In this guide
Interest questions reward two habits: knowing the formulas without hesitation, and using multipliers for compound interest instead of long calculations. With those, most SSC CHSL interest questions take under a minute.
The whole topic rests on one contrast. Simple interest is earned only on the original sum, so it grows by the same amount every year. Compound interest is earned on the sum plus the interest already added, so it grows by the same percentage every year. Keep that picture in mind and the formulas explain themselves.
Simple interest
SI = P × R × T/100 and Amount = P + SI, where P is the principal, R the rate per year and T the time in years.
Why: each year earns R% of P, the same amount every year, so T years earn T times that.
From the one formula you can find any missing value:
- Rate: R = SI × 100/(P × T).
- Time: T = SI × 100/(P × R).
- Principal: P = SI × 100/(R × T).
Doubling and tripling. A sum doubles when SI = P, that is when R × T = 100. At 8%, that takes 100/8 = 12.5 years. It triples when SI = 2P, so R × T = 200.
Compound interest
Amount = P × (1 + R/100)ⁿ for n years, and CI = Amount − P.
Why: each year multiplies the current amount by (1 + R/100). After n years the multiplier has been applied n times, exactly as with successive percentage increases in the percentage guide.
Use fractions for common rates. 10% = 1/10, so each year multiplies by 11/10. Over 2 years, principal : amount = 10² : 11² = 100 : 121. Over 3 years it is 1,000 : 1,331. Likewise 5% gives 20 : 21 per year and 20% gives 5 : 6.
Year-wise interest at 10% on ₹10,000
| Year | Interest that year | SI that year | Amount at year end (CI) |
|---|---|---|---|
| 1 | ₹1,000 | ₹1,000 | ₹11,000 |
| 2 | ₹1,100 | ₹1,000 | ₹12,100 |
| 3 | ₹1,210 | ₹1,000 | ₹13,310 |
The first year's CI and SI are always equal. After that, CI pulls ahead because it earns interest on interest.
Half-yearly and quarterly compounding
Compounding half-yearly means the rate is applied twice a year. Halve the rate and double the number of periods. Quarterly: divide the rate by 4 and multiply the periods by 4.
Why: the bank credits interest every six months at half the annual rate, and the next half-year's interest is earned on that new amount.
The CI–SI difference
- 2 years: difference = P × (R/100)².
- 3 years: difference = P × (R/100)² × (3 + R/100).
Why for 2 years: in the table, year 1 is identical and year 2's CI exceeds SI only by interest on the first year's interest, which is P × R/100 × R/100.
CI doubling rule
If a sum becomes k times itself in n years at compound interest, it becomes k² times in 2n years and k³ times in 3n years. Why: the same multiplier k is applied again for every block of n years.
Seven worked questions
Q1. At what rate of simple interest will ₹5,000 earn ₹1,200 in 4 years?
R = 1,200 × 100/(5,000 × 4) = 6%.
Q2. A sum amounts to ₹5,600 in 2 years and ₹6,400 in 4 years at simple interest. Find the sum and the rate.
The extra 2 years added 800, so SI is 400 a year. P = 5,600 − 800 = ₹4,800. R = 400/4,800 × 100 = 8⅓%.
Q3. Find the CI on ₹12,000 at 10% for 2 years.
12,000 × 121/100 = 14,520. CI = ₹2,520. Year by year: 1,200, then 1,320.
Q4. Find the amount on ₹20,000 at 8% a year, compounded half-yearly, for 1 year.
Rate 4% for 2 periods: 20,000 × 1.04 × 1.04 = ₹21,632. Compounded yearly, it would be ₹21,600.
Q5. The difference between CI and SI on a sum at 5% for 2 years is ₹15. Find the sum.
P × (1/20)² = 15, so P = 15 × 400 = ₹6,000.
Q6. Find the CI and the CI–SI difference on ₹10,000 at 10% for 3 years.
Amount = 13,310, so CI = ₹3,310. SI = 3,000, so the difference is ₹310.
Formula check: 10,000 × (1/10)² × 3.1 = 310.
Q7. A sum doubles in 5 years at compound interest. In how many years will it become 8 times?
8 = 2³, so it takes 3 × 5 = 15 years.
Common mistakes
| Mistake | Fix |
|---|---|
| Using the annual rate for half-yearly compounding | Halve the rate and double the periods |
| Giving the amount when CI is asked | Subtract the principal |
| Applying SI logic to CI | CI grows by a fixed percentage, not a fixed amount |
| Using the 2-year difference formula for 3 years | Multiply by (3 + R/100) as well |
| Assuming doubling time at CI is 100/R | That rule is for SI only |
Practice
- Find the SI on ₹6,500 at 8% for 3 years.
- A sum becomes ₹7,000 in 5 years at 8% simple interest. Find the principal.
- Find the CI on ₹5,000 at 20% for 2 years.
- Find the amount on ₹16,000 at 10% a year, compounded half-yearly, for 1 year.
- The CI–SI difference on a sum at 10% for 2 years is ₹50. Find the sum.
- At what rate of simple interest will a sum triple in 20 years?
- A sum doubles in 4 years at compound interest. In how many years will it become 8 times?
- On ₹10,000 for 2 years, which gives more: 10% simple interest or 9.5% compound interest, and by how much?
Answers:
- ₹1,560. 6,500 × 8 × 3/100.
- ₹5,000. SI is 40% of P in 5 years, so P = 7,000 ÷ 1.4.
- ₹2,200. 5,000 × (6/5)² = 7,200.
- ₹17,640. 16,000 × 1.05².
- ₹5,000. P × (1/10)² = 50.
- 10%. Tripling means SI = 2P, so R × 20 = 200.
- 12 years. 8 = 2³, and 3 × 4 = 12.
- SI, by ₹9.75. SI = ₹2,000. CI = 10,000 × (1.095² − 1) = ₹1,990.25.
What to do next
- Memorise the year-wise table for 10% and rebuild it for 5% and 20%.
- Learn the 2-year and 3-year CI–SI difference formulas with their reasons.
- Solve 20 mixed SI and CI questions, writing each multiplier as a fraction where you can.
- If the multipliers feel slow, go back to percentages and discounts, where the same successive-change logic runs in reverse.
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 Staff Selection Commission website .
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