SSC GD Maths Simplification Tricks By Shubham Sir

SSC GD Maths Simplification Tricks By Shubham Sir is a smart learning resource for aspirants who want to solve mathematics questions quickly and correctly. Simplification (सरलीकरण) is one of the most important topics in competitive exams like SSC GD, Railway Group D, RRB NTPC, and other government tests. Many students lose marks because they do not follow proper calculation steps. With the help of BODMAS rule and algebraic shortcuts, complex expressions can be solved in seconds. 

What is Simplification?

Simplification (सरलीकरण) is a fundamental mathematical process that involves reducing complex expressions to a simpler form. It relies on a specific order of operations to ensure accuracy and consistency. The BODMAS rule serves as the cornerstone for correctly evaluating these expressions, preventing errors that can arise from arbitrary calculation sequences. Mastering simplification is crucial for competitive exams.

Simplification Tricks by BODMAS Rule

Simplification focuses on solving mathematical expressions by following a specific order. The foundational rule for this is BODMAS. This acronym dictates the sequence of operations that must be performed to arrive at the correct answer. Deviating from this sequence will result in an incorrect solution.

The BODMAS Sequence of Operations:

1. B - Bracket: Solve all expressions inside brackets first.

2. O - Of: This operation is treated as multiplication and is performed after brackets.

3. D - Division: Perform all division operations.

4. M - Multiplication: Perform all multiplication operations.

5. A - Addition: Perform all addition operations.

6. S - Subtraction: Finally, perform all subtraction operations.

Hierarchy of Brackets:

When an expression contains multiple nested brackets, they must be solved in a specific order:

1. Line Bracket / Vinculum (e.g., ̅8-5̅)

2. Small Bracket / Parentheses (e.g., ( ... ))

3. Curly Bracket / Braces (e.g., { ... })

4. Square Bracket / Brackets (e.g., [ ... ])

Also Read: SSC Calendar 2026

Worked Examples of Applying BODMAS

The following examples illustrate the step-by-step application of the BODMAS rule.

Example 1: Basic BODMAS with Nested Brackets

Problem: 48 + 25 - [20 - {11 - (16 ÷ 2 × 4)}]

• Step 1 (Innermost Bracket): (16 ÷ 2 × 4)

• 16 ÷ 2 = 8

• 8 × 4 = 32

• Expression becomes: 48 + 25 - [20 - {11 - 32}]

• Step 2 (Curly Bracket): {11 - 32}

• 11 - 32 = -21

• Expression becomes: 48 + 25 - [20 - (-21)]

• Step 3 (Square Bracket): [20 - (-21)]

• 20 - (-21) = 20 + 21 = 41

• Expression becomes: 48 + 25 - 41

• Step 4 (Final Calculation): 48 + 25 - 41

• 73 - 41 = 32

• Answer: 32

Example 2: BODMAS with Fractions and 'Of'

Problem: 4 4/5 ÷ (3/5 of 5) + 4/5 × 3/10 - 1/5

• Step 1 (Convert Mixed Fraction): 4 4/5 = 24/5.

• Step 2 ('Of' operation): (3/5 of 5) = 3.

• Step 3 (Division): 24/5 ÷ 3 = 24/5 × 1/3 = 8/5.

• Step 4 (Multiplication): 4/5 × 3/10 = 12/50 = 6/25.

• Step 5 (Combine): The expression is now 8/5 + 6/25 - 1/5.

• Find LCM (25): (40/25) + (6/25) - (5/25)

• (40 + 6 - 5) / 25 = 41/25.

• Answer: 41/25 or 1 16/25

Example 3: BODMAS with Multiple Operations

Problem: 18 ÷ (3 × 2) × 5 + 72 ÷ 36 × 3 - 4 ÷ 8 × 2

• Part 1: 18 ÷ (3 × 2) × 5 = 18 ÷ 6 × 5 = 3 × 5 = 15.

• Part 2: 72 ÷ 36 × 3 = 2 × 3 = 6.

• Part 3: 4 ÷ 8 × 2 = 0.5 × 2 = 1.

• Combine: 15 + 6 - 1 = 20.

• Answer: 20

Example 4: BODMAS with Line Bracket (Vinculum)

Problem: 10 - [6 - {7 - (6 - ̅8-5̅)}]

• Step 1 (Line Bracket): ̅8-5̅ = 3.

• Step 2 (Smallest Bracket): (6 - 3) = 3.

• Step 3 (Curly Bracket): {7 - 3} = 4.

• Step 4 (Square Bracket): [6 - 4] = 2.

• Step 5 (Final Calculation): 10 - 2 = 8.

• Answer: 8

Example 5: BODMAS with a Complex Sequence

Problem: 2.5 × (144 ÷ 198 × 99)

• Step 1 (Bracketed Part): (144 ÷ 198 × 99)

• 144/198 × 99 = 144/2 = 72.

• Step 2 (Final Multiplication): 2.5 × 72

• (5/2) × 72 = 5 × 36 = 180.

• Answer: 180

Advanced fractional problems also follow these principles, reinforcing multi-step BODMAS application.

Simplification Using Algebraic Formulas

For specific types of structured problems, algebraic formulas offer a significant shortcut, allowing for quicker solutions compared to lengthy arithmetic.

Formula 1: a² + b² + c² - ab - bc - ca

This expression can be simplified using the formula:

1/2 [(a - b)² + (b - c)² + (c - a)²]

• Example A: Given a = 113, b = 115, c = 117.

• a - b = -2 → (a - b)² = 4.

• b - c = -2 → (b - c)² = 4.

• c - a = 4 → (c - a)² = 16.

• Calculation: 1/2 [4 + 4 + 16] = 1/2 [24] = 12.

• Example B (Consecutive Numbers): Given x = 2024, y = 2025, z = 2026.

• x - y = -1 → (x - y)² = 1.

• y - z = -1 → (y - z)² = 1.

• z - x = 2 → (z - x)² = 4.

• Calculation: 1/2 [1 + 1 + 4] = 1/2 [6] = 3.

Formula 2: Difference and Sum of Cubes/Squares

• Expression Type: (a³ - b³) / (a² + ab + b²).

• Since a³ - b³ = (a - b)(a² + ab + b²), the expression simplifies to a - b.

• Example: For a = 1.7 and b = 1.2, the result is 1.7 - 1.2 = 0.5.

• Expression Type: (a² - 2ab + b²) / (a² - b²).

• This simplifies to (a - b)² / [(a - b)(a + b)], which further reduces to (a - b) / (a + b).

• Example: For a = 3.25 and b = 1.75.

• a - b = 1.5

• a + b = 5.0

• Result: 1.5 / 5.0 = 0.3.

Formula 3: (a + b)² - (a - b)²

This expression always simplifies to 4ab.

• Example: (203 + 107)² - (203 - 107)².

• The calculation is 4 × 203 × 107.

• (Memory Tip: When solving multiple-choice questions, consider using shortcuts like checking the unit digit. For 4 × 203 × 107, the unit digit will be 4 × 3 × 7 which is 12 × 7 = 84, so the final answer must end in 4. Also, the result must be divisible by 4, as 4ab is clearly divisible by 4.)

• The full answer is 86,884.