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 \(\frac{x + 2}{x + 3} > 1 \)
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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 \(\frac{1}{5}\left( {\frac{{3x}}{5} + 4} \right) \ge \frac{1}{3}(x - 6)\).
Answer (Detailed Solution Below)
Linear Inequations Question 6 Detailed Solution
Download Solution PDFCALCULATIONS:
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 -
Answer (Detailed Solution Below)
Linear Inequations Question 7 Detailed Solution
Download Solution PDFConcept:
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?
Answer (Detailed Solution Below)
Linear Inequations Question 8 Detailed Solution
Download Solution PDFConcept:
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 \(\frac{2x~+~4}{x~-1}\ge 5\)
Answer (Detailed Solution Below)
Linear Inequations Question 9 Detailed Solution
Download Solution PDFConcept:
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
The solution of \( {1 \over x - 2} < 0\) is :
Answer (Detailed Solution Below)
Linear Inequations Question 10 Detailed Solution
Download Solution PDFConcept:
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
Answer (Detailed Solution Below)
Linear Inequations Question 11 Detailed Solution
Download Solution PDFCalculation:
Here, constraints: x ≤ 40, y ≤ 20 and x, y ≥ 0
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 -
Answer (Detailed Solution Below)
Linear Inequations Question 12 Detailed Solution
Download Solution PDFCALCULATION:
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?
Answer (Detailed Solution Below)
Linear Inequations Question 13 Detailed Solution
Download Solution PDFConcept:
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
Answer (Detailed Solution Below)
Linear Inequations Question 14 Detailed Solution
Download Solution PDFConcept:
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.
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?
Answer (Detailed Solution Below)
Linear Inequations Question 15 Detailed Solution
Download Solution PDFConcept:
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
∴ x < 0 OR x > 12