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How to Solve Simplification Questions: VBODMAS Rules, Tricks & Solved Examples

सरलीकरण (Simplification) के प्रश्न कैसे हल करें: VBODMAS नियम, ट्रिक्स और हल उदाहरण

Simplification (सरलीकरण) is the process of reducing complex arithmetic expressions containing multiple operational symbols, nested brackets, fractions, and powers into a single numerical value. In competitive exams like MP Police Constable, Patwari, Sub-Inspector (SI), SSC GD/CHSL, and Railway RRB NTPC, simplification questions test numerical speed and strict adherence to the VBODMAS hierarchy.

Key Rules & Core Formulas

Understand these foundational properties before solving questions

1The VBODMAS Rule Hierarchy (VBODMAS नियम)

V → B → O → D → M → A → S

Operations must be performed in strict priority order: Vinculum/Bar (रेखा कोष्ठक) → Brackets (कोष्ठक) → Of (का) → Division (भाग) → Multiplication (गुणा) → Addition (जोड़) → Subtraction (घटाव).

'Of' (का) indicates multiplication, but must be resolved before regular division and multiplication.

2Hierarchy of Nested Brackets (कोष्ठकों के खुलने का क्रम)

Line/Vinculum (—) → Round ( ) → Curly { } → Square [ ]

Always simplify nested grouping symbols from the innermost to outermost: first Vinculum bar, second Round brackets, third Curly braces, and fourth Square brackets.

When removing brackets preceded by a minus sign (-), reverse every internal sign: -(a - b + c) = -a + b - c.

3Chained Division Rule (क्रमिक भाग का नियम)

a ÷ b ÷ c ÷ d = a / (b × c × d)

Consecutive division operations without grouping symbols are evaluated strictly from Left to Right, or by converting every division into multiplication of its reciprocal.

For fractions: (a/b) ÷ (c/d) = (a/b) × (d/c).

4Algebraic Simplification Identities (बीजगणितीय सर्वसमिकाएं)

a² - b² = (a - b)(a + b) | (a³ ± b³) / (a² ∓ ab + b²) = a ± b

Competitive exam simplification problems featuring huge numbers or repeating 3-digit decimals are algebraic identities disguised as arithmetic.

(a + b)² = a² + 2ab + b² and (a - b)² = a² - 2ab + b² can also simplify square-root expressions.

5Digital Sum (Digit Root) Shortcut (अंकीय योग विधि)

Digital Sum of LHS = Digital Sum of RHS (Treat 9 as 0)

Sum the digits of every number until a single digit (1-9) is reached. Cast out 9s. The digital sum of the problem statement must equal the digital sum of the correct multiple-choice option.

Eliminates 3 out of 4 exam options in under 5 seconds without carrying out extensive decimal or fraction arithmetic.
Standard Method

Step-by-Step Method to Solve

Follow these universal steps in order to solve any Quantitative Aptitude question accurately:

  1. 1

    Step 1: Eliminate Vinculum (Bar) and innermost Round Brackets

    Scan for numbers topped with a bar line (—) and compute that sub-expression first. Next, evaluate operations inside the innermost round brackets ( ).

    Pro-Tip: A vinculum acts like an invisible bracket: 10 - 4 - 2 with a bar over 4 - 2 equals 10 - (4 - 2) = 10 - 2 = 8.
  2. 2

    Step 2: Resolve all 'Of' (का) and percentage operations

    Convert every 'Of' (का) into multiplication and compute its immediate product as an indivisible unit before performing any adjacent division.

    Pro-Tip: Use percentage symmetry for mental speed: x% of y = y% of x (e.g., 28% of 25 = 25% of 28 = 7).
  3. 3

    Step 3: Process Division and Multiplication from Left to Right

    Convert divisions into reciprocals (÷ x becomes × 1/x) and cancel common factors between numerators and denominators to prevent unwieldy fractions.

    Pro-Tip: Writing division as multiplication by reciprocal eliminates chained division order mistakes.
  4. 4

    Step 4: Group and solve Addition and Subtraction

    Separate positive numbers and negative numbers. Compute the sum of positives and the sum of negatives, then take the difference.

    Pro-Tip: Grouping positives and negatives separately prevents mid-calculation sign errors when working under tight exam timers.
  5. 5

    Step 5: Verify with Unit Digit or Digital Sum before marking

    Perform a 3-second unit digit or digital sum check on the original equation against the four exam options to verify accuracy.

    Pro-Tip: If the four options have different last digits, calculate only the unit digit to select the answer instantly.

Worked Examples & Walkthroughs

Real questions solved step-by-step showing every calculation

1Example 1Method: VBODMAS Nested Brackets Method

Question

Simplify the expression: 25 - [20 - {10 - (7 - (5 - 3))}] (where 5 - 3 has a vinculum bar over it).

Step-by-Step Solution:

Step 1: Evaluate Vinculum (Bar):The vinculum is over 5 - 3: 5 - 3 = 2. The innermost round bracket becomes (7 - 2).
Step 2: Solve inside round brackets ( ):7 - 2 = 5. The expression inside curly braces becomes {10 - 5}.
Step 3: Solve inside curly braces { }:10 - 5 = 5. The expression inside square brackets becomes [20 - 5].
Step 4: Solve inside square brackets [ ]:20 - 5 = 15.
Step 5: Perform final subtraction:25 - 15 = 10.
Final Result:
10
Exam Shortcut / Verification: Work from the innermost core outwards. Notice how signs alternate: 25 - 20 + 10 - 7 + (5 - 3) = 5 + 3 + 2 = 10. Expanding nested brackets algebraically allows instant mental tallying.
2Example 2Method: VBODMAS with Fractions & 'Of' Priority

Question

Evaluate: 3/4 ÷ 9/8 of 4/3 + 2/3 × 1(1/2) - 1/3.

Step-by-Step Solution:

Step 1: Evaluate 'Of' first:'9/8 of 4/3' takes precedence over division: (9/8) × (4/3) = (9 × 4) / (8 × 3) = 36/24 = 3/2.
Step 2: Perform division with reciprocal:Now evaluate 3/4 ÷ 3/2: (3/4) × (2/3) = (3 × 2) / (4 × 3) = 6/12 = 1/2.
Step 3: Convert mixed fraction and multiply:Convert 1(1/2) to improper fraction: 3/2. Now multiply: (2/3) × (3/2) = 1.
Step 4: Combine additions and subtractions:The expression is now: 1/2 + 1 - 1/3. Common denominator of 2 and 3 is 6: 3/6 + 6/6 - 2/6 = (3 + 6 - 2) / 6 = 7/6 = 1(1/6).
Final Result:
7/6 (or 1 1/6)
Exam Shortcut / Verification: The most common trap in competitive exams is doing 3/4 ÷ 9/8 = 2/3 and then multiplying by 4/3. Always isolate 'A of B' as an inseparable bracket (A × B) before applying division!
3Example 3Method: Left-to-Right Chained Division & Percentage Symmetry

Question

Find the value of: 720 ÷ 12 ÷ 3 + 24% of 250 - 5².

Step-by-Step Solution:

Step 1: Chained division from left to right:720 ÷ 12 = 60, then 60 ÷ 3 = 20. Alternatively, 720 / (12 × 3) = 720 / 36 = 20.
Step 2: Percentage shortcut:Use percentage symmetry: 24% of 250 = 250% of 24 = 2.5 × 24 = 60.
Step 3: Evaluate power:5² = 25.
Step 4: Combine terms:20 + 60 - 25 = 80 - 25 = 55.
Final Result:
55
Exam Shortcut / Verification: Remember the chained division rule: a ÷ b ÷ c = a / (b × c). Never compute 12 ÷ 3 first! Also, x% of y = y% of x turns awkward percentages into simple mental arithmetic.
4Example 4Method: Algebraic Identity Substitution

Question

Simplify without manual multiplication: (0.87 × 0.87 × 0.87 + 0.13 × 0.13 × 0.13) / (0.87 × 0.87 - 0.87 × 0.13 + 0.13 × 0.13).

Step-by-Step Solution:

Step 1: Identify algebraic variables:Let a = 0.87 and b = 0.13. Then the expression is: (a³ + b³) / (a² - ab + b²).
Step 2: Apply the sum of cubes identity:Recall the identity: a³ + b³ = (a + b)(a² - ab + b²).
Step 3: Cancel identical terms:[(a + b)(a² - ab + b²)] / (a² - ab + b²) = a + b.
Step 4: Substitute actual values:a + b = 0.87 + 0.13 = 1.00 = 1.
Final Result:
1
Exam Shortcut / Verification: In MP Patwari and SSC exams, questions featuring repeated 3-digit decimal multiplications are 100% based on algebraic identities. Look for (a³ + b³) or (a³ - b³) and write (a + b) or (a - b) directly in 2 seconds!

Common Mistakes to Avoid

Avoid these frequent traps to protect yourself from negative marking

1

Performing regular Division before 'Of' (का)

Why it happens: Students conflate 'Of' with regular multiplication and execute division first following standard left-to-right reading (e.g., calculating 20 ÷ 5 of 2 as 4 of 2 = 8).
Correct approach: In VBODMAS, 'O' (Of) stands before 'D' (Division). Always multiply the numbers joined by 'of' first: 20 ÷ 5 of 2 = 20 ÷ (5 × 2) = 20 ÷ 10 = 2.
2

Evaluating Chained Division from Right to Left

Why it happens: Falsely assuming division is associative like multiplication, leading students to solve a ÷ b ÷ c by computing (b ÷ c) first.
Correct approach: Division is NOT associative. Always evaluate chained division strictly left-to-right, or convert it into a single multiplication of reciprocals: a ÷ b ÷ c = a × (1/b) × (1/c) = a / (b × c).
3

Forgetting sign changes when opening negative brackets

Why it happens: Rushing through scratchwork and failing to distribute the negative sign to all terms inside: writing -(a - b) as -a - b instead of -a + b.
Correct approach: Treat the minus sign outside parentheses as multiplying every inner term by -1: -(x - y + z) = -x + y - z.
4

Treating Vinculum (Bar) as a simple underline or ignoring it

Why it happens: Unfamiliarity with the Vinculum symbol (V in VBODMAS), causing students to overlook it and apply standard bracket rules.
Correct approach: Treat whatever lies directly beneath the bar line as enclosed in parentheses with the highest priority: 8 - 4 - 2 with bar on 4 - 2 means 8 - (4 - 2) = 8 - 2 = 6, NOT (8 - 4) - 2.
Self-Assessment

Try It Yourself (Practice Questions)

Test your understanding on your own first, then reveal the step-by-step solution to check your work:

Practice #1

Simplify the expression: 45 - [38 - {60 ÷ 3 - (6 - 9 ÷ 3)}].

Practice #2

What is the value of: 18 ÷ 3 of 2 × 5 + 16 ÷ 4 × 2 - 3?

Practice #3

Find the value of: (678 × 678 - 322 × 322) / 356.

Quick Revision Summary

  • Always follow the VBODMAS hierarchy: Vinculum (Bar) → Brackets () {} [] → Of → Division & Multiplication (Left to Right) → Addition & Subtraction.
  • 'Of' (का) must be evaluated before division: a ÷ b of c = a ÷ (b × c).
  • Chained division rule: a ÷ b ÷ c ÷ d = a / (b × c × d).
  • Large numbers in multiplication expressions usually simplify instantly via identities like a² - b² = (a - b)(a + b) or a³ ± b³.
  • Use Unit Digit and Digital Sum checks to eliminate incorrect multiple-choice options in under 5 seconds.
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