Simplification, at its core, is about collapsing a messy expression into a single clean number — and doing it without making errors. Approximation is the same idea with a deliberate trade-off: you accept a small rounding error to gain significant speed.
Think of it like navigating Delhi traffic. The exact route might be the fastest on paper, but if you know a reliable shortcut — even one that adds 200 metres — you take it when you're already late. In the exam hall, you're always late.
The governing principle is BODMAS (also written PEMDAS in Western textbooks, but BODMAS is the Indian classroom standard):
x², √x, xⁿHere's what BODMAS is actually saying: multiplication and division are "stronger" operations than addition and subtraction. They bind numbers more tightly. When you write 3 + 4 × 2, the 4 × 2 is a single clump that must be resolved before the +3 can act on it.
The analogy that works: treat multiplication as bolding a word. You read the bold parts first, then fit them into the sentence. 3 + **8** - 5 + **12** = 3 + 8 - 5 + 12 = 18. The addition and subtraction then flow left to right.
A critical misconception to kill early — D before M is not a strict rule. Division and multiplication have equal priority; you work left to right among them. Same for addition and subtraction. The rule is: resolve brackets first, then powers, then handle all × and ÷ left-to-right in one pass, then handle all + and − left-to-right in one pass.
Simplification questions in IBPS PO are typically 1–2 marks each, completable in 30–45 seconds per question if you have the order of operations locked in muscle memory. A mistake here costs you a question you should have banked — that is why this topic gets a dedicated chapter rather than being folded into "miscellaneous".
When brackets nest, resolve from the innermost type outward. The standard nesting order in Indian textbooks is:
‾3+2‾): treated as a bracketExample: [3 + {2 × (5 - 1)}]
(5 - 1) = 4{2 × 4} = 8[3 + 8] = 11In IBPS PO, you'll rarely see all four bracket types simultaneously — but when they appear, the nesting signal is explicit.
After brackets are resolved, handle exponents and roots before anything else.
Key identities to have cold — not derived, just recalled:
aⁿ × aᵐ = aⁿ⁺ᵐaⁿ ÷ aᵐ = aⁿ⁻ᵐ(aⁿ)ᵐ = aⁿˣᵐa⁰ = 1 (for any a ≠ 0)a⁻ⁿ = 1/aⁿ√a × √b = √(ab)√a / √b = √(a/b)Surd simplification — the process of removing irrational terms from denominators (rationalisation):
1/(√a + √b) — multiply numerator and denominator by (√a - √b) to get (√a - √b)/(a - b).
This appears in IBPS PO in the form: "Find the value of 1/(√5 + √3)." The moment you see this form, rationalise immediately — don't try to estimate.
The key fraction operations to automate:
Cross-multiplication for addition/subtraction:
a/b ± c/d = (ad ± bc) / bd
For three fractions: take LCM of denominators, convert all fractions, then operate.
Percentage-fraction equivalence table — burn this in:
| Fraction | Percentage | Fraction | Percentage | |----------|-----------|----------|-----------| | 1/2 | 50% | 1/8 | 12.5% | | 1/3 | 33.33% | 1/9 | 11.11% | | 1/4 | 25% | 1/11 | 9.09% | | 1/5 | 20% | 1/12 | 8.33% | | 1/6 | 16.66% | 1/7 | 14.28% |
This table becomes critical in approximation — when you see 33% of 660, you should see (1/3) × 660 = 220 without doing any multiplication.
Approximation questions give you an expression like:
√(4489) + 73.8% of 1250 - 18.2²
You don't compute exactly. You round to convenient numbers:
√(4489) ≈ √4500 ≈ 67 (since 67² = 4489 exactly, but you check nearby perfect squares)73.8% of 1250 ≈ 75% of 1250 = (3/4) × 1250 = 937.518.2² ≈ 18² = 324Result ≈ 67 + 937.5 - 324 = 680.5 — look at the options and pick the one closest to this value.
The rounding discipline: always round to numbers where a known fraction or perfect square/cube applies. Random rounding gives random errors.
48 ÷ 6 × 2 is NOT 48 ÷ 12 = 4. It is (48 ÷ 6) × 2 = 8 × 2 = 16.
The trap is grouping division and multiplication incorrectly. Work strictly left to right once brackets and powers are done.
Before writing anything, do a single left-to-right eye scan of the expression. Circle every bracket. Resolve innermost first. This pre-scan prevents the most common error — missing a nested bracket mid-calculation.
Standard method (no scan): students often resolve the wrong bracket first, backtrack, lose 30 seconds. With the pre-scan (2 seconds), you resolve in one clean pass.
Micro-example: [4 × {3 + (8 ÷ 2)}] — scan reveals three levels. Pre-circling shows: start with (8 ÷ 2) = 4, then {3 + 4} = 7, then [4 × 7] = 28. One pass, no backtrack.
Step count: without scan — 5 steps with possible error correction; with scan — 4 clean steps.
When you see a percentage of a number, immediately check if the percentage converts to a clean fraction. If yes, multiply by the fraction — it's faster than long multiplication.
When to use: percentage × 3-digit number, where the percentage is in your conversion table.
Micro-example: 37.5% of 480
0.375 × 480 = 180 (decimal multiplication, ~25 seconds)37.5% = 3/8, so (3/8) × 480 = 3 × 60 = 180 (~8 seconds)Speed gain: standard method ~25s vs shortcut ~8s.
For approximating √N where N is not a perfect square:
Find a² (the nearest perfect square below N). Then: √N ≈ a + (N - a²)/(2a).
When to use: approximation questions with a square root in the expression.
Micro-example: √50
7² = 49√50 ≈ 7 + (50 - 49)/(2 × 7) = 7 + 1/14 ≈ 7.077.071. Error < 0.001.Standard approach (trial and error with 7.0, 7.1, 7.2): ~30s. This formula: ~10s.
For expressions with × and ÷ mixed (no brackets), scan left-to-right and never re-order. Write a small arrow above each × or ÷ pointing right to remind yourself: process in sequence.
Prevents the classic trap: 24 ÷ 4 × 3 — wrong instinct groups it as 24 ÷ (4 × 3) = 2; correct answer is (24 ÷ 4) × 3 = 18.
Use in any expression where a ÷ is followed by a ×. The marking takes 1 second and saves a wrong answer (−0.25 marks).
Step comparison: unmarked — 3 steps, ~30% error rate on this pattern; marked — 3 steps, ~2% error rate.
In approximation questions, round up some values and round down others — approximately balance the rounding errors. Don't round everything in the same direction or your answer will be systematically off by a large margin.
When to use: expressions with 3+ terms where each term is rounded.
Micro-example: 199 × 3.02 + 49.8 × 6.1
199 → 200 (up), 3.02 → 3 (down) — net: near zero error on first term49.8 → 50 (up), 6.1 → 6 (down) — net: near zero error on second term600 + 300 = 900 vs actual 601.98 + 303.78 = 905.76Versus rounding everything up: 200 × 3 + 50 × 6 = 600 + 300 = 900 (accidentally balanced here, but the principle saves you when it's not).
The rule: if you round a multiplier up, round the other multiplicand in that pair down.
When you see a simplification question in the exam hall, run this sequence:
For approximation questions, add this decision before step 1:
Time budget: straightforward BODMAS — 25 to 35 seconds. Surd/indices simplification — 45 to 60 seconds. Approximation with 3+ terms — 40 to 55 seconds.
If you've spent more than 60 seconds on a simplification question, mark your best approximation and move on. These questions should be net time savers, not time drains.
Why this question: Tests pure BODMAS with a single bracket and two operations — the most common format in IBPS PO prelims.
Solving path: Bracket first — (30 - 12) = 18. Now the expression is 18 × 2 + 8. Multiplication before addition: 18 × 2 = 36. Final: 36 + 8 = 44. The trap here is doing 30 - 12 × 2 first (ignoring bracket) which gives 30 - 24 = 6, then 6 + 8 = 14 — wrong.
Why this question: Same structure but tests whether you apply multiplication before subtraction after resolving the bracket.
Solving path: (15 + 25) = 40. Expression becomes 40 × 2 - 30. Multiplication first: 40 × 2 = 80. Then: 80 - 30 = 50. Common trap: 40 × (2 - 30) — students sometimes re-bracket incorrectly when writing down intermediate steps.
Why this question: The multiplier is outside the bracket — tests whether you correctly apply the coefficient to the bracket result before handling subtraction.
Solving path: (12 + 8) = 20. Expression: 3 × 20 - 20. Multiplication first: 3 × 20 = 60. Then: 60 - 20 = 40. The option 35 traps students who compute 3 × 12 + 8 - 20 = 36 + 8 - 20 = 24 (forgot to apply bracket first) — eliminate it instantly since it violates B in BODMAS.
Why this question: No brackets — pure order of operations with division before addition and subtraction.
Solving path: Division first: 72 ÷ 9 = 8. Expression becomes 8 + 15 - 6. Left to right: 8 + 15 = 23, then 23 - 6 = 17. The trap answer 21 comes from adding 15 - 6 = 9 first and then computing 72 ÷ 9 + 9 = 8 + 9 = 17 — coincidentally correct here but the wrong reasoning, which fails on harder versions.
Why this question: Tests that you resolve bracket, then handle multiplication and division left-to-right in a single pass.
Solving path: (5 + 3) = 8. Expression: 9 × 8 ÷ 6. Left to right among equal-priority operations: 9 × 8 = 72, then 72 ÷ 6 = 12. The trap: computing 9 × (8 ÷ 6) = 9 × 1.33 = 12 — this accidentally gives the same answer here, but the method is wrong. The correct approach is left-to-right. On a different number set it will fail.
Multiplying before resolving a bracket that contains addition. In 4 × (3 + 5), students sometimes distribute immediately — that is valid algebra but only if done correctly. The faster path is always: resolve bracket first, then multiply. Distributing mid-problem increases error risk.
Treating D before M as a strict rule. 12 ÷ 3 × 4 is not 12 ÷ 12 = 1. It is 4 × 4 = 16. Division does not universally precede multiplication — they share equal priority and are resolved left to right.
Wrong bracket nesting under time pressure. When you see [3 + {2 + (1)}], working from left to right (outer to inner) is the instinct — but it's wrong. Always start from innermost. Under time pressure, this reversal kills accuracy.
Rounding all approximation terms in the same direction. If every term is rounded up, the answer will be consistently higher than the actual value. The answer choice you select will often be the second-highest option, not the highest — but you'll pick the highest. Balance your rounding.
Forgetting a⁰ = 1 and mishandling negative exponents. 2⁻³ = 1/8, not -8. In indices simplification, students frequently subtract the sign and get −8. Write the reciprocal explicitly before continuing.
Misreading the vinculum (overline bar) as a minus sign. In questions that use a bar over a group of digits (common in some IBPS PO formats), the bar indicates that entire group should be computed as a unit first. Treating it as a minus sign changes the answer entirely.