October 26, 2026
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.
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:
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.
| 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.
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.
Solve: |3x − 4| = 11.
Solution:
Case 1:
Case 2:
Find the range of x satisfying |x − 4| ≤ 6.
Solution:
How many real solutions does this equation have?
Solution:
Let t = |x|. Since x² = |x|², the equation becomes:
Therefore, t = 2 or t = 6.
Since t = |x|, the possible values of x are 2, −2, 6 and −6.
Find the minimum value of:
Solution:
The three numbers are 2, 5 and 9. Their median is 5, so the sum of distances is minimised at x = 5.
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:
Solve: |x − 1| + |x − 4| = 5
Solution:
The critical points are x = 1 and x = 4.
Case 1: x ≤ 1
This value satisfies the interval.
Case 2: 1 ≤ x ≤ 4
There is no solution in this interval.
Case 3: x ≥ 4
Find the area enclosed by: 2|x| + 3|y| ≤ 6
Solution:
Put y = 0 to find the x-intercepts:
Put x = 0 to find the y-intercepts:
The region is a rhombus with diagonals of lengths 6 and 4.
Test your understanding with these easy, moderate and advanced modulus IPMAT questions. Try solving them before checking the answer key.
| 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. |
Boost your practice by solving the level-based IPMAT modulus questions for 2027
Q1. What is the value of |−17|?
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.
Q3. Find the value of |−8| + |3|.
Q4. Which interval satisfies |x| < 6?
Q5. Solve the equation |2x| = 10.
Q6. What is the value of |0| + |−4|?
Q7. Solve the inequality |x − 2| ≤ 4.
Correct answer: A. −2 ≤ x ≤ 6
Convert the inequality into −4 ≤ x − 2 ≤ 4. Adding 2 throughout gives −2 ≤ x ≤ 6.
Q1. How many real solutions does x² − 5|x| + 6 = 0 have?
Q2. Find the minimum value of |x − 2| + |x − 7|.
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.
Q3. Solve |x − 1| + |x − 4| = 5.
Q4. Find the number of integer pairs (x, y) satisfying |x| + |y| ≤ 2.
A. 9
Q5. Solve |2x − 3| = |x + 1|.
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.
Q6. Find the minimum value of |x − 1| + |x − 4| + |x − 9|.
Q7. How many distinct real solutions does |x² − 4| = 5 have?
Q8. Solve the inequality |2x + 1| < 5.
Correct answer: A. −3 < x < 2
−5 < 2x + 1 < 5. Subtract 1: −6 < 2x < 4. Divide by 2: −3 < x < 2.
Q1. Find the area of the region defined by |x| + |y| ≤ 4.
Q2. For which values of x does |x − 2| + |x + 2| = 8?
Q3. If |x| = 3, what is the value of x² + |x|?
Q4. How many integer values of x satisfy |x − 3| ≤ 2?
Q5. Find the minimum value of |x − 2| + |x − 5| + |x − 11|.
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.
| 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 |
Before attempting advanced modulus IPMAT questions, revise these fundamental formulas and properties.
For example:
Example: |−3 × 4| = |−3| × |4| = 12.
For b ≠ 0:
This property is useful in questions involving the sum of absolute values and inequalities.
Consider the equation |x| = a.
Example:
For a > 0:
For example:
These rules are the foundation of most basic modulus equations and inequalities.
These questions involve one absolute-value expression and usually require splitting the equation into two cases.
Case 1: 2x − 5 = 7, giving x = 6.
Case 2: 2x − 5 = −7, giving x = −1.
Answer: x = 6 or x = −1.
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.
The critical points are x = 2 and x = −1. Solve the expression separately in the intervals determined by these points.
Inequality questions ask you to find the range of values satisfying a condition.
Consider an expression such as:
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.
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.
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:
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.
You do not need to memorise complicated rules to improve your performance. Follow a systematic approach and build your skills step by step.
Start with the meaning of absolute value. Practise positive and negative numbers until you understand why |−x| = |x|.
Begin with equations containing one modulus sign. Once you are comfortable, move to multiple expressions and inequalities.
For an expression such as |x − 2| + |x − 5|, the critical points are 2 and 5. Divide the number line into three intervals:
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.
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.
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.
Record the questions you get wrong and the reason behind each mistake. Revise this notebook regularly, particularly before IPMAT mock tests and the examination.
The equation |x| = a has no real solution when a is negative and only one solution when a is zero.
The expression |x − 3| equals x − 3 only when x ≥ 3. If x < 3, it equals 3 − x.
When using the interval method, check whether each calculated solution belongs to the interval in which it was obtained.
An equation such as |x| = −4 has no real solution because absolute value cannot be negative.
Minimum-value questions involving distances can often be solved using the number line and median property.
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.
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.
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