Every arithmetic expression is a set of instructions. The problem is that without an agreed-upon sequence, the same expression can give different answers to different people. BODMAS is that agreed-upon sequence — a strict priority order that tells you which operation to perform first, second, and so on.
Here is a useful way to think about it. Imagine you are a factory supervisor, and your workers are the four operations: multiplication, division, addition, subtraction. Left to themselves, each worker would grab the numbers nearest to them. Brackets are like a "do this first — urgent" note. Exponents (Orders) are pre-processed raw material. Multiplication and division are the main assembly line. Addition and subtraction are the final packaging step. If you let packaging happen before assembly, the product comes out wrong. The hierarchy exists so every worker knows exactly when to step in.
The full expansion is:
One critical point that trips up many students: D and M are equal priority, resolved left to right. So are A and S. This is not a strict left-to-right sequence across all six — it is a two-level system where you first clear brackets and of/orders, then handle the DM pair left to right, then handle the AS pair left to right.
SSC MTS uses simplification questions to test whether you apply this rule correctly under time pressure. The expressions look complicated, but they almost always decompose into three or four clean sub-calculations. The only skill is applying the priority order without flinching.
When an expression contains nested brackets, you work from the inside out. The standard convention in Indian textbooks uses three types:
Example: [5 + {3 × (2 + 4)}]
Step 1: Round bracket first → (2 + 4) = 6
Step 2: Curly bracket → {3 × 6} = 18
Step 3: Square bracket → [5 + 18] = 23
There is also the vinculum — a horizontal bar drawn over a group of numbers, like 3̄ + 2̄ meaning (3 + 2). In SSC questions, vinculum appears in old papers and is always resolved before even the round brackets.
"Of" translates directly to multiplication but it sits at the O position in BODMAS, meaning it is processed before regular multiplication and division in many contextual readings. In practice, in SSC papers, treat "of" identically to ×, but resolve it as part of the O-step so it does not get mistakenly pushed to the AS stage.
Look at this: 2 of 3 ÷ 3 × 2
Step 1 (O): 2 of 3 = 6
Now expression: 6 ÷ 3 × 2
Step 2 (DM, left to right): 6 ÷ 3 = 2, then 2 × 2 = 4
If you had read "of" as a plain × and done left-to-right DM without the O-step distinction, you would get the same answer here — but in a nested bracket problem, ignoring the "of" step and treating it as equal-priority multiplication can cause sequencing errors.
When SSC writes a ÷ b of c, it means a ÷ (b × c). This is one of the most-tested traps.
13 ÷ 13 of 13 means 13 ÷ (13 × 13) = 13 ÷ 169 = 1/13
If you incorrectly read it as (13 ÷ 13) of 13 = 1 × 13 = 13, you have made a classic BODMAS error. The "of" belongs to the O-tier and gets processed as part of the divisor.
When expressions involve powers, two rules apply constantly:
aᵐ × aⁿ = aᵐ⁺ⁿ — same base, multiplication → add exponentsaᵐ ÷ aⁿ = aᵐ⁻ⁿ — same base, division → subtract exponentsFor mixed-base products like (2³ × 3⁴) × (2⁵ × 3²), group the same bases:
2³ × 2⁵ = 2⁸ and 3⁴ × 3² = 3⁶, so the answer is 2⁸ × 3⁶.
SSC MTS regularly throws chains like: 28 × 25 ÷ 7 × 15 + 63 × 56 ÷ 72 × 5 + 24 × 7 + 8
The right approach is chunking: identify natural multiplication-division groups separated by the + or − signs, solve each chunk independently, then add them up.
Chunk 1: 28 × 25 ÷ 7 × 15 → (28/7) × 25 × 15 = 4 × 375 = 1500
Chunk 2: 63 × 56 ÷ 72 × 5 → 63 × (56/72) × 5 = 63 × (7/9) × 5 = 7 × 7 × 5 = 245
Chunk 3: 24 × 7 + 8 → 168 + 8 = 176
Total: 1500 + 245 + 176 = 1921
The key insight: addition and subtraction act as natural separators in the expression. Everything between two + or − signs is an independent DM-chain.
Expressions like 0.3 of 12 × (1 − 0.3)² require sequential BODMAS:
Step 1 (B): (1 − 0.3) = 0.7
Step 2 (O): square it → (0.7)² = 0.49
Step 3 (O): 0.3 of 12 = 3.6
Step 4 (M): 3.6 × 0.49 = 1.764
When an expression has a long chain of ×, ÷, +, −, use + and − as natural separators to break the expression into independent DM-chunks. Solve each chunk, then add/subtract the chunk results. Standard method (left to right, no chunking): 8–10 sequential steps with carry errors. Chunking method: 3 parallel sub-calculations + one final addition. Saves 40–50 seconds on chains like the 2022 PYQ above.
Whenever you see a ÷ b of c, write it immediately as a ÷ (bc) before doing anything else. Circle the b of c group. This one rewrite prevents the single most common simplification error in SSC papers. Takes 3 seconds to rewrite and eliminates wrong answers from the options. In the 2023 PYQ, missing this step gives numerator = 13 instead of 1/13 — a 169× error.
For questions involving products of powers (like 2³ × 3⁴ multiplied by 2⁵ × 3²), physically rewrite the expression grouping same bases side by side: (2³ × 2⁵) × (3⁴ × 3²). Then just add exponents within each group. Standard method: multiply out actual numbers (time-consuming, error-prone). Grouping method: 2 additions → answer in one line. Reduces a 90-second calculation to a 10-second one.
In any expression with nested brackets, locate the innermost bracket before you read the full expression. Write its value above the bracket in the margin, then re-read. This prevents the error of accidentally resolving an outer bracket first. Concrete step difference: without this habit you may do 2 extra steps to undo a wrong bracket resolution. With it: solve linearly outward, zero backtracking.
In the final addition-subtraction step, instead of adding all positives first and all negatives first (two passes), keep a running left-to-right total. For 91 − 33 + 7 + 54 − 29 + 22: start at 91, subtract 33 → 58, add 7 → 65, add 54 → 119, subtract 29 → 90, add 22 → 112. One pass, no intermediate paper. Versus two-pass method (sum positives: 91+7+54+22 = 174; sum negatives: 33+29 = 62; 174−62 = 112): same answer but the running total avoids carrying two separate sums.
In the exam hall, use this decision tree the moment you see a simplification question:
Step 1 — Scan for brackets. Are there nested brackets? If yes, identify the innermost one and resolve it first. Continue outward.
Step 2 — Is "of" present? Convert every a ÷ b of c to a ÷ (bc) immediately. This is non-negotiable.
Step 3 — Are there powers or roots? Resolve these next (the O in BODMAS). For index questions, group same bases and add/subtract exponents.
Step 4 — Chunking check. Is the expression a long DM+AS chain? Identify + and − signs as separators. Solve each DM-chunk independently.
Step 5 — Final AS sweep. Add/subtract your chunk results using a left-to-right running total.
Time budget: A well-structured simplification question at MTS level should take 60–90 seconds. If you are past 90 seconds, check whether you missed the "of" trap or forgot to chunk. Do not restart — check only the most recent step you completed.
Elimination shortcut: For answer options that are far apart (like 1821 vs 1921 vs 2101), rough estimation of your largest chunk alone can eliminate two options before you finish the full calculation.
Why this question: Tests pure index laws in a simplification wrapper — common in SSC papers, and students often try to compute actual values (which are huge) instead of recognising the pattern.
Solving path: Write A × B = (2³ × 3⁴) × (2⁵ × 3²). Group same bases: (2³ × 2⁵) × (3⁴ × 3²) = 2⁸ × 3⁶. No actual multiplication needed — just add exponents. Option B matches.
Why this question: The 2023 PYQ is the canonical "of trap" question. The numerator requires you to treat 13 of 13 as 169 (as the denominator of the division), not as a separate multiplication.
Solving path: Numerator: 13 ÷ (13 of 13) + 13 = 13 ÷ 169 + 13 = 1/13 + 13 = 1/13 + 169/13 = 170/13. Denominator: 13 ÷ 13 × 13 + 13 — here no "of", so left-to-right DM: 1 × 13 + 13 = 26. Final: (170/13) ÷ 26 = 170/338 = 85/169. Option B.
Why this question: A long mixed-operation chain — exactly the type where chunking saves time. This is a 2022 question and reflects the style SSC has used consistently.
Solving path: Chunk 1: 28 × 25 ÷ 7 × 15. Simplify 28/7 = 4 first, then 4 × 25 × 15 = 1500. Chunk 2: 63 × 56 ÷ 72 × 5. Simplify 56/72 = 7/9, then 63 × (7/9) × 5 = 7 × 7 × 5 = 245. Chunk 3: 24 × 7 + 8 = 176. Total: 1500 + 245 + 176 = 1921. Option B.
Why this question: Nested "of" with decimals and a bracket containing subtraction — tests three layers of BODMAS in one expression.
Solving path: Step 1 (O): 2 of 3 = 6. Expression becomes 6 ÷ 3 × 2 + {4 × 3 − (5 × 2 + 3)}. Step 2 (innermost bracket): (5 × 2 + 3) = 13. Step 3 (curly bracket): {4 × 3 − 13} = {12 − 13} = −1. Step 4 (DM on remaining): 6 ÷ 3 × 2 = 4. Step 5 (AS): 4 + (−1) = 3. Option C.
Why this question: A 2025 PYQ — current pattern. Short chain with one bracket and "of", representative of what you will actually see on exam day.
Solving path: Identify the two O-tier operations first: 42 ÷ 6 = 7 (division, part of DM-chunk) and 18 of 3 = 54. Expression: 91 − 33 + 7 + 54 − 29 + 22. Running total: 91 − 33 = 58, 58 + 7 = 65, 65 + 54 = 119, 119 − 29 = 90, 90 + 22 = 112. Option A.
Treating "of" as equal to regular multiplication in a division chain. When the expression says a ÷ b of c, the "of" binds to b, making the divisor bc. Students who read left-to-right get (a ÷ b) × c, which is completely different. Always check whether "of" sits immediately after a ÷ sign.
Resolving D before M or M before D without checking left-to-right order. Division and multiplication are equal-priority. 6 ÷ 3 × 2 must go left to right: (6 ÷ 3) × 2 = 4. Doing multiplication first gives 6 ÷ 6 = 1 — wrong.
Skipping the bracket type hierarchy. Students comfortable with round brackets sometimes resolve { } or [ ] before clearing all ( ) inside them. Always clear round brackets first, then curly, then square.
Adding all numbers before handling DM-chunks in a long expression. In 1500 + 245 + 176 this is fine, but in a raw expression like 4 + 3 × 5, adding 4 + 3 first gives 35 instead of the correct 19. The + sign does not mean "add now" if there is an unresolved × nearby.
Computing actual values when only the form matters (index questions). If the question asks for 2³ × 3⁴ × 2⁵ × 3² and the options are all in power form, there is zero reason to compute 8 × 81 × 32 × 9. Recognise that the answer format tells you what method to use.
Sign errors after bracket removal. When a negative sign precedes a bracket — −(10 + 3) — both terms inside flip sign: −10 − 3 = −13. Students often flip only the first term and leave the second unchanged, giving −10 + 3 = −7.