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IPMAT Modulus Questions and Answers PDF Download (2027)

Author : Lalita Vishwakarma

October 26, 2026

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Preparing for the management exam and finding IPMAT modulus questions confusing? You are not alone! Modulus is an important algebra concept that can be tested through absolute-value equations, inequalities, functions and geometry-based problems.

The good news is that modulus questions become much easier when you understand the basic rules and learn how to apply them. This guide covers IPMAT modulus questions with solutions, important formulas, practice questions, previous-year question resources and preparation tips to help you study with confidence.

Whether you are looking for modulus IPMAT questions, a modulus IPMAT question paper, or an IPMAT modulus question paper PDF, start with the concepts and examples below.

IPMAT Modulus Questions 2027: Overview

Modulus, also called absolute value, represents a number's distance from zero on the number line. It is written using two vertical bars, like this: |x|. For example:

|5| = 5 |−5| = 5 |0| = 0

Both 5 and −5 have the same absolute value because they are equally far from zero. In simple words, modulus gives the non-negative distance of a number from zero.

Basic Rules of IPMAT Modulus Questions

Expression Value or Rule
|7| 7
|−9| 9
|0| 0
|x| when x ≥ 0 x
|x| when x < 0 −x

Consider the expression |x − 3|. Its value depends on whether x is greater than or equal to 3 or less than 3.

If x ≥ 3, then |x − 3| = x − 3
If x < 3, then |x − 3| = 3 − x
Remember: Never remove a modulus sign without checking the sign of the expression inside it.

Solved IPMAT Modulus Questions to Practice Now

The following are original practice questions designed to explain common modulus concepts. These also include previous IPMAT questions, so practice them to improve your conceptual clarity and speed.

Question 1: Solve the Modulus Equation

Solve: |3x − 4| = 11.

Solution:

Case 1:

3x − 4 = 11
3x = 15
x = 5

Case 2:

3x − 4 = −11
3x = −7
x = −7/3
Final answer: x = 5 or x = −7/3.

Question 2: Solve a Modulus Inequality

Find the range of x satisfying |x − 4| ≤ 6.

Solution:

−6 ≤ x − 4 ≤ 6
−2 ≤ x ≤ 10
Final answer: −2 ≤ x ≤ 10.

Question 3: Find the Number of Real Solutions

How many real solutions does this equation have?

x² − 8|x| + 12 = 0

Solution:

Let t = |x|. Since x² = |x|², the equation becomes:

t² − 8t + 12 = 0
(t − 2)(t − 6) = 0

Therefore, t = 2 or t = 6.

Since t = |x|, the possible values of x are 2, −2, 6 and −6.

Final answer: 4 distinct real solutions.

Question 4: Find the Minimum Value

Find the minimum value of:

f(x) = |x − 2| + |x − 5| + |x − 9|

Solution:

The three numbers are 2, 5 and 9. Their median is 5, so the sum of distances is minimised at x = 5.

f(5) = |5 − 2| + |5 − 5| + |5 − 9|
= 3 + 0 + 4
= 7
Final answer: 7.

Question 5: Count the Integer Pairs

How many integer pairs (x, y) satisfy: |x − 2| + |y − 3| ≤ 2

Solution:

Let u = x − 2 and v = y − 3. The inequality becomes: |u| + |v| ≤ 2

Count the integer possibilities:

  • When |u| = 0, there are 5 possible values of v.
  • When |u| = 1, there are 2 choices of u and 3 choices of v for each: 6 pairs.
  • When |u| = 2, there are 2 choices of u and 1 choice of v for each: 2 pairs.
5 + 6 + 2 = 13
Final answer: 13 integer pairs.

Question 6: Solve an Equation with Two Modulus Expressions

Solve: |x − 1| + |x − 4| = 5

Solution:

The critical points are x = 1 and x = 4.

Case 1: x ≤ 1

(1 − x) + (4 − x) = 5
5 − 2x = 5
x = 0

This value satisfies the interval.

Case 2: 1 ≤ x ≤ 4

(x − 1) + (4 − x) = 5
3 = 5

There is no solution in this interval.

Case 3: x ≥ 4

(x − 1) + (x − 4) = 5
2x − 5 = 5
x = 5
Final answer: x = 0 or x = 5.

Question 7: Find the Area of a Modulus Region

Find the area enclosed by: 2|x| + 3|y| ≤ 6

Solution:

Put y = 0 to find the x-intercepts:

2|x| = 6 ⇒ x = ±3

Put x = 0 to find the y-intercepts:

3|y| = 6 ⇒ y = ±2

The region is a rhombus with diagonals of lengths 6 and 4.

Area = ½ × d₁ × d₂
= ½ × 6 × 4
= 12 square units
Final answer: 12 square units.

IPMAT Modulus Practice Questions for 2027

Test your understanding with these easy, moderate and advanced modulus IPMAT questions. Try solving them before checking the answer key.

Easy-Level Questions

Q1. Solve the equation |x − 7| = 3.

  1. 4 and 10
  2. 3 and 7
  3. −4 and 10
  4. 4 only

Q2. Find |−15| + |6|.

  1. 9
  2. 21
  3. −9
  4. −21

Q3. Solve the inequality |x| < 4.

  1. x < −4
  2. x > 4
  3. −4 < x < 4
  4. x ≤ 4

Moderate-Level Questions

Q4. How many real solutions does the equation have? x² − 5|x| + 6 = 0

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

Q5. Find the minimum value of: f(x) = |x − 1| + |x − 6|

  1. 0
  2. 5
  3. 6
  4. 7

Q6. Find the number of integer pairs satisfying: |x| + |y| ≤ 1

  1. 3
  2. 4
  3. 5
  4. 6

Advanced-Level Questions

Q7. Solve the equation: |x − 2| + |x + 2| = 8

  1. x = ±2
  2. x = ±4
  3. x = ±6
  4. x = 0 only

Q8. Find the area enclosed by: |x| + |y| ≤ 4

  1. 8 square units
  2. 12 square units
  3. 16 square units
  4. 32 square units

Q9. If a, b and c are distinct real numbers satisfying: |a − b| + |b − c| = |a − c| Which statement must be true?

  1. a is always the largest number.
  2. b lies between a and c.
  3. c is always the smallest number.
  4. a, b and c must be positive.

Q10. Find the minimum value of: f(x) = |x − 2| + |x − 4| + |x − 10|

  1. 6
  2. 8
  3. 10
  4. 12

Answer Key with Explanations

Question Correct Answer Explanation
Q1 A x = 7 ± 3, giving 4 and 10.
Q2 B 15 + 6 = 21.
Q3 C The values lie strictly between −4 and 4.
Q4 D Let t = |x|. Then t² − 5t + 6 = 0, so t = 2 or 3, giving four distinct roots.
Q5 B The minimum sum of distances between 1 and 6 is 5.
Q6 C The five pairs are (0,0), (1,0), (−1,0), (0,1) and (0,−1).
Q7 B The equation holds for x ≤ −2 or x ≥ 2 when x = −4 or x = 4.
Q8 D The rhombus has diagonals of length 8 each. Its area is ½ × 8 × 8 = 32.
Q9 B Equality in the distance triangle inequality means b lies between a and c.
Q10 B The median is 4, giving a minimum of 2 + 0 + 6 = 8.

Level-Based IPMAT Modulus Questions with Answers for 2027

Boost your practice by solving the level-based IPMAT modulus questions for 2027

IPMAT Modulus Questions Easy Level

Q1. What is the value of |−17|?

  • A. −17
  • B. 0
  • C. 17
  • D. 1

Correct answer: C. 17
The modulus of a number represents its non-negative distance from zero. Therefore, |−17| = 17.

Q2. Solve the equation |x − 5| = 3.

  • A. x = 2 or x = 8
  • B. x = 3 or x = 5
  • C. x = −2 or x = 8
  • D. x = 2 only
Show correct answer and explanation

Q3. Find the value of |−8| + |3|.

  • A. 5
  • B. 11
  • C. −5
  • D. −11
Show correct answer and explanation

Q4. Which interval satisfies |x| < 6?

  • A. x < −6
  • B. x > 6
  • C. −6 < x < 6
  • D. −6 ≤ x ≤ 6
Show correct answer and explanation

Q5. Solve the equation |2x| = 10.

  • A. x = 5 only
  • B. x = −5 only
  • C. x = 5 or x = −5
  • D. x = 10 or x = −10
Show correct answer and explanation

Q6. What is the value of |0| + |−4|?

  • A. 0
  • B. 4
  • C. −4
  • D. 8
Show correct answer and explanation

Q7. Solve the inequality |x − 2| ≤ 4.

  • A. −2 ≤ x ≤ 6
  • B. −4 ≤ x ≤ 4
  • C. x ≤ −2 or x ≥ 6
  • D. 2 ≤ x ≤ 4
Show the correct answer and explanation

Correct answer: A. −2 ≤ x ≤ 6

Convert the inequality into −4 ≤ x − 2 ≤ 4. Adding 2 throughout gives −2 ≤ x ≤ 6.

Moderate-Level IPMAT Modulus Questions with Solutions 2027

Q1. How many real solutions does x² − 5|x| + 6 = 0 have?

  • A. 1
  • B. 2
  • C. 3
  • D. 4
Show correct answer and explanation

Q2. Find the minimum value of |x − 2| + |x − 7|.

  • A. 0
  • B. 2
  • C. 5
  • D. 9
Show correct answer and explanation

Correct answer: C. 5

The expression represents the sum of distances from x to 2 and 7. For any x, this sum is at least the distance between 2 and 7, which is 5. Equality holds for every x between 2 and 7.

Moderate · Question 10

Q3. Solve |x − 1| + |x − 4| = 5.

  • A. x = 1 or x = 4
  • B. x = 0 or x = 5
  • C. x = −1 or x = 5
  • D. x = 0 only
Show correct answer and explanation

Q4. Find the number of integer pairs (x, y) satisfying |x| + |y| ≤ 2.

  • A. 9

  • B. 11
  • C. 13
  • D. 15
Show correct answer and explanation

Q5. Solve |2x − 3| = |x + 1|.

  • A. x = 4 or x = 2/3
  • B. x = −4 or x = 2/3
  • C. x = 4 only
  • D. x = −1 or x = 3
Show correct answer and explanation

Correct answer: A. x = 4 or x = 2/3

If |A| = |B|, then A = B or A = −B.

Case 1: 2x − 3 = x + 1, giving x = 4.

Case 2: 2x − 3 = −x − 1, giving 3x = 2 and x = 2/3.

Moderate · Question 13

Q6. Find the minimum value of |x − 1| + |x − 4| + |x − 9|.

  • A. 5
  • B. 8
  • C. 9
  • D. 13
Show correct answer and explanation

Q7. How many distinct real solutions does |x² − 4| = 5 have?

  • A. 0
  • B. 1
  • C. 2
  • D. 4
Show correct answer and explanation

Q8. Solve the inequality |2x + 1| < 5.

  • A. −3 < x < 2
  • B. −2 < x < 3
  • C. x < −3 or x > 2
  • D. −5 < x < 5
Show correct answer and explanation

Correct answer: A. −3 < x < 2

−5 < 2x + 1 < 5. Subtract 1: −6 < 2x < 4. Divide by 2: −3 < x < 2.

Advanced-Level IPMAT Modulus Questions with Answers 2027

Q1. Find the area of the region defined by |x| + |y| ≤ 4.

  • A. 8 square units
  • B. 16 square units
  • C. 24 square units
  • D. 32 square units
Show correct answer and explanation

Q2. For which values of x does |x − 2| + |x + 2| = 8?

  • A. x = −2 or x = 2
  • B. x = −4 or x = 4
  • C. x = −6 or x = 6
  • D. x = 0 only
Show correct answer and explanation

Q3. If |x| = 3, what is the value of x² + |x|?

  • A. 6
  • B. 9
  • C. 12
  • D. 15
Show correct answer and explanation

Q4. How many integer values of x satisfy |x − 3| ≤ 2?

  • A. 3
  • B. 4
  • C. 5
  • D. 6
Show correct answer and explanation

Q5. Find the minimum value of |x − 2| + |x − 5| + |x − 11|.

  • A. 6
  • B. 9
  • C. 12
  • D. 14
Show correct answer and explanation

Correct answer: C. 12

The median of 2, 5 and 11 is 5. Substitute x = 5:

|5 − 2| + |5 − 5| + |5 − 11| = 3 + 0 + 6 = 9.

Correction: The minimum is 9, so the correct option is B.

Quick Answer Key: IPMAT Modulus Questions

Question Answer Question Answer
1 C 11 C
2 A 12 A
3 B 13 B
4 C 14 C
5 C 15 A
6 B 16 D
7 A 17 B
8 D 18 C
9 C 19 C
10 B 20 B
Preparation tip: Complete all 20 questions before checking the answer key. Review the explanations for every incorrect answer and practise the same concept again. These are practice MCQs for IPMAT preparation, not a claim that they appeared in an official previous-year paper.

Important Modulus Formulas for IPMAT 2027

Before attempting advanced modulus IPMAT questions, revise these fundamental formulas and properties.

Formula 1: Definition of Absolute Value

|x| = x, when x ≥ 0
|x| = −x, when x < 0

For example:

|x − 4| = x − 4, when x ≥ 4
|x − 4| = 4 − x, when x < 4

Formula 2: Product Property

|ab| = |a| × |b|

Example: |−3 × 4| = |−3| × |4| = 12.

Formula 3: Quotient Property

For b ≠ 0:

|a / b| = |a| / |b|

Formula 4: Triangle Inequality

|a + b| ≤ |a| + |b|

This property is useful in questions involving the sum of absolute values and inequalities.

Formula 5: Basic Modulus Equation

Consider the equation |x| = a.

  • If a < 0, there is no real solution.
  • If a = 0, the only solution is x = 0.
  • If a > 0, the solutions are x = a and x = −a.

Example:

|x| = 6 ⇒ x = 6 or x = −6

Formula 6: Modulus Inequalities

For a > 0:

|x| < a ⇒ −a < x < a
|x| ≤ a ⇒ −a ≤ x ≤ a
|x| > a ⇒ x < −a or x > a
|x| ≥ a ⇒ x ≤ −a or x ≥ a

For example:

|x − 2| ≤ 3
−3 ≤ x − 2 ≤ 3
−1 ≤ x ≤ 5

These rules are the foundation of most basic modulus equations and inequalities.

Types of IPMAT Modulus Questions in 2027

Type 1: Simple Modulus Equations

These questions involve one absolute-value expression and usually require splitting the equation into two cases.

|2x − 5| = 7

Case 1: 2x − 5 = 7, giving x = 6.

Case 2: 2x − 5 = −7, giving x = −1.

Answer: x = 6 or x = −1.

Type 2: Multiple Modulus Expressions

These questions contain two or more absolute-value expressions. Identify the critical points where the expressions inside the modulus signs become zero, then divide the number line into intervals.

|x − 2| + |x + 1| = 5

The critical points are x = 2 and x = −1. Solve the expression separately in the intervals determined by these points.

Type 3: Modulus Inequalities

Inequality questions ask you to find the range of values satisfying a condition.

|x + 3| < 5
−5 < x + 3 < 5
−8 < x < 2

Type 4: Minimum and Maximum Values

Consider an expression such as:

f(x) = |x − a| + |x − b| + |x − c|

Each term represents a distance on the number line. For three real numbers a, b and c, the sum of distances is minimised when x is the median of those three numbers.

Type 5: Modulus and Coordinate Geometry

These questions combine absolute values with geometric regions, intercepts or integer coordinates.  2|x| + 3|y| ≤ 6

Find the intercepts, sketch the region and use symmetry to calculate the required area.

Is Modulus Important for IPMAT 2027?

Yes, modulus is a useful topic to include in your IPMAT Quantitative Ability preparation. It connects with several algebra concepts and can also be applied to number-line and coordinate-geometry problems.

Focus on understanding these question types:

  • Basic properties of absolute values
  • Modulus equations and inequalities
  • Equations containing two or more modulus expressions
  • Minimum and maximum value problems
  • Number-line and integer-solution questions
  • Coordinate geometry and area-based questions

Some questions can be solved in just a few steps, while others need careful case analysis. Rather than memorising selected questions, learn the underlying methods so you can solve unfamiliar problems. The exact number of modulus questions is not fixed. Use previous-year papers to understand how the concept has been tested, but do not rely on a predicted question count.

How to Prepare for Modulus Questions in IPMAT?

You do not need to memorise complicated rules to improve your performance. Follow a systematic approach and build your skills step by step.

Step 1: Understand the Concept First

Start with the meaning of absolute value. Practise positive and negative numbers until you understand why |−x| = |x|.

Step 2: Practise Equations and Inequalities Separately

Begin with equations containing one modulus sign. Once you are comfortable, move to multiple expressions and inequalities.

Step 3: Master the Interval Method

For an expression such as |x − 2| + |x − 5|, the critical points are 2 and 5. Divide the number line into three intervals:

  • x < 2
  • 2 ≤ x ≤ 5
  • x > 5

Remove each modulus sign according to the sign of its expression in that interval. Solve the resulting equation and retain only the solutions that satisfy the interval.

Step 4: Solve Previous-Year Questions

Previous-year IPMAT questions help you understand how a concept is applied in examination conditions. Review the question structure, calculations and solution method rather than memorising the answer.

Step 5: Practise Under Time Limits

After learning the concepts, attempt a short timed practice set. Review incorrect answers and identify whether your mistake came from a concept gap, a sign error, a calculation error, or poor time management.

Step 6: Maintain a Mistake Notebook

Record the questions you get wrong and the reason behind each mistake. Revise this notebook regularly, particularly before IPMAT mock tests and the examination.

Preparation tip: Accuracy matters more than rushing through questions. First learn to solve them correctly, then work towards improving your speed.

Common Mistakes to Avoid While Solving Modulus Questions

Mistake 1: Assuming Every Equation Has Two Solutions

The equation |x| = a has no real solution when a is negative and only one solution when a is zero.

Mistake 2: Removing Modulus Signs Without Checking the Sign

The expression |x − 3| equals x − 3 only when x ≥ 3. If x < 3, it equals 3 − x.

Mistake 3: Ignoring the Original Interval

When using the interval method, check whether each calculated solution belongs to the interval in which it was obtained.

Mistake 4: Forgetting That Absolute Value Is Non-Negative

An equation such as |x| = −4 has no real solution because absolute value cannot be negative.

Mistake 5: Using Long Calculations Unnecessarily

Minimum-value questions involving distances can often be solved using the number line and median property.

Mistake 6: Practising Only Easy Questions

Once your basics are clear, move on to multiple-modulus equations, inequalities, integer solutions and geometry-based problems.

You can revise the fundamental rules and solve representative questions in one day, particularly if your algebra basics are strong. Advanced questions may require additional practice and revision.

Conclusion

Modulus is a manageable topic once you understand the logic behind absolute values. The key is to recognise when to split an equation into cases, how to handle inequalities and when a number-line or geometric interpretation can simplify the solution.

To prepare effectively, revise the formulas, practise different types of IPMAT modulus questions, solve previous-year questions and review your mistakes regularly. Most importantly, focus on understanding the reasoning rather than memorising solutions. This will help you approach unfamiliar questions with greater confidence and improve your overall Quantitative Ability preparation.

Your next step: Attempt the practice set above, check your answers and then move on to topic-wise previous-year questions to strengthen your IPMAT preparation.

Frequently Asked Questions

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About the Author

Faculty
Lalita Vishwakarma

Content Writer

Lalita Vishwakarma is a professional content writer with over 5+ years of experience covering undergraduate management entrance exams and CUET. She writes about IPMAT Indore, IPMAT Rohtak, JIPMAT, NPAT, SET, CUET UG, and CUET PG, helping students understand admission requirements and prioritise their preparation. Her work includes exam updates, eligibility criteria, application processes, syllabi and preparation strategies. With a research-driven approach and student-friendly writing style, she turns complex exam information into practical guidance, helping aspirants make informed choices and prepare with confidence.... more