Linear Inequations MCQ Quiz - Objective Question with Answer for Linear Inequations - Download Free PDF

Last updated on Jun 30, 2025

Latest Linear Inequations MCQ Objective Questions

Linear Inequations Question 1:

The number of positive integral solutions of the inequation \(\frac{x + 2}{x + 3} > 1 \)

  1. Infinite 
  2. 4
  3. 3
  4. 0

Answer (Detailed Solution Below)

Option 4 : 0

Linear Inequations Question 1 Detailed Solution

Given:

The inequation: (x + 2)/(x + 3) > 1

Formula used:

For a rational inequality of the form (a/b) > 1, we analyze critical points and test values between intervals.

Calculation:

(x + 2)/(x + 3) > 1

⇒ (x + 2) - (x + 3) > 0

⇒ x + 2 - x - 3 > 0

⇒ -1 > 0

This is not possible.

Since the inequality is never satisfied, there are no positive integral solutions.

∴ The correct answer is option (4).

Linear Inequations Question 2:

If \(k<(\sqrt{2}+1)^3 where k is a natural number, the value of k?

  1. 11
  2. 13
  3. 15
  4. 17

Answer (Detailed Solution Below)

Option 2 : 13

Linear Inequations Question 2 Detailed Solution

Calculation:

Given,

The inequality is: \( k < (√{2} + 1)^3 < k + 2 \), where k is a natural number.

We need to compute \((√{2} + 1)^3\)

\( (√{2} + 1)^3 = 2√{2} + 6 + 3√{2} + 1 = 5√{2} + 7 \)

Approximating √ 2≈ 1.414 , we find:

\( 5√{2} + 7 ≈ 5 \times 1.414 + 7 = 7.07 + 7 = 14.07 \)

The inequality becomes:

\( k < 14.07 < k + 2 \)

This implies that k > 12.07 \), so the smallest integer value of k  is:

\( k = 13 \)

∴ The value of k  is 13.

Hence, the correct answer is Option 2. 

Linear Inequations Question 3:

Let x and y be distinct integers where 1 ≤ x ≤ 25 and 1 ≤ y ≤ 25. Then, the number of ways of choosing x and y, such that x + y is divisible by 5, is _____.  

Answer (Detailed Solution Below) 120

Linear Inequations Question 3 Detailed Solution

Calculation: 

x + y = 5λ

Cases :

\( \begin{array}{|c|c|c|} \hline \mathbf{x} & \mathbf{y} & \text{Number of ways} \\ \hline 5\lambda & 5\lambda & 20 \\ 5\lambda+1 & 5\lambda+4 & 25 \\ 5\lambda+2 & 5\lambda+3 & 25 \\ 5\lambda+3 & 5\lambda+2 & 25 \\ 5\lambda+4 & 5\lambda+1 & 25 \\ \hline \end{array} \)

Total = 120

Hence, the correct answer is 120. 

Linear Inequations Question 4:

The statement (p ∧ (~q)) ⇒ (p ⇒ (~q)) is 

  1. equivalent to (~p) ∨ (~q) 
  2. a tautology 
  3. equivalent to p ∨ q 
  4. a contradiction

Answer (Detailed Solution Below)

Option 2 : a tautology 

Linear Inequations Question 4 Detailed Solution

Calculation: 

(p ∧ ~ q) → (p → ~ q) 

≡ (~ (p ∧ ~q)) ∨ (~ p ∨ ~q)

≡ (~ (p ∨ ~q)) ∨ (~ p ∨ ~q)

≡ ~ p ∨ t ≡ t

∴ The statement (p ∧ (~q)) ⇒ (p ⇒ (~q)) is tautology 

Hence, the correct answer is Option 2. 

Linear Inequations Question 5:

Number of integral values of x satisfying x2 - 4x - 21 > 0 and x2 - 9x + 8 < 0 is 

  1. one 
  2. two 
  3. many 
  4. nil 

Answer (Detailed Solution Below)

Option 4 : nil 

Linear Inequations Question 5 Detailed Solution

Calculation

x² - 4x - 21 > 0:

(x-7)(x+3) > 0.

Solutions: x < -3 or x > 7.

x² - 9x + 8 < 0:

(x-1)(x-8) < 0.

Solutions: 1 < x < 8.

Combine solutions: x < -3 or x > 7, and 1 < x < 8.

 

Intersection is 7 < x < 8.

Integral values between 7 and 8: None.

Therefore, no integral values satisfy both inequalities.

Hence option 4 is correct

Top Linear Inequations MCQ Objective Questions

Write the solution of inequality \(\frac{1}{5}\left( {\frac{{3x}}{5} + 4} \right) \ge \frac{1}{3}(x - 6)\).

  1. \(x \le \frac{{105}}{8}\)
  2. \(x \ge \frac{{105}}{8}\)
  3. \(x \ge 120\)
  4. \(x \le 120\)

Answer (Detailed Solution Below)

Option 1 : \(x \le \frac{{105}}{8}\)

Linear Inequations Question 6 Detailed Solution

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CALCULATIONS:

Given \(\frac{1}{5}\left( {\frac{{3x}}{5} + 4} \right) \ge \frac{1}{3}(x - 6)\)

\(\begin{array}{l} \Rightarrow 3\left( {\frac{{3x}}{5} + 4} \right) \ge 5(x - 6)\\ \Rightarrow \left( {\frac{{9x}}{5} + 12} \right) \ge 5x - 30\\ \Rightarrow \left( {30 + 12} \right) \ge - \frac{{9x}}{5} + 5x\\ \Rightarrow 42 \ge \frac{{ - 9x + 25x}}{5}\\ \Rightarrow 42 \ge \frac{{16x}}{5}\\ \Rightarrow \frac{{42 \times 5}}{{16}} \ge x\\ \Rightarrow x \le \frac{{105}}{8} \end{array}\)

Therefore, option (1) is the correct answer.

If |2x - 3| < |x + 5| then x belongs to - 

  1. (-3, 5)
  2. (5, 9)
  3. \(\left(-\dfrac{2}{3}, 8\right)\)
  4. \(\left(-8, \dfrac{2}{3}\right)\)

Answer (Detailed Solution Below)

Option 3 : \(\left(-\dfrac{2}{3}, 8\right)\)

Linear Inequations Question 7 Detailed Solution

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Concept:

Comparison using inequality:

For any two real numbers \(a\) and \(b\) if \(\rm |a| < |b|\) then \(\rm a^2 < b^2\).

Calculation:

Use the given inequality and proceed as follows:

\(\rm|2x-3| < |x+5|\\ (2x-3)^2 < (x+5)^2\\ 4x^2+9-12x < x^2+25+10x\\ 3x^2-22x-16 < 0\\ 3x^2-24x+2x-16 < 0\\ 3x(x-8)+2(x-8) < 0\\ (3x+2)(x-8) < 0 \)

Therefore, we conclude that \(\left(-\dfrac{2}{3}, 8\right)\).

What is the solution set for 0 < − x/2 < 3?

  1. (−6, 6)
  2. (−6, 0)
  3. (0, 6)
  4. (−∞, −6)

Answer (Detailed Solution Below)

Option 2 : (−6, 0)

Linear Inequations Question 8 Detailed Solution

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Concept:

Rules for Operations on Inequalities:

  • Adding the same number to each side of an inequality does not change the direction of the inequality symbol.
  • Subtracting the same number from each side of an inequality does not change the direction of the inequality symbol.
  • Multiplying each side of an inequality by a positive number does not change the direction of the inequality symbol.
  • Multiplying each side of an inequality by a negative number reverses the direction of the inequality symbol.
  • Dividing each side of an inequality by a positive number does not change the direction of the inequality symbol.
  • Dividing each side of an inequality by a negative number reverses the direction of the inequality symbol.

 

Calculation

Given: 0 < − x/2 < 3

Multiply by 2 in above inequality (Here 2 is a positive number so the direction of the inequality does not change)

⇒ 0 < −x < 6

⇒ −6 < x < 0

∴ x lies in (−6, 0)

Solve the following in equations \(\frac{2x~+~4}{x~-1}\ge 5\)

  1. (1, 2]
  2. (1, 3)
  3. (1, 3]
  4. (1, 4]

Answer (Detailed Solution Below)

Option 3 : (1, 3]

Linear Inequations Question 9 Detailed Solution

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Concept:

Rules for Operations on Inequalities:

  • Adding the same number to each side of an inequality does not change the direction of the inequality symbol.
  • Subtracting the same number from each side of an inequality does not change the direction of the inequality symbol.
  • Multiplying each side of an inequality by a positive number does not change the direction of the inequality symbol.
  • Multiplying each side of an inequality by a negative number reverses the direction of the inequality symbol.
  • Dividing each side of an inequality by a positive number does not change the direction of the inequality symbol.
  • Dividing each side of an inequality by a negative number reverses the direction of the inequality symbol.

 

Calculation:

Given:

\(\frac{2x+~4}{x-1}\ge 5\)

\(\Rightarrow \frac{2x+~4}{x-1}-5\ge 0\)

\(\Rightarrow ~\frac{2x+4-5\left( x-1 \right)}{x-1}\ge 0\)

\(\Rightarrow ~\frac{2x+4-5x+5}{x-1}\ge 0\)

\(\Rightarrow \frac{9-3x}{x-1}\ge 0\)

\(\Rightarrow ~\frac{-3~\left( x-3 \right)}{x-1}\ge 0\)

Multiplying each side of an inequality by a negative number (-1/3) reverses the direction of the inequality symbol.

\(\Rightarrow ~\frac{~\left( x-3 \right)}{x-1}\le 0\), Here x – 1 ≠ 0 ⇒ x ≠ 1

F1 A.K 24.4.2 Pallavi D3

Hence the solution set of the given in equations is (1, 3]

The solution of \( {1 \over x - 2} < 0\) is :

  1. x  ∈ (−∞ ,2)
  2. x  ∈ (∞,2)
  3. x  ∈ (2,∞)
  4. x  ∈ (2,∞)

Answer (Detailed Solution Below)

Option 1 : x  ∈ (−∞ ,2)

Linear Inequations Question 10 Detailed Solution

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Concept:

Solution Set: The set of all possible values of x.

When (a > 0 and b < 0) or (a < 0 and b > 0) then \(\rm \frac a b <0\)

When (a > 0 and b > 0) or (a < 0 and b < 0) then \(\rm \frac a b >0\)

Calculation:

Here, \( {1 \over x - 2} < 0\)

So, x - 2 < 0

⇒ x < 2

∴ x ∈ (-∞ , 2)

The Minimum value of P = 6x + 16y subject to constraints x ≤ 40, y ≤ 20 and x, y ≥ 0 is

  1. 240
  2. 320
  3. 0
  4. None of these

Answer (Detailed Solution Below)

Option 3 : 0

Linear Inequations Question 11 Detailed Solution

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Calculation:

Here, constraints: x ≤ 40, y ≤  20 and x, y ≥ 0 

F1 Aman 29.9.20 Pallavi D4

 

We get minimum value of P, only when x = 0 and y = 0

So, P = 6(0) + 16(0) = 0

Hence, option (3) is correct.

The solution of the inequality \(\frac{x}{4} > \frac{x}{2} + 1\)will be - 

  1. x > 4
  2. x > -4
  3. x < -4
  4. -4 < x >4

Answer (Detailed Solution Below)

Option 3 : x < -4

Linear Inequations Question 12 Detailed Solution

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CALCULATION:

Given \(\frac{x}{4} > \frac{x}{2} + 1\)

\( \Rightarrow \frac{x}{4} - \frac{x}{2} > 1\)

\(\Rightarrow \frac{{x - 2x}}{4} > 1\)

\( \Rightarrow \frac{{ - x}}{4} > 1\)

\( \Rightarrow \frac{{ - x}}{4} > 1\)

\( \Rightarrow - x > 4\)

\( \Rightarrow x < - 4\)

Therefore option (3) is the correct answer.

If |x2 - 3x + 2| > x2 - 3x + 2, then which one of the following is correct?

  1. x ≤ 1 or x ≥ 2
  2. 1 ≤ x ≤ 2
  3. 1 < x < 2
  4. x is any real value except 3 and 4

Answer (Detailed Solution Below)

Option 3 : 1 < x < 2

Linear Inequations Question 13 Detailed Solution

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Concept:

If f (x) = |x|, then \(f\left( x \right) = \;\left\{ {\begin{array}{*{20}{c}} {x\;,\;x \ge 0}\\ { - x,\;x < 0} \end{array}} \right.\)

Calculation:

Given: |x2 - 3x + 2| > x2 - 3x + 2

Case-1: If x ≥ 0

⇒ x2 - 3x + 2 > x2 - 3x + 2

∴ No real value of x satisfies the above equation.

Case-2:- If x < 0

⇒ - (x2 - 3x + 2) > x2 - 3x + 2

⇒ x2 - 3x + 2 < 0

⇒ (x - 1) (x - 2) < 0

⇒ 1 < x < 2.

The constraints –x + y ≤ 1, −x + 3y ≤ 9 and x, y ≥ 0 defines on

  1. Bounded feasible space
  2. Unbounded feasible space
  3. Both unbounded and bounded feasible space
  4. None of the above

Answer (Detailed Solution Below)

Option 2 : Unbounded feasible space

Linear Inequations Question 14 Detailed Solution

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Concept:

First, graph the "equals" line, then shade in the correct area.

There are three steps:

  • Rearrange the equation so "y" is on the left and everything else on the right.
  • Plot the "y = " line (make it a solid line for y ≤ or y ≥, and a dashed line for y < or y >)
  • Shade above the line for a "greater than" (y > or y ≥) or below the line for a "less than" (y < or y ≤).

 

Calculation:

Given: The constraints –x + y ≤ 1, −x + 3y ≤ 9 and x, y ≥ 0

–x + y ≤ 1

⇒ y ≤ 1 + x

Hence shade below the line.

−x + 3y ≤ 9

3y ≤ 9 + x

⇒ y ≤ 3 + x/3

Hence shade below the line.

F1 A.K 24.4.2 Pallavi D1

We can see through above graph region is unbounded feasible space

If \(\frac{4}{x}<\frac{1}{3}\), what is the possible range of values for x?

  1. x < 3 OR x > 4
  2. x < 0 OR x > 12
  3. x < 3 OR x > 0
  4. x < 0 OR x > 3

Answer (Detailed Solution Below)

Option 2 : x < 0 OR x > 12

Linear Inequations Question 15 Detailed Solution

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Concept:

Rules for Operations on Inequalities:

  • Adding the same number to each side of an inequality does not change the direction of the inequality symbol.
  • Subtracting the same number from each side of an inequality does not change the direction of the inequality symbol.
  • Multiplying each side of an inequality by a positive number does not change the direction of the inequality symbol.
  • Multiplying each side of an inequality by a negative number reverses the direction of the inequality symbol.
  • Dividing each side of an inequality by a positive number does not change the direction of the inequality symbol.
  • Dividing each side of an inequality by a negative number reverses the direction of the inequality symbol.

 

Calculation:

Given:

\(\frac{4}{x}<\frac{1}{3}\)

\(\Rightarrow \frac{4}{x}-~\frac{1}{3}<0\)

\(\Rightarrow \frac{12-x}{3x}<0\)

\(\Rightarrow \frac{-~\left( x-12 \right)}{3x}<0\)

Multiplying each side of an inequality by a negative number reverses the direction of the inequality symbol.

\(\Rightarrow \frac{\left( x-12 \right)}{3x}>0\), Here x ≠ 0

F1 A.K 24.4.2 Pallavi D4

∴ x < 0 OR x > 12

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