1. BODMAS order
Brackets → Of / Orders (powers, roots) → Division and Multiplication (left to right) → Addition and Subtraction (left to right). Brackets are solved in the order ( ) → { } → [ ]. A bar (vinculum) is solved first of all.
2. Algebraic identities that save time
| Identity | Use |
|---|---|
| (a + b)² = a² + 2ab + b² | squares like 102² |
| (a − b)² = a² − 2ab + b² | 98² |
| a² − b² = (a + b)(a − b) | 67² − 33² = 100 × 34 |
| (a + b)² − (a − b)² = 4ab | ASO 2021 type |
| (a + b)² + (a − b)² = 2(a² + b²) | |
| a³ + b³ = (a + b)(a² − ab + b²) | a³ + b³ ÷ (a² − ab + b²) = a + b |
| a³ − b³ = (a − b)(a² + ab + b²) | |
| If a + b + c = 0, a³ + b³ + c³ = 3abc |
3. Solved examples
(ASO 2021 type) [(799 + 267)² − (799 − 267)²] ÷ (799 × 267) = 4ab/ab = 4.
(0.7³ + 0.3³) ÷ (0.7² − 0.21 + 0.3²) = 0.7 + 0.3 = 1.
48 ÷ 4 × 3 + 2 = 12 × 3 + 2 = 38 (divide and multiply left to right).
(ASO 2025) Hens 50, goats 45, camels 8 and k keepers. Feet are 224 more than heads. Each hen/keeper adds 1, each goat/camel adds 3: 50 + 159 + k = 224 → k = 15.
4. Approximation tip
When options are far apart, round numbers (e.g. 49.8 ≈ 50, 19.97 ≈ 20) and compute quickly.
Solved examples by question type
Each type that comes in the ASO paper, worked step by step. Try it yourself first, then read the solution.
Basic level
Example 1. Simplify: 43 + 4 × 4 − 21
(a) 38 (b) -25 (c) 42 (d) 167
Answer: (a) 38
Solution: BODMAS: multiply first: 4×4 = 16. Then 43 + 16 − 21 = 38.
Example 2. 1.2 × 0.03 is equal to:
(a) 3.60 × 10^-2 (b) 3.60 × 10^-4 (c) 3.60 × 10^-1 (d) 3.60 × 10^-3
Answer: (a) 3.60 × 10^-2
Solution: 1.2 × 0.03 = 0.036 = 3.60 × 10^-2. Count decimal places carefully.
Example 3. 0.2 × 0.05 is equal to:
(a) 1.00 × 10^-1 (b) 1.00 × 10^-3 (c) 1.00 × 10^-2 (d) 1.00 × 10^-4
Answer: (c) 1.00 × 10^-2
Solution: 0.2 × 0.05 = 0.01 = 1.00 × 10^-2. Count decimal places carefully.
Example 4. 90% of 1000 + 70% of 880 = ?
(a) 1,492 (b) 1,516 (c) 1,511 (d) 1,526
Answer: (b) 1,516
Solution: 90% of 1000 = 900; 70% of 880 = 616. Sum = 1,516.
Example 5. 40% of 580 + 60% of 340 = ?
(a) 431 (b) 484 (c) 436 (d) 446
Answer: (c) 436
Solution: 40% of 580 = 232; 60% of 340 = 204. Sum = 436.
Example 6. 21 × 7 ÷ 3 + 159 − 89 = ?
(a) -199 (b) 122 (c) 119 (d) 297
Answer: (c) 119
Solution: Divide/multiply first: 21 × 7 ÷ 3 = 49. Then 49 + 159 − 89 = 119.
ASO exam level
Example 7. Simplify: [(958 + 42)² − (958 − 42)²] ÷ (958 × 42)
(a) 8 (b) 4 (c) 1 (d) 2
Answer: (b) 4
Solution: (a+b)² − (a−b)² = 4ab. So 4ab/ab = 4. No calculation needed.
Example 8. Simplify: (0.7³ + 0.3³) ÷ (0.7² − 0.7×0.3 + 0.3²)
(a) 0.4 (b) 0.21 (c) 1 (d) 2
Answer: (c) 1
Solution: a³ + b³ = (a + b)(a² − ab + b²). So the answer is a + b = 1.
Example 9. What is 3/5 of 5/8 of 640?
(a) 480 (b) 240 (c) 784 (d) 400
Answer: (b) 240
Solution: 3/5 × 5/8 × 640 = 240.
Example 10. What is 1/4 of 4/5 of 300?
(a) 240 (b) 315 (c) 60 (d) 120
Answer: (c) 60
Solution: 1/4 × 4/5 × 300 = 60.
Example 11. 0.3̄ + 0.2̄ = ?
(a) 5/99 (b) 5/9 (c) 5/10 (d) 0.5
Answer: (b) 5/9
Solution: 0.3̄ = 3/9 and 0.2̄ = 2/9. Sum = 5/9.
Example 12. 4/5 of 6/7 of 980 = ?
(a) 672 (b) 1,344 (c) 784 (d) 840
Answer: (a) 672
Solution: 980 × 4/5 × 6/7 = 672.
Example 13. (616 + 26)² − (616 − 26)² = ?
(a) 64,068 (b) 64,064 (c) 32,032 (d) 32,708
Answer: (b) 64,064
Solution: (x + y)² − (x − y)² = 4xy = 4 × 616 × 26 = 64,064.
Example 14. (583 + 88)² − (583 − 88)² = ?
(a) 1,02,608 (b) 1,10,352 (c) 2,05,216 (d) 2,05,220
Answer: (c) 2,05,216
Solution: (x + y)² − (x − y)² = 4xy = 4 × 583 × 88 = 2,05,216.
Advanced level
Example 15. In a caravan there are 60 hens, 20 goats and 10 camels with some keepers. If the total number of feet is 224 more than the number of heads, how many keepers are there?
(a) 73 (b) 76 (c) 148 (d) 74
Answer: (d) 74
Solution: Each hen/keeper adds 1 (2 feet − 1 head); each goat/camel adds 3. 60×1 + (20+10)×3 + k×1 = 224 → k = 74.
Example 16. Find a number such that when 15 is subtracted from 7 times the number, the result is 10 more than 2 times the number.
(a) 10 (b) 6 (c) 5 (d) 2.14
Answer: (c) 5
Solution: 7x − 15 = 2x + 10 → 5x = 25 → x = 5.
Example 17. If a + b = 20 and ab = 75, then a² + b² = ?
(a) 250 (b) 550 (c) 400 (d) 325
Answer: (a) 250
Solution: a² + b² = (a + b)² − 2ab = 400 − 150 = 250.
Example 18. If a + b = 53 and ab = 696, then a² + b² = ?
(a) 2,113 (b) 2,809 (c) 1,417 (d) 4,201
Answer: (c) 1,417
Solution: a² + b² = (a + b)² − 2ab = 2809 − 1392 = 1417.
Topic-wise practice for Simplification and BODMAS (Quant): BODMAS, squares and cubes, continued fractions, approximation. Useful for OPSC ASO, OSSC, OSSSC and other Odisha exams.
Question types covered
- BODMAS expressions
- Squares, square roots, cubes and cube roots in expressions
- Continued fractions
- Expressions using algebraic identities
- Approximation and finding the nearest value
- Missing number in an expression