Interest is the price of money borrowed or lent. If you borrow ₹1000 from a bank for a year at 10%, you owe ₹100 extra at the end — that ₹100 is interest. Simple and Compound Interest are two different systems of calculating this charge.
Simple Interest (SI) is the straightforward version. Every year, interest is calculated only on the original amount (the principal). If you borrow ₹1000 at 10% for 3 years, you pay ₹100 each year — same amount, every year — totalling ₹300. The principal stays fixed as the reference point throughout.
Compound Interest (CI) is the reinvestment version. After each period, the interest earned gets added to the principal, and the next period's interest is calculated on this new, larger amount. So in year 1 you earn ₹100 on ₹1000. In year 2 you earn interest on ₹1100. In year 3 you earn interest on ₹1210. The amount snowballs.
Here's the analogy that makes this stick: think of SI as a farmer who harvests the same field every year without improving it — same yield, every time. CI is a farmer who ploughs the harvest back into the land — each year's yield is larger than the last.
For SSC GD, about 80-90% of SI/CI questions are pure calculation questions. Very few are conceptual traps. That means your priority is speed with the formulas — not deeper theory. If you know the four key formulas cold and can spot the "doubles in N years" and "CI-SI difference" patterns, you will clear this topic in under 60 seconds per question.
One important note: in SI questions, whenever you see the word "amount", remember — Amount = Principal + Interest. A very common error is treating Amount as Interest. Don't.
Simple Interest:
Where:
P = Principal (original sum)R = Rate of interest per annum (%)T = Time in yearsCompound Interest (compounded annually):
For SSC GD, you will almost always encounter CI compounded annually. Half-yearly compounding (rare) uses R/2 and 2T.
This is one of the most frequently tested SI patterns. Here's the logic:
If a sum doubles in N years at SI, the interest earned equals the principal. So:
Now, for the sum to treble (become 3 times), interest must equal 2P:
So the answer is always 2N. If it doubles in 10 years, it trebles in 20 years. If it doubles in 8 years, it trebles in 16 years. You don't need to find R at all.
General rule: If a sum becomes k times, the time required = (k-1) × N where N is the doubling time.
For 2 years, the difference between CI and SI is:
This formula is extremely useful when the question gives you the difference and asks for P or R.
For 3 years, the difference is:
The 3-year formula is rarely needed for SSC GD — the 2-year version covers 90% of exam questions.
When a person invests different sums at different rates (or borrows at one rate and lends at another), treat each investment separately and then combine.
Profit in borrow-and-lend problems:
Where R₂ is the lending rate and R₁ is the borrowing rate. This is clean and quick — you only need the rate difference, not two separate calculations.
If Amount and Time are given:
R = (SI × 100) / (P × T)This is the most common "find the rate" structure on SSC GD. The calculation is always clean — if you're getting a messy decimal, recheck your SI value.
| Scenario | Key Equation |
|---|---|
| Sum doubles in N years (SI) | R = 100/N |
| Sum trebles in ? years (SI) | Time = 2N |
| Sum becomes k times (SI) | Time = (k-1) × N |
| CI - SI for 2 years | PR²/10000 |
| Borrow-lend profit | P(R₂ - R₁)T/100 |
When a sum doubles in N years at SI, it trebles in exactly 2N years — no rate calculation needed.
Why: Doubling means SI = P (one unit of interest in N years). Trebling means SI = 2P (two units), so time = 2N.
Worked example: "Doubles in 10 years, trebles in ?" — Answer: 20 years. Solved in 3 seconds.
Standard method (find R, then solve for trebling time): ~45 seconds. This pattern: 3 seconds.
For 2 years: CI - SI = PR²/10000. Memorise this as "P times R-squared divided by ten-thousand".
When to use: Any question that says "difference between CI and SI for 2 years is ₹X" — plug directly and solve for P or R.
Worked example: Difference = ₹225, R = 15%. P × 225/10000 = 225 → P = 10000. Done in 15 seconds.
Standard method (compute CI and SI separately, then subtract): 5-6 calculation steps, ~90 seconds.
For borrow-and-lend problems, skip computing two interest amounts. Directly use: Profit = P × (R₂ - R₁) × T / 100.
When to use: "Borrowed at X%, lent at Y%, profit after T years" — one shot calculation.
Worked example: Borrowed ₹8000 at 15%, lent at 18%, 3 years. Profit = 8000 × 3 × 3/100 = ₹720. One step instead of two.
Standard method (calculate SI paid and SI received separately): 2 calculations + subtraction = ~50 seconds. This: ~15 seconds.
Before every SI question, mark whether the question gives/asks for "Amount" or "Interest". Write A = P + SI at the top.
When to use: Any question where "amount" appears — especially when the question gives Amount and asks for SI or vice versa.
Worked example: If Amount = ₹25800, Principal = ₹15000, then SI = ₹10800 (not ₹25800). Misreading Amount as SI is the single most common error in this topic. Catching it takes 2 seconds; fixing it after a wrong answer costs you the question.
This trick prevents ~40% of careless errors in SI questions — not a speed gain, but a score protection move.
For small rates and short periods, use approximate CI via the binomial shortcut.
(1 + R/100)² ≈ 1 + 2R/100 + R²/10000
The 2R/100 part is the SI component. The R²/10000 part is the extra CI earns over SI.
When to use: Verifying CI answers quickly when T = 2 and R is a round number (10%, 15%, 20%).
Worked example: CI on ₹10000 at 10% for 2 years. SI portion = ₹2000. Extra = 10000 × 100/10000 = ₹100. So CI = ₹2100. No need to compute (1.1)² in full.
Standard method (compute 1.1² = 1.21, multiply by 10000): ~35 seconds. This: ~12 seconds.
Read the question and identify the type within 5 seconds:
Step 1 — Classify: Is it SI only, CI only, or SI vs CI (difference)?
Step 2 — SI questions: Check if it's "doubles/trebles" pattern → use 2N rule. Otherwise, identify what's unknown (P, R, T, or SI) and apply SI = PRT/100 directly.
Step 3 — CI questions: Check if it's the 2-year CI-SI difference type → use PR²/10000. Otherwise, apply A = P(1 + R/100)^T. For T=2, expand mentally; for T=3, calculate step-by-step.
Step 4 — Two-investment / borrow-lend: Use rate-difference formula P(R₂ - R₁)T/100 for profit. For two separate investments, compute SI for each and add.
Step 5 — Sanity check: Did you use Amount where SI was needed? Is the answer in the option range? If CI, is it always greater than SI for same P, R, T?
Target: 45-60 seconds per question. If you're past 90 seconds, pick the closest option and move.
Why this question: The classic "doubles → trebles" question that appears almost every year in some form. Tests whether you know the shortcut or waste time finding R.
Solving path: If it doubles in 10 years, the rate R = 100/10 = 10%. To treble, interest needed = 2P. Time = (2P × 100)/(P × 10) = 20 years. Or, directly: trebling time = 2 × doubling time = 2 × 10 = 20 years. Answer: 20 years.
Why this question: The borrow-and-lend profit structure — one of the most common real-world SI setups on SSC GD. Tests if you know to subtract interests rather than add them.
Solving path: Rate difference = 18% - 15% = 3%. Profit = 8000 × 3 × 3/100 = 72000/100 = ₹720. One step using the rate-difference trick. Answer: ₹720.
Why this question: Find Time given Amount and Rate — the standard rearrangement of the SI formula. Tests your ability to extract SI from Amount correctly.
Solving path: SI = Amount - Principal = 25800 - 15000 = ₹10800. Now T = (SI × 100)/(P × R) = (10800 × 100)/(15000 × 12) = 1080000/180000 = 6 years. Answer: 6 years.
Why this question: Two-part question — first find Rate from given data, then apply it to a new scenario. Tests systematic approach.
Solving path: SI for ₹6000 over 2 years = 7200 - 6000 = ₹1200. Rate = (1200 × 100)/(6000 × 2) = 10%. Now apply to new case: SI on ₹9000 at 10% for 3 years = (9000 × 10 × 3)/100 = ₹2700. Answer: ₹2700.
Why this question: The CI-SI difference formula at its cleanest — given difference and rate, find principal. This is a direct formula application question.
Solving path: Use CI - SI = PR²/10000. Given: 225 = P × (15)²/10000 = P × 225/10000. So P = 225 × 10000/225 = ₹10000. Answer: ₹10000.
Treating Amount as Interest. The question says "amounts to ₹25800" — that is the total amount, not the interest. SI = Amount - Principal. If you skip this subtraction and plug 25800 directly into the SI formula, every subsequent calculation is wrong.
Forgetting the "doubles → trebles = 2N" rule and solving the long way. Students find R = 10%, then set up a fresh equation for trebling. That works but costs 40 extra seconds. Know the pattern.
Using the CI formula when the question says Simple Interest. The word "interest" alone does not mean CI. Look for "compound" or "compounded" explicitly. Default is always SI unless stated otherwise.
Getting the CI-SI difference formula sign wrong. CI is always greater than or equal to SI for the same P, R, T (with T > 1 year). If your CI - SI is coming negative, you've made an error somewhere — stop and recheck, do not continue.
Confusing rate per annum with rate per period for compounding. For half-yearly compounding, R becomes R/2 and T becomes 2T. For SSC GD this is rare, but if you see "compounded half-yearly", apply the adjustment before plugging into the formula.
In two-investment problems, adding both interests instead of subtracting. In borrow-and-lend, one interest is a cost (paid to bank) and one is income (received from friend). Profit = income - cost. Students in a hurry add both, which gives a wrong inflated answer that often appears as a trap option.