In this guide
Interest is the one arithmetic topic that is also your future job. A clerk at the counter explains fixed deposit maturity, loan interest and recurring deposits to customers, and the exam tests the same ideas. Clerk papers commonly include an interest question among the arithmetic word problems, and interest figures sometimes appear inside DI sets.
The topic rests on two formulas. Everything else is a shortcut you can prove from them in one line, which is worth doing once so you trust the shortcuts in the hall.
Simple interest
SI = P × R × T ÷ 100, and Amount = P + SI.
Here P is the principal, R the rate per cent per year, and T the time in years. Simple interest is charged only on the original principal, so the same amount of interest is added every year. That is why SI grows in a straight line with time.
Two useful consequences:
- A sum doubles under SI when SI = P, which happens when R × T = 100. At 12.5% it takes 8 years.
- A sum becomes n times when SI = (n − 1)P, so R × T = 100(n − 1).
Compound interest
Amount = P × (1 + R/100)ᵀ, and CI = Amount − P.
Under compounding, each year's interest is added to the principal and earns interest itself. So the amount is multiplied by the same factor every year, which is why the formula has a power.
Effective rates worth memorising
| Rate | 2 years (CI as % of P) | 3 years (CI as % of P) |
|---|---|---|
| 5% | 10.25% | 15.7625% |
| 10% | 21% | 33.1% |
| 20% | 44% | 72.8% |
The 2-year figure comes straight from the successive-change rule: R + R + R²/100. For 10%, that is 10 + 10 + 1 = 21%.
The CI–SI difference
- 2 years: CI − SI = P × (R/100)².
- 3 years: CI − SI = P × (R/100)² × (3 + R/100).
Why the 2-year rule works: in year one, CI and SI are equal. In year two, CI also earns interest on the first year's interest, which is P × R/100. Interest on that at R% is P × (R/100)². That is the whole difference.
Half-yearly and quarterly compounding
- Half-yearly: use R/2 per period and 2T periods.
- Quarterly: use R/4 per period and 4T periods.
More frequent compounding always gives slightly more interest, because interest starts earning interest sooner.
Doubling under CI
If a sum doubles in n years under CI, it becomes 4 times in 2n years and 8 times in 3n years. The multiplier over each n-year block is 2, and blocks multiply: 2 × 2 = 4, 2 × 2 × 2 = 8.
Worked examples
Example 1. Find the SI on ₹7,500 at 6% a year for 2 years 8 months.
- Time = 2 8/12 = 8/3 years.
- SI = 7,500 × 6 × 8/3 ÷ 100 = 7,500 × 16 ÷ 100 = ₹1,200.
Example 2. Find the CI on ₹10,000 at 10% a year for 3 years.
- Amount = 10,000 × 1.1 × 1.1 × 1.1 = 13,310.
- CI = ₹3,310. Or use the table: 33.1% of 10,000.
Example 3. Find CI − SI on ₹10,000 at 10% for 2 years and for 3 years.
- 2 years: 10,000 × (0.1)² = ₹100. Check: CI 2,100, SI 2,000.
- 3 years: 10,000 × 0.01 × 3.1 = ₹310. Check: CI 3,310, SI 3,000.
Example 4. A sum doubles in 8 years at SI. What is the rate, and when will it triple?
- SI = P in 8 years, so R × 8 = 100 and R = 12.5%.
- Tripling needs SI = 2P, so R × T = 200 and T = 16 years.
Example 5. Find the CI on ₹8,000 at 10% a year, compounded half-yearly, for 1 year.
- Rate per half-year = 5%; periods = 2.
- Amount = 8,000 × 1.05 × 1.05 = 8,820.
- CI = ₹820. With yearly compounding it would be ₹800, so half-yearly earns ₹20 more.
Example 6. A sum amounts to ₹4,840 in 2 years at 10% CI. Find the principal.
- P × 1.21 = 4,840.
- P = 4,840 ÷ 1.21 = ₹4,000.
Common mistakes
- Using months as years. Convert: 8 months is 8/12 = 2/3 year.
- Forgetting to halve the rate when compounding half-yearly, or halving the rate but not doubling the periods.
- Giving the amount when the question asks for interest, or the reverse.
- Applying the 2-year difference rule to 3 years.
- Mixing up the doubling rules. Under SI, a sum that doubles in 8 years becomes 3 times (not 4 times) in 16 years. The "4 times in 2n years" rule is for CI only.
| Question asks | Use |
|---|---|
| Interest only | SI or CI, not the amount |
| Amount | Principal plus interest |
| Rate from doubling (SI) | R × T = 100 |
| CI − SI, 2 years | P × (R/100)² |
Practice
Set a 5-minute timer.
- SI on ₹6,000 at 5% for 4 years?
- CI on ₹5,000 at 20% for 2 years?
- CI − SI on ₹20,000 at 5% for 2 years?
- A sum triples in 10 years at SI. Find the rate.
- ₹4,000 amounts to ₹4,960 in 3 years at SI. Find the rate.
- CI on ₹16,000 at 10% a year, compounded half-yearly, for 1 year?
- The difference between CI and SI on a sum for 2 years at 8% is ₹32. Find the sum.
- A sum amounts to ₹2,420 in 2 years and ₹2,662 in 3 years at CI. Find the rate and the principal.
Answers:
- ₹1,200. 6,000 × 5 × 4 ÷ 100.
- ₹2,200. 5,000 × 1.44 = 7,200; minus 5,000. Or 44% of 5,000.
- ₹50. 20,000 × (0.05)² = 20,000 × 0.0025.
- 20%. SI = 2P, so R × 10 = 200.
- 8%. SI = 960; R = 960 × 100 ÷ (4,000 × 3).
- ₹1,640. 5% for 2 half-years: 16,000 × 1.1025 = 17,640.
- ₹5,000. 32 ÷ (0.08)² = 32 ÷ 0.0064.
- 10%; ₹2,000. The third year's interest is 2,662 − 2,420 = 242, which is 10% of 2,420. Then P = 2,420 ÷ 1.21 = 2,000.
What to do next
- Memorise the effective-rate table for 5%, 10% and 20%.
- Prove the 2-year CI–SI rule once on paper, then use it without doubt.
- Do eight mixed SI and CI questions a day for a week.
- Revise percentage for successive changes, and read banking basics for how these ideas appear in the job.
A note on dates and numbers. Exam patterns, vacancies and schedules change from year to year. Always confirm the current details in the latest notification on the Institute of Banking Personnel Selection website .
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