Arithmetic word problems are the backbone of SBI PO's quantitative aptitude section. The exam doesn't test your ability to recall formulas in isolation — it tests whether you can read a messy real-world scenario, extract the mathematical structure underneath, and solve it before time bleeds away.
The core skill is translation. Every word problem is a sentence written in English that encodes an equation or a set of equations. "A and B together finish work in 12 days" means 1/a + 1/b = 1/12. "He marks up by 40%" means MP = 1.4 × CP. The moment you write that equation, the problem is 60% done.
Here is the mental model that separates fast solvers from slow ones: think in rates and multipliers, not amounts. Instead of tracking rupees and percentages separately, compress them. A 40% markup followed by a 10% discount is not calculated step-by-step — it is 1.4 × 0.9 = 1.26 in one line. A pipe filling at 1/12 of a tank per hour and a leak draining at 1/60 per hour net out to 1/12 - 1/60 = 4/60 = 1/15 per hour. Rates add and subtract cleanly.
Think of it like mixing currents in a wire. If two currents flow in opposite directions on the same path, the net current is the difference. Pipes, work-rates, and speeds all behave identically once you frame them as rates.
Arithmetic word problems on SBI PO cluster into a handful of recurring structures:
Master these seven families and you have covered the vast majority of what SBI PO throws at you.
Every CP-MP-SP problem is a chain of multiplications. Memorise these compressed forms:
r%: MP = CP × (1 + r/100)d%: SP = MP × (1 - d/100)d1% and d2%: effective discount factor = (1 - d1/100)(1 - d2/100)Look — successive discounts are not additive. A 10% + 15% discount is not 25%. It is 0.9 × 0.85 = 0.765, i.e., 23.5% effective discount. Confusing this is one of the most common wrong answers in SBI PO Prelims.
For the "find CP from SP with markup + two discounts" pattern: write SP as CP × (markup factor) × (discount1 factor) × (discount2 factor) and solve for CP directly. No working backwards step-by-step required.
SI is clean: SI = PRT/100. CI involves P(1 + R/100)^n - P. The key structural insight: the interest earned in the 3rd year of CI equals the amount at the end of 2 years multiplied by the rate. This is because each year's interest is earned on that year's principal, which already includes prior interest.
So if you are given amounts at year 2 and year 3, the rate is just:
R = (A3 - A2) / A2 × 100
No need to set up a full compound interest equation. The difference between consecutive CI amounts directly gives you the rate.
The textbook approach (add the fractions 1/a + 1/b) is fine, but the LCM method is faster for three-person problems:
For the "A+B, B+C, C+A" structure specifically: add all three equations to get 2(A+B+C) rate, halve it to get (A+B+C) rate, then subtract each pair to isolate individuals.
Alligation is the fastest tool for mixture problems where you know two "strengths" and want a target strength. Draw the cross:
Strength 1 (S1) Strength 2 (S2)
\ /
Mean (M)
/ \
(S2 - M) (M - S1)
The ratio of quantities is (S2 - M) : (M - S1). This works for profit percentages, concentrations, prices — anything that behaves as a weighted average.
Average problems often look complex but reduce to one equation: new sum = new average × new count. When a new member is added or removed, compute the change in total sum, which equals the difference between the new member's value and the old average, multiplied by... no, simpler: new member's value = (new average × new count) - (old average × old count).
Structurally identical to work-time. Filling pipes have positive rates, leaks have negative rates. Net rate = sum of all filling rates minus sum of all leaking rates. Time to fill = 1 / net rate.
One trap: the problem says the pipe fills in 12 hours "alone", but asks for the time WITH the leak. The leak's rate is found by subtraction: leak rate = fill rate alone - fill rate with leak.
Problems like "add x liters of water to change ratio from A:B to C:D" always have the same structure: the quantity of the component you are NOT adding stays fixed. Set up one equation with one unknown. Cross-multiply, solve. Done.
When CI amounts are given for year n and year n+1, the rate is simply the ratio of the increment to the earlier amount. If amount grows from ₹4840 to ₹5324, rate = (5324 - 4840)/4840 × 100 = 484/4840 × 100 = 10%. Standard method: set up P(1+r/100)^n equations and divide — 4 steps. This pattern: 1 step. Time saving: roughly 50 seconds vs 10 seconds.
For two successive discounts d1 and d2, final price factor = (1 - d1/100)(1 - d2/100). For 10% + 15%: 0.9 × 0.85 = 0.765. Effective discount = 23.5%. Standard method: apply first discount, take new price, apply second discount — 6 arithmetic operations. This method: 1 multiplication. Saves 30+ seconds on chain-discount problems.
For (A+B) = 1/p, (B+C) = 1/q, (C+A) = 1/r: sum all three to get 2(A+B+C) = 1/p + 1/q + 1/r. Halve it to get (A+B+C). Then A alone = (A+B+C) - (B+C) = half-sum minus 1/q. This is 3 fraction additions and 1 subtraction — standard algebraic elimination of three variables takes 8-10 steps. This method: 4 steps.
Draw the alligation cross instead of writing algebra. Place S1 top-left, S2 top-right, mean in center. Cross-differences give the ratio. For 8% and 18% profit with 14% mean: cross differences are (18-14)=4 and (14-8)=6. Ratio = 4:6 = 2:3. Quantity at 18% = 3/5 × 1000 = 600 kg. Standard algebra (two equations, two unknowns): 5 steps. Alligation cross: 2 steps.
When one person is added to a group and the average changes, the new person's value = new total sum - old total sum. New sum = new average × new count. Old sum = old average × old count. For the teacher problem: 21×19 - 20×18 = 399 - 360 = 39. No equation needed, no variable introduced. Standard method: set up equation with unknown x — 3 steps. Direct difference: 2 multiplications and a subtraction, under 15 seconds.
In the exam hall, follow this decision tree for every arithmetic word problem:
Step 1 — Identify the family. Read the first two lines. Does it mention CP/MP/SP, profit, discount? Go to multiplier chain. Does it say "invested at interest"? Go to SI/CI. Does it say "complete work", "fill tank", "pipe"? Go to rate addition.
Step 2 — Write the compressed form immediately. Don't narrate, don't paraphrase. Write the equation in multiplier or rate form in under 20 seconds.
Step 3 — Check if alligation applies. Any problem where two things mix to produce a known average — profits, concentrations, prices — use the alligation cross. It halves your work time.
Step 4 — Plug answer options if algebra feels slow. On SBI PO, four options are given. For problems like "find CP given SP", plug the options directly into the formula. Two wrong options will fail immediately. This is not guessing — it is structured verification that often takes under 30 seconds.
Step 5 — Sanity-check units. If the question asks for "days", your answer should not be a fraction unless the options show fractions. If something feels off in 3 seconds of checking, reread the question — the most common error is misreading "A alone" versus "A and B together".
Why this question: Three-person work problems are a recurring SBI PO pattern. The key is recognising the "add all three, halve, subtract" structure before you start.
Solving path: Let 1/a, 1/b, 1/c be individual rates. You have 1/a + 1/b = 1/12, 1/b + 1/c = 1/15, 1/c + 1/a = 1/20. Add all three: 2(1/a + 1/b + 1/c) = 1/12 + 1/15 + 1/20. Find LCM(12,15,20) = 60. Sum = 5/60 + 4/60 + 3/60 = 12/60 = 1/5. So 1/a + 1/b + 1/c = 1/10. Now 1/a = 1/10 - (1/b + 1/c) = 1/10 - 1/15 = 3/30 - 2/30 = 1/30. A alone takes 30 days.
Why this question: Markup + double discount is a classic SBI PO Prelims structure. The multiplier chain collapses this to one line.
Solving path: Let CP = x. MP = 1.4x. After 10% discount: 0.9 × 1.4x = 1.26x. After 15% discount: 0.85 × 1.26x = 1.071x. Given SP = 1071, so 1.071x = 1071, giving x = ₹1000. Note: you could also plug options. If CP = ₹1000, then SP = 1000 × 1.4 × 0.9 × 0.85 = 1071. Confirmed in under 20 seconds.
Why this question: Alligation on profit percentages — looks like algebra but alligation cross cuts it to two arithmetic operations.
Solving path: Two groups: one at 8% profit, one at 18% profit, blended at 14%. Alligation cross: difference from mean for 8% side = 14 - 8 = 6; for 18% side = 18 - 14 = 4. Ratio of (8%-group):(18%-group) = 4:6 = 2:3. Total = 5 parts = 1000 kg. 18%-group = (3/5) × 1000 = 600 kg.
Why this question: CI rate from consecutive year amounts — use the direct pattern, not the full formula.
Solving path: Amount in year 3 minus amount in year 2 = interest on year-2 amount for 1 year. Interest = 5324 - 4840 = ₹484. This is earned on principal of ₹4840. Rate = (484/4840) × 100 = 10%. Done in two arithmetic steps. Then verify: P × (1.1)² = 4840 gives P = 4840/1.21 = ₹4000, and 4000 × 1.331 = ₹5324. Consistent.
Why this question: Mixture ratio problem — fixed component trick avoids setting up two simultaneous equations.
Solving path: Initial milk = (7/8) × 80 = 70 litres. Initial water = (1/8) × 80 = 10 litres. Adding x litres of water, milk stays at 70. New ratio: 70 : (10 + x) = 7 : 3. Cross multiply: 70 × 3 = 7 × (10 + x), so 210 = 70 + 7x, giving x = 20 litres.
Why this question: Pipes and cisterns with a leak — the leak rate is found by subtraction of the two known fill rates.
Solving path: Fill rate without leak = 1/12 tank/hour. Fill rate with leak = 1/15 tank/hour. Leak rate = 1/12 - 1/15 = 5/60 - 4/60 = 1/60 tank/hour. Time for leak to empty full tank = 60 hours.
Treating successive discounts as additive. A 10% + 15% discount is not 25%. Always multiply the factors: (1 - 0.10)(1 - 0.15) = 0.765, giving 23.5% net discount. This single error eliminates tens of thousands of candidates every year in SBI PO Prelims.
Adding percentages across different bases. "8% profit on one batch and 18% on another" cannot be averaged arithmetically unless the batches are equal in cost. The correct method is alligation or weighted average with actual cost values.
Forgetting to halve the sum in three-person work problems. When you add all three pair equations, you get 2(A + B + C), not (A + B + C). Forgetting to divide by 2 produces an answer exactly double the correct rate, leading to a time exactly half the correct answer.
In CI problems, computing interest instead of the CI formula difference. The question asks for CI, not SI. If given the CI and asked for SI (or vice versa), use the direct formulas — do not subtract SI from CI by computing both from scratch unless you have time to spare.
In average problems, using old count instead of new count. After adding the teacher to 20 students, the new count is 21, not 20. Using 20 × 19 instead of 21 × 19 for the new total sum is a very common error, particularly under exam-hall time pressure.
In ratio-mixture problems, changing the component you should keep fixed. When you add water to milk, the milk quantity stays constant. Set up the equation with the constant component on one side of the ratio, not the changing component. Mixing this up leads to an algebraically solvable but semantically wrong equation that gives a plausible-looking wrong answer.