Theory & Concepts

SSC CGL Inequalities Questions, Formulas & Tricks

Get comprehensive theory, expert shortcuts, and hand-picked practice questions for Inequalities specifically designed for the SSC CGL 2025-26 pattern.

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20 min readDifficulty: Easy-Intermediate

Inequalities test your ability to handle ranges of numbers rather than exact values. In SSC CGL, you will often face questions combining linear, quadratic, and modulus inequalities. The most critical mistake candidates make is forgetting to flip the sign.

Learning path

  • The Negative Flip Rule
  • Quadratic & Wavy Curve
  • Modulus Properties
  • 10 Exam-Level Examples

1. The Negative Flip Rule

Treat inequalities exactly like normal equations (you can add or subtract numbers from both sides safely). But there is one strict exception.

Positive Operations

Adding, subtracting, or multiplying/dividing by a Positive number does not change the sign.

2x>10  ⟹  x>52x > 10 \implies x > 5

The Danger Zone

Multiplying or dividing by a Negative number instantly FLIPS the inequality sign.

−2x>10  ⟹  x<−5-2x > 10 \implies x < -5

2. Quadratic & Wavy Curve Method

When you have a quadratic inequality like x2−5x+6<0x^2 - 5x + 6 < 0, factorize it first into (x−2)(x−3)<0(x-2)(x-3) < 0.

The "Less Than" Sandwich

If (x−a)(x−b)<0(x-a)(x-b) < 0 (where a<ba < b):

xx is sandwiched between the roots: a<x<ba < x < b.

Example: (x−2)(x−3)<0  ⟹  2<x<3(x-2)(x-3) < 0 \implies 2 < x < 3.

The "Greater Than" Split

If (x−a)(x−b)>0(x-a)(x-b) > 0:

xx lies outside the roots: x<ax < a OR x>bx > b.

Example: (x−2)(x−3)>0  ⟹  x<2 or x>3(x-2)(x-3) > 0 \implies x < 2 \text{ or } x > 3.

3. Modulus Properties

The absolute value ∣x∣|x| represents the distance from 0. Inequalities with modulus follow exact translation rules.

Distance Less Than

∣x∣<a|x| < a

Translates directly to: −a<x<a-a < x < a

Example: ∣x∣<5  ⟹  −5<x<5|x| < 5 \implies -5 < x < 5

Distance Greater Than

∣x∣>a|x| > a

Translates to: x<−a OR x>ax < -a \text{ OR } x > a

Example: ∣x∣>5  ⟹  x<−5 OR x>5|x| > 5 \implies x < -5 \text{ OR } x > 5

4. 10 Solved examples

Question 01Exam Pattern

Solve for \( x \): \( 3x - 5 > 10 \).

x > 5
x < 5
x > 15
x < -5
Correct answer: a) x > 5

Solution

Step 1: Treat it like an equation. Add 5 to both sides.
Step 2: \( 3x > 10 + 5 \).
Step 3: \( 3x > 15 \).
Final calculation: Divide by 3 (positive, so don't flip). \( x > 5 \).
Question 02Exam Pattern

Solve for \( x \): \( -2x + 4 \ge 12 \).

x ≥ -4
x ≤ -4
x ≥ 4
x ≤ 4
Correct answer: b) x ≤ -4

Solution

Step 1: Subtract 4 from both sides: \( -2x \ge 12 - 4 \).
Step 2: \( -2x \ge 8 \).
Step 3: Divide by -2. Since we are dividing by a NEGATIVE number, we must FLIP the sign.
Final calculation: \( x \le -4 \).
Question 03Exam Pattern

Solve: \( 5x - 3 \le 2x + 9 \).

x ≤ 3
x ≤ 4
x ≥ 4
x ≥ 3
Correct answer: b) x ≤ 4

Solution

Step 1: Move variables to one side: \( 5x - 2x \le 9 + 3 \).
Step 2: Simplify: \( 3x \le 12 \).
Final calculation: Divide by 3. \( x \le 4 \).
Question 04Exam Pattern

Solve: \( |x - 3| < 5 \).

-2 < x < 8
-8 < x < 2
2 < x < 8
x < -2 or x > 8
Correct answer: a) -2 < x < 8

Solution

Step 1: Use the 'Distance Less Than' rule. Remove modulus by sandwiching.
Step 2: \( -5 < x - 3 < 5 \).
Step 3: Add 3 to all three sections to isolate \( x \).
Final calculation: \( -5 + 3 < x < 5 + 3 \implies -2 < x < 8 \).
Question 05Exam Pattern

Solve: \( |2x + 1| \ge 7 \).

-4 ≤ x ≤ 3
x ≤ -4 or x ≥ 3
x ≤ -3 or x ≥ 4
-3 ≤ x ≤ 4
Correct answer: b) x ≤ -4 or x ≥ 3

Solution

Step 1: Use the 'Distance Greater Than' rule. It splits into two inequalities.
Step 2: Positive branch: \( 2x + 1 \ge 7 \).
Step 3: Negative branch: \( 2x + 1 \le -7 \).
Step 4: Solve Positive: \( 2x \ge 6 \implies x \ge 3 \).
Step 5: Solve Negative: \( 2x \le -8 \implies x \le -4 \).
Final calculation: \( x \le -4 \text{ or } x \ge 3 \).
Question 06Exam Pattern

Find the range of \( x \) for \( x^2 - 5x + 6 < 0 \).

x < 2 or x > 3
2 < x < 3
-3 < x < -2
x < -3 or x > -2
Correct answer: b) 2 < x < 3

Solution

Step 1: Factorize the quadratic: \( (x - 2)(x - 3) < 0 \).
Step 2: Identify the roots: 2 and 3.
Step 3: Since it is 'Less Than' (< 0), the answer is sandwiched between the roots.
Final calculation: \( 2 < x < 3 \).
Question 07Exam Pattern

Find the range of \( x \) for \( x^2 - x - 12 \ge 0 \).

-3 ≤ x ≤ 4
x ≤ -3 or x ≥ 4
x ≤ -4 or x ≥ 3
-4 ≤ x ≤ 3
Correct answer: b) x ≤ -3 or x ≥ 4

Solution

Step 1: Factorize: \( x^2 - 4x + 3x - 12 \ge 0 \implies (x - 4)(x + 3) \ge 0 \).
Step 2: Identify roots: 4 and -3.
Step 3: Since it is 'Greater Than' (≥ 0), the answer points outward away from the roots.
Final calculation: \( x \le -3 \text{ or } x \ge 4 \).
Question 08Exam Pattern

Solve: \( \frac{x-2}{x+3} > 0 \).

-3 < x < 2
x < -3 or x > 2
x < -2 or x > 3
-2 < x < 3
Correct answer: b) x < -3 or x > 2

Solution

Step 1: The wavy curve method applies to fractions just like products.
Step 2: The critical points (roots of numerator and denominator) are 2 and -3.
Step 3: Because the fraction is 'Greater Than' 0, the solution lies outside the critical points.
Final calculation: \( x < -3 \text{ or } x > 2 \). (Note: \( x \neq -3 \) because denominator cannot be 0).
Question 09Exam Pattern

If \( -5 \le 2x - 1 \le 9 \), find the range of \( x \).

-2 ≤ x ≤ 5
-3 ≤ x ≤ 4
-2 ≤ x ≤ 4
-3 ≤ x ≤ 5
Correct answer: a) -2 ≤ x ≤ 5

Solution

Step 1: We must isolate \( x \) in the middle. Start by adding 1 to all three parts.
Step 2: \( -5 + 1 \le 2x - 1 + 1 \le 9 + 1 \).
Step 3: \( -4 \le 2x \le 10 \).
Step 4: Divide all parts by 2.
Final calculation: \( -2 \le x \le 5 \).
Question 010Exam Pattern

Solve the system: \( 2x > 4 \) and \( x - 5 \le 0 \).

2 < x ≤ 5
2 ≤ x < 5
x > 2
x ≤ 5
Correct answer: a) 2 < x ≤ 5

Solution

Step 1: Solve the first inequality: \( 2x > 4 \implies x > 2 \).
Step 2: Solve the second inequality: \( x - 5 \le 0 \implies x \le 5 \).
Step 3: Combine the conditions. \( x \) must be greater than 2 AND less than or equal to 5.
Final calculation: \( 2 < x \le 5 \).