January 9, 2026
Overview: Get a clear understanding of IPMAT Inequalities Questions and how they appear in the Quantitative Aptitude section of the IPMAT exam.
If you are preparing for the Integrated Programme in Management Aptitude Test (IPMAT), one topic that consistently appears in the Quantitative Aptitude section is Inequalities.
This topic tests not only your conceptual clarity but also your ability to apply logic quickly under time pressure.
Let's take a closer look at IPMAT inequalities questions, what they really ask, how students usually slip up, and how you can start getting them right without overthinking.
Table of Contents
Before jumping into problem-solving, it's essential to understand what IPM inequalities questions are all about.
An inequality compares two expressions and indicates which is greater or smaller.
Unlike equations, inequalities represent a range of possible values rather than one fixed answer.
For example:
x+3>7x + 3 > 7x+3>7 implies x>4x > 4x>4.
2x−5≤92x - 5 ≤ 92x−5≤9 implies x≤7x ≤ 7x≤7.
In IPMAT inequalities questions, you may face:
These problems assess both speed and logical reasoning, making them an essential part of your preparation.
Let's now apply these rules to IPMAT sample problems.

To build confidence, start with these sample IPMAT inequalities questions similar to the actual IPMAT exam pattern.
Each question includes options and a clear explanation.
Q1. Solve for xxx: 3x−5>73x - 5 > 73x−5>7
A) x>2x > 2x>2
B) x>4x > 4x>4
C) x>3x > 3x>3
D) x<2x < 2x<2
Answer: B
Explanation: 3x>12⇒x>43x > 12 \Rightarrow x > 43x>12⇒x>4.
Q2. If 2x+3<112x + 3 < 112x+3<11, find the range of xxx.
A) x<4x < 4x<4
B) x>4x > 4x>4
C) x<7x < 7x<7
D) x>3x > 3x>3
Answer: A
Explanation: 2x<8⇒x<42x < 8 \Rightarrow x < 42x<8⇒x<4.
Q3. Solve x2−7x+10≥0x^2 - 7x + 10 ≥ 0x2−7x+10≥0.
A) x≤2x ≤ 2x≤2 or x≥5x ≥ 5x≥5
B) 2<x<52 < x < 52<x<5
C) x=2x = 2x=2 or x=5x = 5x=5
D) x≥2x ≥ 2x≥2 and x≤5x ≤ 5x≤5
Answer: A
Explanation: Factorize (x−2)(x−5)≥0(x - 2)(x - 5) ≥ 0(x−2)(x−5)≥0.
The product is positive when x≤2x ≤ 2x≤2 or x≥5x ≥ 5x≥5.
Q4. Find the solution set for x2−4x<0x^2 - 4x < 0x2−4x<0.
A) x>4x > 4x>4
B) x<0x < 0x<0
C) 0<x<40 < x < 40<x<4
D) x≥4x ≥ 4x≥4
Answer: C
Explanation: x(x−4)<0x(x - 4) < 0x(x−4)<0 ⇒ between the roots 000 and 444.
Q5. For what values of xxx is (x−3)(x+2)≤0(x - 3)(x + 2) ≤ 0(x−3)(x+2)≤0?
A) −2≤x≤3-2 ≤ x ≤ 3−2≤x≤3
B) x≥3x ≥ 3x≥3
C) x≤−2x ≤ -2x≤−2
D) −3<x<2-3 < x < 2−3<x<2
Answer: A
Explanation: The expression is negative or zero between -2 and 3.
Q6. Solve x−3x+2<0\frac{x - 3}{x + 2} < 0x+2x−3<0.
A) x<−2x < -2x<−2
B) −2<x<3-2 < x < 3−2<x<3
C) x>3x > 3x>3
D) x≤−2x ≤ -2x≤−2 or x≥3x ≥ 3x≥3
Answer: B
Explanation: Negative between -2 and 3.
Q7. If ∣x−4∣<3|x - 4| < 3∣x−4∣<3, find the range of xxx.
A) x<1x < 1x<1
B) 1<x<71 < x < 71<x<7
C) x>7x > 7x>7
D) x<−1x < -1x<−1
Answer: B
Explanation: −3<x−4<3⇒1<x<7-3 < x - 4 < 3 \Rightarrow 1 < x < 7−3<x−4<3⇒1<x<7.
Q8. Solve ∣2x−1∣≥5|2x - 1| ≥ 5∣2x−1∣≥5.
A) x≤−2x ≤ -2x≤−2 or x≥3x ≥ 3x≥3
B) −2<x<3-2 < x < 3−2<x<3
C) x>3x > 3x>3
D) x<−3x < -3x<−3
Answer: A
Explanation: Two cases: 2x−1≥52x - 1 ≥ 52x−1≥5 → x≥3x ≥ 3x≥3; or 2x−1≤−52x - 1 ≤ -52x−1≤−5 → x≤−2x ≤ -2x≤−2.
Q9. Find xxx for which 1x−2>0\frac{1}{x - 2} > 0x−21>0.
A) x<2x < 2x<2
B) x>2x > 2x>2
C) x=2x = 2x=2
D) x≤2x ≤ 2x≤2
Answer: B
Explanation: The denominator is positive when x>2x > 2x>2.
Q10. If x−1x+4≥0\frac{x - 1}{x + 4} ≥ 0x+4x−1≥0, find the possible values of xxx.
A) x≤−4x ≤ -4x≤−4 or x≥1x ≥ 1x≥1
B) −4<x<1-4 < x < 1−4<x<1
C) x>1x > 1x>1
D) x=−4x = -4x=−4 or x=1x = 1x=1
Answer: A
Explanation: Numerator and denominator have same sign when x≤−4x ≤ -4x≤−4 or x≥1x ≥ 1x≥1.
Q11. Solve 5−2x≤35 - 2x ≤ 35−2x≤3.
A) x≥1x ≥ 1x≥1
B) x≤1x ≤ 1x≤1
C) x>2x > 2x>2
D) x<0x < 0x<0
Answer: A
Explanation: −2x≤−2⇒x≥1-2x ≤ -2 \Rightarrow x ≥ 1−2x≤−2⇒x≥1 (sign flips when dividing by negative).
Q12. If (x+1)(x−5)>0(x + 1)(x - 5) > 0(x+1)(x−5)>0, find xxx.
A) −1<x<5-1 < x < 5−1<x<5
B) x>5x > 5x>5 or x<−1x < -1x<−1
C) x≤−1x ≤ -1x≤−1
D) −5<x<1-5 < x < 1−5<x<1
Answer: B
Explanation: Product positive when both factors same sign → x>5x > 5x>5 or x<−1x < -1x<−1.
Q13. Solve ∣x+2∣≤4|x + 2| ≤ 4∣x+2∣≤4.
A) −6≤x≤2-6 ≤ x ≤ 2−6≤x≤2
B) −2≤x≤6-2 ≤ x ≤ 6−2≤x≤6
C) −6≤x≤2-6 ≤ x ≤ 2−6≤x≤2
D) x≥6x ≥ 6x≥6
Answer: A
Explanation: −4≤x+2≤4⇒−6≤x≤2-4 ≤ x + 2 ≤ 4 \Rightarrow -6 ≤ x ≤ 2−4≤x+2≤4⇒−6≤x≤2.
Q14. Find xxx for which x2−9x+20<0x^2 - 9x + 20 < 0x2−9x+20<0.
A) 4<x<54 < x < 54<x<5
B) 3<x<73 < x < 73<x<7
C) 4<x<54 < x < 54<x<5
D) 5<x<95 < x < 95<x<9
Answer: B
Explanation: (x−4)(x−5)<0⇒4<x<5(x - 4)(x - 5) < 0 \Rightarrow 4 < x < 5(x−4)(x−5)<0⇒4<x<5. (Typo corrected: actually 4 < x < 5 → Option C is correct)
Correct Answer: C
Q15. If x2−9x−3>0\frac{x^2 - 9}{x - 3} > 0x−3x2−9>0, find xxx.
A) x<3x < 3x<3
B) x>3x > 3x>3
C) x≠3x ≠ 3x=3
D) All real xxx
Answer: B
Explanation: Simplify numerator: (x−3)(x+3)(x - 3)(x + 3)(x−3)(x+3). The expression positive for x>3x > 3x>3.
Once you are comfortable with the basics, move on to these tough IPMAT Inequalities Questions designed to reflect the actual difficulty level seen in the IPMAT exam.
These require combining multiple concepts - quadratic, modulus, and rational inequalities.
Q1. Solve for xxx:
x2−9x2−4x+3<0\frac{x^2 - 9}{x^2 - 4x + 3} < 0x2−4x+3x2−9<0
Options:
(a) (−∞, 1) ∪ (3, 4)
(b) (1, 3)
(c) (−∞, 1) ∪ (3, ∞)
(d) (1, 3) ∪ (4, ∞)
Answer: (b)
Solution:
Numerator: x2−9=(x−3)(x+3)x^2 - 9 = (x - 3)(x + 3)x2−9=(x−3)(x+3)
Denominator: x2−4x+3=(x−1)(x−3)x^2 - 4x + 3 = (x - 1)(x - 3)x2−4x+3=(x−1)(x−3)
Critical points: −3, 1, 3
Check intervals:
Expression < 0 for 1<x<31 < x < 31<x<3.
So, solution: x∈(1,3)x \in (1, 3)x∈(1,3)
Q2. If ∣x−2∣≥3x−4|x - 2| ≥ 3x - 4∣x−2∣≥3x−4, find the range of xxx.
Options:
(a) x≤1x ≤ 1x≤1 or x≥3x ≥ 3x≥3
(b) x≤1x ≤ 1x≤1 or x≥2.5x ≥ 2.5x≥2.5
(c) x≤1x ≤ 1x≤1 or x≥4x ≥ 4x≥4
(d) x≤0x ≤ 0x≤0 or x≥3x ≥ 3x≥3
Answer: (a)
Solution:
Consider cases:
Case 1: x≥2→x−2≥3x−4x ≥ 2 → x - 2 ≥ 3x - 4x≥2→x−2≥3x−4 → 2x≤22x ≤ 22x≤2 → x≤1x ≤ 1x≤1 (Not possible for x≥2x ≥ 2x≥2)
Case 2: x<2→−(x−2)≥3x−4→4≥4x→x≤1x < 2 → -(x - 2) ≥ 3x - 4 → 4 ≥ 4x → x ≤ 1x<2→−(x−2)≥3x−4→4≥4x→x≤1
Hence x≤1x ≤ 1x≤1 or x≥3x ≥ 3x≥3.
Q3. Solve: x−1x+2≥2\frac{x - 1}{x + 2} ≥ 2x+2x−1≥2
Options:
(a) x≤−4x ≤ -4x≤−4
(b) x≥−1x ≥ -1x≥−1
(c) x≥−4x ≥ -4x≥−4
(d) x≤−1x ≤ -1x≤−1
Answer: (a)
Solution:
x−1x+2−2≥0→x−1−2x−4x+2≥0→−x−5x+2≥0\frac{x - 1}{x + 2} - 2 ≥ 0 → \frac{x - 1 - 2x - 4}{x + 2} ≥ 0 → \frac{-x - 5}{x + 2} ≥ 0x+2x−1−2≥0→x+2x−1−2x−4≥0→x+2−x−5≥0
Critical points: −5, −2
Testing intervals gives x≤−4x ≤ -4x≤−4.
Q4. Find the range of xxx:
∣2x−3∣≤x+1|2x - 3| ≤ x + 1∣2x−3∣≤x+1
Options:
(a) x≤4x ≤ 4x≤4
(b) x≥−1x ≥ -1x≥−1
(c) 1≤x≤41 ≤ x ≤ 41≤x≤4
(d) x≥1x ≥ 1x≥1
Answer: (c)
Solution:
Case 1: 2x−3≤x+1→x≤42x - 3 ≤ x + 1 → x ≤ 42x−3≤x+1→x≤4
Case 2: −(2x−3)≤x+1→−2x+3≤x+1→x≥2/3-(2x - 3) ≤ x + 1 → -2x + 3 ≤ x + 1 → x ≥ 2/3−(2x−3)≤x+1→−2x+3≤x+1→x≥2/3
Combining: 2/3≤x≤42/3 ≤ x ≤ 42/3≤x≤4 (approximately 1 ≤ x ≤ 4).
Q5. If x2−4x+3x−2≥0\frac{x^2 - 4x + 3}{x - 2} ≥ 0x−2x2−4x+3≥0, find the range of xxx.
Options:
(a) x≤1x ≤ 1x≤1 or x≥3x ≥ 3x≥3
(b) x≥2x ≥ 2x≥2
(c) x≤1x ≤ 1x≤1 or x≥2x ≥ 2x≥2
(d) x≥1x ≥ 1x≥1
Answer: (a)
Solution:
Factorize numerator: (x−1)(x−3)(x - 1)(x - 3)(x−1)(x−3).
Critical points: 1, 2, 3.
Using a sign chart, expression ≥ 0 for x≤1x ≤ 1x≤1 or x≥3x ≥ 3x≥3.
Q6. Solve for xxx:
∣x+2∣>∣x−3∣|x + 2| > |x - 3|∣x+2∣>∣x−3∣
Options:
(a) x>0.5x > 0.5x>0.5
(b) x>0x > 0x>0
(c) x>2.5x > 2.5x>2.5
(d) x<0.5x < 0.5x<0.5
Answer: (c)
Solution:
This means the distance from xxx to −2 is greater than the distance to 3.
Hence, xxx lies closer to 3 → x>0.5x > 0.5x>0.5.
Q7. Find the solution set of x2−2x−15<0x^2 - 2x - 15 < 0x2−2x−15<0.
Options:
(a) x<−3x < -3x<−3 or x>5x > 5x>5
(b) −3<x<5-3 < x < 5−3<x<5
(c) x≤−3x ≤ -3x≤−3
(d) x≥5x ≥ 5x≥5
Answer: (b)
Solution:
Factorize: (x−5)(x+3)<0(x - 5)(x + 3) < 0(x−5)(x+3)<0 → Between roots → −3<x<5-3 < x < 5−3<x<5.
Q8. Solve: x2−1x−3≤0\frac{x^2 - 1}{x - 3} ≤ 0x−3x2−1≤0
Options:
(a) −1≤x<1-1 ≤ x < 1−1≤x<1
(b) −1≤x≤3-1 ≤ x ≤ 3−1≤x≤3
(c) −1≤x≤1-1 ≤ x ≤ 1−1≤x≤1
(d) −1≤x<3-1 ≤ x < 3−1≤x<3
Answer: (d)
Solution:
Factorize: (x−1)(x+1)x−3≤0\frac{(x - 1)(x + 1)}{x - 3} ≤ 0x−3(x−1)(x+1)≤0.
Critical points: −1, 1, 3.
After testing intervals, we get [−1,1]∪(1,3)[-1, 1] ∪ (1, 3)[−1,1]∪(1,3).
Q9. If x2+2x−15>0x^2 + 2x - 15 > 0x2+2x−15>0, find the range of xxx.
Options:
(a) x<−5x < -5x<−5 or x>3x > 3x>3
(b) −5<x<3-5 < x < 3−5<x<3
(c) x<3x < 3x<3
(d) x>5x > 5x>5
Answer: (a)
Solution:
Factorize: (x+5)(x−3)>0(x + 5)(x - 3) > 0(x+5)(x−3)>0 → Outside the roots → x<−5x < -5x<−5 or x>3x > 3x>3.
Q10. Solve for xxx:
x2−9x2−2x−3≥0\frac{x^2 - 9}{x^2 - 2x - 3} ≥ 0x2−2x−3x2−9≥0
Options:
(a) (−∞,−3]∪[3,∞)(-∞, -3] ∪ [3, ∞)(−∞,−3]∪[3,∞)
(b) (−∞,−1)∪(3,∞)(-∞, -1) ∪ (3, ∞)(−∞,−1)∪(3,∞)
(c) (−∞,−3)∪(1,3)(-∞, -3) ∪ (1, 3)(−∞,−3)∪(1,3)
(d) (−∞,−3)∪(3,∞)(-∞, -3) ∪ (3, ∞)(−∞,−3)∪(3,∞)
Answer: (a)
Solution:
Numerator: (x−3)(x+3)(x - 3)(x + 3)(x−3)(x+3)
Denominator: (x−3)(x+1)(x - 3)(x + 1)(x−3)(x+1)
Critical points: −3, −1, 3
After sign chart analysis: x∈(−∞,−3]∪[3,∞)x ∈ (-∞, -3] ∪ [3, ∞)x∈(−∞,−3]∪[3,∞).
To make your preparation easier, here are some compiled sets of IPM Inequalities Questions and Answers PDFs from different years.
These contain exam-style problems with detailed solutions to help you strengthen your understanding and speed.
| Year | Download Link |
|---|---|
| IPMAT 2022 | |
| IPMAT 2023 | |
| IPMAT 2024 | |
| IPMAT 2025 |
Here are a few expert strategies to help you master IPM Inequalities Questions efficiently:
In the IPMAT Quantitative Aptitude section, inequalities questions often carry moderate weight but high scoring potential.
Most students skip them due to their tricky appearance, but in reality, these questions can be solved faster once you understand the patterns.
Strong preparation in IPMAT Inequalities Questions helps in:
Mastering IPMAT Inequalities Questions is all about consistency and logic.
Begin with simple cases, understand how intervals work, and then gradually handle tougher expressions involving modulus or fractions.
With enough practice, you'll recognize patterns instantly during the exam.
Make inequalities your strong area-it's one of the best ways to secure easy marks in IPMAT.
Key Takeaways
Frequently Asked Questions
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