Simultaneous Linear Equations — Multiple Choice Questions

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

1. x = 1, y = 1

2. x = 1, y = -1

3. x = -1, y = 1

4. 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.

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

1. x = 4, y = 6

2. x = 4, y = -6

3. x = -4, y = 6

4. 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.

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

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

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

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

4. 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.

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

1. x = 12, y = 8

2. x = 10, y = 8

3. x = 8, y = 12

4. 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.

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

1. 25

2. 13

3. $\dfrac{1}{13}$

4. $-\dfrac{1}{13}$

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.

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

1. x = 0, y = 0

2. x = 0, y = 1

3. x = 1, y = 0

4. 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.

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

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

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

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

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

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.

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 :

1. ₹ 2,750

2. ₹ 2,705

3. ₹ 2,075

4. ₹ 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.

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 :

1. 90

2. 18

3. 81

4. 54

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.

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 :

1. 10 years

2. 20 years

3. 30 years

4. 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.

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 :

1. one-rupee coins = 100 and fifty-paisa coins = 80

2. one-rupee coins = 80 and fifty-paisa coins = 100

3. one-rupee coins = 110 and fifty-paisa coins = 80

4. 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.

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 :

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

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

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

4. 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.

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 :

1. speed of boat = 3 kmph, speed of current = 2 kmph

2. speed of boat = 6 kmph, speed of current = 4 kmph

3. speed of boat = 9 kmph, speed of current = 2 kmph

4. 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.

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 :

1. boys = 4, girls = 6

2. boys = 6, girls = 4

3. boys = 7, girls = 3

4. 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.

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 :

1. x = 40°, y = 20°

2. x = 20°, y = 40°

3. x = 45°, y = 15°

4. 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.