Mathematical Operations questions in Reasoning belong to a very specific trap category. The examiner takes the four standard arithmetic operations — addition, subtraction, multiplication, division — and disguises them using letters, symbols, or even swaps one for another. Your job: decode the disguise, rewrite the expression in true mathematical form, apply BODMAS, and pick the answer.
Here's the thing most students miss: the question does not test your arithmetic. It tests whether you can hold the substitution table in your head for 30 seconds without accidentally using the original symbol's meaning.
Think of it like a uniform swap. Imagine four constables named +, -, ×, ÷. The question says: today, constable + will wear constable ×'s uniform. Now you see someone in +'s uniform — you do not treat them as addition. You look at the order sheet (the question), find what uniform that constable was assigned, and respond accordingly.
Three sub-types appear in UP Police Constable papers:
Type 1 — Letter/Symbol Substitution. A key like "T means ×, U means −, V means ÷, W means +" is given. You rewrite the expression and calculate.
Type 2 — Operation Swap. The four standard operators are cyclically or randomly reassigned. "+ means −, − means ×" etc. The expression is written with the original symbols, but you must treat them as their coded counterparts.
Type 3 — Custom Operator Definition. A new symbol is defined with a formula, like A % B = A + B − 1. You plug in numbers and evaluate using the given formula, not any standard operation.
Type 4 — Number Swapping to Balance Equations. An equation is given that does not balance. You must try swapping pairs of numbers from the options until it does.
The analogy: Type 1 and Type 2 are translation problems. Type 3 is a formula-substitution problem (like अनुपात with a custom rule). Type 4 is a trial-and-verification problem.
All four types appear in recent UP Police Constable papers. You need to be fast and systematic — guessing costs you here because all options are numerically plausible.
This is the most straightforward type. The question gives you a decoding table. Your only task: replace every coded letter/symbol with its real operation, then apply BODMAS faithfully.
Example: T = ×, U = −, V = ÷, W = +
Expression: (56 V 4) W (26 T 5)
Step 1 — Translate: (56 ÷ 4) + (26 × 5)
Step 2 — Evaluate brackets first (BODMAS):
56 ÷ 4 = 1426 × 5 = 130Step 3 — Combine: 14 + 130 = 144
Done. The critical discipline: never skip the translation step and work directly on the coded expression. That is how students pick 144 for the wrong reason or arrive at the wrong answer entirely.
Here the expression is written using standard symbols, but those symbols mean something different for this question. This is slightly trickier because your brain has 12+ years of conditioning to read + as addition.
The rule: Read the expression left to right. Every time you encounter a symbol, pause, check the swap table, replace it mentally, then proceed.
Example: + means −, − means ×, × means +, ÷ means +
Expression: 16 × 4 ÷ 4 + 14 − 2
Step 1 — Translate each operator:
× → −÷ → ++ → ×− → ÷Translated expression: 16 − 4 + 4 × 14 ÷ 2
Step 2 — Apply BODMAS to the translated expression:
4 × 14 = 5656 ÷ 2 = 2816 − 4 + 28 = 40Answer: 40.
Common failure mode: Students apply BODMAS to the original expression and then try to swap. Wrong order. Swap first, then BODMAS.
The question defines a new operation using a formula. Example: A % B = A + B − 1. You receive 10 % 2 and must find the value.
Straightforward: identify A = 10, B = 2, substitute into formula.
10 + 2 − 1 = 11
When the definition contains standard operations inside it (like A * B = A + B × 4), apply BODMAS to the definition's right-hand side after substituting values.
Example: A * B = A + B × 4
For 16 * 20: substitute A = 16, B = 20.
= 16 + 20 × 4 = 16 + 80 = 96
Note that B × 4 is multiplication and happens before the addition. The definition's formula obeys BODMAS.
When custom operators are nested (one result feeds into another definition), evaluate the innermost operation first — treat it exactly like nested brackets.
You receive an equation that does not hold. You must find which two numbers, when swapped, make it correct.
Systematic approach:
Example: 14 × 8 ÷ 12 + 64 − 2 = 108
Option C says swap 2 and 12. New expression: 14 × 8 ÷ 2 + 64 − 12
Evaluate:
14 × 8 = 112112 ÷ 2 = 5656 + 64 = 120120 − 12 = 108 ✓This matches. Option C is correct.
Speed tip: start with options that involve numbers appearing in places where their swap would most dramatically change the result (typically denominators or subtracted values).
A partially written equation has blank slots where operators need to go. You insert operators from the given option and verify.
Example: (12 * 7 * 6) * 13 * 6 — fill in * slots.
Option D says: ×, −, ÷, =
Substituting: (12 × 7 − 6) ÷ 13 = 6
12 × 7 = 8484 − 6 = 7878 ÷ 13 = 6 ✓Here, you are testing each option sequentially. With only 4 options, this is a pure substitution game — no derivation needed.
The single most important rule: in Type 2 (operation swap), write down the translated expression on paper before touching a single number. Students who jump into calculation using the original symbols lose 30-40 seconds backtracking. The physical act of rewriting takes 5 seconds and eliminates the most common error. Standard method (mentally tracking swaps while calculating): 60s, 40% error rate. Translation-first method: 20s, near-zero error rate.
After translating (Type 1 or Type 2), your brain still sees the original expression. Lock onto the translated version. A quick drill: cover the original expression with your hand or finger after writing the translated version. Work only from the translated line. This eliminates the two-line confusion that causes students to mix up operator precedence from the wrong version. Saves approximately 3 steps of re-checking.
In Type 4 (number-swap to balance), do not analyse which number "should" logically move. Instead, go straight to option A, perform the swap, evaluate the expression, check against the target. If wrong, move to option B. With 4 options and expressions that take 10-15 seconds each to evaluate, worst case is 60 seconds. Most students waste 90 seconds trying to reason analytically before trying options. Pure substitution wins here: saves 30+ seconds vs analytical guessing.
When a custom operator formula has two variables (like A # B = A − B ÷ 3), label which number in the expression is A and which is B before substituting. Write A=35, B=96 next to the expression. Then substitute mechanically: 35 − 96 ÷ 3. This prevents the common mistake of putting B in A's position, which happens when students evaluate under time pressure. Two extra seconds of labelling saves 30 seconds of wrong-answer recovery.
In sign-insertion questions, the = sign must appear at a specific slot. If the question has 4 blanks and the answer must be a true equation, any option where = is not placed logically (e.g., = appears before all operations are resolved) can be eliminated immediately. In most UP Police Constable questions, this eliminates 1-2 options before calculation, reducing your work from 4 substitutions to 2-3. Time saving: 15-20 seconds per question.
In the exam hall, classify the question in the first 5 seconds:
Is a decoding table given (letters like T, U, V or symbols like @)? → Yes: Type 1 or Type 3. Check if a formula is given (Type 3) or just a one-to-one replacement (Type 1). Translate, then BODMAS.
Are standard operators (+, −, ×, ÷) redefined as other standard operators? → Yes: Type 2. Write the translated expression on a separate line. Apply BODMAS only to the translated version.
Is a full equation given that does not balance, and you must swap two numbers? → Yes: Type 4. Go straight to option substitution. Start with option A, check, move to B if wrong.
Are blank slots (*) in an expression you must fill with operators? → Yes: Type 5 (sign insertion). Eliminate options where = is placed illogically. Try remaining options via substitution.
For nested custom operators: always evaluate innermost first. Write intermediate results clearly. Do not carry two live values in your head simultaneously.
Time budget: 45-60 seconds per question maximum. If you have not arrived at an answer in 50 seconds, mark your best guess and move on — these questions do not reward extended calculation.
Why this question: Direct Type 1 substitution — the bread-and-butter UP Police format with a decoding table and a two-bracket expression. Tests pure translation discipline.
Solving path: Identify the table: T=×, U=−, V=÷, W=+. Translate: (56 ÷ 4) + (26 × 5). Evaluate brackets: 14 and 130. Sum: 144. Total time: under 25 seconds.
Why this question: The classic simple single-symbol substitution. Tests whether you can hold one definition in memory and apply it without overthinking.
Solving path: @ means ÷. So 15 @ 5 = 15 ÷ 5 = 3. Done in under 10 seconds. If you see a question this simple, do not second-guess yourself into a wrong answer.
Why this question: Type 3 — custom operator with a formula. Tests formula substitution with BODMAS inside the definition.
Solving path: Formula: A % B = A + B − 1. Identify A = 10, B = 2. Substitute: 10 + 2 − 1 = 11. Answer: 11. The trap here is that % looks like a percentage operator — ignore that instinct entirely.
Why this question: Type 5 — sign insertion into a balanced equation. Tests systematic option-by-option substitution with BODMAS.
Solving path: Try option D: ×, −, ÷, =. Expression becomes (12 × 7 − 6) ÷ 13 = 6. Evaluate: 84 − 6 = 78, 78 ÷ 13 = 6. Matches. Note that the bracket in the original expression forces multiplication and subtraction to resolve first — respect the bracket even in these substitution problems.
Why this question: Type 4 — number swapping to balance. Tests the plug-and-check method under time pressure.
Solving path: Original: 14 × 8 ÷ 12 + 64 − 2 = 108. Try option C (swap 2 and 12): 14 × 8 ÷ 2 + 64 − 12. Step by step: 14 × 8 = 112, 112 ÷ 2 = 56, 56 + 64 = 120, 120 − 12 = 108. Confirmed. The key insight: swapping the denominator (12 → 2) dramatically changes the division result, which is the most leveraged position in this expression.
Why this question: Type 2 — operation swap with BODMAS applied to the translated expression. The most important type to get right because the brain's instinct to use original symbols is strongest here.
Solving path: Swap table: +→×, ×→−, −→÷, ÷→+. Original: 16 × 4 ÷ 4 + 14 − 2. Translate operator by operator: ×→−, ÷→+, +→×, −→÷. Translated: 16 − 4 + 4 × 14 ÷ 2. Now BODMAS on the translated version: 4 × 14 = 56, 56 ÷ 2 = 28, 16 − 4 = 12, 12 + 28 = 40. Answer: 40.
Applying BODMAS before translating (Type 2). Students simplify 16 × 4 ÷ 4 using the original operators first, then try to apply the swap. This is backwards. The translation must happen before any calculation.
Reversing A and B in custom operator formulas. When the formula is A # B = A − B ÷ 3, students sometimes compute B − A ÷ 3 under time pressure. Always write A = (first number), B = (second number) explicitly before substituting.
Forgetting BODMAS inside custom operator definitions. In A * B = A + B × 4, the B × 4 must be evaluated before adding A. Students who read left-to-right compute (A + B) × 4 = 144 instead of A + (B × 4) = 96.
In number-swapping, swapping operators instead of numbers. The question asks which two numbers to swap. Students sometimes swap the positions of operation signs instead. Read the question carefully — the operators stay fixed, only the numbers move.
In sign insertion, ignoring brackets. When the expression has (12 * 7 * 6) * 13 * 6, the bracket forces the operations inside it to be evaluated first regardless of which operators fill the slots. Students who ignore the bracket order get wrong intermediate values.
Treating the = sign as just another blank. In sign insertion questions, the = is the final sign and defines what the expression must equal. If an option places the = sign mid-expression in a way that leaves numbers orphaned on the right, that option is structurally impossible — eliminate it before calculating.