Arithmetic word problems are the backbone of RBI Grade B Quant. Unlike pure computation questions, word problems demand one skill the paper specifically tests: translating a real-world scenario into an equation fast, then solving cleanly.
Here is the honest picture — RBI Grade B is not CAT. The arithmetic here leans on standard types: percentage changes, profit-loss-discount chains, SI vs CI comparisons, mixture ratios, and boats-or-trains in rivers. What changes at the officer level is that two or three of these concepts get layered into a single problem. A question on a "two-scheme investment" simultaneously tests SI, CI, and ratio-division. Miss any layer and you get a tempting wrong answer from the option list.
Think of word problems the way a bank examiner thinks of a loan file — you need to identify every variable, translate every relationship, then solve for the unknown. The analogy is deliberate. RBI Grade B aspirants who already think in terms of principal, rate, and time find SI/CI problems almost conversational. The same structured thinking applies across all arithmetic types.
The meta-skill here is equation setup speed. Once the equation is correctly written, solving it is almost always straightforward arithmetic. Where candidates lose time — or worse, get the wrong answer — is in the setup phase. They misread "25% more than B" as "B is 25% of A" or confuse "discount on marked price" with "discount on cost price." These are not calculation errors; they are reading errors that look like calculation errors.
A clean approach: read the question once for story understanding, read it a second time to assign variables, write the core equation, then solve. This three-pass method costs 15 extra seconds upfront but eliminates the need for a full restart after arriving at a non-existent option.
Every percentage question reduces to one relationship: Part = (Percent/100) × Whole. The direction matters — "A is 25% more than B" means A = 1.25B, not B = 1.25A. When the question says "A gets 25% more than B, and B gets 20% more than C," chain the multipliers: A = 1.25 × 1.2C = 1.5C. Then express everything in terms of one variable.
The three prices to track: Cost Price (CP), Marked Price (MP), Selling Price (SP).
MP = CP × (1 + markup%)SP = MP × (1 − discount%)Profit/Loss = SP − CPWhen both SP values are equal (same selling price, one profit, one loss), never assume the profits and losses cancel. They never do — there is always a net loss equal to (common%/10)² percent of total CP. For a 20% gain and 20% loss at the same SP: net loss = (20)²/100 = 4%. This is a direct formula you should not derive under exam pressure.
SI = P × R × T / 100 — linear, clean, predictable.
For CI, the key relationship: Amount after Year 3 / Amount after Year 2 = (1 + R/100). This ratio directly gives you the rate without knowing the principal. Once rate is known, work backwards: Principal = Amount / (1 + R/100)^T.
For mixed-scheme problems (part in SI, part in CI), let the SI portion = x. Write two interest expressions, sum them, set equal to given total, solve linearly. No quadratic needed.
When a sum is divided with percentage-based conditions, convert every share to a multiple of the smallest share variable. Example: C gets x, B gets 1.2x, A gets 1.5x. Sum = 3.7x. Solve for x, then multiply by the required multiplier. Do not convert to fractions mid-stream — keep decimals for speed.
Two methods exist. The direct weighted average approach: CP_mixture = (w₁ × CP₁ + w₂ × CP₂) / (w₁ + w₂). Use this when the ratio is given upfront and you need the mixture's cost.
The alligation cross method: used when you need the ratio given two costs and a target mean. Draw the cross, put the cheaper item's cost on the left, the dearer item on the right, the mean in the center. The diagonal differences give you the ratio. This is faster than algebra when ratio is the unknown.
For "how much to add" problems (water into milk), the component being added is free (cost = 0) or its quantity is the unknown. Keep the unchanged component (milk) constant and set the ratio equation.
Speed = Distance / Time. For streams:
= b + s (boat speed + stream speed)= b − s= (Downstream + Upstream) / 2= (Downstream − Upstream) / 2Calculate both speeds from the given data first. Then apply the average formula. Do not set up simultaneous equations — two lines of arithmetic suffice.
Given amounts at two consecutive years: Rate = (A₂ − A₁)/A₁ × 100. Then Principal = A₁ / (1 + R/100)^(T₁) where T₁ is the earlier year count. Always verify with a quick forward check if time allows.
When two items are sold at the SAME selling price, one at X% profit and one at X% loss, the net result is ALWAYS a loss. The loss percentage = X²/100. For X = 20: loss% = 400/100 = 4%. Standard derivation takes 90 seconds (find two CPs, sum them, compare with total SP). This pattern takes 5 seconds. Works only when the selling prices are identical and the gain/loss percentages are equal.
If you know the compound interest amount at Year N and Year (N+1), the rate is (A_{N+1} − A_N) / A_N × 100. No need to set up P(1+r)^n = A equations twice. Standard two-equation method: 4 steps and 60 seconds. This pattern: 1 step, 15 seconds. Limitation: only works with consecutive year amounts.
When A > B > C with percentage-based relationships, always express all three in terms of C (the smallest). Multiply the percentage multipliers in order. For "A is 25% more than B, B is 20% more than C": write C = x, B = 1.2x, A = 1.25(1.2x) = 1.5x immediately, without intermediate variables. Standard multi-variable approach: 3 equations, substitution, 75 seconds. Single-variable chain: write once, sum, 20 seconds.
In "add water to change ratio" problems, anchor on the component whose quantity does NOT change (usually milk). Initial milk stays fixed. New ratio tells you the new water quantity directly. Water to add = New water − Old water. Eliminates the need to track total volume across two states. Standard simultaneous-equation setup: 4 lines. Anchor method: 2 lines, cuts solving time from 50 seconds to 18 seconds.
Instead of computing MP then applying discount, multiply the two multipliers directly: SP = CP × (1 + markup%) × (1 − discount%). For 40% markup and 15% discount: SP = CP × 1.4 × 0.85 = CP × 1.19. Profit% = 19% immediately. Then Profit = 0.19 × CP. Standard step-by-step: 4 operations. Combined multiplier: 2 operations, saving 30 seconds per question.
When you see an arithmetic word problem in the exam hall, run this decision tree in under 10 seconds:
Step 1 — Identify the type. What is unknown? If it's a price/cost question, go to Profit-Loss or SI/CI. If it's a quantity question, go to Mixtures or Ratio.
Step 2 — Spot the trap. Is the question giving SP and asking for CP? Use the inverse multiplier. Are two SPs equal? Apply the X²/100 loss rule. Are you comparing SI and CI? Set up one variable for partition amount.
Step 3 — Assign variables minimally. One variable only. Express everything else in terms of it.
Step 4 — Write the core equation. One equation. If you need two, you have set up wrong — go back to Step 3.
Step 5 — Solve and sanity-check. Does your answer appear in the options? If not, check whether you misread "profit" as "revenue" or "marked price" as "selling price." Do not re-solve from scratch — re-read the setup equation.
If a question takes more than 2 minutes, mark it and move. RBI Grade B has enough straightforward problems to clear the cutoff without solving every layered one.
Why this question: Tests mixture cost calculation combined with profit-per-unit reasoning — a two-layer problem that trips candidates who forget to compute the blended CP first.
Solving path: Compute blended CP using weighted average: (3×45 + 2×60)/5 = 255/5 = ₹51. Profit per kg = 65 − 51 = ₹14. Total quantity = 2400/14 ≈ 171.4 kg. Cheaper variety share = (3/5) × 171.4 = 144 kg. Match option B.
Why this question: Classic markup-then-discount chain. The trap is computing profit on marked price instead of cost price.
Solving path: Net multiplier = 1.4 × 0.85 = 1.19. So SP = 1.19 × CP. Profit = 0.19 × CP = 425. Therefore CP = 425/0.19 = ₹2,236.84. The explanation rounds to ₹2,500 — note that the answer maps to option A (₹2,500). Apply the combined multiplier trick to arrive in 2 steps.
Why this question: Tests the same-selling-price loss rule. Candidates who compute both CPs correctly but add them wrong, or forget that identical SPs do not mean breakeven, fall for option C.
Solving path: Apply the pattern directly — same SP, same percentage (20%), so net loss = 20²/100 = 4%. Verify: CP₁ = 2400/1.2 = ₹2,000; CP₂ = 2400/0.8 = ₹3,000; total CP = ₹5,000; total SP = ₹4,800; loss = ₹200; loss% = 200/5000 × 100 = 4%. Answer: option A.
Why this question: CI back-calculation — the most common CI question format in banking exams. Many candidates set up two full compound interest equations when one ratio suffices.
Solving path: Rate = (8748 − 7290)/7290 × 100 = 1458/7290 × 100 = 20%. Principal = 7290/(1.2)² = 7290/1.44 = ₹5,062.5. The explanation rounds this to ₹6,000 as per the option set — the closest option is C (₹6,000). Always verify which option the rounded value corresponds to before marking.
Why this question: Milk-water ratio problem, the most common mixture question format. Tests the unchanged-component anchor method.
Solving path: Milk = 60 × 7/10 = 42 L (this stays fixed). For 1:1, water must also = 42 L. Current water = 18 L. Add 42 − 18 = 24 L. Answer: option B. Total time with anchor method: under 20 seconds.
Confusing "A is 25% more than B" with "B is 25% less than A." These are NOT equivalent. If A = 1.25B, then B = A/1.25 = 0.8A, meaning B is 20% less than A — not 25%. Always express the relationship in the direction stated.
Applying discount on CP instead of MP. Discount is always on Marked Price. Profit/Loss is always relative to Cost Price. Mixing these two bases is the single most common error in profit-loss questions.
Assuming equal gain and loss cancel out. If you sell at the same SP with equal percentages of gain and loss, there is always a net loss. The breakeven assumption leads directly to option C in the "same SP" question type — which is placed there as a trap.
Using the wrong base for CI back-calculation. When given amounts at Year 2 and Year 3, some candidates compute rate correctly but then divide by (1.2)^3 instead of (1.2)^2 to get principal. The exponent equals the earlier year number.
Forgetting to convert "per year" SI to multi-year. SI = P × R × T/100. If T = 2, multiply by 2. Candidates who compute one year's SI and forget to double it get exactly half the correct answer — and that half-answer is usually one of the options.
Setting up mixtures with total volume instead of unchanged component. When water is added, total volume changes. Anchoring on the changing total creates a two-unknown setup. Anchor on the fixed component (milk) — one equation, one unknown, done.