = -5

⇒ -x = -5

⇒ x = 5.

Substituting value of x in equation 1,

⇒ 103x + 51y = 617

⇒ 103(5) + 51y = 617

⇒ 515 + 51y = 617

⇒ 51y = 617 - 515

⇒ 51y = 102

⇒ y = $\dfrac{102}{51}$ = 2.

Hence, x = 5, y = 2.

Question 22

Solve the following simultaneous equations:

23x − 29y = 98, 29x − 23y = 110

Answer

Given,

Equations :

23x − 29y = 98 .........(1),

29x − 23y = 110 ........(2)

Adding equations 1 and 2,

⇒ 23x − 29y + (29x − 23y) = 98 + 110

⇒ 23x − 29y + 29x − 23y = 98 + 110

⇒ 52x − 52y = 208

⇒ 52(x - y) = 208

⇒ (x - y) = $\dfrac{208}{52}$

⇒ x - y = 4 .........(3)

Subtracting equation 2 from 1,

⇒ 23x − 29y - (29x − 23y) = 98 - 110

⇒ 23x − 29y - 29x + 23y = 98 - 110

⇒ -6x - 6y = -12

⇒ -6(x + y) = -12

⇒ x + y = $\dfrac{-12}{-6}$

⇒ x + y = 2 .......(4)

Adding equation 4 and 5,

⇒ x + y + x - y = 2 + 4

⇒ 2x = 6

⇒ $x = \dfrac{6}{2}$

⇒ x = 3.

Substituting value of x in equation 1,

⇒ 23x − 29y = 98

⇒ 23(3) − 29y = 98

⇒ 69 − 29y = 98

⇒ −29y = 98 − 69

⇒ −29y = 29

⇒ y = $\dfrac{29}{-29}$ = -1.

Hence, x = 3, y = -1.

Question 23

Solve the following simultaneous equations:

$\dfrac{a}{x} - \dfrac{b}{y} = 0, \dfrac{ab^2}{x} + \dfrac{a^2 b}{y} = (a^2 + b^2)$

Answer

Substituting $\dfrac{1}{x} = m \text{ and } \dfrac{1}{y} = n$ in $\dfrac{a}{x} - \dfrac{b}{y} = 0$ ,

⇒ am - bn = 0

⇒ am = bn

⇒ m = $\dfrac{bn}{a}$ ........(1)

Substituting $\dfrac{1}{x} = m \text{ and } \dfrac{1}{y} = n$ in $\dfrac{ab^2}{x} + \dfrac{a^2b}{y} = (a^2 + b^2)$

⇒ ab²m + a²bn = a² + b² .........(2)

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

⇒ ab² $\Big(\dfrac{bn}{a}\Big)$ + a²bn = a² + b²

⇒ b³n + a²bn = a² + b²

⇒ bn(b² + a²) = a² + b²

⇒ bn = $\dfrac{(a^2 + b^2)}{(a^2 + b^2)}$

⇒ bn = 1

⇒ n = $\dfrac{1}{b}$

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

⇒ m = $\dfrac{b}{a} \times \dfrac{1}{b}$

⇒ m = $\dfrac{1}{a}$

$\Rightarrow \dfrac{1}{x} = m \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{1}{a} \\[1em] \Rightarrow x = a \\[1em] \Rightarrow \dfrac{1}{y} = n \\[1em] \Rightarrow \dfrac{1}{y} = \dfrac{1}{b} \\[1em] \Rightarrow y = b.$

Hence, x = a, y = b.

Question 24

If 2x + y = 32 and 3x + 4y = 68, find the value of $\dfrac{x}{y}$ .

Answer

Given,

Equations :

2x + y = 32 .........(1),

3x + 4y = 68 .........(2)

Multiplying equation (1) by 4, we get :

⇒ 4(2x + y) = 32 × 4

⇒ 8x + 4y = 128 .........(3)

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

⇒ 8x + 4y - (3x + 4y) = 128 - 68

⇒ 8x + 4y - 3x - 4y = 60

⇒ 5x = 60

⇒ x = $\dfrac{60}{5}$

⇒ x = 12.

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

⇒ 3x + 4y = 68

⇒ 3 × 12 + 4y = 68

⇒ 36 + 4y = 68

⇒ 4y = 68 - 36

⇒ 4y = 32

⇒ y = $\dfrac{32}{4}$

⇒ y = 8.

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

$\Rightarrow \dfrac{x}{y} = \dfrac{12}{8} = \dfrac{3}{2}$ .

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

Question 25

The sides of an equilateral triangle are (x + 3y) cm, (3x + 2y − 2) cm and $\Big(4x + \dfrac{y}{2} + 1\Big)$ cm. Find the length of each side.

Answer

Given,

In an equilateral triangle, all three sides are equal.

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

Solving L.H.S of the above equation, we get :

⇒ x + 3y = 3x + 2y - 2

⇒ x - 3x + 3y - 2y = -2

⇒ -2x + y = -2

⇒ y = 2x - 2 .....(1)

Solving R.H.S of the above equation, we get :

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

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

⇒ 3(2x - 2) - 2x = 6

⇒ 6x - 6 - 2x = 6

⇒ 4x - 6 = 6

⇒ 4x = 6 + 6

⇒ 4x = 12

⇒ x = $\dfrac{12}{4}$

⇒ x = 3.

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

⇒ y = 2(3) - 2

⇒ y = 6 - 2

⇒ y = 4.

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

⇒ x + 3y = 3 + 3(4) = 3 + 12 = 15 cm.

Since, all sides are equal.

$\Rightarrow \dfrac{x}{(2) \times (10) - (1) \times (25)} = \dfrac{y}{(25) \times (2) - (10) \times (3)} = \dfrac{1}{(3) \times (1) - (2) \times (2)} \\[1em] \Rightarrow \dfrac{x}{20 - 25} = \dfrac{y}{50 - 30} = \dfrac{1}{3 - 4} \\[1em] \Rightarrow \dfrac{x}{-5} = \dfrac{y}{20} = \dfrac{1}{-1} \\[1em] \Rightarrow \dfrac{x}{-5} = \dfrac{1}{-1} \text{ and } \dfrac{y}{20} = \dfrac{1}{-1} \\[1em] \Rightarrow x = \dfrac{-5}{-1} \text{ and } y = \dfrac{20}{-1} \\[1em] \Rightarrow x = 5 \text{ and } y = -20.$

Hence, x = 5 and y = -20.

Question 11

Solve the following system of equations by using the method of cross multiplication:

$\dfrac{5}{x} - \dfrac{4}{y} + 2 = 0, \dfrac{2}{x} + \dfrac{3}{y} = 13$ , (x ≠ 0, y ≠ 0)

Answer

Substituting $\dfrac{1}{x}= a, \dfrac{1}{y} = b$ in $\dfrac{5}{x} - \dfrac{4}{y} + 2 = 0$ , we get:

⇒ 5a - 4b + 2 = 0 ..........(1)

Substituting $\dfrac{1}{x} = a, \dfrac{1}{y} = b$ in $\dfrac{2}{x} + \dfrac{3}{y} = 13$ , we get :

⇒ 2a + 3b = 13

⇒ 2a + 3b - 13 = 0 ........(2)

Applying cross-multiplication method for solving equations (1) and (2), we get :

$\Rightarrow \dfrac{a}{(-4) \times (-13) - (3) \times (2)} = \dfrac{b}{(2) \times (2) - (-13) \times (5)} = \dfrac{1}{(5) \times (3) - (2) \times (-4)} \\[1em] \Rightarrow \dfrac{a}{52 - 6} = \dfrac{b}{4 + 65} = \dfrac{1}{15 + 8} \\[1em] \Rightarrow \dfrac{a}{46} = \dfrac{b}{69} = \dfrac{1}{23} \\[1em] \Rightarrow \dfrac{a}{46} = \dfrac{1}{23} \text{ and } \dfrac{b}{69} = \dfrac{1}{23} \\[1em] \Rightarrow a = \dfrac{46}{23} \text{ and } b = \dfrac{69}{23} \\[1em] \Rightarrow a = 2 \text{ and } b = 3.$

Now we have a = 2 and b = 3,

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

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

Question 12

Solve the following system of equations by using the method of cross multiplication:

$\dfrac{1}{x} + \dfrac{1}{y} = 7, \dfrac{2}{x} + \dfrac{3}{y} = 17$ (x ≠ 0, y ≠ 0)

Answer

Substituting $\dfrac{1}{x}= a, \dfrac{1}{y} = b$ in $\dfrac{1}{x} + \dfrac{1}{y} = 7$ , we get :

⇒ a + b = 7

⇒ a + b - 7 = 0 .........(1)

Substituting $\dfrac{1}{x} = a, \dfrac{1}{y} = b$ in $\dfrac{2}{x} + \dfrac{3}{y} = 17$ , we get :

⇒ 2a + 3b = 17

⇒ 2a + 3b - 17 = 0 .........(2)

Applying cross-multiplication method for solving equations (1) and (2), we get :

$\Rightarrow \dfrac{a}{(1) \times (-17) - (3) \times (-7)} = \dfrac{b}{(-7) \times (2) - (-17) \times (1)} = \dfrac{1}{(1) \times (3) - (2) \times (1)} \\[1em] \Rightarrow \dfrac{a}{(-17) + 21} = \dfrac{b}{-14 + 17} = \dfrac{1}{3 - 2} \\[1em] \Rightarrow \dfrac{a}{4} = \dfrac{b}{3} = \dfrac{1}{1} \\[1em] \Rightarrow \dfrac{a}{4} = 1 \text{ and } \dfrac{b}{3} = \dfrac{1}{1} \\[1em] \Rightarrow a = \dfrac{4}{1} \text{ and } b = \dfrac{3}{1} \\[1em] \Rightarrow a = 4 \text{ and } b = 3.$

Now we have a = 4 and b = 3,

$\Rightarrow \dfrac{1}{x} = a \\[1em] \Rightarrow \dfrac{1}{x} = 4 \\[1em] \Rightarrow x = \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{1}{y} = b \\[1em] \Rightarrow \dfrac{1}{y} = 3 \\[1em] \Rightarrow y = \dfrac{1}{3}.$

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

Question 13

Solve the following system of equations by using the method of cross multiplication:

$\dfrac{10}{x + y} + \dfrac{2}{x - y} = 4, \dfrac{15}{x + y} - \dfrac{5}{x - y} + 2 = 0$ , where x ≠ -y and x ≠ y

Answer

Substituting $\dfrac{1}{x + y}= a, \dfrac{1}{x - y} = b$ in $\dfrac{10}{x + y} + \dfrac{2}{x - y} = 4$ , we get:

⇒ 10a + 2b = 4

⇒ 10a + 2b - 4 = 0 .....(1)

Substituting $\dfrac{1}{x + y}= a, \dfrac{1}{x - y} = b$ in $\dfrac{15}{x + y} - \dfrac{5}{x - y} + 2 = 0$ , we get:

⇒ 15a - 5b + 2 = 0 ....(2)

Applying cross-multiplication method for solving equations (1) and (2), we get :

$\Rightarrow \dfrac{a}{(2) \times (2) - (-5) \times (-4)} = \dfrac{b}{(-4) \times (15) - (2) \times (10)} = \dfrac{1}{(10) \times (-5) - (15) \times (2)} \\[1em] \Rightarrow \dfrac{a}{(4) - 20} = \dfrac{b}{-60 - 20} = \dfrac{1}{-50 - 30} \\[1em] \Rightarrow \dfrac{a}{-16} = \dfrac{b}{-80} = \dfrac{1}{-80} \\[1em] \Rightarrow \dfrac{a}{-16} = \dfrac{1}{-80} \text{ and } \dfrac{b}{-80} = \dfrac{1}{-80} \\[1em] \Rightarrow a = \dfrac{-16}{-80} \text{ and } b = \dfrac{-80}{-80} \\[1em] \Rightarrow a = \dfrac{1}{5} \text{ and } b = 1.$

Now we have a = $\dfrac{1}{5}$ and b = 1,

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

On adding equations (3) and (4) we get,

⇒ (x + y) + (x - y) = 5 + 1

⇒ 2x = 6

⇒ x = $\dfrac{6}{2} = 3$ .

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

⇒ x - y = 1

⇒ 3 - y = 1

⇒ y = 3 - 1 = 2.

Hence, x = 3, y = 2.

Question 14

Solve the following system of equations by using the method of cross multiplication:

$\dfrac{5}{x + 1} - \dfrac{2}{y - 1} = \dfrac{1}{2},\dfrac{10}{x + 1} + \dfrac{2}{y - 1} = \dfrac{5}{2}$ , where x ≠ -1 and x ≠ 1.

Answer

$\dfrac{5}{x + 1} - \dfrac{2}{y - 1} = \dfrac{1}{2},\dfrac{10}{x + 1} + \dfrac{2}{y - 1} = \dfrac{5}{2}$

Substituting $\dfrac{1}{x + 1}= u, \dfrac{1}{y - 1} = v$ in $\dfrac{5}{x + 1} - \dfrac{2}{y - 1} = \dfrac{1}{2}$ , we get:

⇒ 5u - 2v = $\dfrac{1}{2}$

⇒ 5u - 2v - $\dfrac{1}{2}$ = 0 ....(1)

Substituting $\dfrac{1}{x + 1}= u, \dfrac{1}{y - 1} = v$ in $\dfrac{10}{x + 1} + \dfrac{2}{y - 1} = \dfrac{5}{2}$ , we get:

⇒ 10u + 2v = $\dfrac{5}{2}$

⇒ 10u + 2v - $\dfrac{5}{2}$ = 0 ....(2)

Multiply equation (1) and (2) by 2, we get,

⇒ $2\Big(5u - 2v - \dfrac{1}{2}\Big)$ = 0

⇒ 10u - 4v - 1 = 0 .......(3)

⇒ $2\Big(10u + 2v - \dfrac{5}{2}\Big)$ = 0

⇒ 20u + 4v - 5 = 0 .........(4)

Applying cross-multiplication method for solving equations (3) and (4), we get :

$\Rightarrow \dfrac{u}{(-4) \times (-5) - (4) \times (-1)} = \dfrac{v}{(-1) \times (20) - (-5) \times (10)} = \dfrac{1}{(10) \times (4) - (20) \times (-4)} \\[1em] \Rightarrow \dfrac{u}{(20) + 4} = \dfrac{v}{-20 + 50} = \dfrac{1}{40 + 80} \\[1em] \Rightarrow \dfrac{u}{24} = \dfrac{v}{30} = \dfrac{1}{120} \\[1em] \Rightarrow \dfrac{u}{24} = \dfrac{1}{120} \text{ and } \dfrac{v}{30} = \dfrac{1}{120} \\[1em] \Rightarrow u = \dfrac{24}{120} \text{ and } v = \dfrac{30}{120} \\[1em] \Rightarrow u = \dfrac{1}{5} \text{ and } v = \dfrac{1}{4}.$

Now we have u = $\dfrac{1}{5} \text{ and } v = \dfrac{1}{4}$ ,

$\Rightarrow \dfrac{1}{x + 1} = u \\[1em] \Rightarrow \dfrac{1}{x + 1} = \dfrac{1}{5} \\[1em] \Rightarrow x + 1 = 5 \\[1em] \Rightarrow x = 5 - 1 = 4 \\[1em] \Rightarrow \dfrac{1}{y - 1} = v \\[1em] \Rightarrow \dfrac{1}{y - 1} = \dfrac{1}{4} \\[1em] \Rightarrow y - 1 = 4 \\[1em] \Rightarrow y = 4 + 1 = 5.$

Hence, x = 4, y = 5.

Exercise 5C

Question 1

The sum of two numbers is 53 and their difference is 25. Find the numbers.

Answer

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

Given,

Sum of numbers = 53

⇒ x + y = 53 .......(1)

Difference of numbers = 25

⇒ x - y = 25 ......(2)

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

⇒ (x + y) + (x - y) = 53 + 25

⇒ x + x + y - y = 78

⇒ 2x = 78

⇒ x = $\dfrac{78}{2}$ = 39.

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

⇒ x + y = 53

⇒ 39 + y = 53

⇒ y = 53 - 39 = 14.

Hence, the numbers are 39 and 14.

Question 2

The sum of two numbers exceeds thrice the smaller by 2. If the difference between them is 19, find the numbers.

Answer

Let the numbers be x and y, such that x < y.

Given,

The sum of two numbers exceeds thrice the smaller by 2.

∴ (x + y) - 3x = 2

⇒ x - 3x + y = 2

⇒ y - 2x = 2 ......(1)

Given,

Difference between the numbers = 19

⇒ y - x = 19 ......(2)

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

⇒ y - x - (y - 2x) = 19 - 2

⇒ y - x - y + 2x = 19 - 2

⇒ x = 17.

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

⇒ y - 2x = 2

⇒ y - 2 × 17 = 2

⇒ y - 34 = 2

⇒ y = 36.

Hence, the numbers are 17 and 36.

Question 3

The sum of two numbers is 51. If the larger is doubled and the smaller is tripled, the difference is 12. Find the numbers.

Answer

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

Given,

Sum of numbers = 51

⇒ x + y = 51

⇒ x = 51 - y ......(1)

Given,

Difference of numbers when larger is doubled and the smaller is tripled = 12

⇒ 2x - 3y = 12 ......(2)

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

⇒ 2(51 - y) - 3y = 12

⇒ 102 - 2y - 3y = 12

⇒ -5y = 12 - 102

⇒ -5y = -90

⇒ y = $\dfrac{90}{5} = 18$ .

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

⇒ x = 51 - y

⇒ x = 51 - 18

⇒ x = 33.

Hence, the numbers are 33 and 18.

Question 4

Find two numbers such that the sum of twice the first and thrice the second is 103 and four times the first exceeds seven times the second by 11.

Answer

Let two numbers be x and y.

Given,

Sum of twice the first and thrice the second = 103

⇒ 2x + 3y = 103

⇒ 2x = 103 - 3y

⇒ x = $\Big(\dfrac{103 - 3y}{2}\Big)$ ......(1)

Given,

Four times the first number exceeds seven times the second by 11.

⇒ 4x - 7y = 11

⇒ 4x = 7y + 11 .....(2)

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

⇒ $4\Big(\dfrac{103 - 3y}{2}\Big)$ = 7y + 11

⇒ 2(103 - 3y) = 7y + 11

⇒ 206 - 6y = 7y + 11

⇒ 206 - 11 = 7y + 6y

⇒ 195 = 13y

⇒ y = $\dfrac{195}{13} = 15$ .

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

$\Rightarrow x = \Big(\dfrac{103 - 3y}{2}\Big) \\[1em] \Rightarrow x = \Big(\dfrac{103 - 3 \times 15}{2}\Big) \\[1em] \Rightarrow x = \Big(\dfrac{103 - 45}{2}\Big) \\[1em] \Rightarrow x = \Big(\dfrac{58}{2}\Big) = 29.$

Hence, the numbers are 29 and 15.

Question 5

Find two numbers such that the sum of thrice the first and the second is 142 and four times the first exceeds the second by 138.

Answer

Let two numbers be x and y.

Given,

Sum of thrice the first and the second = 142

⇒ 3x + y = 142

⇒ y = 142 - 3x .....(1)

Given,

Four times the first exceeds the second by 138.

⇒ 4x - y = 138

⇒ 4x = y + 138 ......(2)

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

⇒ 4x = 142 - 3x + 138

⇒ 4x + 3x = 142 + 138

⇒ 7x = 280

⇒ x = $\dfrac{280}{7} = 40$ .

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

⇒ y = 142 - 3x

⇒ y = 142 - 3(40)

⇒ y = 142 - 120

⇒ y = 22.

Hence, the numbers are 40 and 22.

Question 6

Of the two numbers, 4 times the smaller one is less than 3 times the larger one by 6. Also, the sum of the numbers is larger than 6 times their difference by 5. Find the numbers.

Answer

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

Given,

4 times the smaller one is less than 3 times the larger one by 6.

⇒ 3x - 4y = 6

⇒ 4y = 3x - 6

⇒ y = $\Big(\dfrac{3x - 6}{4}\Big)$ .............(1)

Given,

Sum of the numbers is larger than 6 times their difference by 5.

⇒ x + y - 6(x - y) = 5

⇒ x + y - 6x + 6y = 5

⇒ x - 6x + y + 6y = 5

⇒ 7y - 5x = 5 ...........(2)

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

$\Rightarrow 7\Big(\dfrac{3x - 6}{4}\Big) - 5x = 5 \\[1em] \Rightarrow \Big(\dfrac{21x - 42}{4}\Big) - 5x = 5 \\[1em] \Rightarrow \Big(\dfrac{21x - 42 - 20x}{4}\Big) = 5 \\[1em] \Rightarrow \dfrac{x - 42}{4} = 5 \\[1em] \Rightarrow x - 42 = 20 \\[1em] \Rightarrow x = 20 + 42 \\[1em] \Rightarrow x = 62.$

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

$\Rightarrow y = \Big(\dfrac{3x - 6}{4}\Big) \\[1em] \Rightarrow y = \Big(\dfrac{3 \times 62 - 6}{4}\Big) \\[1em] \Rightarrow y = \Big(\dfrac{186 - 6}{4}\Big) \\[1em] \Rightarrow y = \Big(\dfrac{180}{4}\Big) = 45.$

Hence, the numbers are 62 and 45.

Question 7

If from twice the greater number of the two numbers, 45 is subtracted, the result is the other number. If from twice the smaller number, 21 is subtracted, the result is the greater number. Find the numbers.

Answer

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

Given,

If from twice the greater number of the two numbers, 45 is subtracted, the result is the other number.

⇒ 2x - 45 = y .....(1)

Given,

If from twice the smaller number, 21 is subtracted, the result is the greater number.

⇒ 2y - 21 = x .....(2)

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

⇒ 2y - 21 = x

⇒ 2(2x - 45) - 21 = x

⇒ 4x - 90 - 21 = x

⇒ 4x - x - 111 = 0

⇒ 3x = 111

⇒ x = $\dfrac{111}{3} = 37.$

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

⇒ 2x - 45 = y

⇒ 2(37) - 45 = y

⇒ 74 - 45 = y

⇒ y = 29.

Hence, the numbers are 37 and 29.

Question 8

If three times the larger of the two numbers is divided by the smaller, then the quotient is 4 and remainder is 5. If 6 times the smaller is divided by the larger, the quotient is 4 and the remainder is 2. Find the numbers.

Answer

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

Given,

Three times the larger divided by the smaller gives quotient 4.

⇒ 3x = 4y + 5

⇒ x = $\dfrac{4y + 5}{3}$ .....(1)

Given,

Six times the smaller divided by the larger gives quotient 4 and remainder 2.

⇒ 6y = 4x + 2 .....(2)

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

$\Rightarrow 6y = 4 \Big(\dfrac{4y + 5}{3}\Big) + 2 \\[1em] \Rightarrow 6y = \Big(\dfrac{16y + 20}{3}\Big) + 2 \\[1em] \Rightarrow 6y = \Big(\dfrac{16y + 20 + 6}{3}\Big) \\[1em] \Rightarrow 6y = \Big(\dfrac{16y + 26}{3}\Big) \\[1em] \Rightarrow 6y \times 3 = 16y + 26 \\[1em] \Rightarrow 18y = 16y + 26 \\[1em] \Rightarrow 18y - 16y = 26 \\[1em] \Rightarrow 2y = 26 \\[1em] \Rightarrow y = \dfrac{26}{2} \\[1em] \Rightarrow y = 13.$

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

$\Rightarrow x = \dfrac{4y + 5}{3} \\[1em] \Rightarrow x = \dfrac{4 \times 13 + 5}{3} \\[1em] \Rightarrow x = \dfrac{52 + 5}{3} \\[1em] \Rightarrow x = \dfrac{57}{3} = 19.$

Hence, the numbers are 19 and 13.

Question 9

If 2 is added to each of two given numbers, then their ratio becomes 1 : 2. However, if 4 is subtracted from each of the given numbers, the ratio becomes 5 : 11. Find the numbers.

Answer

Let two numbers be x and y.

Given,

If 2 is added to each, the ratio becomes 1 : 2.

⇒ $\dfrac{x + 2}{y + 2} = \dfrac{1}{2}$

⇒ 2(x + 2) = y + 2

⇒ 2x + 4 = y + 2

⇒ y = 2x + 4 - 2

⇒ y = 2x + 2 ....(1)

Given,

If 4 is subtracted from each number, the ratio becomes 5 : 11.

⇒ $\dfrac{x - 4}{y - 4} = \dfrac{5}{11}$

⇒ 11(x - 4) = 5(y - 4)

⇒ 11x - 44 = 5y - 20

⇒ 11x - 5y = -20 + 44

⇒ 11x - 5y = 24 ....(2)

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

⇒ 11x - 5(2x + 2) = 24

⇒ 11x - 10x - 10 = 24

⇒ x - 10 = 24

⇒ x = 24 + 10

⇒ x = 34.

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

⇒ y = 2x + 2

⇒ y = 2 × 34 + 2

⇒ y = 68 + 2

⇒ y = 70.

Hence, the numbers are 34 and 70.

Question 10

The difference between two numbers is 12 and the difference between their squares is 456. Find the numbers.

Answer

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

Given,

Difference between two numbers = 12.

⇒ x - y = 12 .........(1)

Given,

Difference between their squares = 456

⇒ x² - y² = 456

By identity,

⇒ x² - y² = (x + y)(x - y)

⇒ (x + y)(x - y) = 456

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

⇒ (x + y)(12) = 456

⇒ (x + y) = $\dfrac{456}{12}$

⇒ x + y = 38 ........(2)

Adding equations (1) and (2),

⇒ x + y + x - y = 38 + 12

⇒ 2x = 50

⇒ x = $\dfrac{50}{2}$

⇒ x = 25.

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

⇒ x - y = 12

⇒ 25 - y = 12

⇒ y = 25 - 12

⇒ y = 13.

Hence, the numbers are 25 and 13.

Question 11

Find the fraction which becomes $\dfrac{1}{2}$ when its numerator is increased by 6 and is equal to $\dfrac{1}{3}$ when its denominator is increased by 7.

Answer

Let the numerator be x and denominator be y.

Thus, fraction = $\dfrac{x}{y}$

Given,

The fraction becomes $\dfrac{1}{2}$ when its numerator is increased by 6.

⇒ $\dfrac{x + 6}{y} = \dfrac{1}{2}$

⇒ 2(x + 6) = y

⇒ 2x + 12 = y ..........(1)

Given,

The fraction becomes $\dfrac{1}{3}$ when its denominator is increased by 7.

⇒ $\dfrac{x}{y + 7} = \dfrac{1}{3}$

⇒ 3x = y + 7 .......(2)

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

⇒ 3x = (2x + 12) + 7

⇒ 3x = 2x + 19

⇒ 3x - 2x = 19

⇒ x = 19.

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

⇒ 2 × 19 + 12 = y

⇒ 38 + 12 = y

⇒ y = 50.

Hence, the fraction = $\dfrac{19}{50}$ .

Question 12

A fraction becomes $\dfrac{1}{2}$ when 1 is subtracted from its numerator and 1 is added to its denominator; it becomes $\dfrac{1}{3}$ when 6 is subtracted from its numerator and 1 from its denominator. Find the original fraction.

Answer

Let the numerator be x and denominator be y.

Thus, fraction = $\dfrac{x}{y}$

Given,

The fraction becomes $\dfrac{1}{2}$ when 1 is subtracted from the numerator and 1 is added to the denominator,

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

⇒ 2(x - 1) = y + 1

⇒ 2x - 2 = y + 1

⇒ y = 2x - 2 - 1

⇒ y = 2x - 3 .........(1)

Given,

The fraction becomes $\dfrac{1}{3}$ when 6 is subtracted from the numerator and 1 from the denominator,

⇒ $\dfrac{x - 6}{y - 1} = \dfrac{1}{3}$

⇒ 3(x - 6) = y - 1

⇒ 3x - 18 = y - 1

⇒ y = 3x - 18 + 1

⇒ y = 3x - 17 .........(2)

From equations (1) in (2), we get :

⇒ 3x - 17 = 2x - 3

⇒ 3x - 2x = -3 + 17

⇒ 3x - 2x = 14

⇒ x = 14.

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

⇒ y = 2(14) - 3 = 28 - 3 = 25.

Fraction = $\dfrac{x}{y} = \dfrac{14}{25}$ .

Hence, the fraction = $\dfrac{14}{25}$ .

Question 13

The denominator of a fraction is greater than its numerator by 9. If 7 is subtracted from both, its numerator and denominator, the fraction becomes $\dfrac{2}{3}$ . Find the original fraction.

Answer

Let the numerator be x.

Given,

The denominator of a fraction is greater than its numerator by 9.

⇒ Denominator = x + 9

Thus, fraction = $\dfrac{x}{x + 9}$

Given,

The fraction becomes $\dfrac{2}{3}$ , when 7 is subtracted from both numerator and denominator,

⇒ $\dfrac{x - 7}{(x + 9) - 7} = \dfrac{2}{3}$

⇒ $\dfrac{x - 7}{x + 2} = \dfrac{2}{3}$

⇒ 3(x - 7) = 2(x + 2)

⇒ 3x - 21 = 2x + 4

⇒ 3x - 2x = 4 + 21

⇒ x = 25.

The denominator is (x + 9) = 25 + 9 = 34.

Fraction = $\dfrac{x}{y} = \dfrac{25}{34}$ .

Hence, the fraction = $\dfrac{25}{34}$ .

Question 14

A number consists of two digits, the difference of whose digits is 3. If 4 times the number equals 7 times the number obtained by reversing its digits, find the number.

[Hint. Original number is greater than the number obtained by reversing its digits. ∴ In original number, ten's digit greater than unit's digit.]

Answer

Let the tens and unit digits of required number be x and y. x > y (according to hint)

Given,

Difference of digits of the number is 3.

⇒ x - y = 3

⇒ x = y + 3 .....(1)

Original number = 10x + y

Number obtained by reversing the digits = (10y + x)

Given,

4 times the number equals 7 times the number obtained by reversing its digits.

⇒ 4(10x + y) = 7(10y + x)

⇒ 40x + 4y = 70y + 7x

⇒ 40x - 7x + 4y - 70y = 0

⇒ 33x - 66y = 0 .....(2)

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

⇒ 33(y + 3) - 66y = 0

⇒ 33y + 33 × 3 - 66y = 0

⇒ 99 - 33y = 0

⇒ 33y = 99

⇒ y = $\dfrac{99}{33}$

⇒ y = 3.

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

⇒ x = y + 3

⇒ x = 3 + 3

⇒ x = 6.

Original number = 10x + y

= 10 × 6 + 3

= 63.

Hence, the number is 63.

Question 15

A number consists of two digits, the difference of whose digits is 5. If 8 times the number is equal to 3 times the number obtained by reversing the digits, find the number.

Answer

According to question,

Original number is smaller than the number obtained by reversing its digits.

∴ In original number, unit's digit greater than ten's digit.

Let the ten's and unit's digit of required number be x and y respectively y > x.

Given,

⇒ y - x = 5

⇒ y = x + 5 .....(1)

Number obtained by reversing the digits = (10y + x)

Given,

8 times the number is equal to 3 times the number obtained by reversing the digits.

⇒ 8(10x + y) = 3(10y + x)

⇒ 80x + 8y = 30y + 3x

⇒ 80x - 3x + 8y - 30y = 0

⇒ 77x - 22y = 0 .....(2)

Substituting the value of x from equation (2) in 77x - 22y = 0, we get:

⇒ 77x - 22y = 0

⇒ 77x - 22 × (x + 5) = 0

⇒ 77x - 22x - 110 = 0

⇒ 55x - 110 = 0

⇒ 55x = 110

⇒ x = $\dfrac{110}{55}$

⇒ x = 2.

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

⇒ y = x + 5

⇒ y = 2 + 5

⇒ y = 7.

Number = (10x + y) = 10 × 2 + 7 = 27.

Hence, the number is 27.

Question 16

The result of dividing a two-digit number by the number with its digits reversed is $\Big(1\dfrac{3}{4}\Big)$ . If the sum of the digits is 12, find the number.

Answer

Let the ten's and unit's digits of required number be x and y respectively.

Given,

Sum of the digits of the number is 12.

⇒ x + y = 12

⇒ x = 12 - y .....(1)

Original number = 10x + y

Number obtained by reversing the digits = 10y + x

Given,

On dividing the number by the number with its digits reversed, the result is $\Big(1\dfrac{3}{4}\Big)$ .

$\therefore \Big(\dfrac{10x + y}{10y + x}\Big) = \Big(1\dfrac{3}{4}\Big) \\[1em] \Rightarrow \Big(\dfrac{10x + y}{10y + x}\Big) = \dfrac{7}{4} \\[1em] \Rightarrow (10x + y) \times 4 = (10y + x) \times 7 \\[1em] \Rightarrow 40x + 4y = 70y + 7x \\[1em] \Rightarrow 40x - 7x + 4y - 70y = 0 \\[1em] \Rightarrow 33x - 66y = 0 \text{.....(2)}$

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

⇒ 33(12 - y) - 66y = 0

⇒ 396 - 33y - 66y = 0

⇒ 396 - 99y = 0

⇒ 99y = 396

⇒ y = $\dfrac{396}{99}$

⇒ y = 4.

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

⇒ x = 12 - y

⇒ x = 12 - 4

⇒ x = 8.

Original number = (10x + y)

= 10 × 8 + 4

= 84.

Hence, the number is 84.

Question 17

When a two-digit number is divided by the sum of its digits, the quotient is 8. On diminishing the ten’s digit by three times the unit’s digit, the remainder obtained is 1. Find the number.

Answer

Let the ten's and unit's digit of required number be x and y respectively.

Number = 10x + y

Given,

On dividing the number by the sum of its digits, the quotient is 8.

$\Rightarrow \Big(\dfrac{10x + y}{x + y}\Big) =8 \\[1em] \Rightarrow 10x + y = 8(x + y) \\[1em] \Rightarrow 10x + y = 8x + 8y \\[1em] \Rightarrow 10x - 8x + y - 8y = 0 \\[1em] \Rightarrow 2x - 7y = 0 \text{ .....(1)}$

Given,

On diminishing the ten’s digit by three times the unit’s digit, the remainder obtained is 1.

⇒ x - 3y = 1

⇒ x = 3y + 1 .........(2)

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

⇒ 2(3y + 1) - 7y = 0

⇒ 6y + 2 - 7y = 0

⇒ 2 - y = 0

⇒ y = 2.

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

⇒ x = 3 × 2 + 1

⇒ x = 6 + 1

⇒ x = 7.

Number = 10x + y

= 10 × 7 + 2

= 72.

Hence, the number is 72.

Question 18

A number of two digits exceeds four times the sum of its digits by 6, and the number is increased by 9 on reversing its digits. Find the number.

Answer

Let the ten's and unit's digit of required number be x and y respectively.

Number = 10x + y

Given,

A number of two digits exceeds four times the sum of its digits by 6.

⇒ 10x + y - 4(x + y) = 6

⇒ 10x + y - 4x - 4y = 6

⇒ 6x - 3y = 6 .....(1)

Number obtained by reversing the digits = 10y + x

Given,

Number is increased by 9 on reversing its digits.

⇒ 10y + x = 10x + y + 9

⇒ 10y - y + x - 10x = 9

⇒ 9y - 9x = 9

⇒ y - x = 1

⇒ x = y - 1 .....(2)

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

⇒ 6(y - 1) - 3y = 6

⇒ 6y - 6 - 3y = 6

⇒ 3y - 6 = 6

⇒ 3y = 6 + 6

⇒ 3y = 12

⇒ y = $\dfrac{12}{3}$

⇒ y = 4.

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

⇒ x = 4 - 1

⇒ x = 3.

Number = 10x + y

= 10 × 3 + 4

= 34.

Hence, the number is 34.

Question 19

The sum of the digits of a two-digit number is 12. If the digits are reversed, the new number is 12 less than twice the original number. Find the original number.

Answer

Let the ten's and unit's digits of required number be x and y.

Number = 10x + y

Given,

Sum of digits = 12.

⇒ x + y = 12

⇒ x = 12 - y .....(1)

Given,

Number obtained by reversing the digits = (10y + x)

If the digits are reversed, the new number is 12 less than twice the original number.

⇒ (10y + x) = 2(10x + y) - 12

⇒ 10y + x = 20x + 2y - 12

⇒ 10y - 2y + x - 20x = -12

⇒ 8y - 19x = - 12 .....(2)

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

⇒ 8y - 19(12 - y) = -12

⇒ 8y - 228 + 19y = -12

⇒ 27y = -12 + 228

⇒ 27y = 216

⇒ y = $\dfrac{216}{27}$

⇒ y = 8.

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

⇒ x = 12 - 8

⇒ x = 4.

The number is,

⇒ (10x + y) = 10 × 4 + 8 = 48.

Hence, the number is 48.

Question 20

If 11 pens and 19 pencils together cost ₹ 502, while 19 pens and 11 pencils together cost ₹ 758, how much do 3 pens and 6 pencils cost together?

Answer

Let the cost of a pen be ₹ x and ₹ y be the cost of a pencil.

Given,

11 pens and 19 pencils together cost ₹ 502.

⇒ 11x + 19y = 502 ........(1)

Given,

19 pens and 11 pencils together cost ₹ 758.

⇒ 19x + 11y = 758 .........(2)

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

⇒ 19x + 11y - (11x + 19y) = 758 - 502

⇒ 19x - 11x + 11y - 19y = 256

⇒ 8x - 8y = 256

⇒ 8(x - y) = 256

⇒ x - y = $\dfrac{256}{8}$

⇒ x - y = 32

⇒ x = 32 + y .....(3)

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

⇒ 11x + 19y = 502

⇒ 11(32 + y) + 19y = 502

⇒ 352 + 11y + 19y = 502

⇒ 30y = 502 - 352

⇒ 30y = 150

⇒ y = $\dfrac{150}{30}$

⇒ y = ₹ 5.

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

⇒ x = 32 + y

⇒ x = 32 + 5

⇒ x = ₹ 37.

The amount to buy the 3 pens and 6 pencils is,

⇒ 3x + 6y

⇒ 3 × 37 + 6 × 5

⇒ 111 + 30

⇒ ₹ 141.

Hence, 3 pens and 6 pencils costs ₹ 141.

Question 21

5 kg sugar and 7 kg rice together cost ₹ 258, while 7 kg sugar and 5 kg rice together cost ₹ 246. Find the total cost of 8 kg sugar and 10 kg rice.

Answer

Let cost of sugar be ₹ x/kg and cost of rice be ₹ y/kg.

Given,

5 kg sugar and 7 kg rice together cost ₹ 258,

⇒ 5x + 7y = 258

⇒ 7y = 258 - 5x

⇒ y = $\dfrac{258 - 5x}{7}$ .....(1)

Given,

7 kg sugar and 5 kg rice together cost ₹ 246,

⇒ 7x + 5y = 246 .....(2)

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

$\Rightarrow 7x + 5\Big(\dfrac{258 - 5x}{7}\Big) = 246 \\[1em] \Rightarrow \Big(\dfrac{49x + 1290 - 25x}{7}\Big) = 246 \\[1em] \Rightarrow 24x + 1290 = 246 \times 7 \\[1em] \Rightarrow 24x + 1290 = 1722 \\[1em] \Rightarrow 24x = 1722 - 1290 \\[1em] \Rightarrow x = \dfrac{432}{24} = 18.$

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

$\Rightarrow y = \dfrac{258 - 5 \times 18}{7} \\[1em] \Rightarrow y = \dfrac{258 - 90}{7} \\[1em] \Rightarrow y = \dfrac{168}{7} \\[1em] \Rightarrow y = 24.$

Total cost of 8 kg sugar and 10 kg rice = 8x + 10y

= 8 × 18 + 10 × 24

= 144 + 240 = ₹ 384.

Hence, the total cost of 8 kg sugar and 10 kg rice = ₹ 384.

Question 22

One year ago a man was four times as old as his son. After 6 years, his age exceeds twice his son’s age by 9 years. Find their present ages.

Answer

Let the present age of man be x years and his son be y years.

Given,

One year ago a man was four times as old as his son,

⇒ x - 1 = 4(y - 1)

⇒ x - 1 = 4y - 4

⇒ x = 4y - 4 + 1

⇒ x = 4y - 3 ....(1)

Given,

After 6 years, the man's age will exceed twice his son's age by 9,

⇒ x + 6 = 2(y + 6) + 9

⇒ x + 6 = 2y + 12 + 9

⇒ x = 2y + 12 + 9 - 6

⇒ x = 2y + 15 .....(2)

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

⇒ 4y - 3 = 2y + 15

⇒ 4y - 2y = 15 + 3

⇒ 2y = 18

⇒ y = $\dfrac{18}{2}$ = 9.

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

⇒ x = 4y - 3

⇒ x = 4(9) - 3

⇒ x = 36 - 3 = 33.

Hence, man's present age = 33 years and son's present age = 9 years.

Question 23

5 years ago, A was thrice as old as B and 10 years later A shall be twice as old as B. What are the present ages of A and B?

Answer

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

Given,

5 years ago, A was thrice as old as B.

⇒ x - 5 = 3(y - 5)

⇒ x - 5 = 3y - 15

⇒ x = 3y - 15 + 5

⇒ x = 3y - 10 ...,,,,.(1)

Given,

10 years later, A shall be twice as old as B.

⇒ x + 10 = 2(y + 10)

⇒ x + 10 = 2y + 20

⇒ x = 2y + 20 - 10

⇒ x = 2y + 10 .........(2)

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

⇒ 3y - 10 = 2y + 10

⇒ 3y - 2y = 10 + 10

⇒ y = 20.

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

⇒ x = 3y - 10

⇒ x = 3(20) - 10

⇒ x = 60 - 10 = 50.

Hence, A's present age = 50 years and B's present age = 20 years.

Question 24

The monthly incomes of A and B are in the ratio 7 : 5 and their expenditures are in the ratio 3 : 2. If each saves ₹ 1500 per month, find their monthly incomes.

Answer

Let monthly income of A be 7x and B be 5x.

Let monthly expenditure of A be 3y and B be 2y.

Given,

Both A and B save ₹ 1500 per month.

For A :

⇒ 7x - 3y = 1500

⇒ 7x = 1500 + 3y

⇒ x = $\dfrac{1500 + 3y}{7}$ .....(1)

For B :

⇒ 5x - 2y = 1500 ......(2)

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

$\Rightarrow 5x - 2y = 1500 \\[1em] \Rightarrow 5\Big(\dfrac{1500 + 3y}{7}\Big) - 2y = 1500 \\[1em] \Rightarrow \Big(\dfrac{7500 + 15y}{7}\Big) - 2y = 1500 \\[1em] \Rightarrow \Big(\dfrac{7500 + 15y - 14y}{7}\Big) = 1500 \\[1em] \Rightarrow 7500 + y = 10500 \\[1em] \Rightarrow y = 10500 - 7500 \\[1em] \Rightarrow y = 3000.$

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

$\Rightarrow x = \dfrac{1500 + 3y}{7} \\[1em] \Rightarrow x = \dfrac{1500 + 3 \times 3000}{7} \\[1em] \Rightarrow x = \dfrac{1500 + 9000}{7} \\[1em] \Rightarrow x = \dfrac{10500}{7} = 1500.$

A's income = ₹ 7x = ₹ 7 × 1500 = ₹ 10,500,

B's income = ₹ 5x = ₹ 5 × 1500 = ₹ 7,500.

Hence, A's income = ₹ 10,500, B's income = ₹ 7,500.

Question 25

A 90% acid solution is mixed with a 97% acid solution to obtain 21 litres of a 95% solution. Find the quantity of each the solutions to get the resultant mixture.

Answer

Let x litres be the quantity of the 90% acid solution and y litres be the quantity of the 97% acid solution.

Given,

Total volume = 21 litres

⇒ x + y = 21

⇒ x = 21 - y ....(1)

Als,

⇒ 90% of x + 97% of y = 95% of 21

⇒ 0.90x + 0.97y = 0.95 × 21

⇒ 0.90x + 0.97y = 19.95 ....(2)

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

⇒ 0.90(21 - y) + 0.97y = 19.95

⇒ 18.9 - 0.90y + 0.97y = 19.95

⇒ 18.9 + 0.07y = 19.95

⇒ 0.07y = 19.95 - 18.9

⇒ 0.07y = 1.05

⇒ y = $\dfrac{1.05}{0.07} = 15$ .

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

⇒ x = 21 - y

⇒ x = 21 - 15

⇒ x = 6.

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

Question 26

There are two examination halls A and B. If 12 pupils are sent from A to B, the number of pupils in each room becomes the same. If 11 pupils are sent from B to A, then the number of pupils in A is double their number in B. Find the number of pupils in each room.

Answer

Let x and y be the initial number of pupils in examination hall A and B respectively.

Given,

If 12 pupils are sent from A to B, the number of pupils in each room becomes the same.

⇒ x - 12 = y + 12

⇒ x = y + 12 + 12

⇒ x = 24 + y .....(1)

Given,

If 11 pupils are sent from B to A, then the number of pupils in A is double their number in B.

⇒ x + 11 = 2(y - 11)

⇒ x + 11 = 2y - 22

⇒ x - 2y = -22 - 11

⇒ x - 2y = -33 ........(2)

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

⇒ 24 + y - 2y = -33

⇒ 24 - y = -33

⇒ y = 24 + 33

⇒ y = 57.

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

⇒ x = 24 + y

⇒ x = 24 + 57

⇒ x = 81.

Hence, no. of pupils in examination hall A = 81 and examination hall B = 57.

Question 27

A and B each have a certain number of marbles. A says to B, “If you give 30 to me, I will have twice as many as left with you.” B replies, “If you give me 10, I will have thrice as many as left with you.” How many marbles does each have?

Answer

Let the number of marbles A has be x, and the number of marbles B has be y.

After B gives 30 marbles to A,

⇒ x + 30 = 2(y - 30)

⇒ x + 30 = 2y - 60

⇒ x = 2y - 60 - 30

⇒ x = 2y - 90 .......(1)

After A gives 10 marbles to B,

⇒ y + 10 = 3(x - 10)

⇒ y + 10 = 3x - 30

⇒ y = 3x - 30 - 10

⇒ y = 3x - 40 ........(2)

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

⇒ y = 3(2y - 90) - 40

⇒ y = 6y - 270 - 40

⇒ y = 6y - 310

⇒ y - 6y = -310

⇒ -5y = -310

⇒ y = $\dfrac{-310}{-5} = 62$ .

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

⇒ x = 2y - 90

⇒ x = 2(62) - 90

⇒ x = 124 - 90

⇒ x = 34.

Hence, A has 34 marbles and B has 62 marbles.

Question 28

The present age of a man is 3 years more than three times the age of his son. Three years hence, the man’s age will be 10 years more than twice the age of his son. Determine their present ages.

Answer

Let x be present age of the man and y be the present age of son.

Given,

The present age of a man is 3 years more than three times the age of his son,

⇒ x = 3y + 3 ........(1)

Given,

Three years hence, the man’s age will be 10 years more than twice the age of his son.

⇒ x + 3 = 2(y + 3) + 10

⇒ x + 3 = 2y + 6 + 10

⇒ x = 2y + 16 - 3

⇒ x = 2y + 13 ........(2)

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

⇒ 3y + 3 = 2y + 13

⇒ 3y - 2y = 13 - 3

⇒ y = 10.

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

⇒ x = 3y + 3

⇒ x = 3(10) + 3

⇒ x = 30 + 3

⇒ x = 33.

Hence, the present age of son = 10 years and that of man = 33 years.

Question 29

The length of a room exceeds its breadth by 3 meters. If the length is increased by 3 m and breadth is decreased by 2 meters, the area remains the same. Find the length and breadth of the room.

Answer

Let x meters be the length of a room and y meters be the breadth of the room.

Given,

The length of a room exceeds its breadth by 3 m,

⇒ x = y + 3 ....(1)

Given,

If length is increased by 3 and breadth decreased by 2, area remains the same,

⇒ (x + 3)(y - 2) = xy

⇒ (xy - 2x + 3y - 6) = xy

⇒ xy - xy = 2x - 3y + 6

⇒ 2x - 3y + 6 = 0 ....(2)

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

⇒ 2(y + 3) - 3y + 6 = 0

⇒ 2y + 6 - 3y + 6 = 0

⇒ -y + 12 = 0

⇒ y = 12.

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

⇒ x = y + 3

⇒ x = 12 + 3

⇒ x = 15.

Hence, length and breadth of the room are 15 m and 12 m respectively.

Question 30

The area of a rectangle gets reduced by 8 m², if its length is reduced by 5 m and breadth increased by 3 m. If we increase the length by 3 m and breadth by 2 m, the area is increased by 74 m². Find the length and breadth of the rectangle.

Answer

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

Given,

If the length is reduced by 5 m and breadth is increased by 3 m, then the area reduces by 8 m².

⇒ (x - 5)(y + 3) = xy - 8

⇒ (xy + 3x - 5y - 15) = xy - 8

⇒ 3x - 5y - 15 + 8 = xy - xy

⇒ 3x - 5y - 7 = 0

⇒ 3x = 5y + 7

⇒ x = $\dfrac{5y + 7}{3}$ ........(1)

Given,

If the length is increased by 3 m and breadth by 2 m, then the area increases by 74 m².

⇒ (x + 3)(y + 2) = xy + 74

⇒ (xy + 2x + 3y + 6) = xy + 74

⇒ 2x + 3y + 6 - 74 = xy - xy

⇒ 2x + 3y - 68 = 0 .......(2)

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

$\Rightarrow 2 \Big(\dfrac{5y + 7}{3}\Big) + 3y - 68 = 0 \\[1em] \Rightarrow \Big(\dfrac{10y + 14}{3}\Big) + 3y = 68 \\[1em] \Rightarrow \Big(\dfrac{10y + 14 + 9y}{3}\Big) = 68 \\[1em] \Rightarrow 19y + 14 = 68 \times 3 \\[1em] \Rightarrow 19y + 14 = 204 \\[1em] \Rightarrow 19y = 204 - 14 \\[1em] \Rightarrow 19y = 190 \\[1em] \Rightarrow y = \dfrac{190}{19} = 10.$

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

$\Rightarrow x = \dfrac{5y + 7}{3} \\[1em] \Rightarrow x = \dfrac{5 \times 10 + 7}{3} \\[1em] \Rightarrow x = \dfrac{57}{3} = 19.$

Hence, the length and breadth of rectangles are 19 m and 10 m respectively.

Question 31

A motorboat takes 6 hours to cover 100 km downstream and 30 km upstream. If the motorboat goes 75 km downstream and returns back to its starting point in 8 hours, find the speed of the motorboat in still water and the rate of the stream.

Answer

Let x km/hr be the speed of motorboat in still water and y km/hr be the speed of stream.

Downstream speed = x + y km/h

Upstream speed = x - y km/h

Given,

Time $= \dfrac{\text{Distance}}{\text{Speed}}$

Motorboat takes 6 hours to cover 100 km downstream and 30 km upstream.

$\Rightarrow \dfrac{100}{x + y} + \dfrac{30}{x - y} = 6$ .........(1)

Given,

Motorboat goes 75 km downstream and returns back to its starting point in 8 hours.

$\Rightarrow \dfrac{75}{x + y} + \dfrac{75}{x - y} = 8$ ....(2)

Substituting, u = $\dfrac{1}{x + y}$ , v = $\dfrac{1}{x - y}$ in equation (1), we get :

⇒ 100u + 30v = 6

⇒ 10(10u + 3v) = 6

⇒ (10u + 3v) = $\dfrac{6}{10}$ ........(3)

Substituting, u = $\dfrac{1}{x + y}$ , v = $\dfrac{1}{x - y}$ in equation (2), we get :

⇒ 75u + 75v = 8

⇒ 75(u + v) = 8

⇒ u + v = $\dfrac{8}{75}$

⇒ u = $\dfrac{8}{75} - v$ ........(4)

Substituting value of u from equation (4) in (3), we get :

$\Rightarrow 10 \Big(\dfrac{8}{75} - v\Big) + 3v = \dfrac{6}{10} \\[1em] \Rightarrow \dfrac{80}{75} - 10v + 3v = \dfrac{6}{10} \\[1em] \Rightarrow \dfrac{80}{75} - 7v = \dfrac{6}{10} \\[1em] \Rightarrow 7v = \dfrac{80}{75} - \dfrac{6}{10} \\[1em] \Rightarrow 7v = \dfrac{160 - 90}{150} \\[1em] \Rightarrow 7v = \dfrac{70}{150} \\[1em] \Rightarrow v = \dfrac{7}{15 \times 7} \\[1em] \Rightarrow v = \dfrac{1}{15}.$

Substituting value of v in equation (4), we get :

$\Rightarrow u = \dfrac{8}{75} - v \\[1em] \Rightarrow u = \dfrac{8}{75} - \dfrac{1}{15} \\[1em] \Rightarrow u = \dfrac{8 - 5}{75} \\[1em] \Rightarrow u = \dfrac{3}{75} \\[1em] \Rightarrow u = \dfrac{1}{25}.$

Since,

$\Rightarrow u = \dfrac{1}{x + y} \\[1em] \Rightarrow \dfrac{1}{25} = \dfrac{1}{x + y}$

⇒ x + y = 25 .......(5)

$\Rightarrow v = \dfrac{1}{x - y} \\[1em] \Rightarrow \dfrac{1}{15} = \dfrac{1}{x - y}$

⇒ x - y = 15 ........(6)

Adding equations (5) and (6),

⇒ x + y + x - y = 25 + 15

⇒ 2x = 40

⇒ x = $\dfrac{40}{2} = 20$ .

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

⇒ x - y = 15

⇒ 20 - y = 15

⇒ 20 - 15 = y

⇒ y = 5.

Hence, the speed of the motorboat in still water = 20 km/hr and the speed of the stream = 5 km/hr.

Question 32

A man sold a chair and a table for ₹ 2,178, thereby making a profit of 12% on the chair and 16% on the table. By selling them for ₹ 2,154, he gains 16% on the chair and 12% on the table. Find the cost price of each.

Answer

Let x be the cost of chair and y be the cost of table.

Given,

When sold for ₹ 2,178, he makes profit of 12% on the chair and 16% on the table.

$\therefore x + \dfrac{12}{100}x + y + \dfrac{16}{100}y = 2178 \\[1em] \Rightarrow \dfrac{100x + 12x}{100} + \dfrac{100y + 16y}{100} = 2178 \\[1em] \Rightarrow \dfrac{112x}{100} + \dfrac{116y}{100} = 2178 \\[1em] \Rightarrow 112x + 116y = 2178 \times 100 \\[1em] \Rightarrow 112x + 116y = 217800\\[1em] \Rightarrow 4(28x + 29y) = 217800 \\[1em] \Rightarrow 28x + 29y = \dfrac{217800}{4} \\[1em] \Rightarrow 28x + 29y = 54450 \text{ ....(1) }$

Given,

When sold for ₹ 2,154, he makes profit of 16% on the chair and 12% on the table.

$\therefore x + \dfrac{16}{100}x + y + \dfrac{12}{100}y = 2154 \\[1em] \Rightarrow \dfrac{100x + 16x}{100} + \dfrac{100y + 12}{100} = 2154 \\[1em] \Rightarrow \dfrac{116x}{100} + \dfrac{112y}{100} = 2154 \\[1em] \Rightarrow 116x + 112y = 2154 \times 100 \\[1em] \Rightarrow 116x + 112y = 215400 \\[1em] \Rightarrow 4(29x + 28y) = 215400 \\[1em] \Rightarrow 29x + 28y = \dfrac{215400}{4} \\[1em] \Rightarrow 29x + 28y = 53850 \text{ ........(2) }$

Subtracting equation eqn 1 from 2, we get :

⇒ 29x + 28y - (28x + 29y) = 53850 - 54450

⇒ 29x + 28y - 28x - 29y = -600

⇒ x - y = -600

⇒ x = y - 600 ....(3)

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

⇒ 28x + 29y = 54450

⇒ 28(y - 600) + 29y = 54450

⇒ 28y - 16800 + 29y = 54450

⇒ 57y = 54450 + 16800

⇒ 57y = 71250

⇒ y = $\dfrac{71250}{57}$

⇒ y = ₹ 1,250

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

⇒ 29x + 28 × 1250 = 53850

⇒ 29x + 35000 = 53850

⇒ 29x = 53850 - 35000

⇒ 29x = 18850

⇒ x = $\dfrac{18850}{29}$

⇒ x = ₹ 650

Hence, cost price of table = ₹ 1,250 and cost price of chair = ₹ 650.

Question 33

A man travels 600 km partly by train and partly by car. If he covers 120 km by train and the rest by car, it takes him 8 hours. But if he travels 200 km by train and the rest by car, he takes 20 minutes longer. Find the speed of the car and that of the train.

Answer

Let x km/hr be the speed of train and y km/hr be the speed of car.

$Time = \dfrac{\text{Distance}}{\text{Speed}}$

Given,

120 km by train, rest (600 - 120 = 480 km) by car takes 8 hours.

$\Rightarrow \dfrac{120}{x} + \dfrac{480}{y} = 8$ ........(1)

Given,

200 km by train, rest (600 - 200 = 400 km) by car takes 8 hours 20 mins = $8 + \dfrac{20}{60} = 8 + \dfrac{1}{3} = \dfrac{25}{3}$ hours.

$\Rightarrow \dfrac{200}{x} + \dfrac{400}{y} = \dfrac{25}{3}$ ..........(2)

Substituting, u = $\dfrac{1}{x}$ , v = $\dfrac{1}{y}$ in equation (1), we get :

⇒ 120u + 480v = 8

⇒ 120u + 480v - 8 = 0

⇒ 8(15u + 60v - 1) = 0

⇒ 15u + 60v - 1 = 0

⇒ 15u = 1 - 60v

⇒ u = $\dfrac{1 - 60v}{15}$ ..........(3)

Substituting, u = $\dfrac{1}{x}$ , v = $\dfrac{1}{y}$ in equation (2), we get :

⇒ 200u + 400v = $\dfrac{25}{3}$ .........(4)

Substituting value of u from equation (3) in equation (4), we get :

$\Rightarrow 200 \Big(\dfrac{1 - 60v}{15}\Big) + 400v = \dfrac{25}{3} \\[1em] \Rightarrow \Big(\dfrac{200 - 12000v}{15}\Big) + 400v = \dfrac{25}{3} \\[1em] \Rightarrow \Big(\dfrac{200 - 12000v + 6000v}{15}\Big) = \dfrac{25}{3} \\[1em] \Rightarrow 200 - 6000v = \dfrac{25}{3} \times 15 \\[1em] \Rightarrow 200 - 6000v = 25 \times 5 \\[1em] \Rightarrow 200 - 6000v = 125 \\[1em] \Rightarrow 6000v = 200 - 125 \\[1em] \Rightarrow 6000v = 75 \\[1em] \Rightarrow v = \dfrac{75}{6000} \\[1em] \Rightarrow v = \dfrac{1}{80}.$

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

$\Rightarrow u = \dfrac{1 - 60 \times \dfrac{1}{80}}{15} \\[1em] \Rightarrow u = \dfrac{1 - \dfrac{3}{4}}{15} \\[1em] \Rightarrow u = \dfrac{\dfrac{1}{4}}{15} \\[1em] \Rightarrow u = \dfrac{1}{60}.$

Substituting,

⇒ u = $\dfrac{1}{x}$

⇒ $\dfrac{1}{60} = \dfrac{1}{x}$

⇒ x = 60.

⇒ v = $\dfrac{1}{y}$

⇒ $\dfrac{1}{80} = \dfrac{1}{y}$

⇒ y = 80.

Hence, speed of train = 60 km/h, speed of car = 80 km/h.

Question 34

6 men and 8 boys can finish a piece of work in 14 days while 8 men and 12 boys can do it in 10 days. Find the time taken by one man alone and by one boy alone to finish the work.

Answer

Lets assume that one man takes x days to do work and y days for one boy.

So, the amount of work done by 1 man in 1 day = $\dfrac{1}{x}$

So, the amount of work done by 1 boy in 1 day = $\dfrac{1}{y}$

Given,

6 men and 8 boys finish the work in 14 days,

$\Rightarrow \dfrac{6}{x} + \dfrac{8}{y} = \dfrac{1}{14} \text{ ....(1) }$

Given,

8 men and 12 boys finish the same work in 10 days,

$\Rightarrow \dfrac{8}{x} + \dfrac{12}{y} = \dfrac{1}{10}$ ....(2)

Multiply equation (1) by 2, we get:

$\Rightarrow 2 \Big(\dfrac{6}{x} + \dfrac{8}{y}\Big) = \dfrac{1}{14} \times 2 \\[1em] \Rightarrow \dfrac{12}{x} + \dfrac{16}{y} = \dfrac{1}{7} \text{ ....(3) }$

Multiply equation (2) by $\dfrac{3}{2}$ , we get:

$\Rightarrow \dfrac{3}{2} \Big(\dfrac{8}{x} + \dfrac{12}{y}\Big) = \dfrac{1}{10} \times \dfrac{3}{2} \\[1em] \Rightarrow \dfrac{12}{x} + \dfrac{18}{y}= \dfrac{3}{20} \text{ ....(4) }$

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

$\Rightarrow \Big(\dfrac{12}{x} + \dfrac{18}{y}\Big) - \Big(\dfrac{12}{x} + \dfrac{16}{y}\Big) = \dfrac{3}{20} - \dfrac{1}{7} \\[1em] \Rightarrow \Big(\dfrac{12}{x} + \dfrac{18}{y} - \dfrac{12}{x} - \dfrac{16}{y}\Big) = \dfrac{21}{140} - \dfrac{20}{140} \\[1em] \Rightarrow \dfrac{18}{y} - \dfrac{16}{y} = \dfrac{1}{140} \\[1em] \Rightarrow \dfrac{2}{y} = \dfrac{1}{140} \\[1em] \Rightarrow y = 2 \times 140 = 280.$

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

$\Rightarrow \dfrac{6}{x} + \dfrac{8}{y} = \dfrac{1}{14} \\[1em] \Rightarrow \dfrac{6}{x} + \dfrac{8}{280} = \dfrac{1}{14} \\[1em] \Rightarrow \dfrac{6}{x} = \dfrac{1}{14} - \dfrac{8}{280} \\[1em] \Rightarrow \dfrac{6}{x} = \dfrac{20}{280} - \dfrac{8}{280} \\[1em] \Rightarrow \dfrac{6}{x} = \dfrac{12}{280} \\[1em] \Rightarrow \dfrac{6 \times 280}{12} = x \\[1em] \Rightarrow x = 140.$

Hence, one man can finish the work in = 140 days and one boy finish the work in = 280 days .

Question 35

A lady has 25-P and 50-P coins in her purse. If in all she has 80 coins totalling ₹ 25, how many coins of each kind does she have ?

Answer

Let Number of 25-P coins be x and number of 50-P coins be y.

Given,

Total number of coins = 80,

⇒ x + y = 80

⇒ x = 80 - y .........(1)

Given,

Total value = ₹ 25 = 2500 paise,

⇒ 25x + 50y = 2500 .......(2)

Substituting value of x from equation (1) in 25x + 50y = 2500, we get :

⇒ 25x + 50y = 2500

⇒ 25(80 - y) + 50y = 2500

⇒ 2000 - 25y + 50y = 2500

⇒ -25y + 50y = 2500 - 2000

⇒ 25y = 500

⇒ y = $\dfrac{500}{25}$

⇒ y = 20.

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

⇒ x = 80 - y

⇒ x = 80 - 20

⇒ x = 60.

Hence, number of 25-P coins = 60 and number of 50-P coins = 20.

Question 36

A and B together can do a piece of work in 6 days. If A’s one day’s work is $1\dfrac{1}{2}$ times the one day's work of B, find how many days, each alone can finish the work.

Answer

Let A's one day work be x and B's one day work be y.

According to first condition given in the problem,

A works $1\dfrac{1}{2}$ times of B,

$\Rightarrow x = 1\dfrac{1}{2}y$

$\Rightarrow x = \dfrac{3}{2}y$

⇒ 2x = 3y

⇒ 2x - 3y = 0 ......(1)

Also given, A and B together can do a piece of work in 6 days.

$\therefore x + y = \dfrac{1}{6}$

⇒ 6(x + y) = 1

⇒ 6x + 6y = 1 ...(2)

Multiplying (1) by 2 we get,

⇒ 2(2x - 3y) = 2 × 0

⇒ 4x - 6y = 0 ...(3)

Adding equations (2) and (3) we get,

⇒ 6x + 6y + 4x - 6y = 1 + 0

⇒ 10x = 1

⇒ x = $\dfrac{1}{10}$

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

⇒ 2 × $\dfrac{1}{10}$ - 3y = 0

⇒ $\dfrac{1}{5}$ - 3y = 0

⇒ 3y = $\dfrac{1}{5}$

⇒ y = $\dfrac{1}{15}$

Since, A's one day work is x and B's one day work is y, so A can do complete work in $\dfrac{1}{x}$ and B can do work in $\dfrac{1}{y}$ days.

$\dfrac{1}{x} = \dfrac{1}{\dfrac{1}{10}}$ = 10 days

$\dfrac{1}{y} = \dfrac{1}{\dfrac{1}{15}}$ = 15 days

Hence, A can finish the work in 10 days while B can finish the work in 15 days.

Multiple Choice Questions

Question 1

The solution of the simultaneous equations 3x - 2y = 5 and x + 2y = -1 is :

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

Answer

Given,

Equations: 3x - 2y = 5, x + 2y = -1

⇒ 3x - 2y = 5

⇒ 3x - 5 = 2y

⇒ y = $\dfrac{3x - 5}{2}$ ....(1)

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

⇒ x + $2\Big(\dfrac{3x - 5}{2}\Big)$ = -1

⇒ x + 3x - 5 = -1

⇒ 4x = -1 + 5

⇒ 4x = 4

⇒ x = $\dfrac{4}{4}$

⇒ x = 1.

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

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

Hence, option 2 is the correct option.

Question 2

The solution of the simultaneous equations $\dfrac{x}{2} - \dfrac{y}{3} = 0$ and $\dfrac{3x}{2} + \dfrac{2y}{3} + 10 = 0$ is :

x = 4, y = 6
x = 4, y = -6
x = -4, y = 6
x = -4, y = -6

Answer

Given,

Equations: $\dfrac{x}{2} - \dfrac{y}{3} = 0, \dfrac{3x}{2} + \dfrac{2y}{3} + 10 = 0$

Solving equation $\dfrac{x}{2} - \dfrac{y}{3} = 0$ ,

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

$\dfrac{3x}{2} + \dfrac{2y}{3} + 10 = 0 \text{ ....(2)}$

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

$\Rightarrow \dfrac{3\Big(\dfrac{2y}{3}\Big)}{2} + \dfrac{2y}{3} + 10 = 0 \\[1em] \Rightarrow \dfrac{2y}{2} + \dfrac{2y}{3} + 10 = 0 \\[1em] \Rightarrow \dfrac{6y + 4y + 60}{6} = 0 \\[1em] \Rightarrow 10y + 60 = 0 \\[1em] \Rightarrow 10y = -60 \\[1em] \Rightarrow y = \dfrac{-60}{10} = -6.$

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

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

Hence, option 4 is the correct option.

Question 3

The solution of the simultaneous equations $2x + \dfrac{1}{y} = 0$ and $3x + \dfrac{1}{2y} = -2$ is :

x = 1, y = $\dfrac{1}{2}$
x = 1, y = $-\dfrac{1}{2}$
x = -1, y = $\dfrac{1}{2}$
x = -1, y = $-\dfrac12$

Answer

Given,

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

Solving equation 1,

⇒ $2x + \dfrac{1}{y} = 0$

⇒ 2x = $-\dfrac{1}{y}$

⇒ x = $\dfrac{-1}{2y}$ .......(1)

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

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

Substituting y = $\dfrac{1}{2}$ in equation (1), we get :

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

Hence, option 3 is the correct option.

Question 4

If $\dfrac{x}{6} + 6 = y, \dfrac{3x}{4} = 1 + y$ , then :

x = 12, y = 8
x = 10, y = 8
x = 8, y = 12
x = 12, y = 6

Answer

Given,

Equations: $\dfrac{x}{6} + 6 = y, \dfrac{3x}{4} = 1 + y$

Solving first equation,

⇒ $\dfrac{x}{6} + 6 = y$

⇒ $\dfrac{x + 36}{6} = y$

⇒ x + 36 = 6y

⇒ x = 6y - 36 .......(1)

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

$\Rightarrow \dfrac{3(6y - 36)}{4} = 1 + y \\[1em] \Rightarrow 18y - 108 = 4(1 + y) \\[1em] \Rightarrow 18y - 108 = 4 + 4y \\[1em] \Rightarrow 18y - 4y = 4 + 108 \\[1em] \Rightarrow 14y = 112 \\[1em] \Rightarrow y = \dfrac{112}{14} = 8.$

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

⇒ x = 6y - 36

⇒ x = 6(8) - 36

⇒ x = 48 - 36 = 12.

Hence, option 1 is the correct option.

Question 5

If x and y are real numbers and (2x - 1) ² + (3y - 1) ² = 0, then $\Big(\dfrac{1}{x^2} + \dfrac{1}{y^2}\Big) =$

25
13
$\dfrac{1}{13}$
$-\dfrac{1}{13}$

Answer

Given,

⇒ (2x - 1)² + (3y - 1)² = 0

⇒ (2x - 1)² = 0 and (3y - 1)² = 0

⇒ (2x - 1) = 0 and 3y - 1 = 0

⇒ 2x = 1 and 3y = 1

⇒ x = $\dfrac{1}{2}$ and y = $\dfrac{1}{3}$ .

Substituting value of x and y in $\Big(\dfrac{1}{x^2} + \dfrac{1}{y^2}\Big)$ , we get :

$\Rightarrow \Big(\dfrac{1}{x^2} + \dfrac{1}{y^2}\Big) \\[1em] \Rightarrow \dfrac{1}{\Big(\dfrac{1}{2}\Big)^2} + \dfrac{1}{\Big(\dfrac{1}{3}\Big)^2} \\[1em] \Rightarrow \dfrac{1}{\dfrac{1}{4}} + \dfrac{1}{\dfrac{1}{9}} \\[1em] \Rightarrow 4 + 9 \\[1em] \Rightarrow 13.$

Hence, option 2 is the correct option.

Question 6

The solution of $\sqrt{5}x - \sqrt{7}y = 0$ and $\sqrt{3}y + \sqrt{13}x = 0$ is :

x = 0, y = 0
x = 0, y = 1
x = 1, y = 0
x = 1, y = -1

Answer

Given,

Equations: $\sqrt{5}x - \sqrt{7}y = 0, \sqrt{3}y + \sqrt{13}x = 0$

Solving first equation,

$\Rightarrow \sqrt{5}x - \sqrt{7}y = 0 \\[1em] \Rightarrow \sqrt{5}x = \sqrt{7}y \\[1em] \Rightarrow x = \dfrac{\sqrt{7}}{\sqrt{5}}y \text{ ....(1)}$

Substituting value of x from equation (1) in $\sqrt{3}y + \sqrt{13}x = 0$ , we get :

$\Rightarrow \sqrt{3}y + \sqrt{13}x = 0 \\[1em] \Rightarrow \sqrt{3}y + \sqrt{13} \times \dfrac{\sqrt{7}}{\sqrt{5}}y = 0 \\[1em] \Rightarrow y \Big(\sqrt{3} + \dfrac{\sqrt{91}}{\sqrt{5}}\Big) = 0.$

Since, $\Big(\sqrt{3} + \dfrac{\sqrt{91}}{\sqrt{5}}\Big)$ is not equal to zero thus, y = 0.

Substituting value of y = 0 in equation (1), we get :

$\Rightarrow x = \dfrac{\sqrt{7}}{\sqrt{5}}y = \dfrac{\sqrt{7}}{\sqrt{5}} \times 0$ = 0.

Hence, option 1 is the correct option.

Question 7

The solution of 0.4x + 3y = 1.2 and 7x - 2y = $\dfrac{17}{6}$ is :

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

Answer

Given,

Equations: 0.4x + 3y = 1.2 and 7x - 2y = $\dfrac{17}{6}$

Solving first equation,

⇒ 0.4x + 3y = 1.2

Multiplying both sides of the equation by 10,

⇒ 10(0.4x + 3y) = 10 × 1.2

⇒ 4x + 30y = 12

⇒ 4x = 12 - 30y

⇒ x = $\dfrac{12 - 30y}{4}$ .......(1)

⇒ 7x - 2y = $\dfrac{17}{6}$ .......(2)

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

$\Rightarrow 7 \Big(\dfrac{12 - 30y}{4}\Big) - 2y = \dfrac{17}{6} \\[1em] \Rightarrow \dfrac{84 - 210y}{4} - 2y = \dfrac{17}{6} \\[1em] \Rightarrow \dfrac{84 - 210y - 8y}{4} = \dfrac{17}{6} \\[1em] \Rightarrow 84 - 218y = 4 \times \dfrac{17}{6} \\[1em] \Rightarrow 84 - 218y = 2 \times \dfrac{17}{3} \\[1em] \Rightarrow 3(84 - 218y) = 17 \times 2 \\[1em] \Rightarrow 252 - 654y = 34 \\[1em] \Rightarrow 654y = 252 - 34 \\[1em] \Rightarrow 654y = 218 \\[1em] \Rightarrow y = \dfrac{218}{654} \\[1em] \Rightarrow y = \dfrac{1}{3}.$

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

$\Rightarrow x = \dfrac{12 - 30y}{4} \\[1em] \Rightarrow x = \dfrac{12 - 30 \times \dfrac{1}{3}}{4} \\[1em] \Rightarrow x = \dfrac{12 - 10}{4} \\[1em] \Rightarrow x = \dfrac{2}{4} = \dfrac{1}{2}.$

Hence, option 2 is the correct option.

Question 8

2 tables and 3 chairs together cost ₹ 1,075 and 3 tables and 8 chairs together cost ₹ 1,875. The cost of 4 tables and 5 chairs together will be :

₹ 2,750
₹ 2,705
₹ 2,075
₹ 2,057

Answer

Let ₹ x be the cost of table and ₹ y be cost of the chair.

Given,

2 tables and 3 chairs together cost ₹ 1,075.

⇒ 2x + 3y = 1075 .......(1)

3 tables and 8 chairs together cost ₹ 1,875.

⇒ 3x + 8y = 1875 .......(2)

Multiplying equation (1) by 3, we get :

⇒ 3(2x + 3y) = 1075 × 3

⇒ 6x + 9y = 3225 ......(3)

Multiplying equation (2) by 2,

⇒ 2(3x + 8y) = 1875 × 2

⇒ 6x + 16y = 3750 ....(4)

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

⇒ (6x + 16y) - (6x + 9y) = 3750 - 3225

⇒ 6x + 16y - 6x - 9y = 525

⇒ 7y = 525

⇒ y = $\dfrac{525}{7}$ = ₹ 75.

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

⇒ 2x + 3(75) = 1075

⇒ 2x + 225 = 1075

⇒ 2x = 1075 - 225

⇒ 2x = 850

⇒ x = $\dfrac{850}{2}$ = ₹ 425.

The cost of 4 tables and 5 chairs,

⇒ 4x + 5y = 4 × 425 + 5 × 75

= 1700 + 375 = ₹ 2,075.

Hence, option 3 is the correct option.

Question 9

The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Then the original number is :

Answer

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

Number = 10x + y,

Given,

Sum of the digits of a two-digit = 9

⇒ x + y = 9 ....(1)

Given,

Nine times the number is twice the number obtained by reversing the order of the digits,

⇒ 9(10x + y) = 2(10y + x)

⇒ 90x + 9y = 20y + 2x

⇒ 90x - 2x = 20y - 9y

⇒ 88x = 11y

⇒ y = $\dfrac{88}{11}x$

⇒ y = 8x ....(2)

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

⇒ x + 8x = 9

⇒ 9x = 9

⇒ x = $\dfrac{9}{9}$ = 1.

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

⇒ y = 8x

⇒ y = 8(1) = 8.

Number = 10x + y = 10(1) + 8 = 18.

Hence, option 2 is the correct option.

Question 10

Five years ago, Bharat was thrice as old as Rajat. Ten years later, Bharat will be twice as old as Rajat. The difference between their present ages is :

10 years
20 years
30 years
35 years

Answer

Let x be Bharat's present age and y be Rajat's present age,

Given,

Five years ago, Bharat was thrice as old as Rajat.

⇒ x - 5 = 3(y - 5)

⇒ x - 5 = 3y - 15

⇒ x = 3y - 15 + 5

⇒ x = 3y - 10 ....(1)

Given,

Ten years later, Bharat will be twice as old as Rajat,

⇒ x + 10 = 2(y + 10)

⇒ x + 10 = 2y + 20

⇒ x = 2y + 20 - 10

⇒ x = 2y + 10 ....(2)

Substituting value of x from equation (1) in x = 2y + 10, we get :

⇒ 3y - 10 = 2y + 10

⇒ 3y - 2y = 10 + 10

⇒ y = 20 years.

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

⇒ x = 3y - 10

⇒ x = 3(20) - 10

⇒ x = 60 - 10

⇒ x = 50 years.

The difference between their present ages = x - y = 50 - 20 = 30 years.

Hence, option 3 is the correct option.

Question 11

A bag contains some one-rupee coins and some fifty-paisa coins. The total amount is ₹ 140. If half of the one-rupee coins are replaced by fifty-paisa coins, then the amount becomes ₹ 115. The coins of each type in the bag initially, were :

one-rupee coins = 100 and fifty-paisa coins = 80
one-rupee coins = 80 and fifty-paisa coins = 100
one-rupee coins = 110 and fifty-paisa coins = 80
one-rupee coins = 70 and fifty-paisa coins = 90

Answer

Let x be the number of one rupee coins and y be the number of 50 paisa coins in the bag initially.

Given,

Initial total amount = ₹ 140.

⇒ x + 0.5y = 140

⇒ x = 140 - 0.5y .......(1)

Given,

After replacing half of the 1-rupee coins with 50-paisa coins, the amount becomes ₹ 115.

⇒ $\dfrac{x}{2} \times 1$ + 0.5y + 0.5 $\Big(\dfrac{x}{2}\Big)$ = 115

⇒ 0.5x + 0.5y + 0.25x = 115

⇒ 0.75x + 0.5y = 115 ........(2)

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

⇒ 0.75(140 - 0.5y) + 0.5y = 115

⇒ 105 - 0.375y + 0.5y = 115

⇒ 0.125y = 115 - 105

⇒ 0.125y = 10

⇒ y = $\dfrac{10}{0.125}$ = 80.

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

⇒ x = 140 - 0.5y

⇒ x = 140 - 0.5(80)

⇒ x = 140 - 40

⇒ x = 100.

∴ Number of one rupee coins = 100 and number of fifty paisa coins = 80.

Hence, option 1 is the correct option.

Question 12

X takes 3 hours more than Y to walk a distance of 30 km, but if X doubles his race, he is able to be ahead of Y by $1\dfrac{1}{2}$ hours, then the speed of their walking will be :

X’s speed = $\dfrac{10}{3}$ km/hr, Y’s speed = 5 km/hr
X’s speed = 5 km/hr, Y’s speed = $\dfrac{10}{3}$ km/hr
X’s speed = 10 km/hr, Y’s speed = $\dfrac{5}{3}$ km/hr
X’s speed = $\dfrac{5}{3}$ km/hr, Y’s speed = 10 km/hr

Answer

Let X's speed and Y's speed be x km/hr and y km/hr respectively.

Time = $\dfrac{\text{Distance}}{\text{Speed}}$

Given,

X takes 3 hours more than Y to walk 30 km.

⇒ $\dfrac{30}{x} = \dfrac{30}{y} + 3$

⇒ $\dfrac{30}{y} = \dfrac{30}{x} - 3$ .........(1)

Given,

If X doubles his race, he is able to be ahead of Y by $1\dfrac{1}{2}$ hours.

⇒ $\dfrac{30}{2x} = \dfrac{30}{y} - \dfrac{3}{2}$

⇒ $\dfrac{30}{y} = \dfrac{30}{2x} + \dfrac{3}{2}$ .........(2)

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

$\Rightarrow \dfrac{30}{x} - 3 = \dfrac{30}{2x} + \dfrac{3}{2} \\[1em] \Rightarrow \dfrac{30}{x} - \dfrac{30}{2x} = \dfrac{3}{2} + 3 \\[1em] \Rightarrow \dfrac{60 - 30}{2x} = \dfrac{3 + 6}{2} \\[1em] \Rightarrow \dfrac{30}{2x} = \dfrac{9}{2} \\[1em] \Rightarrow x = \dfrac{30}{9} = \dfrac{10}{3} \text{ km/hr}.$

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

$\Rightarrow \dfrac{30}{y} = \dfrac{30}{\dfrac{10}{3}} - 3 \\[1em] \Rightarrow \dfrac{30}{y} = \dfrac{90}{10} - 3 \\[1em] \Rightarrow \dfrac{30}{y} = 9 - 3 \\[1em] \Rightarrow \dfrac{30}{y} = 6 \\[1em] \Rightarrow y = \dfrac{30}{6} = 5 \text{ km/hr}.$

Hence, option 1 is the correct option.

Question 13

A boat takes 10 hours to go 44 km downstream and 30 km upstream. Again, the same boat takes 13 hours to go 55 km downstream and 40 km upstream. The speed of the boat and the current will be :

speed of boat = 3 kmph, speed of current = 2 kmph
speed of boat = 6 kmph, speed of current = 4 kmph
speed of boat = 9 kmph, speed of current = 2 kmph
speed of boat = 8 kmph, speed of current = 3 kmph

Answer

Let x be the speed of the boat in still water and y be the speed of current,

Downstream speed = (x + y) km/hr

Upstream speed = (x - y) km/hr

Time = $\dfrac{\text{Distance}}{\text{Speed}}$

Given,

It takes 10 hours to go 44 km downstream and 30 km upstream.

⇒ $\dfrac{44}{x + y} + \dfrac{30}{x - y} = 10$ .........(1)

Given,

It takes 13 hours to go 55 km downstream and 40 km upstream.

⇒ $\dfrac{55}{x + y} + \dfrac{40}{x - y} = 13$ ........(2)

Substituting $\dfrac{1}{x + y} = u, \dfrac{1}{x - y} = v$ , in equation (1),

⇒ 44u + 30v = 10 ....(3)

Substituting $\dfrac{1}{x + y} = u, \dfrac{1}{x - y} = v$ , in equation (2),

⇒ 55u + 40v = 13 ....(4)

Multiply equation (3) by 4, we get :

⇒ 4(44u + 30v = 10)

⇒ 176u + 120v = 40 ....(5)

Multiply equation (4) by 3, we get :

⇒ 3(55u + 40v = 13)

⇒ 165u + 120v = 39 ....(6)

Subtracting equation (5) from equation (6), we get :

⇒ (165u + 120v) - (176u + 120v) = 39 - 40

⇒ (165u + 120v - 176u - 120v) = -1

⇒ -11u = -1

⇒ u = $\dfrac{-1}{-11} = \dfrac{1}{11}$ .

Substituting value of u in equation (3), we get :

⇒ 44u + 30v = 10

⇒ $44 \times \dfrac{1}{11}$ + 30v = 10

⇒ 4 + 30v = 10

⇒ 30v = 10 - 4

⇒ 30v = 6

⇒ v = $\dfrac{6}{30} = \dfrac{1}{5}$ .

$\Rightarrow \dfrac{1}{x + y} = u \\[1em] \Rightarrow \dfrac{1}{x + y} = \dfrac{1}{11} \\[1em] \Rightarrow x + y = 11 \text{ ........(7) } \\[1em] \Rightarrow \dfrac{1}{x - y} = v \\[1em] \Rightarrow \dfrac{1}{x - y} = \dfrac{1}{5} \\[1em] \Rightarrow x - y = 5 \text{ .........(8) }$

Adding equations (7) and (8) we get,

⇒ x + y + x - y = 11 + 5

⇒ 2x = 16

⇒ x = $\dfrac{16}{2} = 8$ .

Substituting value of x in equation (8),

⇒ x - y = 5

⇒ 8 - y = 5

⇒ 8 - 5 = y

⇒ y = 3.

The speed of the boat in still water is 8 km/hr and the speed of the current is 3 km/hr.

Hence, option 4 is the correct option.

Question 14

42 mangoes are to be distributed among some boys and girls. If each boy is given 3 mangoes, then each girl gets 6 mangoes; and if each boy gets 5 mangoes, then each girl gets 3 mangoes. The number of boys and girls will be :

boys = 4, girls = 6
boys = 6, girls = 4
boys = 7, girls = 3
boys = 3, girls = 7

Answer

Let x be the number of boys and y be the number of girls,

Given,

Case 1:

If each boy is given 3 mangoes, then each girl gets 6 mangoes.

⇒ 3x + 6y = 42 .......(1)

Case 2:

If each boy is given 5 mangoes, then each girl gets 3 mangoes.

⇒ 5x + 3y = 42 ....(2)

Multiply equation by 2 we get,

⇒ 2(5x + 3y = 42)

⇒ 10x + 6y = 84 ......(3)

Subtracting equation (1) from (3), we get:

⇒ 10x + 6y - (3x + 6y) = 84 - 42

⇒ 10x + 6y - 3x - 6y = 84 - 42

⇒ 7x = 42

⇒ x = $\dfrac{42}{7}$ = 6.

Substituting value of x in equation 1, we get :

⇒ 3x + 6y = 42

⇒ 3(6) + 6y = 42

⇒ 18 + 6y = 42

⇒ 6y = 42 - 18

⇒ 6y = 24

⇒ y = $\dfrac{24}{6}$ = 4.

Thus, no. of boys = 6, no. of girls = 4.

Hence, option 2 is the correct option.

Question 15

If ∠A = 2x°, ∠B = (6y + 10)°, ∠C = (2x + y)° and ∠D = (x + 10)° are the angles of a quadrilateral, then the values of x and y will be :

x = 40°, y = 20°
x = 20°, y = 40°
x = 45°, y = 15°
x = 15°, y = 45°

Answer

We know that,

Sum of all interior angles of a quadrilateral = 360°.

⇒ ∠A + ∠B + ∠C + ∠D = 360°

⇒ 2x° + 6y° + 10° + 2x° + y° + x° + 10° = 360°

⇒ 5x° + 7y° + 20° = 360°

⇒ 5x° + 7y° = 340°

Substituting x = 40°, y = 20° in L.H.S. of the above equation, we get :

⇒ 5 × 40° + 7 × 20°

⇒ 200° + 140°

⇒ 340°.

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

Solution : x = 40°, y = 20°.

Hence, option 1 is the correct option.

Case Study Based Questions

Question 1

Case Study I Ritesh bought a new well-furnished two-bedroom flat in a society. The layout of the flat is shown in the figure alongside. The builder claims that the areas of the two bedrooms and the kitchen together is 95 square metres. All the dimensions in the figure are in metre (m).

Ritesh bought a new well-furnished two-bedroom flat in a society. The layout of the flat is shown in the figure alongside. The builder claims that the areas of the two bedrooms and the kitchen together is 95 square metres. All the dimensions in the figure are in metre (m). R.S. Aggarwal Mathematics Solutions ICSE Class 9.

Study the above information and answer the following questions.

Which of the following pair of linear equations represent the given situation?
(a) x + y = 19, 2x + y = 13
(b) x + y = 13, 2x + y = 19
(c) x − y = 19, 2x − y = 13
(d) x + y = 13, 2x − y = 19
The perimeter of the outer boundary of the layout is:
(a) 54 m
(b) 27 m
(c) 50 m
(d) 52 m
Total area of bedroom 1 and kitchen is:
(a) 60 m²
(b) 70 m²
(c) 65 m²
(d) 95 m²
The area of the living room is:
(a) 50 m²
(b) 60 m²
(c) 70 m²
(d) 75 m²
The cost of laying tiles on the floor of the kitchen at the rate of ₹200 per sq m is:
(a) ₹ 7,000
(b) ₹ 6,000
(c) ₹ 5,200
(d) ₹ 5,000

Answer

1. Given,

From picture,

x = Length of Bedroom 1, y = Length of kitchen

Length of bathroom = 2

Equations:

Length of rectangular flat = x + y + 2

In rectangle, opposites sides are equal.

⇒ x + y + 2 = 15

⇒ x + y = 15 - 2

⇒ x + y = 13 ....(1)

Area = length × breadth

From figure,

Area of each Bedroom = 5 × x

Area of Kitchen = 5 × y

Given,

Area of two bedrooms and kitchen is 95.

⇒ 2 × (5 × x) + 5 × y = 95

⇒ 10x + 5y = 95

⇒ 5(2x + y) = 95

⇒ 2x + y = $\dfrac{95}{5}$

⇒ 2x + y = 19

⇒ 2x + y = 19 ....(2)

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

⇒ 2x + y - (x + y) = 19 - 13

⇒ x = 6 m.

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

⇒ 6 + y = 13

⇒ y = 13 - 6 = 7 m.

Hence, Option (b) is the correct option.

2. From figure,

⇒ Breadth = 15 m

⇒ Length = 5 + 2 + 5 = 12 m

Perimeter = 2(l + b)

= 2(12 + 15)

= 2 × 27

= 54 m.

Hence, Option (a) is the correct option.

3. Given,

Area of bedroom1 = l × b

= x × 5

= 6 × 5

= 30 m².

Area of kitchen = l × b

= y × 5

= 7 × 5

= 35 m².

Area of Bedroom 1 + Kitchen = 30 + 35 = 65 m².

Hence, option (c) is the correct option.

4. From figure,

Area of bedroom 2 = l × b = 5 × x = 5 × 6 = 30 m².

Area of bedroom2 + living room = l × b

= 15 × 7

= 105 m².

Area of living room = 105 - area of bedroom2 = 105 - 30 = 75 m².

Hence, option (d) is the correct option.

5. Given,

Area of kitchen = l × b

= y × 5

= 7 × 5

= 35 m².

Cost of laying tiles on the floor of the kitchen at the rate of ₹ 200 per sq m = 35 × 200 = ₹ 7,000.

Hence, Option (a) is the correct option.

Question 2

Case Study II
There are two mobile-phone companies – P and Q, that offer different plans. Company P charges a monthly fee of ₹ 40 plus ₹ 0.5 per minute of talk time. Company Q charges a monthly service fee of ₹ 30 plus ₹ 1 per minute of talk time.

There are two mobile-phone companies – P and Q, that offer different plans. Company P charges a monthly fee of ₹ 40 plus ₹ 0.5 per minute of talk time. Company Q charges a monthly service fee of ₹ 30 plus ₹ 1 per minute of talk time. R.S. Aggarwal Mathematics Solutions ICSE Class 9.

Based on this information answer the following questions:

The linear equation which expresses the plan of company P is :
(a) y = 0.5x + 40
(b) y = 40x + 0.5
(c) y = x + 40.5
(d) y = 40 − 0.5x
The linear equation which expresses the plan of company Q is:
(a) y = 30x + 1
(b) y = x + 30
(c) y = x − 30
(d) y = 30x − 1
How many minutes of talk time would yield equal expenditure from both companies?
(a) 10 minutes
(b) 15 minutes
(c) 20 minutes
(d) 25 minutes
Manisha took the plan of company P and used 400 minutes of talk time. She spent:
(a) ₹240
(b) ₹430
(c) ₹220
(d) ₹215
If in a month, Anurag wants to use only 300 minutes of talk time, then which company’s plan is better for him?
(a) Company P
(b) Company Q
(c) Both offer the same plan
(d) Can’t be determined

Answer

1. Given,

Let x = number of minutes, and y = total cost in ₹

Company P charges: ₹40/month + ₹0.5/min talk time

⇒ y = 0.5x + 40.

Hence, Option (a) is the correct option.

2. Given,

Company Q charges: ₹30/month + ₹1/min talk time

⇒ y = x + 30

Hence, Option (b) is the correct option.

3. If the expenditure is equal, then equating the values of y from part (1) and (2)

⇒ 0.5x + 40 = x + 30

⇒ 40 - 30 = x - 0.5x

⇒ 10 = 0.5x

⇒ x = $\dfrac{10}{0.5}$ = 20 minutes.

Hence, Option (c) is the correct option.

4. Given,

Manisha used 400 minutes on Company P. Calculating, the money spent by her,

⇒ y = 0.5x + 40

⇒ y = 0.5(400) + 40

⇒ y = 200 + 40

⇒ y = ₹ 240.

Hence, option (a) is the correct option.

5. Given,

Anurag uses 300 minutes,

Cost if he took Company P's plan.

⇒ y = 0.5x + 40

⇒ y = 0.5(300) + 40

⇒ y = 150 + 40

⇒ y = ₹ 190

Cost if he took Company Q's plan.

⇒ y = 1x + 30

⇒ y = 300 + 30

⇒ y = ₹ 330.

Thus, Anurag gets benefit if he uses company P's plan.

Hence, option (a) is the correct option.

Question 3

Case Study III
Tanusha went to a bank to withdraw money. She asked the cashier to give her ₹ 100 and ₹ 500 rupee notes only. The cashier agreed. Tanusha got x, ₹ 100-rupee notes and y, 500-rupee notes.

Based on this information, answer the following questions.

If Tanusha withdrew ₹ 15,000, then the above information can be represented by the linear equation:
(a) x + 5y = 150
(b) 5x + y = 150
(c) x + 5y + 150 = 0
(d) x + y = 150
If she got 54 notes in all, then the above information can be represented by the linear equation:
(a) 100x + 500y = 54
(b) x + y = 54
(c) 500x + 100y = 54
(d) 100x + y = 54
If Tanusha withdraws ₹16 000, then which combination of notes might she get?
(a) ₹500 notes = 30, ₹100 notes = 20
(b) ₹500 notes = 25, ₹100 notes = 25
(c) ₹500 notes = 20, ₹100 notes = 30
(d) ₹500 notes = 30, ₹100 notes = 10
If she gets twenty 500-rupee notes and twenty-five 100-rupee notes, then the amount she withdraws is:
(a) ₹10,000
(b) ₹11,000
(c) ₹12,000
(d) ₹12,500
Can Tanusha withdraw ₹10,050 under the given conditions?
(a) yes
(b) no
(c) can’t say anything
(d) none of these

Answer

1. Tanusha got x, ₹ 100-rupee notes and y, 500-rupee notes and withdraws ₹ 15,000.

⇒ 100x + 500y = 15000

⇒ 100(x + 5y) = 15000

⇒ x + 5y = $\dfrac{15000}{100}$

⇒ x + 5y = 150.

Hence, Option (a) is the correct option.

2. If she received x number ₹ 100 notes and y number of ₹ 500 notes and total notes are 54.

⇒ x + y = 54

Hence, Option (b) is the correct option.

3. If she withdaws ₹ 16,000 and gets x, ₹ 100 notes and y, 500 notes then.

⇒ 100x + 500y = 16000

⇒ 100(x + 5y) = 16000

⇒ x + 5y = $\dfrac{16000}{100}$

⇒ x + 5y = 160.

Substituting y = 30 and x = 10 in L.H.S. of the above equation, we get :

⇒ 10 + 5(30) = 10 + 150 = 160 = R.H.S.

Thus, no. of ₹ 100 notes = 10 and no. of ₹ 500 notes = 30.

Hence, Option (d) is the correct option.

4. Given,

She received 25 number ₹ 100 notes and 20 number of ₹ 500 notes,

Amount withdrawn = 100 × 25 + 500 × 20

= 2,500 + 10,000

= ₹ 12,500.

Hence, Option (d) is the correct option.

5. Let x be the number of ₹100 notes and y be the number of ₹500 notes.

The total amount is 100x + 500y.

Can Tanusha withdraw ₹ 10,050.

This means we need to check if the equation 100x + 500y = 10050 has integer solutions for x and y.

The value of x ₹100 notes is 100x, which is a multiple of 100.

The value of y ₹500 notes is 500y, which is also a multiple of 100.

The sum of these two values, the total amount withdrawn, must also be a multiple of 100.

Since, ₹ 10,050 is not a multiple of 100.

Therefore, it is not possible to form the amount ₹ 10,050 using only ₹ 100 and ₹ 500 notes.

Hence, Option (b) is the correct option.

Assertion Reason Type Questions

Question 1

Assertion(A): If 8x + 7y = 37 and 7x + 8y = 38, then x = -2, y = 3.

Reason(R): ax + by = c and bx + ay = d is not simultaneous linear equations in two variables.

A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false

Answer

Given,

Equations: 8x + 7y = 37 and 7x + 8y = 38

⇒ 8x + 7y = 37

⇒ 8x = 37 - 7y

⇒ x = $\dfrac{37 - 7y}{8}$ .....(1)

Substituting value of x from equation (1) in 7x + 8y = 38, we get :

$\Rightarrow 7\Big(\dfrac{37 - 7y}{8}\Big) + 8y = 38 \\[1em] \Rightarrow \dfrac{259 - 49y}{8} + 8y = 38 \\[1em] \Rightarrow \dfrac{259 - 49y + 64y}{8} = 38\\[1em] \Rightarrow 259 + 15y = 38 \times 8 \\[1em] \Rightarrow 259 + 15y = 304 \\[1em] \Rightarrow 15y = 304 - 259 \\[1em] \Rightarrow 15y = 45\\[1em] \Rightarrow y = \dfrac{45}{15} = 3.$

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

$\Rightarrow x = \dfrac{37 - 7y}{8} \\[1em] \Rightarrow x = \dfrac{37 - 7 \times 3}{8} \\[1em] \Rightarrow x = \dfrac{37 - 21}{8} \\[1em] \Rightarrow x = \dfrac{16}{8} = 2.$

x = 2 and y = 3.

∴ Assertion (A) is false.

⇒ ax + by = c and bx + ay = d are simultaneous linear equations in two variables x and y.

∴ Reason (R) is false.

Hence, option 4 is the correct option.

Question 2

Assertion(A): $\dfrac{2}{m} + \dfrac{3}{m} = 0 \text{ and } \dfrac{2}{3m} + \dfrac{2}{n} = \dfrac{1}{6}$ is a pair of simultaneous linear equations.

Reason(R): An equation of the form ax + by + c = 0, a ≠ 0, b ≠ 0 is called linear equations in two variables.

A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false

Answer

Given,

Equations: $\dfrac{2}{m} + \dfrac{3}{m} = 0 \text{ and } \dfrac{2}{3m} + \dfrac{2}{n} = \dfrac{1}{6}$

$\Rightarrow \dfrac{2}{m} + \dfrac{3}{m} = 0 \\[1em] \Rightarrow \dfrac{5}{m} = 0 \\[1em] \Rightarrow 5 = 0 \times m \\[1em] \Rightarrow 5 \ne 0 .$

Since 5 is not equal to 0, this equation has no solution for m. We cannot find values for m and n that satisfy both equations simultaneously.

$\Rightarrow \dfrac{2}{3m} + \dfrac{2}{n} = \dfrac{1}{6}$ Again, this is not linear in variables m and n, because the variables are in denominators.

Assertion (A) is false.

An equation of the form ax + by + c = 0, a ≠ 0, b ≠ 0, is called a linear equation in two variables.

Reason(R) is true.

A is false, R is true

Hence, option 2 is the correct option.

Question 3

Assertion(A): A pair of linear equations in two variables cannot have more than one solution.

Reason(R): If we solve a pair of linear equations in two variables, first by elimination method and then by cross multiplication method, then in some cases the two solutions so obtained may be different.

A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false

Answer

A pair of linear equations in two variables cannot have more than one solution. When the pair is consistent and independent, it has exactly one unique solution.

∴ Assertion (A) is true.

If we solve a pair of linear equations in two variables, first by elimination method and then by cross multiplication method, then in some cases the two solutions so obtained may be different.

This is false statement because both methods give same final answers.

∴ Reason(R) is false.

A is true, R is false.

Hence, option 1 is the correct option.

Competency Focused Questions

Question 1

If 0.4x + 0.3y = 2.3 and 2.5x - 2y = -5, then the value of xy is :

Answer

Given,

0.4x + 0.3y = 2.3 and 2.5x - 2y = -5

Solving first equation,

⇒ 0.4x + 0.3y = 2.3

⇒ 10(0.4x + 0.3y = 2.3) [Multiplying both sides by 10]

⇒ 4x + 3y = 23

⇒ 4x = 23 - 3y

⇒ x = $\dfrac{23 - 3y}{4}$ ....(1)

⇒ 2.5x - 2y = -5 ....(2)

Substituting value of x from equation (1) in 2.5x - 2y = -5, we get :

$\Rightarrow 2.5 \Big(\dfrac{23 - 3y}{4}\Big) - 2y = -5 \\[1em] \Rightarrow \dfrac{57.5 - 7.5y}{4} - 2y = -5 \\[1em] \Rightarrow \dfrac{57.5 - 7.5y - 8y}{4} = -5 \\[1em] \Rightarrow 57.5 - 15.5y = -5 \times 4 \\[1em] \Rightarrow 57.5 - 15.5y = -20 \\[1em] \Rightarrow -15.5y = -20 - 57.5 \\[1em] \Rightarrow -15.5y = -77.5 \\[1em] \Rightarrow y = \dfrac{-77.5}{-15.5} = 5.$

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

$\Rightarrow x = \dfrac{23 - 3y}{4} \\[1em] \Rightarrow x = \dfrac{23 - 3 \times 5}{4} \\[1em] \Rightarrow x = \dfrac{23 - 15}{4} \\[1em] \Rightarrow x = \dfrac{8}{4} = 2.$

x = 2 and y = 5.

xy = 10.

Hence, option 1 is the correct option.

Question 2

If ax - by = a² + b² and $\dfrac{x + y}{2} = a$ , then the value of x - y is :

Answer

Given,

Equations: ax - by = a² + b² and $\dfrac{x + y}{2} = a$

Solving equation $\dfrac{x + y}{2} = a$ ,

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

⇒ ax - by = a² + b² .......(2)

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

⇒ ax - by = a² + b²

⇒ ax - b(2a - x) = a² + b²

⇒ ax - 2ab + bx = a² + b²

⇒ ax + bx = a² + b² + 2ab

⇒ x(a + b) = a² + b² + 2ab

⇒ x(a + b) = (a + b)²

⇒ x = $\dfrac{(a + b)^2}{a + b}$

⇒ x = (a + b).

Substituting value of x in equation (1),

⇒ y = 2a - x

⇒ y = 2a - (a + b)

⇒ y = 2a - a - b

⇒ y = a - b

Now,

⇒ x - y = (a + b) - (a - b)

⇒ x - y = (a + b - a + b)

⇒ x - y = 2b.

Hence, option 4 is the correct option.

Question 3

If 8x + 9y = 42xy and 2x + 3y = 12xy, then the value of $\dfrac{1}{xy}$ is :

1
4
6
$\dfrac{1}{6}$

Answer

Given,

Equations:

⇒ 8x + 9y = 42xy

⇒ 2x + 3y = 12xy

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

$\Rightarrow \dfrac{8x + 9y}{xy} = \dfrac{42xy}{xy} \\[1em] \Rightarrow \dfrac{8x}{xy} + \dfrac{9y}{xy} = 42 \\[1em] \Rightarrow \dfrac{8}{y} + \dfrac{9}{x} = 42 \text{ .......(1)}$

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

$\Rightarrow \dfrac{2x + 3y}{xy} = \dfrac{12xy}{xy} \\[1em] \Rightarrow \dfrac{2x}{xy} + \dfrac{3y}{xy} = 12 \\[1em] \Rightarrow \dfrac{2}{y} + \dfrac{3}{x} = 12.$

Multiplying both sides of the above equation by 4, we get :

$\Rightarrow 4\Big(\dfrac{2}{y} + \dfrac{3}{x}\Big) = 12 \times 4 \\[1em] \Rightarrow \dfrac{8}{y} + \dfrac{12}{x} = 48 \text{ .......(2)}$

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

$\Rightarrow \dfrac{8}{y} + \dfrac{12}{x} - \Big(\dfrac{8}{y} + \dfrac{9}{x}\Big) = 48 - 42 \\[1em] \Rightarrow \dfrac{8}{y} + \dfrac{12}{x} - \dfrac{8}{y} - \dfrac{9}{x} = 6 \\[1em] \Rightarrow \dfrac{12}{x} - \dfrac{9}{x} = 6 \\[1em] \Rightarrow \dfrac{12 - 9}{x} = 6 \\[1em] \Rightarrow \dfrac{3}{x} = 6 \\[1em] \Rightarrow x = \dfrac{3}{6} = \dfrac{1}{2}.$

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

$\Rightarrow \dfrac{8}{y} + \dfrac{9}{x} = 42 \\[1em] \Rightarrow \dfrac{8}{y} + \dfrac{9}{\dfrac{1}{2}} = 42 \\[1em] \Rightarrow \dfrac{8}{y} + 18 = 42 \\[1em] \Rightarrow \dfrac{8}{y} = 42 - 18 \\[1em] \Rightarrow \dfrac{8}{y} = 24 \\[1em] \Rightarrow y = \dfrac{8}{24} = \dfrac{1}{3}.$

Calculating xy,

xy = $\dfrac{1}{2} \times \dfrac{1}{3} = \dfrac{1}{6}$ .

$\therefore \dfrac{1}{xy} = 6$

Hence, option 3 is the correct option.

Question 4

A shopkeeper sold a table and a chair for ₹ 1,050, thereby making a profit of 10% on the table and 25% on the chair. If he had taken a profit of 25% on the table and 10% on the chair, then he would have got ₹ 1,065. What is the cost price of 1 table and 1 chair.

Answer

Let cost price of the table be ₹ x and cost price of the chair be ₹ y.

According to case 1 :

⇒ Profit on table = 10%

Selling Price of table = Cost price (1 + Profit%) = $x \Big(1 + \dfrac{10}{100}\Big)$ = ₹ x × 1.10

⇒ Profit on chair = 25%

Selling Price of chair = Cost price (1 + Profit%) = $y \Big(1 + \dfrac{25}{100}\Big)$ = ₹ y × 1.25

⇒ 1.10x + 1.25y = 1050

Multiply the equation by 100,

⇒ 100(1.10x + 1.25y) = 100 × 1050

⇒ 110x + 125y = 105000

⇒ 5(22x + 25y) = 5 × 21000

⇒ 22x + 25y = 21000

⇒ 22x = 21000 - 25y

⇒ x = $\dfrac{21000 - 25y}{22}$ ......(1)

According to case 2 :

⇒ Profit on table = 25%

Selling Price of table = Cost price (1 + Profit%) = $x \Big(1 + \dfrac{25}{100}\Big)$ = ₹ x × 1.25

⇒ Profit on chair = 10%

Selling Price of chair = Cost price (1 + Profit%) = $y \Big(1 + \dfrac{10}{100}\Big)$ = ₹ y × 1.10

⇒ 1.25x + 1.10y = 1065

Multiply the equation by 100,

⇒ 100(1.25x + 1.10y) = 1065 × 100

⇒ 125x + 110y = 106500

⇒ 5(25x + 22y) = 5 × 21300

⇒ 25x + 22y = 21300 .......(2)

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

$\Rightarrow 25 \Big(\dfrac{21000 - 25y}{22}\Big) + 22y = 21300 \\[1em] \Rightarrow \dfrac{25(21000 - 25y) + 484y}{22} = 21300 \\[1em] \Rightarrow 525000 - 625y + 484y = 468600 \\[1em] \Rightarrow -141y = 468600 - 525000 \\[1em] \Rightarrow -141y = -56400 \\[1em] \Rightarrow y = \dfrac{-56400}{-141} = 400.$

Substituting value of y in equation 1, we get :

$\Rightarrow x = \dfrac{21000 - 25y}{22} \\[1em] \Rightarrow x = \dfrac{21000 - 25 \times 400}{22} \\[1em] \Rightarrow x = \dfrac{21000 - 10000}{22} \\[1em] \Rightarrow x = \dfrac{11000}{22} = 500.$

Hence, cost Price of Table = ₹ 500 and cost Price of Chair = ₹ 400.

Question 5

A boatman rowing at the rate of 5 km/hr in still water takes thrice as much time in going 40 km upstream as in going 40 km downstream. What is the speed of the stream?

Answer

Let x be speed of the stream.

Given,

Speed of boat in still water = 5 km/hr.

Speed of boat in upstream = (5 - x) km/hr

Speed of boat in downstream = (5 + x) km/hr

By formula,

Time = $\dfrac{\text{Distance}}{\text{Speed}}$

Given,

The boatman takes thrice as much time in going 40 km upstream as in going 40 km downstream.

$\therefore \dfrac{40}{5 - x} = 3 \times \dfrac{40}{5 + x} \\[1em] \Rightarrow \dfrac{1}{5 - x} = \dfrac{1}{5 + x} \\[1em] \Rightarrow 5 + x = 3(5 - x) \\[1em] \Rightarrow 5 + x = 15 - 3x \\[1em] \Rightarrow x + 3x = 15 - 3 \\[1em] \Rightarrow 4x = 10 \\[1em] \Rightarrow x = \dfrac{10}{4} \\[1em] \Rightarrow x = 2.5 \text{ km/hr }$

Hence, the speed of stream = 2.5 km/hr.

Question 6

A shopkeeper buys pens and pencils at ₹ 5 and ₹ 1 per price respectively. For every two pens, he buys three pencils. He sold pens and pencils at 12% and 10% profit respectively. If his total sale is ₹ 725, then find the number of pens and pencils sold by him.

Answer

Let x be the number of pens sold and y be the number of pencils sold.

Given,

Cost Price (CP) of 1 pen = ₹ 5

Profit on pens = 12%

⇒ SP of 1 pen = CP of pen + (Profit % of CP)

⇒ SP of 1 pen = 5 + $\dfrac{12}{100} \times 5$ = 5 + 0.12 × 5 = 5 + 0.60 = ₹ 5.60

Given,

Cost Price (CP) of 1 pencil = ₹ 1

Profit on pencils = 10%

SP of 1 pencil = CP of pencil + (Profit % of CP)

SP of 1 pencil = 1 + $\dfrac{10}{100}$ × 1 = 1 + 0.10 = ₹ 1.10

For every two pens, he buys three pencils,

This means the ratio of pens to pencils is x : y = 2 : 3.

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

⇒ y = $\dfrac{3}{2}x$ .......(1)

Given,

Total sale is ₹ 725,

⇒ x × 5.60 + y × 1.10 = 725

⇒ 5.60x + 1.10y = 725 ......(2)

Substitute the expression for y from equation (1) in (2), we get :

⇒ 5.60x + 1.10 $\times \dfrac{3}{2}x$ = 725

⇒ 5.60x + 1.10 × 1.5 = 725

⇒ 5.60x + 1.65x = 725

⇒ 7.25x = 725

⇒ y = $\dfrac{725}{7.25}$ = 100.

Substituting value of y in equation 1 we get,

⇒ y = $\dfrac{3}{2} \times 100$

⇒ y = 3 × 50

⇒ y = 150.

Hence, the number of pens sold = 100 and number of pencils sold = 150.

Question 7

The angles of a triangle in ascending order are x, y and z. If y - x = z - y = 10°, then find the angles of the triangle.

Answer

Given,

The three angles of the triangle in ascending order are x, y, and z.

Given,

⇒ y − x = 10°

⇒ y = 10° + x ....(1)

Given,

⇒ z − y = 10° ....(2)

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

⇒ z − y = 10°

⇒ z - (10° + x) = 10°

⇒ z = 10° + 10° + x

⇒ z = 20° + x.

So, the three angles of the triangle in terms of x are:

First angle: x

Second angle: x + 10°

Third angle: x + 20°

We know that,

The sum of the angles in any triangle is always 180°.

⇒ x + y + z = 180°

⇒ x + (x + 10°) + (x + 20°) = 180°

⇒ 3x + 30° = 180°

⇒ 3x = 180° − 30°

⇒ 3x = 150°

⇒ x = $\dfrac{150°}{3}$

⇒ x = 50°.

Substituting the value of x,

⇒ y = x + 10° = 50° + 10° = 60°

⇒ z = x + 20° = 50° + 20° = 70°.

Hence, x = 50°, y = 60°, z = 70°.

Chapter 5

Simultaneous Linear Equations

Class - 9 RS Aggarwal Mathematics Solutions

Exercise 5A

Question 1

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

Question 23

Question 24

Question 25

Exercise 5B

Question 1

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

Exercise 5C

Question 1

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

Question 23

Question 24

Question 25

Question 26

Question 27

Question 28

Question 29

Question 30

Question 31

Question 32

Question 33

Question 34

Question 35