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.
Use these properties to reorder or regroup terms to make mental calculation easier. For example, 27 + 13 + 73 = (27 + 73) + 13 = 100 + 13 = 113.
Follow the sequence below when evaluating an expression:
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
Example: Original price = ₹800, new price = ₹920. Percentage increase = $\frac{920 - 800}{800} \times 100 = \frac{120}{800} \times 100 = 15\%$.
Convert between fraction, decimal and percentage using these relations:

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.

| Number | Cube |
|---|---|
| 1 | 1 |
| 2 | 8 |
| 3 | 27 |
| 4 | 64 |
| 5 | 125 |
| 6 | 216 |
| 7 | 343 |
| 8 | 512 |
| 9 | 729 |
| 10 | 1000 |
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.
Apply these identities to expand or factor expressions quickly. Example: 102² = (100 + 2)² = 10000 + 400 + 4 = 10404.
| Number | Rule |
|---|---|
| 2 | Last digit is even (0, 2, 4, 6, 8) |
| 3 | Sum of digits divisible by 3 |
| 4 | Last two digits divisible by 4 |
| 5 | Ends with 0 or 5 |
| 6 | Divisible by both 2 and 3 |
| 8 | Last three digits divisible by 8 |
| 9 | Sum of digits divisible by 9 |
| 10 | Ends with 0 |
| 11 | Difference 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.
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.
| 1. What are the most important formulas I need to memorize for IBPS PO simplification questions? | ![]() |
| 2. How do I simplify expressions with multiple brackets and operators without making silly mistakes? | ![]() |
| 3. Why do approximation methods matter in Bank Exams when I could just calculate exact answers? | ![]() |
| 4. What's the difference between simplification formulas and algebraic identities I should know for Bank Exams? | ![]() |
| 5. Are there any common calculation traps in simplification that trick IBPS PO candidates? | ![]() |