Linear Inequations — Multiple Choice Questions

Which of the following is not a linear inequation?

1. 3x - 8 > 5 + 2x

2. $\dfrac{5}{2}x - 3 \le \dfrac{9}{4}x + 12$

3. $4x - \dfrac{7}{3} \ge 13x + 8$

4. $6x + 13 \lt \dfrac{4}{x} - 3$

Answer

Solving, option 4 :

$\Rightarrow 6x + 13 \lt \dfrac{4}{x} - 3 \\[1em] \Rightarrow 6x + 13 \lt \dfrac{4 - 3x}{x} \\[1em]$

Case 1 : If x is positive,

$\Rightarrow x (6x + 13) \lt 4 - 3x \\[1em] \Rightarrow 6x^2+ 13x \lt 4 - 3x \\[1em] \Rightarrow 6x^2+ 13x + 3x \lt 4 \\[1em] \Rightarrow 6x^2+ 16x \lt 4 \\[1em]$

Case 2 : If x is negative,

$\Rightarrow x(6x + 13) \gt 4 - 3x \\[1em] \Rightarrow 6x^2 + 13x \gt 4 - 3x \\[1em] \Rightarrow 6x^2 + 13x + 3x \gt 4 \\[1em] \Rightarrow 6x^2 + 16x \gt 4 \\[1em]$

Since, the highest power of x is 2.

$\therefore 6x + 13 \lt \dfrac{4}{x} - 3$ is not linear.

Hence, Option 4 is the correct option.

Which of the following is a linear inequation?

1. x² + 3 ≥ 2x - 7

2. $\dfrac{7}{(x - 1)} \lt 3x + 4$

3. $\dfrac{7}{3}x - 9 \le 5 + \dfrac{4}{7}x$

4. $5 - \dfrac{9}{x} \gt 3x + 4$

Answer

$\Rightarrow \dfrac{7}{3}x - 9 \le 5 + \dfrac{4}{7}x$

In the above inequation the highest power of x will be 1 so it is a linear inequation.

Hence, Option 3 is the correct option.

Which of the following is not a general form of a linear inequation?

1. ax + b > c

2. $\dfrac{a}{x}$ + b ≥ c

3. ax + b ≤ c

4. ax² + b < c

Answer

ax² + b < c is not a general form of a linear inequation, as here the hightest power of x is 2, while in linear inequation the highest power should be 1.

Hence, Option 4 is the correct option.

Which of the following is reducible to a linear inequation?

1. 7 - x² ≤ 5x + 3

2. $\dfrac{3}{x} - 8 \gt 4$

3. $4 - \dfrac{2}{x} \ge \dfrac{1}{x^2} - 6$

4. $11 - x \le 5x ^ 2 + 6$

Answer

Solving option 2,

$\Rightarrow \dfrac{3}{x} - 8 \gt 4 \\[1em] \Rightarrow \dfrac{3 - 8x}{x} \gt 4 \\[1em] \Rightarrow 3 - 8x \gt 4x \\[1em] \Rightarrow 4x + 8x \gt 3 \\[1em] \Rightarrow 12x \gt 3.$

The above is a linear inequation.

Hence, Option 2 is the correct option.

Which of following is not reducible to a linear inequation ?

1. $3 - \dfrac{7}{x} \lt 5$

2. $8 + \dfrac{3}{x} \ge \dfrac{5}{x} - 4$

3. $\dfrac{6}{x} - 8 \gt \dfrac{4}{x} + 3x$

4. $\dfrac{3}{x - 1} - 7 \lt 9$

Answer

Solving option 3,

$\Rightarrow \dfrac{6}{x} - 8 \gt \dfrac{4}{x} + 3x\\[1em] \Rightarrow \dfrac{6 - 8x}{x} \gt \dfrac{4 + 3x^2}{x} \\[1em]$

Case 1 : If x is positive,

$\Rightarrow 6 - 8x \gt 4 + 3x^2 \\[1em] \Rightarrow 4 + 3x^2 \lt 6 - 8x \\[1em] \Rightarrow 3x^2 + 8x \lt 6 - 4 \\[1em] \Rightarrow 3x^2 + 8x \lt 2 \\[1em]$

Case 2 : If x is negative,

$\Rightarrow 6 - 8x \lt 4 + 3x^2 \\[1em] \Rightarrow 4 + 3x^2 \gt 6 - 8x \\[1em] \Rightarrow 3x^2 + 8x \gt 6 - 4 \\[1em] \Rightarrow 3x^2 + 8x \gt 2 \\[1em]$

Since, the highest power of x is 2 in both the cases,

$\dfrac{6}{x} - 8 \gt \dfrac{4}{x} + 3x$ is not reducible to linear inequation.

Hence, Option 3 is the correct option.

Which of the following is not true for the linear inequations ?

1. Adding the same number to each side of an inequation does not change the inequality.

2. Multiplying each side of an inequation by the same positive number reverses the inequality.

3. Multiplying each side of an inequation by the same negative number reverses the inequality.

4. Dividing each side of an inequation by the same positive number does not change the inequality.

Answer

We know that,

Multiplying each side of an inequation by the same negative number reverses the inequality, but there is no change on multiplying with positive number.

Hence, Option 2 is the correct option.

Which of the following is not true?

1. p > q ⇔ q > p

2. p < q ⇔ q > p

3. p ≥ q ⇔ q ≤ p

4. p ≤ q ⇔ q ≥ p

Answer

The greater than relation > is asymmetric, meaning :

If p > q, then q < p.

It is not the case in, p > q ⇔ q > p.

So, p > q ⇔ q > p it is incorrect

Hence, Option 1 is the correct option.

If -4x > 8y; then

1. x > 2y

2. x > -2y

3. x < -2y

4. x < 2y

Answer

Solving,

⇒ -4x > 8y

⇒ 4x < -8y

⇒ x < $-\dfrac{8y}{4}$

⇒ x < -2y.

Hence, Option 3 is the correct option.

Which of the following is the solution set of x ≤ 7 when the replacement set is the set of natural numbers?

1. {1, 2, 3, 4, 5, 6, 7}

2. {0, 1, 2, 3, 4, 5, 6, 7}

3. {1, 2, 3, 4, 5, 6}

4. {0, 1, 2, 3, 4, 5, 6}

Answer

The natural numbers start at 1.

Since, x ≤ 7

Solution set = {1, 2, 3, 4, 5, 6, 7}

Hence, Option 1 is the correct option.

Which of the following is the solution set of x < 0 when the replacement set is the set of whole numbers?

1. {0}

2. {....-3, -2, -1, }

3. {.......,-3, -2, -1, 0}

4. Null set

Answer

Whole numbers include 0 and all positive integers, but not any negative numbers.

For x < 0 and x ∈ W

Solution set will be a null set.

Hence, Option 4 is the correct option.

Which of the following is the solution set of x ≤ 3 when the replacement set is the set of integers?

1. {0, 1, 2, 3}

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

3. {...........,-2,-1, 0, 1, 2, 3}

4. {.....,-2,-1, 1, 2, 3 }

Answer

For x ≤ 3 and x ∈ I

Solution set = {3, 2, 1, 0, -1, -2, .........}

Hence, Option 3 is the correct option.

Which of the following statements is not true ? (r is a positive integer)

1. If p < q then p - r < q - r

2. If p ≥ q then -pr ≥ -qr

3. If p > q, then $\dfrac{p}{r} \gt \dfrac{q}{r}$

4. If p ≤ q then p + r ≤ q + r

Answer

⇒ p ≥ q

Since, r is positive so -r will be negative.

Multiplying by -r on both sides we get,

⇒ -pr ≤ -qr (As on multiplying by negative number the sign reverses.)

∴ -pr ≥ -qr is not true.

Hence, Option 2 is the correct option.

Which of the following statements is true? (r is a positive integer)

1. If p ≥ q, then $-\dfrac{p}{r} \le -\dfrac{q}{r}$

2. If p ≤ q then pr ≥ qr

3. If p > q, then -pr > -qr

4. If p < q then p - r > q - r

Answer

⇒ p ≥ q

Since, r is positive so -r will be negative.

Dividing by -r on both sides we get,

$\Rightarrow -\dfrac{p}{r} \le - \dfrac{q}{r}$ (As on dividing by negative number the sign reverses.)

$\therefore -\dfrac{p}{r} \le -\dfrac{q}{r}$ is true.

Hence, Option 1 is the correct option.

Graphical representation of following inequation on the number line is {x : -3 < x < 2, x ∈ I}

1.

2.

3.

4.

Answer

For -3 < x < 2 and x ∈ I

Solution set = {-2, -1, 0, 1}

Hence, Option 3 is the correct option.

Which of the following is the correct graphical representation of {x : x < 1, x ∈ I} on the number line?

1.

2.

3.

4.

Answer

For, x < 1 and x ∈ I

Solution set = {0, -1, -2, ...........}

Hence, Option 1 is the correct option.

Which of the following is the correct graphical representation of {x : -4 < x ≤ 2, x ∈ R} on the number line?

1.

2.

3.

4.

Answer

Given,

-4 < x ≤ 2 and x ∈ R

It contains all the numbers between -4 and 2 excluding -4 and including 2.

Hence, Option 3 is the correct option.

The solution set of 4x - 3 ≤ 5, where x ∈ N, is :

1. {1}

2. {1, 2}

3. {0, 1, 3}

4. none of these

Answer

Given,

⇒ 4x - 3 ≤ 5

⇒ 4x ≤ 5 + 3

⇒ 4x ≤ 8

⇒ x ≤ $\dfrac{8}{4}$

⇒ x ≤ 2

Since, x ∈ N

⇒ Solution set = {1, 2}

Hence, Option 2 is the correct option.

The solution set of 3x + 1 > 5x - 7, where x ∈ R, is :

1. {x : x < 4, x ∈ R}

2. {x : x > 4, x ∈ R}

3. {x : x < 8, x ∈ R}

4. {x : x > -4, x ∈ R}

Answer

Given,

⇒ 3x + 1 > 5x - 7

⇒ 5x - 3x < 7 + 1

⇒ 2x < 8

⇒ x < $\dfrac{8}{2}$

⇒ x < 4

Since, x ∈ R

⇒ Solution set = {x : x < 4, x ∈ R}

Hence, Option 1 is the correct option.

The solution set of 4x - 9 ≥ 7, where x ∈ {1, 2, 3, 4, 5, 6, 7, 8}, is :

1. {1, 2, 3, 4}

2. {4, 5, 6, 7, 8}

3. {1, 2, 3}

4. {5, 6, 7, 8}

Answer

⇒ 4x - 9 ≥ 7

⇒ 4x ≥ 7 + 9

⇒ 4x ≥ 16

⇒ x ≥ $\dfrac{16}{4}$

⇒ x ≥ 4

Since, x ∈ {1, 2, 3, 4, 5, 6, 7, 8}

Solution set = {4, 5, 6, 7, 8}.

Hence, Option 2 is the correct option.

Find the values of x in the inequation 3x - 2 > 9x - 16, where x ∈ I.

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

2. {2, 3, 4, 5, ......}

3. {......, -2, -1, 0, 1, 2}

4. {......, -2, -1, 0, 1, 2, 3}

Answer

Given,

⇒ 3x - 2 > 9x - 16

⇒ 9x - 16 < 3x - 2

⇒ 9x - 3x < -2 + 16

⇒ 6x < 14

⇒ x < $\dfrac{14}{6}$

⇒ x < 2.33

Since, x ∈ I

⇒ Solution set = {2, 1, 0, -1, ......}

Hence, Option 3 is the correct option.

The largest value of x for which, 3(x - 2) ≤ 6 - x, where x ∈ W, is :

1. 3

2. 4

3. 6

4. none of these

Answer

Given,

⇒ 3(x - 2) ≤ 6 - x

⇒ 3x - 6 ≤ 6 - x

⇒ 3x + x ≤ 6 + 6

⇒ 4x ≤ 12

⇒ x ≤ $\dfrac{12}{4}$

⇒ x ≤ 3

So, the largest whole number value of x = 3.

Hence, Option 1 is the correct option.

If x is a negative integer, then find the solution set of 3 + 2(x + 1) > -1.

1. {-3, -2, -1}

2. {-2, -1}

3. {-1}

4. none of these

Answer

Given,

⇒ 3 + 2(x + 1) > -1

⇒ 3 + 2x + 2 > -1

⇒ 2x + 5 > -1

⇒ 2x > -1 - 5

⇒ 2x > -6

Dividing 2 on both sides we get,

⇒ x > -3

Since, x is a negative integer

⇒ Solution set = {-2, -1}.

Hence, Option 2 is the correct option.

Find the smallest value of x in the following inequation.

3(x + 4) ≤ 5(x - 1) + 4 and x ∈ N

1. 5

2. 6

3. 7

4. 8

Answer

⇒ 3(x + 4) ≤ 5(x - 1) + 4

⇒ 3x + 12 ≤ 5x - 5 + 4

⇒ 3x + 12 ≤ 5x - 1

⇒ 5x - 3x ≥ 12 + 1

⇒ 2x ≥ 13

⇒ x ≥ $\dfrac{13}{2}$

⇒ x ≥ $6\dfrac{1}{2}$

Since, x ∈ N.

The smallest value of x is 7.

Hence, Option 3 is the correct option.

What is the smallest value of x in the following inequation?

20 - 5x < 5(x + 8) and x ∈ I

1. -1

2. 0

3. 1

4. cannot be determined

Answer

⇒ 20 - 5x < 5(x + 8)

⇒ 20 - 5x < 5x + 40

⇒ 5x + 40 > 20 - 5x

⇒ 5x + 5x > 20 - 40

⇒ 10x > 20 - 40

⇒ 10x > -20

⇒ x > $-\dfrac{20}{10}$

⇒ x > -2

Since, x ∈ I and x > -2

⇒ Smallest value of x in following inequation = -1.

Hence, Option 1 is the correct option.

If -3 ≤ -4x + 5 and x ∈ W, then the solution set is :

1. {0, 1, 2}

2. {1, 2}

3. {2, 3, 4, 5}

4. {....., -3, -2, -1, 0, 1, 2, 3, ......}

Answer

Solving,

⇒ -3 ≤ -4x + 5

⇒ 4x ≤ 5 + 3

⇒ 4x ≤ 8

⇒ x ≤ $\dfrac{8}{4}$

⇒ x ≤ 2.

Since, x ≤ 2 and x ∈ W.

∴ Solution set = {0, 1, 2}.

Hence, Option 3 is the correct option.

Given, x + 2 ≤ $\dfrac{x}{3}$ + 3 and x is a prime number. The solution set for x is :

1. ∅

2. {0}

3. {1}

4. {0, 1}

Answer

Solving the given equation :

$\Rightarrow x + 2 \le \dfrac{x}{3} + 3 \\[1em] \Rightarrow x - \dfrac{x}{3} \le 3 - 2 \\[1em] \Rightarrow \dfrac{3x - x}{3} \le 1 \\[1em] \Rightarrow \dfrac{2x}{3} \le 1 \\[1em] \Rightarrow x \le \dfrac{3}{2} \\[1em] \Rightarrow x \le 1.5$

Since, x is a prime number less than 1.5

Solution set is empty.

Hence, option 1 is the correct option.

If 2x - 15 > 4x + 9, then:

1. x < -12

2. x < 12

3. x > −12

4. x > 12

Answer

Given,

⇒ 2x - 15 > 4x + 9

⇒ -15 - 9 > 4x - 2x

⇒ -24 > 2x

⇒ $\dfrac{-24}{2}$ > x

⇒ -12 > x

⇒ x < -12.

Hence, option 1 is the correct option.

The solution set for 0 < $-\dfrac{x}{3}$ < 2, x ∈ Z, is :

1. {-5, -4, -3, -2, -1}

2. {-6, -5, -4, -3, -2, -1}

3. {-5, -4, -3, -2, -1, 0}

4. {-6, -5, -4, -3, -2, -1, 0}

Answer

Given,

0 < $-\dfrac{x}{3}$ < 2

Solving L.H.S. of equation :

⇒ 0 < $-\dfrac{x}{3}$

⇒ $\dfrac{x}{3}$ < 0

⇒ x < 0 ... (1)

Solving R.H.S. of equation :

⇒ $-\dfrac{x}{3}$ < 2

⇒ -x < 2

× 3

⇒ -x < 6

⇒ x > -6 .... (2)

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

-6 < x < 0 and x ∈ Z.

Solution set for x = {-5, -4, -3, -2, -1}.

Hence, option 1 is the correct option.

What is the solution set for the inequation represented by the following number line?

1. {x ∈ R : -3 < x ≤ 4}

2. {x ∈ R : -3 < x < 4}

3. {x ∈ R : -3 ≤ x < 4}

4. {x ∈ R : -3 ≤ x ≤ 4}

Answer

The number line represents a solution set = {x ∈ R : -3 < x ≤ 4}

Hence, Option 1 is the correct option.

The solution set of the inequation x - 3 ≥ -5, x ∈ R is :

1. {x : x > -2, x ∈ R}

2. {x : x ≤ -2, x ∈ R}

3. {x : x ≥ -2, x ∈ R}

4. {-2, -1, 0, 1, 2}

Answer

Given,

⇒ x - 3 ≥ -5

⇒ x ≥ -5 + 3

⇒ x ≥ -2

⇒ Solution set = {x : x ≥ -2, x ∈ R}

Hence, Option 3 is the correct option.

Find the greatest integer which is such that if 7 is added to its double, then the resulting number is greater than three times the integer.

1. 7

2. 8

3. 6

4. none of these

Answer

Let the integer be x.

According to question,

⇒ 7 + 2x > 3x

⇒ 7 > 3x - 2x

⇒ 7 > x

⇒ x < 7

Let's try x = 6,

⇒ 7 + 2x

⇒ 7 + 2(6)

⇒ 19, which is greater than 3 times the number (i.e. 3x or 18).

Hence, Option 3 is the correct option.

The solution set for the inequation 2x + 4 ≤ 14, x ∈ W is :

1. {1, 2, 3, 4, 5}

2. {0, 1, 2, 3, 4, 5}

3. {1, 2, 3, 4}

4. {0, 1, 2, 3, 4}

Answer

Given,

⇒ 2x + 4 ≤ 14

⇒ 2x ≤ 14 - 4

⇒ 2x ≤ 10

⇒ x ≤ $\dfrac{10}{2}$

⇒ x ≤ 5

Since, x ∈ W

⇒ Solution set = {0, 1, 2, 3, 4, 5}.

Hence, Option 2 is the correct option.