Linear Equations and Inequalities in One Variable — Multiple Choice Questions

Which of the following is not a linear equation in one variable?

1. $3x + 2 = 0$

2. $2y - 4 = y$

3. $x + 2y = 7$

4. $2(x - 3) + 7 = 0$

Answer

We know that,

A linear equation in one variable is of the type $ax + b = 0$, where $a \ne 0$ and it contains only one variable.

The equation $x + 2y = 7$ contains two variables $x$ and $y$.

So, it is not a linear equation in one variable.

Hence, option 3 is the correct option.

The solution of the equation $\dfrac{2}{3}x + 1 = \dfrac{15}{9}$ is

1. 1

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

3. 2

4. $\dfrac{2}{3}$

Answer

Solving,

$\Rightarrow \dfrac{2}{3}x + 1 = \dfrac{15}{9} \\[1em] \Rightarrow \dfrac{2}{3}x = \dfrac{15}{9} - 1 \\[1em] \Rightarrow \dfrac{2}{3}x = \dfrac{15 - 9}{9} \\[1em] \Rightarrow \dfrac{2}{3}x = \dfrac{6}{9} \\[1em] \Rightarrow \dfrac{2}{3}x = \dfrac{2}{3} \\[1em] \Rightarrow x = 1$

Hence, option 1 is the correct option.

The solution of the equation $4z + 3 = 6 + 2z$ is

1. 1

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

3. 2

4. 3

Answer

Solving,

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

Hence, option 2 is the correct option.

The solution of the equation $\dfrac{3x}{5} + 1 = \dfrac{4x}{15} + 7$ is

1. 12

2. 14

3. 16

4. 18

Answer

$\dfrac{3x}{5} + 1 = \dfrac{4x}{15} + 7$

L.C.M. of 5 and 15 = 15

Multiplying both sides by 15, we get:

$\Rightarrow \Big[\dfrac{3x}{5} + 1\Big] \times 15 = \Big[\dfrac{4x}{15} + 7\Big] \times 15 \\[1em] \Rightarrow \dfrac{3x}{5} \times 15 + 1 \times 15 = \dfrac{4x}{15} \times 15 + 7 \times 15 \\[1em] \Rightarrow 9x + 15 = 4x + 105 \\[1em] \Rightarrow 9x - 4x = 105 - 15 \\[1em] \Rightarrow 5x = 90 \\[1em] \Rightarrow x = \dfrac{90}{5} \\[1em] \Rightarrow x = 18$

Hence, option 4 is the correct option.

The solution of the equation $\dfrac{x}{2} - \dfrac{1}{5} = \dfrac{x}{3} + \dfrac{1}{4}$ is

1. 2.7

2. 1.8

3. 2.9

4. 1.7

Answer

$\dfrac{x}{2} - \dfrac{1}{5} = \dfrac{x}{3} + \dfrac{1}{4}$

L.C.M. of 2, 5, 3 and 4 = 60

Multiplying both sides by 60, we get:

$\Rightarrow \Big[\dfrac{x}{2} - \dfrac{1}{5}\Big] \times 60 = \Big[\dfrac{x}{3} + \dfrac{1}{4}\Big] \times 60 \\[1em] \Rightarrow \dfrac{x}{2} \times 60 - \dfrac{1}{5} \times 60 = \dfrac{x}{3} \times 60 + \dfrac{1}{4} \times 60 \\[1em] \Rightarrow 30x - 12 = 20x + 15 \\[1em] \Rightarrow 30x - 20x = 15 + 12 \\[1em] \Rightarrow 10x = 27 \\[1em] \Rightarrow x = \dfrac{27}{10} \\[1em] \Rightarrow x = 2.7$

Hence, option 1 is the correct option.

The solution of the equation $\dfrac{8x - 3}{3x} = 2$ is

1. $\dfrac{1}{2}$

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

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

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

Answer

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

By cross multiplication, we get

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

Hence, option 3 is the correct option.

If we subtract $\dfrac{1}{2}$ from a number and multiply the result by $\dfrac{1}{2}$, we get $\dfrac{1}{8}$, then the number is

1. $\dfrac{1}{2}$

2. $\dfrac{3}{4}$

3. $\dfrac{1}{4}$

4. none of these.

Answer

Let the number be $x$.

According to the given condition,

$\Rightarrow \left(x - \dfrac{1}{2}\right) \times \dfrac{1}{2} = \dfrac{1}{8} \\[1em] \Rightarrow x - \dfrac{1}{2} = \dfrac{1}{8} \times 2 \\[1em] \Rightarrow x - \dfrac{1}{2} = \dfrac{1}{4} \\[1em] \Rightarrow x = \dfrac{1}{4} + \dfrac{1}{2} \\[1em] \Rightarrow x = \dfrac{1 + 2}{4} \\[1em] \Rightarrow x = \dfrac{3}{4}$

Hence, option 2 is the correct option.

Fifteen years from now Ravi's age will be four times his present age. What is Ravi's present age?

1. 4 years

2. 5 years

3. 6 years

4. 3 years

Answer

Let Ravi's present age be $x$ years.

Fifteen years from now, Ravi's age = $(x + 15)$ years

According to the given condition,

$\Rightarrow x + 15 = 4x \\[1em] \Rightarrow 15 = 4x - x \\[1em] \Rightarrow 3x = 15 \\[1em] \Rightarrow x = \dfrac{15}{3} \\[1em] \Rightarrow x = 5$

Hence, option 2 is the correct option.

If the sum of three consecutive integers is 51, then the largest integer is

1. 16

2. 17

3. 18

4. 19

Answer

Let the three consecutive integers be $x, x + 1$ and $x + 2$.

Given,

Sum of three consecutive integers is 51.

$\Rightarrow x + (x + 1) + (x + 2) = 51 \\[1em] \Rightarrow 3x + 3 = 51 \\[1em] \Rightarrow 3x = 51 - 3 \\[1em] \Rightarrow 3x = 48 \\[1em] \Rightarrow x = \dfrac{48}{3} \\[1em] \Rightarrow x = 16$

So, the largest integer = 16 + 2 = 18

Hence, option 3 is the correct option.

If the perimeter of a rectangle is 13 cm and its width is $2\dfrac{3}{4}$ cm, then its length is

1. $2\dfrac{3}{4}$ cm

2. $3\dfrac{3}{4}$ cm

3. $4\dfrac{3}{4}$ cm

4. $5\dfrac{3}{4}$ cm

Answer

Let the length of the rectangle be $x$ cm.

Given, width = $2\dfrac{3}{4}$ cm = $\dfrac{11}{4}$ cm

We know that,

Perimeter of a rectangle = 2(length + width)

According to the given condition,

$\Rightarrow 2\left(x + \dfrac{11}{4}\right) = 13 \\[1em] \Rightarrow x + \dfrac{11}{4} = \dfrac{13}{2} \\[1em] \Rightarrow x = \dfrac{13}{2} - \dfrac{11}{4} \\[1em] \Rightarrow x = \dfrac{26 - 11}{4} \\[1em] \Rightarrow x = \dfrac{15}{4} = 3\dfrac{3}{4}$

Hence, option 2 is the correct option.

What should be added to twice the rational number $\dfrac{-7}{3}$ to get $\dfrac{3}{7}$?

1. $\dfrac{58}{21}$

2. $\dfrac{29}{21}$

3. $\dfrac{89}{21}$

4. $\dfrac{107}{21}$

Answer

Let the required number be $x$.

Twice the rational number $\dfrac{-7}{3} = 2 \times \dfrac{-7}{3} = \dfrac{-14}{3}$

According to the given condition,

$\Rightarrow x + \left(\dfrac{-14}{3}\right) = \dfrac{3}{7} \\[1em] \Rightarrow x = \dfrac{3}{7} + \dfrac{14}{3} \\[1em] \Rightarrow x = \dfrac{9 + 98}{21} \\[1em] \Rightarrow x = \dfrac{107}{21}$

Hence, option 4 is the correct option.

Sum of digits of a two digit number is 8. If the number obtained by reversing the digits is 18 more than the original number, then the original number is

1. 35

2. 53

3. 26

4. 62

Answer

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

So, the original number = $10y + x$

On reversing the digits, the new number = $10x + y$

Given, sum of the digits = 8

⇒ $x + y = 8 \quad \text{...(1)}$

According to the given condition,

$\Rightarrow (10x + y) - (10y + x) = 18 \\[1em] \Rightarrow 9x - 9y = 18 \\[1em] \Rightarrow x - y = \dfrac{18}{9} \\[1em] \Rightarrow x - y = 2 \quad \text{...(2)}$

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

$\Rightarrow 2x = 10 \\[1em] \Rightarrow x = 5$

Substituting $x = 5$ in equation (1), we get $y = 8 - 5 = 3$

So, the original number = 10 × 3 + 5 = 35

Hence, option 1 is the correct option.

Arjun is twice as old as Shriya. If five years ago his age was three times Shriya's age, then Arjun's present age is

1. 10 years

2. 15 years

3. 20 years

4. 25 years

Answer

Let Shriya's present age be $x$ years.

Then, Arjun's present age = $2x$ years

Five years ago, Shriya's age = $(x - 5)$ years and Arjun's age = $(2x - 5)$ years

According to the given condition,

$\Rightarrow 2x - 5 = 3(x - 5) \\[1em] \Rightarrow 2x - 5 = 3x - 15 \\[1em] \Rightarrow 15 - 5 = 3x - 2x \\[1em] \Rightarrow x = 10$

So, Arjun's present age = 2 × 10 = 20 years

Hence, option 3 is the correct option.

If the replacement set is {-5, -3, -1, 0, 1, 3}, then the solution set of the inequation $-3 \lt x \lt 3$ is

1. {-2, -1, 0, 1, 2}

2. {-1, 0, 1, 2}

3. {-3, -1, 0, 1, 3}

4. {-1, 0, 1}

Answer

$- 3 \lt x \lt 3$

∵ The replacement set = {-5, -3, -1, 0, 1, 3}.

The elements of the replacement set which are greater than - 3 and less than 3 are - 1, 0 and 1.

So, solution set = {-1, 0, 1}

Hence, option 4 is the correct option.