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Simplification: Important Formula

Simplification and approximation are crucial topics in quantitative aptitude, commonly tested in competitive examinations. Mastering these formulas and techniques helps you reduce calculation time and improve accuracy in problem solving.

1. Basic arithmetic operations

Addition and subtraction

  • Commutative property of addition: a + b = b + a
  • Commutative property of subtraction: a - b ≠ b - a (subtraction is not commutative)
  • Associative property of addition: (a + b) + c = a + (b + c)

Use these properties to reorder or regroup terms to make mental calculation easier. For example, 27 + 13 + 73 = (27 + 73) + 13 = 100 + 13 = 113.

Multiplication and division

  • Commutative property of multiplication: a × b = b × a
  • Division is not commutative: a ÷ b ≠ b ÷ a
  • Associative property of multiplication: (a × b) × c = a × (b × c)
  • Distributive property: a × (b + c) = a × b + a × c and a × (b - c) = a × b - a × c Apply the distributive property to simplify expressions before calculating. Example: 25 × 16 = 25 × (10 + 6) = 250 + 150 = 400.

2. Order of operations (BODMAS)

Follow the sequence below when evaluating an expression:

  • B - Brackets ( ), [ ], { }
  • O - Orders (powers and roots)
  • D - Division ÷
  • M - Multiplication ×
  • A - Addition +
  • S - Subtraction -

Example: Evaluate 5 + 3 × (8 ÷ 2)²

Follow the order strictly:

3 × (8 ÷ 2)²

8 ÷ 2 = 4

(8 ÷ 2)² = 4² = 16

3 × 16 = 48

5 + 48 = 53

3. Percentage formulas

  • x% of y = $\left(\frac{x}{100}\right) \times y$
  • y is what % of x = $ \left(\frac{y}{x}\right) \times 100$
  • Percentage increase/decrease = $\left( \frac{\text{New value} - \text{Original value}}{\text{Original value}} \right) \times 100$

Example: Original price = ₹800, new price = ₹920. Percentage increase = $\frac{920 - 800}{800} \times 100 = \frac{120}{800} \times 100 = 15\%$.

4. Fraction, decimal and percentage conversions

Convert between fraction, decimal and percentage using these relations:

  • Fraction to decimal: divide numerator by denominator $(e.g., \frac{3}{4} = 0.75)$
  • Decimal to percentage: multiply by 100 (e.g., 0.75 = 75%)
  • Percentage to fraction: $x\% = \frac{x}{100}$  (simplify if possible)
4. Fraction, decimal and percentage conversions

5. Approximation techniques

  • Rounding off numbers:Look at the digit immediately after the required place.
    • If that digit is ≥ 5, round up the last retained digit by 1.
    • If that digit is < 5, keep the last retained digit unchanged (round down).
  • Estimating sums/differences: Round each addend to a convenient value and then add/subtract. Example: 498 + 372 ≈ 500 + 370 = 870.
  • Estimating products: Round factors to near tens or hundreds. Example: 49 × 31 ≈ 50 × 30 = 1500.
  • Use significant figures when required to present answers with appropriate precision.

6. Squares and cubes-shortcuts and tables

Squares (useful up to 30)

Memorise squares of small numbers and use patterns: (a ± b)² = a² ± 2ab + b². Examples: 25² = (20 + 5)² = 400 + 200 + 25 = 625; 29² = (30 - 1)² = 900 - 60 + 1 = 841.

Squares (useful up to 30)

Cubes (1-10)

NumberCube
11
28
327
464
5125
6216
7343
8512
9729
101000

7. Square-root and cube-root approximations

  • √2 ≈ 1.414
  • √3 ≈ 1.732
  • √5 ≈ 2.236
  • √10 ≈ 3.162
  • Cube root of 2 ≈ 1.26
  • Cube root of 3 ≈ 1.442

Use these approximations to quickly estimate expressions involving roots. For example, √50 ≈ √(49) = 7 (so √50 ≈ 7.071), which is close enough for many competitive questions where exact precision is not required.

8. Algebraic identities to simplify expressions

  • (a + b)² = a² + 2ab + b²
  • (a - b)² = a² - 2ab + b²
  • a² - b² = (a + b)(a - b)
  • (a + b)³ = a³ + 3a²b + 3ab² + b³
  • (a - b)³ = a³ - 3a²b + 3ab² - b³

Apply these identities to expand or factor expressions quickly. Example: 102² = (100 + 2)² = 10000 + 400 + 4 = 10404.

9. Divisibility rules (quick checks)

NumberRule
2Last digit is even (0, 2, 4, 6, 8)
3Sum of digits divisible by 3
4Last two digits divisible by 4
5Ends with 0 or 5
6Divisible by both 2 and 3
8Last three digits divisible by 8
9Sum of digits divisible by 9
10Ends with 0
11Difference between sum of digits in odd places and sum of digits in even places is divisible by 11

Use divisibility rules to test factors quickly before attempting long division or factorisation.


Final tips

  • Practice mental arithmetic daily to increase speed and reduce dependence on paper or calculator.
  • Use approximation when exact values are not necessary or when it helps eliminate options quickly in multiple-choice questions.
  • Memorise key squares, cubes and common roots (up to at least 30² and 10³) to save time in calculations.
  • Use algebraic identities and properties to transform expressions into easier forms before computing.
  • When solving stepped calculations, write intermediate results clearly to avoid careless errors.

This guide presents essential formulas and techniques for simplification and approximation. Regular practice of these methods-together with timed problem solving-will improve both speed and accuracy.

The document Simplification: Important Formula is a part of the Bank Exams Course IBPS PO Prelims & Mains Preparation.
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FAQs on Simplification: Important Formula

1. What are the most important formulas I need to memorize for IBPS PO simplification questions?
Ans. Essential simplification formulas include BODMAS/PEMDAS order of operations, percentage calculations (percentage = part/whole × 100), profit-loss ratios, average formulas, and ratio-proportion relationships. Students should also master fraction-to-decimal conversions and power/exponent rules. These core formulas appear repeatedly across IBPS PO prelims and are foundational for tackling complex numerical expressions quickly within time constraints.
2. How do I simplify expressions with multiple brackets and operators without making silly mistakes?
Ans. Apply BODMAS strictly: solve brackets first (innermost to outermost), then exponents, division and multiplication (left to right), and finally addition and subtraction. Breaking complex simplification problems into smaller chunks reduces calculation errors. Rewrite expressions step-by-step rather than attempting mental math. Practising with mind maps and visual flashcards on operator precedence helps reinforce the sequence and builds accuracy for competitive exams.
3. Why do approximation methods matter in Bank Exams when I could just calculate exact answers?
Ans. Approximation techniques dramatically cut solving time-critical when handling dozens of simplification questions under strict time limits in IBPS PO exams. Rounding decimals, using nearest whole numbers, and recognising fraction patterns lets candidates solve in 30-45 seconds instead of minutes. This strategy doesn't sacrifice accuracy; it prioritises efficiency, allowing more questions to be attempted and maximising overall scores in the quantitative aptitude section.
4. What's the difference between simplification formulas and algebraic identities I should know for Bank Exams?
Ans. Simplification formulas focus on computational shortcuts and order-of-operations rules (like percentage tricks and ratio methods), whereas algebraic identities are mathematical truths like (a+b)² = a² + 2ab + b². Both matter for IBPS PO: formulas accelerate direct calculations, while identities help factorise and restructure complex expressions. Understanding when to apply each prevents confusion and improves problem-solving speed during competitive exams.
5. Are there any common calculation traps in simplification that trick IBPS PO candidates?
Ans. Yes-sign errors with negative numbers, incorrect operator precedence (adding before multiplying), and premature rounding are frequent mistakes. Many candidates forget bracket hierarchy rules or miscalculate percentages by using wrong base values. Studying important formula notes alongside previous year question solutions helps identify these pitfalls. Regular practice with worksheets and MCQ tests strengthens awareness of exam-specific traps and builds confidence for the actual Bank Exams.
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