Theory & Concepts

SSC CGL Simplification (BODMAS) Questions, Tricks & Formulas

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

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25 min readDifficulty: Easy-Intermediate

Simplification questions are the most scoring section in the SSC CGL Quant paper. However, examiners intentionally design them to trigger silly mistakes. Mastering the strict order of operations (V-BODMAS) and algebraic identity substitutions is crucial for maximizing your speed and accuracy.

Learning path

  • The V-BODMAS Rule
  • Algebraic Identities Substitutions
  • Arithmetic Series Shortcuts
  • 10 Exam-Level Examples

1. The V-BODMAS Rule

In mathematical expressions with multiple operations, you must strictly follow the V-BODMAS priority order to avoid incorrect results.

Priority Order

V: Vinculum (Bar Bracket a−b‾\overline{a-b})

B: Brackets ( (),{},[](), \{\}, [] )

O: Of (Acts as Multiplication, but calculated BEFORE Division)

Standard Operations

D: Division (÷\div)

M: Multiplication (×\times)

A: Addition (++)

S: Subtraction (−-)

2. Algebraic Identity Substitutions

SSC frequently asks complex-looking decimal calculation questions that are secretly basic algebraic formulas.

The Sum of Cubes Trap

If you see large identical numbers being multiplied three times and added, look for this exact pattern:

a3+b3a2−ab+b2=a+b\frac{a^3 + b^3}{a^2 - ab + b^2} = a + b
a3−b3a2+ab+b2=a−b\frac{a^3 - b^3}{a^2 + ab + b^2} = a - b

Just add or subtract the two base numbers directly. Don't multiply them out!

3. Telescoping & Arithmetic Series

These are continuous fractions or sum sequences that cancel each other out wonderfully.

Sum of Natural Numbers

1+2+3⋯+n=n(n+1)21 + 2 + 3 \dots + n = \frac{n(n+1)}{2}

Use this to quickly sum numbers 1 through N.

Telescoping Cancellation

(1−12)(1−13)⋯=1n(1 - \frac{1}{2})(1 - \frac{1}{3}) \dots = \frac{1}{n}

Notice how numerators and denominators cancel out diagonally.

4. 10 Solved examples

Question 01Exam Pattern

Find the value of \( 5 \times 2 \div 2 \text{ of } 5 \).

5
1
25
10
Correct answer: b) 1

Solution

Step 1: According to BODMAS, 'Of' is evaluated before Division.
Step 2: Calculate 'Of': \( 2 \text{ of } 5 = 2 \times 5 = 10 \).
Step 3: The expression becomes \( 5 \times 2 \div 10 \).
Step 4: Now perform Division: \( 2 \div 10 = 0.2 \).
Final calculation: Perform Multiplication: \( 5 \times 0.2 = 1 \).
Question 02Exam Pattern

Find the value of \( [(4 \times 3) + 2] \div 7 - 1 \).

2
1
3
0
Correct answer: b) 1

Solution

Step 1: Solve the innermost bracket first: \( (4 \times 3) = 12 \).
Step 2: Solve the square bracket: \( [12 + 2] = 14 \).
Step 3: The expression becomes \( 14 \div 7 - 1 \).
Step 4: Perform Division: \( 14 \div 7 = 2 \).
Final calculation: Perform Subtraction: \( 2 - 1 = 1 \).
Question 03Exam Pattern

Simplify: \( \frac{0.87 \times 0.87 \times 0.87 + 0.13 \times 0.13 \times 0.13}{0.87 \times 0.87 - 0.87 \times 0.13 + 0.13 \times 0.13} \).

0.74
1
1.13
0.87
Correct answer: b) 1

Solution

Step 1: Let \( a = 0.87 \) and \( b = 0.13 \).
Step 2: The numerator is \( a^3 + b^3 \).
Step 3: The denominator is \( a^2 - ab + b^2 \).
Step 4: Using the identity: \( \frac{a^3 + b^3}{a^2 - ab + b^2} = a + b \).
Final calculation: \( a + b = 0.87 + 0.13 = 1 \).
Question 04Exam Pattern

Find the value of \( 108 \div 36 \text{ of } \frac{1}{4} + \frac{2}{5} \times 3\frac{1}{4} \).

12.5
13.3
14.2
15.1
Correct answer: b) 13.3

Solution

Step 1: Convert the mixed fraction: \( 3\frac{1}{4} = \frac{13}{4} \).
Step 2: Solve 'Of' first: \( 36 \text{ of } \frac{1}{4} = 36 \times \frac{1}{4} = 9 \).
Step 3: Perform Division: \( 108 \div 9 = 12 \).
Step 4: Perform Multiplication: \( \frac{2}{5} \times \frac{13}{4} = \frac{26}{20} = \frac{13}{10} = 1.3 \).
Final calculation: Perform Addition: \( 12 + 1.3 = 13.3 \).
Question 05Exam Pattern

Simplify: \( \frac{(0.05)^2 + (0.41)^2 + (0.073)^2}{(0.005)^2 + (0.041)^2 + (0.0073)^2} \).

10
100
1000
0.1
Correct answer: b) 100

Solution

Step 1: Notice that each term in the denominator is exactly \( \frac{1}{10} \) of the corresponding term in the numerator.
Step 2: For example, \( (0.005)^2 = (0.05 / 10)^2 = \frac{(0.05)^2}{100} \).
Step 3: Factor out \( \frac{1}{100} \) from the entire denominator.
Step 4: The expression becomes \( \frac{\text{Numerator}}{\frac{1}{100} \times \text{Numerator}} \).
Final calculation: The Numerators cancel out. \( 1 \div (1/100) = 100 \).
Question 06Exam Pattern

Find the value of \( (1 - \frac{1}{3})(1 - \frac{1}{4})(1 - \frac{1}{5}) \dots (1 - \frac{1}{n}) \).

\( \frac{1}{n} \)
\( \frac{2}{n} \)
\( \frac{2}{n-1} \)
\( \frac{1}{n(n-1)} \)
Correct answer: b) \( \frac{2}{n} \)

Solution

Step 1: Simplify each term inside the brackets.
Step 2: \( (1 - 1/3) = 2/3 \).
Step 3: \( (1 - 1/4) = 3/4 \).
Step 4: The series becomes \( \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \dots \frac{n-1}{n} \).
Final calculation: Every numerator cancels with the previous denominator. Only the first numerator (2) and the last denominator (n) are left. Result = \( \frac{2}{n} \).
Question 07Exam Pattern

Find the sum of all natural numbers from 1 to 100.

5000
5050
5150
5500
Correct answer: b) 5050

Solution

Step 1: The sequence is an arithmetic progression: \( 1 + 2 + 3 + \dots + 100 \).
Step 2: Use the formula for the sum of the first N natural numbers: \( S_n = \frac{n(n+1)}{2} \).
Step 3: Here, \( n = 100 \).
Final calculation: \( S = \frac{100 \times 101}{2} = 50 \times 101 = 5050 \).
Question 08Exam Pattern

Simplify: \( 999\frac{995}{999} \times 999 \).

998995
998996
999996
998000
Correct answer: b) 998996

Solution

Step 1: Expand the mixed fraction: \( (999 + \frac{995}{999}) \times 999 \).
Step 2: Distribute the multiplication: \( 999 \times 999 + 995 \).
Step 3: Rewrite \( 999^2 \) as \( (1000 - 1)^2 \).
Step 4: \( (1000 - 1)^2 = 1000000 - 2000 + 1 = 998001 \).
Final calculation: Add 995: \( 998001 + 995 = 998996 \).
Question 09Exam Pattern

If \( a+b+c = 0 \), find the value of \( \frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} \).

0
1
2
3
Correct answer: d) 3

Solution

Step 1: Take the LCM of the denominators, which is \( abc \).
Step 2: The expression becomes \( \frac{a^3 + b^3 + c^3}{abc} \).
Step 3: We know the standard algebraic identity: If \( a+b+c = 0 \), then \( a^3 + b^3 + c^3 = 3abc \).
Final calculation: Substitute \( 3abc \) into the numerator: \( \frac{3abc}{abc} = 3 \).
Question 010Exam Pattern

If \( x + \frac{1}{x} = 5 \), find the value of \( x^2 + \frac{1}{x^2} \).

23
25
27
21
Correct answer: a) 23

Solution

Step 1: We are given \( x + \frac{1}{x} = 5 \).
Step 2: Square both sides: \( (x + \frac{1}{x})^2 = 5^2 \).
Step 3: Expand the left side using \( (a+b)^2 = a^2 + b^2 + 2ab \).
Step 4: \( x^2 + \frac{1}{x^2} + 2(x)(\frac{1}{x}) = 25 \).
Step 5: The \( x \) terms in the middle cancel out, leaving: \( x^2 + \frac{1}{x^2} + 2 = 25 \).
Final calculation: Subtract 2 from both sides: \( 25 - 2 = 23 \).

5. Strategy errors to avoid

!

The "Of" Trap

Never treat "Of" as just a multiplication sign. While mathematically similar, it has HIGHER priority than Division. If you see \\( 10 \\div 5 \\text{ of } 2 \\), the result is 1, not 4!

!

Neglecting the Bar

The Vinculum (bar) over numbers acts as the strongest bracket. Solve terms under the bar first, regardless of what operations are outside.

Master the next modules

Approximation & RoundingSurds & IndicesAlgebraic Identities