Inequalities, Modulus and Absolute Value Previous Year Questions (PYQs) for CAT: 14+ Solved Questions with Step-by-Step Solutions

Solve 14+ Inequalities, Modulus and Absolute Value previous year questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Inequalities, Modulus and Absolute Value

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Chapter roadmap

Inequalities, Modulus and Absolute Value

🧭 Modulus Equations and Absolute Value Cases

Learn how absolute value behaves as distance, how to split cases, and how CAT hides simple equations inside modulus symbols.

9 direct CAT PYQs | strongest topic in this chapter

🔢 Inequalities with Integers and Intervals

Convert inequality conditions into clean intervals and count integer solutions carefully.

4 direct CAT PYQs

📈 Rational and Polynomial Inequalities

Use critical points, sign charts, and interval testing to solve higher-level inequality questions.

4 direct CAT PYQs

By the end, you should be able to split expressions by intervals instead of guessing signs randomly.

Topic Hero: Modulus Equations and Absolute Value Cases

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Algebra → Inequalities, Modulus and Absolute Value → Topic 1

Modulus Equations and Absolute Value Cases

Every modulus question asks: “Which side of the breakpoint are we on?”

✅ Meaning of

∣ x ∣

as distance from zero

✅ Split cases using breakpoints

✅ Solve equations with multiple modulus terms

✅ Use geometry shortcuts for distance-sum equations

✅ Count integer and real solutions safely

Inequalities, Modulus and Absolute Value: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Ability MCQ

The largest real value of a for which the equation $∣ x + a ∣ + ∣ x - 1∣ = 2$ has an infinite number of solutions for x is

A.
-1
B.
0
C.
1
D.
2

Correct Answer:

Step-by-Step Solution

Key idea: Distance interpretation of absolute value. Recognisable because it equates a sum of two absolute values to a constant and asks for infinite solutions.

Step 1: Interpret geometrically.

The equation $∣ x + a ∣ + ∣ x - 1∣ = 2$ can be rewritten as:

$∣ x - (- a) ∣ + ∣ x - 1∣ = 2$

The expression $∣ x - (- a) ∣ + ∣ x - 1∣$ represents the sum of the distances from a point $x$ on the number line to two fixed points: $P = - a$ and $Q = 1$ .

The sum of distances from $x$ to $P$ and $Q$ is minimum when $x$ lies on the line segment between $P$ and $Q$ . The minimum value is exactly the distance between $P$ and $Q$ , which is $∣ - a - 1∣ = ∣ a + 1∣$ .

For any $x$ outside the segment $P Q$ , the sum of distances is strictly greater than $∣ a + 1∣$ .

Step 2: Condition for infinite solutions.

The equation states that the sum of distances is a constant, 2.

For the equation to have an infinite number of solutions, the entire line segment between $P$ and $Q$ must satisfy the equation.

This happens if and only if the minimum sum of distances equals the constant 2.

Therefore, the distance between $P$ and $Q$ must be exactly 2.

$∣ a + 1∣ = 2$

Step 3: Solve for $a$ .

$a + 1 = 2 ⟹ a = 1$

$a + 1 = - 2 ⟹ a = - 3$

Step 4: Find the largest real value.

The possible values for $a$ are 1 and -3. The largest is 1.

Answer: 1

Question 2 · Quantitative Ability MCQ

Let $0 \leq a \leq x \leq 100$ and $f (x) = ∣ x - a ∣ + ∣ x - 100∣ + ∣ x - a - 50∣$ . Then the maximum value of f(x) becomes 100 when a is equal to

A.
25
B.
100
C.
50
D.
0

Correct Answer:

Step-by-Step Solution

Key idea: Maximizing a sum of absolute values on an interval. Recognisable because it asks for the maximum value of a modulus function over a restricted domain.

Step 1: Simplify the known absolute values using the given bounds.

Given $0 \leq a \leq x \leq 100$ .

Since $x \geq a$ , we have $x - a \geq 0$ , so $∣ x - a ∣ = x - a$ .

Since $x \leq 100$ , we have $x - 100 \leq 0$ , so $∣ x - 100∣ = 100 - x$ .

Step 2: Substitute these back into $f (x)$ .

$f (x) = (x - a) + (100 - x) + ∣ x - a - 50∣$

$f (x) = 100 - a + ∣ x - (a + 50) ∣$

Step 3: Maximize $f (x)$ on the interval $[a, 100]$ .

The term $∣ x - (a + 50) ∣$ is a V-shaped function, so its maximum on a closed interval must occur at one of the endpoints, $x = a$ or $x = 100$ .

At $x = a$ : $∣ a - (a + 50) ∣ = ∣ - 50∣ = 50$ .

At $x = 100$ : $∣100 - (a + 50) ∣ = ∣50 - a ∣$ .

The maximum of $f (x)$ is $100 - a + max (50, ∣50 - a ∣)$ .

Step 4: Set the maximum value to 100 and solve for $a$ .

We are given that the maximum value is 100:

$100 - a + max (50, ∣50 - a ∣) = 100$

$max (50, ∣50 - a ∣) = a$

Since the maximum is at least 50, we must have $a \geq 50$ .

If $a \geq 50$ , then $∣50 - a ∣ = a - 50$ .

So the condition becomes $max (50, a - 50) = a$ .

Since $a \geq 50$ , $a - 50$ is non-negative, but it cannot equal $a$ (as $- 50 \neq = 0$ ).

Thus, the maximum must be 50, which gives $50 = a$ .

Let's check $a = 50$ : $max (50, ∣50 - 50∣) = max (50, 0) = 50$ . This perfectly matches $a = 50$ .

Answer: 50

Question 3 · Quantitative Ability NAT

The number of distinct integer solutions $(x, y)$ of the equation $∣ x + y ∣ + ∣ x - y ∣ = 2$ , is

Correct Answer:

Step-by-Step Solution

This is a modulus counting question, recognisable by the sum of two absolute values \ $|x+y| + |x-y|\$ set equal to a constant.

Step 1: Apply the absolute value identity.

There is a powerful identity for any real numbers \ $x\$ and \ $y\$ :

\ $|x + y| + |x - y| = 2 \cdot \max(|x|, |y|)\$

Why this works: If \ $|x| \geq |y|\$ , then \ $x+y\$ and \ $x-y\$ have the same sign as \ $x\$ (or one is zero). Their absolute values sum to \ $|x+y| + |x-y| = |x+y + x-y| = |2x| = 2|x| = 2\max(|x|, |y|)\$ . (A similar argument holds if \ $|y| > |x|\$ .)

Step 2: Simplify the given equation.

\ $2 \cdot \max(|x|, |y|) = 2\$

\ $\max(|x|, |y|) = 1\$

Step 3: Interpret the condition geometrically.

\ $\max(|x|, |y|) = 1\$ means that both \ $|x| \leq 1\$ and \ $|y| \leq 1\$ , and <b>at least one of them equals 1</b>.

For integers, this means:

\ $x \in \\{-1, 0, 1\\} \quad \text{and} \quad y \in \\{-1, 0, 1\\}\$

Step 4: Count the valid pairs.

Total pairs with \ $x \in \\{-1, 0, 1\\}\$ and \ $y \in \\{-1, 0, 1\\}\$ : \ $3 \times 3 = 9\$ .

But we must exclude the pair where \ $\max(|x|, |y|) = 0\$ , which is \ $(0, 0)\$ .

Check: \ $|0+0| + |0-0| = 0 \neq 2\$ . So \ $(0,0)\$ is invalid.

Valid pairs: \ $9 - 1 = 8\$ .

Step 5: List them for verification.

\ $(1, 0), (1, 1), (1, -1), (0, 1), (0, -1), (-1, 0), (-1, 1), (-1, -1)\$

Quick check: \ $(1,1): |2| + |0| = 2\$ ✓; \ $(0,1): |1| + |-1| = 2\$ ✓; all others follow similarly.

Answer: 8

Question 4 · Quantitative Ability NAT

The number of distinct real values of x, satisfying the equation $max {x, 2} - min {x, 2} = ∣ x + 2∣ - ∣ x - 2∣$ , is

Correct Answer:

Step-by-Step Solution

This is a max-min and absolute value question, recognisable because the left side uses max and min functions while the right side uses absolute values.

Step 1: Simplify the left-hand side.

For any two real numbers, $max (a, b) - min (a, b) = ∣ a - b ∣$ .

So: $max (x, 2) - min (x, 2) = ∣ x - 2∣$ .

Step 2: Rewrite the equation.

$∣ x - 2∣ = ∣ x + 2∣ - ∣ x - 2∣$

Adding $∣ x - 2∣$ to both sides:

$2∣ x - 2∣ = ∣ x + 2∣$

Step 3: Split into cases based on breakpoints.

The breakpoints are $x = - 2$ and $x = 2$ .

Case 1: $x \geq 2$

$∣ x - 2∣ = x - 2$ and $∣ x + 2∣ = x + 2$ .

$2 (x - 2) = x + 2 ⟹ 2 x - 4 = x + 2 ⟹ x = 6$

Check: $6 \geq 2$ ✓.

Case 2: $- 2 \leq x < 2$

$∣ x - 2∣ = 2 - x$ and $∣ x + 2∣ = x + 2$ .

$2 (2 - x) = x + 2 ⟹ 4 - 2 x = x + 2 ⟹ 3 x = 2 ⟹ x = \frac{2}{3}$

Check: $- 2 \leq \frac{2}{3} < 2$ ✓.

Case 3: $x < - 2$

$∣ x - 2∣ = 2 - x$ and $∣ x + 2∣ = - (x + 2)$ .

$2 (2 - x) = - (x + 2) ⟹ 4 - 2 x = - x - 2 ⟹ x = 6$

But $6 \neq < - 2$ , so no solution in this case.

Step 4: Count valid solutions.

We found $x = 6$ and $x = \frac{2}{3}$ . Both are distinct real values.

Answer: 2

Question 5 · Quantitative Ability MCQ

For a real number x the condition $∣3 x - 20∣ + ∣3 x - 40∣ = 20$ necessarily holds if

A.
$10 < x < 15$
B.
$9 < x < 14$
C.
$7 < x < 12$
D.
$6 < x < 11$

Correct Answer:

Step-by-Step Solution

This is a sum-of-distances (absolute value) question — recognisable because it's a sum of two modulus terms, $∣3 x - 20∣ + ∣3 x - 40∣$ , set equal to a constant. The key idea is to read $∣ y - a ∣ + ∣ y - b ∣$ as "the sum of distances from $y$ to $a$ and from $y$ to $b$ on the number line."

Step 1 — Substitute $y = 3 x$ to simplify.

$∣ y - 20∣ + ∣ y - 40∣ = 20$

Step 2 — Use the distance interpretation. The minimum possible value of $∣ y - 20∣ + ∣ y - 40∣$ is the distance between 20 and 40, which is $40 - 20 = 20$ . This minimum is achieved for every $y$ between 20 and 40 (inclusive) — not just one point, because for any point between two fixed points, the two distances always add up to exactly the distance between the fixed points.

Step 3 — Since the equation asks for the sum to equal exactly 20 (the minimum), the solution set is precisely:

$20 \leq y \leq 40$

Step 4 — Convert back to x using $y = 3 x$ .

$20 \leq 3 x \leq 40 ⟹ \frac{20}{3} \leq x \leq \frac{40}{3}$

$6.67 \leq x \leq 13.33 (approximately)$

Step 5 — This is the exact (tightest) solution set. Now check what "necessarily holds" means. The question asks which interval is such that every x in that interval is guaranteed to satisfy the equation — that is, the interval must lie entirely inside the true solution set $[\frac{20}{3}, \frac{40}{3}]$ , not the other way round.

Step 6 — Test each option against $[6.67, 13.33]$ :

$10 < x < 15$ : contains x=14, which is outside 13.33 — fails.
$9 < x < 14$ : contains x=13.5, outside 13.33 — fails.
$7 < x < 12$ : entirely within $[6.67, 13.33]$ since $6.67 < 7$ and $12 < 13.33$ — holds for every x in this range.
$6 < x < 11$ : contains x=6.5, which is below 6.67 — fails.

Trap to avoid: It's easy to think the question wants the widest matching interval or to test only whether the interval overlaps the solution set, rather than checking it is fully contained in the solution set.

Final answer: $7 < x < 12$ (option C).

Inequalities, Modulus and Absolute Value Previous Year Questions (PYQs) for CAT: 14+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Inequalities, Modulus and Absolute Value

Inequalities, Modulus and Absolute Value

Topic Hero: Modulus Equations and Absolute Value Cases

Modulus Equations and Absolute Value Cases

Inequalities, Modulus and Absolute Value: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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