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

Last updated on Apr 16, 2025

Linear Inequalities is a less familiar concept and its questions may feel tricky to some people. Practice Linear Inequalities MCQs Quiz and improve your accuracy of arriving at the correct answer. This page has listed a range of Linear Inequalities objective questions. We have made the solutions and explanations to all these questions available for our candidates’ reference. Also find tips, strategies and insights for solving Linear Inequalities question answers in lesser time with lesser efforts.

Latest Linear Inequalities MCQ Objective Questions

Linear Inequalities Question 1:

(x − 2)(x − 4)(x − 6)(x − 8) ... (x − 20) ≤ 0. How many integer values of x satisfy the given equation?

Answer (Detailed Solution Below) 15

Linear Inequalities Question 1 Detailed Solution

x = 2, 4, 6, 8, ... , 20 satisfy

(x − 2)(x − 4)(x − 6) ... (x − 20) = 0.

Also, for (x − 2)(x − 4)(x − 6) ... (x − 20) < 0, odd number of terms should be negative.

For x > 20 and x < 2 all terms will be positive.

For x = 19, x − 20 < 0, while all other terms are positive.

Thus, x = 19 is one of the solutions.

For x = 17, x − 20 < 0 and x − 18 < 0, while all other terms are positive.

But then (x − 2)(x − 4) ... (x − 20) > 0, hence x = 17 is not the solution.

For x = 15, three terms are negative.

Hence, x = 15 is one of the solutions.

Similarly, x = 11, 7, 3 are the other solutions.

∴ The required number of solutions = 10 + 5 = 15.

Linear Inequalities Question 2:

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 \( x^2 - 11x + |a| = 0 \) has integer roots. How many possible integer values can 'a' take?

Answer (Detailed Solution Below) 11

Linear Inequalities Question 2 Detailed Solution

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To begin, the discriminant should be a perfect square. Let the discriminant be represented by \( D \).

From the quadratic formula:

\( \frac{-11 \pm \sqrt{D}}{2} \)

we can see that the numerator must be an even number for the roots to be integers.

This means that the discriminant should be a perfect square and specifically the square of an odd number. (This is the condition for odd + odd = even in the numerator.)

The discriminant equation is:

\( D = 121 - 4|a| = 121 - 4|a| \)

Since \( 4|a| \) cannot be negative, we conclude that \( D \) can take values such as 121, 81, 49, 25, 9, and 1.

The value of \( |a| \) can be 0, 10, 18, 24, 28, or 30.

Hence, \( a \) can take the values 0, \( \pm 10 \), \( \pm 18 \), \( \pm 24 \), \( \pm 28 \), and \( \pm 30 \).

Linear Inequalities Question 3:

Find the number of integral values of x for which

(5x - 1) < (x + 1)2 < (7x – 3)

  1. 1
  2. 2
  3. 3
  4. 4
  5. 5

Answer (Detailed Solution Below)

Option 1 : 1

Linear Inequalities Question 3 Detailed Solution

Calculation:

Taking

⇒ (x + 1)> (5x - 1)

⇒ x2 – 3x + 2 > 0

⇒ (x – 1 )( x – 2 ) > 0

X < 1 or x > 2        ……(i)

Taking

⇒ (x + 1)2 < (7x – 3)

⇒ x2 – 5x + 4 > 0

⇒ (x – 1 )( x – 4 ) > 0

⇒ 1 < x < 4         ……(ii)

Combining (1) and (2) we get 2 < x < 4

Hence, x will only take one integer i.e. 3

Linear Inequalities Question 4:

The value of x satisfies the inequality lx - 1| + lx - 2| ≥ 4 if

  1. \(x \in (- \infty, - \frac{1}{2}] \cup [\frac{7}{2}, \infty)\)
  2. \(x \in (- \infty, - \frac{1}{2}) \cup (\frac{7}{2}, \infty)\)
  3. \(x \in [-\frac{1}{2}, \frac{7}{2}]\)
  4. \(x \in (-\frac{1}{2}, \frac{7}{2}]\)
  5. \(x \in (\frac{1}{2}, \frac{7}{2}]\)

Answer (Detailed Solution Below)

Option 1 : \(x \in (- \infty, - \frac{1}{2}] \cup [\frac{7}{2}, \infty)\)

Linear Inequalities Question 4 Detailed Solution

Concept:

 

Calculation :

Case 1: x2

 

Simplifying

⇒ 

case 2: 

⇒ 

Simplifying

⇒ 

this is contradiction

case 3:  x   1 

⇒ 

Simplifying

⇒ 

From Cases 1 and 3, we get

⇒ 

Hence option 1 is correct

Linear Inequalities Question 5:

If a, b and c are the lengths of the sides of a triangle then the range of the values of \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\) is

  1. [1, 2)
  2. [3/2, 3]
  3. [3/2, 2)
  4. [1, 3)

Answer (Detailed Solution Below)

Option 3 : [3/2, 2)

Linear Inequalities Question 5 Detailed Solution

We know that,

AM (b + c, c + a, a + b) ≥ HM (b + c, c + a, a + b)

\(\Rightarrow \frac{b+c+c+a+a+b}{3} \geq \frac{3}{\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}}\)

\(\Rightarrow \frac{2}{9}(a+b+c) \geq \frac{1}{\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}}\)

\(\Rightarrow \frac{9}{2(a+b+c)} \leq \frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\)

\(\Rightarrow \frac{9}{2} \leq \frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b} \)

\(\Rightarrow \frac{9}{2} \leq \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+3 \)

\(\Rightarrow \frac{3}{2} \leq \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+3 \)

\(\Rightarrow \frac{3}{2} \leq \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\) .........(1)

Now, since a, b and c are the lengths of a triangle b + c > a

2b + 2c > a + b + c

\(\frac{1}{2(b+c)}<\frac{1}{a+b+c} \)

\(\frac{a}{b+c}<\frac{2 a}{a+b+c}\)

Similarly, we have, \(\frac{b}{c+a}<\frac{2 b}{a+b+c}\)

\(\frac{c}{a+b}<\frac{2 c}{a+b+c}\)

Adding these inequalities, we get 

\(\frac{\mathrm{a}}{\mathrm{~b}+\mathrm{c}}+\frac{\mathrm{b}}{\mathrm{c}+\mathrm{a}}+\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}<2\) .........(2)

Therefore, From (A) and (B), we haved

\(\frac{3}{2} \leq \frac{a}{b+c}+\frac{b}{b+c}+\frac{c}{a+b}<2\)

Top Linear Inequalities MCQ Objective Questions

Calculate the least whole number, which when subtracted from both the terms of the ratio 5 : 6 gives a ratio less than 17 : 22.

  1. 5
  2. 3
  3. 2
  4. 4

Answer (Detailed Solution Below)

Option 3 : 2

Linear Inequalities Question 6 Detailed Solution

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Given:

Initial ratio = 5 ∶ 6

Final ratio should be less than 17 ∶ 22

Calculation:

Let the least whole number that is needed to be subtracted be a.

According to the question,

(5 - a)/(6 - a) < 17/22

⇒ 5 × 22 - 22a < 17 × 6 - 17a 

⇒ 110 - 22a < 102 - 17a 

⇒ 110 - 102 < - 17a + 22a 

⇒ 8 < 5a 

⇒ 8/5 = 1.6 < a 

∴ The least whole number must be 2.

If a2 - b2 = 88, a - b = 4 then find the value of ab.

  1. 88
  2. 2√26
  3. 117
  4. 22

Answer (Detailed Solution Below)

Option 3 : 117

Linear Inequalities Question 7 Detailed Solution

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Given:

a2 - b2 = 88

a - b = 4

Formula used:

a2 - b2 = (a - b)(a + b)

Calculation:

a - b = 4      ----(1)

(a - b)(a + b) = 88

⇒ 4 × (a + b) = 88

⇒ a + b = 88/4

⇒ a + b = 22      ----(2)

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

⇒ a - b + a + b = 4 + 22

2a = 26

⇒ a = 13

Put the value of a in equation (2), we get

13 + b = 22

⇒ b = 9

value of ab = 13 × 9

⇒ ab = 117

∴ The value of ab is 117.

If \(x = 8 - 2\sqrt {15}\), then find the value of \({\left( {\frac{{{\rm{x}} + {\rm{}}1}}{{\sqrt {\rm{x}} }}} \right)^2}\)

  1. \(15 - \frac{{3\sqrt {15} }}{2}\)
  2. \(17 - \frac{{3\sqrt {15} }}{2}\)
  3. \(14 - \frac{{5\sqrt {15} }}{2}\)
  4. \(12 - \frac{{3\sqrt {15} }}{2}\)

Answer (Detailed Solution Below)

Option 4 : \(12 - \frac{{3\sqrt {15} }}{2}\)

Linear Inequalities Question 8 Detailed Solution

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Given:

 x = 8 – 2√(15)

Calculation:

 x = (√5)2 + (√3)2 – 2√(15)

⇒ x = (√5 - √3)2

⇒ √x = √5 - √3

And 1/√x = (√5 + √3)/2

According to question,

\({\left( {\sqrt x + \frac{1}{{\sqrt x }}} \right)^2} = {\left[ {\sqrt 5 - \sqrt 3 + \left( {\frac{{\sqrt 5 + \sqrt 3 \;}}{2}} \right)} \right]^2}\)

\(\Rightarrow {\left[ {\frac{{3\sqrt 5 - \sqrt 3 }}{2}} \right]^2}\)

\(\Rightarrow \left[ {\frac{{45 + 3 - 6\sqrt {15} }}{4}} \right]\)

\(\Rightarrow \frac{{48 - 6\sqrt {15} }}{4}\)

⇒ \(12 - \frac{{3\sqrt {15} }}{2}\)

Consider the following inequalities:

1. \(\frac{a^2 - b^2 }{a^2 + b^2 } > \frac{a - b}{a + b} \) where a > b > 0

2. \(\frac{a^3 + b^3 }{a^2 + b^2 } > \frac{a^2 + b^2}{a + b} \) only when a > b > 0

Which of the above is/are correct?

  1. 1 only
  2. 2 only
  3. Both 1 and 2
  4. Neither 1 nor 2

Answer (Detailed Solution Below)

Option 1 : 1 only

Linear Inequalities Question 9 Detailed Solution

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Calculation:

Statement:1    \(\frac{a^2 - b^2 }{a^2 + b^2 } > \frac{a - b}{a + b} \) where a > b > 0

Let a = 2 and b = 1

⇒ \(\frac{2^2 - 1^2 }{2^2 + 1^2 } > \frac{2 - 1}{2 + 1} \)

⇒ \(\frac{3 }{5 } > \frac{1}{3} \) which is correct.

Hence, statement 1 is correct.

Statemnt:2   \(\frac{a^3 + b^3 }{a^2 + b^2 } > \frac{a^2 + b^2}{a + b} \) only when a > b > 0

Let a = 2 and b = 1

⇒ \(\frac{2^3 + 1^3 }{2^2 + 1^2 } > \frac{2^2 + 1^2}{2 + 1} \)

⇒ \(\frac{9 }{5 } > \frac{5}{3} \) Which is also correct.

This inequality also holds for a = 1 and b = 2 i.e.

Statement 2 is correct for b > a > 0 as well.

But according to statement 2, inequality will be correct only if a > b > 0

So, statement 2 is incorrect.

Hence, statement (1) is correct but (2) statement (2) is incorrect.

Mistake PointsPlease note the language of  the second statement

"only when a > b > 0"

We have proved thet, statement is correct for 

a = 2 & b = 1 (i.e. a > b > 0) and b = 2 & a = 1 (i.e. b > a > 0)

Hence, due to only word, statement 2 will become incorrect.

If x - 1/x = 10, then x3 - 1/x3 = ?

  1. 970
  2. 1000
  3. 1030
  4. 1060

Answer (Detailed Solution Below)

Option 3 : 1030

Linear Inequalities Question 10 Detailed Solution

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Given:

x - 1/x = 10

Formula:

(a - b)3 = a3 - b3 - 3ab(a - b)

Calculation:

(x - 1/x)3 = x3 - 1/x3 - 3(x)(1/x)(x - 1/x)

⇒ 103 = x3 - 1/x3 - 3 × 10

⇒  x3 - 1/x3 = 1000 + 30

⇒  x3 - 1/x3 = 1030

∴ The value of x3 - 1/x3 is 1030.

If x - 1/x = n,

x3 - 1/x3 = n3 + 3n

Here, n = 10

x3 - 1/x3 = 103 + 3 × 10

⇒ x3 - 1/x3 = 1000 + 30

⇒  x3 - 1/x3 = 1030

∴ The value of x3 - 1/x3 is 1030.

If a + b = 5 and 2a – b = 4. Find the relation between a and b.

  1. a > b
  2. a < b
  3. a = b
  4. No relation

Answer (Detailed Solution Below)

Option 1 : a > b

Linear Inequalities Question 11 Detailed Solution

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a + b = 5      ----(1)

2a – b = 4      ----(2)

Adding Eq. (1) and (2), we get

⇒ a + b + 2a – b = 5 + 4

⇒ 3a = 9

⇒ a = 3

Putting a = 3 in eq. (1) we get b = 2

∴ a > b

The equations ax + 9y = 1 and 9y - x - 1 = 0 represent the same line if a =

  1. 1
  2. -1
  3. 9
  4. None of these

Answer (Detailed Solution Below)

Option 2 : -1

Linear Inequalities Question 12 Detailed Solution

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Given:

Equation1 = ax + 9y = 1

Equation2 = 9y - x - 1 = 0 

Concept used:

If linear equations are a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0. Here, the equations have an infinite number of solutions, if

a1/a2 = b1/b2 = c1/c2

Calculation:

We have equations,

ax + 9y = 1 

⇒ ax + 9y - 1 = 0

and, 9y - x - 1 = 0

⇒ x - 9y + 1 = 0

Here, a1 = a, b1 = 9, c1 = -1

and, a2 = 1, b2 = -9, c2 = 1

As we know that 

a1/a2 = b1/b2 = c1/c2

⇒ a/1 = 9/-9 = -1/1

⇒ a = -1 = -1

⇒ a = -1

∴ The value of a is -1.

Alternate Method

ax + 9y = 1    ----(i)

9y - x - 1 = 0    ----(ii)

From equation (ii) we get,

-x + 9y = 1    ----(iii)

On comparing eq(iii) and eq (i)

We get, a = -1

For what value of 'a', does the inequality 9a - a2 \(\leq\) 17a + 15 holds

  1. -2
  2. -5
  3. -6
  4. All the above

Answer (Detailed Solution Below)

Option 4 : All the above

Linear Inequalities Question 13 Detailed Solution

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Calculation: 

Given: 9a - a2 \(\leq\) 17a + 15

On rearranging

-a2 + 9a \(\leq\) 17a + 15

Shifting the sign

a2 - 9a \(≥\) - 17a - 15

a2 - 9a + 17a + 15 \(≥\) 0

a2 + 8a + 15 \(≥\) 0

a2 + 5a + 3a + 15 \(≥\) 0

a(a+ 5) + 3(a + 5)  \(≥\) 0

(a + 3)(a + 5)  \(≥\) 0

So, -3 and -5 are the roots of the equation.

Now look at the diagram given below,

F1 Shubham Madhuri 24.03.2021 D1

We see that all the numbers less than -5 and all the numbers greater than -3 will give us positive result. While numbers between -5 and -3 will give us negative results

So, all the above value holds for the above equation.

Solve the inequality: \(\rm \begin {vmatrix}\frac{2}{x-4} \end{vmatrix}\) > 1, x ≠ 4

  1. ϵ (4, 6)
  2. ϵ 4
  3. ϵ (-6, 2)
  4. ϵ 0

Answer (Detailed Solution Below)

Option 1 : x ϵ (4, 6)

Linear Inequalities Question 14 Detailed Solution

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Calculation-

\(\rm \begin {vmatrix}\frac{2}{x-4} \end{vmatrix}\) > 1 ⇒\(\rm \begin {vmatrix}\frac{x-4}{2} \end{vmatrix}\) < 1 ⇒|x - 4| < 2

⇒ -2 < x - 4 < 2 ⇒ 2 < x < 6 

As x ≠ 4 hence x ϵ (2, 4) ∪ (4, 6) 

Since, (4, 6) is lying in the interval for all the acceptable values of x ϵ (2, 4) ∪ (4, 6) so this will be the best option.  

∴ The solution of the equality \(\rm \begin {vmatrix}\frac{2}{x-4} \end{vmatrix}\) > 1 is x ϵ (4, 6).

If 2 (3x + 5) > 4x - 5 < 3x + 2; then x can take which of the following values?

  1. 8
  2. 6
  3. -8
  4. -10

Answer (Detailed Solution Below)

Option 2 : 6

Linear Inequalities Question 15 Detailed Solution

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2 (3x + 5) > 4x - 5 < 3x + 2

2 (3x + 5) > 4x - 5

⇒ 6x + 10 > 4x – 5

⇒ 2x > -15

⇒ x > -15/2

4x - 5 < 3x + 2

⇒ x < 7

So, -15/2 < x < 7

∴ The required value of x = 6.
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