Simple Interest (SI) and Compound Interest (CI) differ in one fundamental way: with SI, the interest is always calculated on the original principal. With CI, the interest earned each period gets added to the principal, and the next period's interest is calculated on this new, larger amount. Interest earns interest. That is the whole story.
Here is an analogy that sticks. Imagine you lend ₹100 to a friend at 10% per year. If it is SI, he pays you ₹10 at the end of every year — flat. After 3 years you get ₹30 interest total. Now imagine instead you reinvest that ₹10 into the same deal each year. Year 2's interest is 10% of ₹110, not ₹100. Year 3's interest is 10% of ₹121. After 3 years you get ₹33.10 instead of ₹30. That extra ₹3.10 is the power of compounding.
For SSC MTS, the questions on compound interest generally fall into four types:
The math level stays accessible — you will rarely see rates like 7.3% or times beyond 3 years. The trick is knowing which formula fits which situation and executing the arithmetic cleanly. Most errors in this topic are not conceptual — they are calculation slips, especially when dealing with fractions like 12.5% = 1/8.
The Amount A after n years on principal P at rate R% per annum compounded annually:
Compound Interest = Amount − Principal:
This is the only formula you need to know cold. Everything else is a variation.
If you know A, R, and n, and need P:
Look at how the PYQ about Geeta uses this: Amount = ₹27,783, R = 5%, n = 3.
21³ = 9261, 20³ = 8000. So P = 27783 × 8000/9261. Now 27783 ÷ 9261 = 3, so P = 3 × 8000 = ₹24,000. Clean and fast.
The key habit here: always check if the fraction simplifies before cubing. 100/105 = 20/21 — much friendlier to cube.
For n = 2 at rate R%:
Or more usefully, the effective rate for 2 years:
For R = 20%: 2(20) + (400/100) = 40 + 4 = 44%. So CI on ₹1200 = 1200 × 44/100 = ₹528. This is the single fastest approach for 2-year CI problems — use it every time for annual compounding.
This is where aspirants lose marks by mixing up inputs.
Rule: When the compounding period changes, adjust both rate and time:
| Compounding | New Rate per Period | New Number of Periods | |-------------|--------------------|-----------------------| | Half-yearly | R/2 | 2n | | Quarterly | R/4 | 4n |
So for 10% p.a. compounded half-yearly over 18 months:
Amount = 1200 × (1.05)³ = 1200 × 1.157625 = ₹1389.15
Do not make the mistake of using 10% and 1.5 years — that gives the wrong answer every time.
If a sum grows by a certain factor in n years:
If a sum becomes 1.331 times in 3 years: (1 + R/100)³ = 1.331. Recognise 1.331 = 1.1³, so R = 10%.
For finding rate when you have amounts at two consecutive time points: divide the larger by the smaller. The ratio is exactly (1 + R/100). For example, if Amount at year 3 is ₹3149.28 and at year 2 is ₹2916, then 3149.28/2916 = 1.08, so R = 8%.
These are more involved but completely manageable. Steps:
The common error is forgetting to compound first, then subtract. Never subtract the payment from the principal and then compound.
| Rate | Fraction form | |------|--------------| | 10% | 1/10 | | 12.5% | 1/8 | | 20% | 1/5 | | 25% | 1/4 | | 5% | 1/20 |
Practise CI calculations using fractions. 1/8 is far easier to square or cube than 0.125.
For 2-year CI, never expand (1 + R/100)² term by term. Instead use: Effective % = 2R + R²/100. For R = 20%: 40 + 4 = 44%. Apply to principal directly. Standard method (expanding bracket): ~40 seconds, 4 steps. This shortcut: ~10 seconds, 1 step. Works for any rate — test it with R = 10%: 20 + 1 = 21%. ₹1000 at 10% for 2 years → CI = ₹210. Correct.
When the question asks for Principal given Amount, rate, and time — multiply Amount by (100/(100+R))ⁿ. Before computing, always reduce 100/(100+R) to lowest terms: 100/105 = 20/21, 100/120 = 5/6, 100/110 = 10/11. Cubing 20/21 is 8000/9261; cubing 10/11 is 1000/1331. These are manageable. Cubing 0.9524 (the decimal form of 100/105) takes twice as long and invites rounding errors. Fraction method: 3 steps. Decimal method: 6 steps with rounding risk.
The SSC loves using perfect powers as growth factors: 1.331 = 1.1³, 1.728 = 1.2³, 1.125 = (9/8)², 1.44 = (1.2)². Build a small mental library: 1.21 = 1.1², 1.44 = 1.2², 1.69 = 1.3², 1.331 = 1.1³, 1.728 = 1.2³. When a "sum becomes X times in n years" question appears, check your library first. Finding the answer from the library: 5 seconds. Solving the equation algebraically: 60+ seconds.
Half-yearly compounding: halve the rate, double the time. Quarterly: quarter the rate, quadruple the time. Write it as a ratio: (Rate, Time) → (R/2, 2T) for half-yearly. This is the only adjustment. If you forget whether to halve or double something, ask: "Does reducing compounding period increase or decrease each period's rate?" It decreases it. So rate per period goes down, periods go up — halve rate, double time. This mental check prevents swapping the operations — a common error that takes 3 marks in one shot.
12.5% = 1/8. So P × (1 + 12.5/100)² = P × (9/8)²= P × 81/64. For the PYQ with P = ₹1,32,000: 1,32,000 × 81/64. Split: 1,32,000 ÷ 64 = 2062.5, then 2062.5 × 81 = 1,67,062.50. Using decimal 1.125²: you must first compute 1.265625, then multiply — far messier. Fraction approach: 2 steps. Decimal approach: 4 steps with a non-terminating intermediate value.
When you see a CI problem in the exam hall, run through this decision tree:
Step 1 — Identify what's unknown: Amount? Principal? Rate? CI only?
Step 2 — Check the compounding period: Annually? Half-yearly? Quarterly? If not annual, adjust rate and time immediately before doing anything else.
Step 3 — Check the time period:
n = 1: CI = Simple Interest. Use P × R/100.n = 2: Use effective rate shortcut 2R + R²/100, then multiply by principal directly.n = 3: Look for clean fractions. Convert rate to fraction (e.g., 5% = 1/20, so multiplier = 21/20). Cube the fraction.Step 4 — For "find principal" questions: Multiply the given amount by (100/(100+R))ⁿ. Reduce fraction first.
Step 5 — For "find rate" questions: If two consecutive year amounts are given, divide bigger by smaller. If a multiplier is given, match it against your mental library of perfect powers.
Step 6 — For partial repayment: Compute full amount first, then subtract payment. Never subtract first.
Why this question: Tests reverse CI — finding principal from amount. Many students try to guess or work forwards from each option. The structured approach is faster.
Solving path: Amount = ₹27,783, R = 5%, n = 3. Principal = 27783 × (100/105)³ = 27783 × (20/21)³. Simplify: 27783 ÷ 9261 = 3. So P = 3 × 8000 = ₹24,000. Time taken with this approach: under 30 seconds.
Why this question: Combines CI calculation with a real-world "partial payment + asset" setup. Tests whether you compute Amount first before doing any subtraction.
Solving path: 12.5% = 1/8, so multiplier = 9/8. Amount = 1,32,000 × (9/8)² = 1,32,000 × 81/64. 1,32,000 ÷ 64 = 2062.5. 2062.5 × 81 = 1,67,062.50. Scooter cost = 1,67,062.50 − 1,07,062.50 = ₹60,000.
Why this question: Classic 2-year CI — perfect for the effective rate shortcut. Options are close together, so estimation won't work. Exact method needed.
Solving path: Effective rate = 2(20) + (20)²/100 = 40 + 4 = 44%. CI = 1200 × 44/100 = ₹528. Done in 2 steps.
Why this question: Half-yearly compounding with a non-integer year (18 months). Exactly the type that causes errors when aspirants forget to convert time.
Solving path: Half-yearly adjustment: rate = 10/2 = 5% per period, periods = 18/6 = 3. Amount = 1200 × (1.05)³. (1.05)³ = 1.157625. Amount = 1200 × 1.157625 = ₹1389.15.
Why this question: Tests whether you can reverse-engineer the rate from a growth multiplier. Quick if you recognise the cube.
Solving path: (1 + R/100)³ = 1.331. Recognise 1.331 = (1.1)³. Therefore 1 + R/100 = 1.1, giving R = 10%.
Why this question: Partial repayment after 2 years — tests the sequence of operations.
Solving path: Amount = 50,000 × (1.2)² = 50,000 × 1.44 = ₹72,000. Paid ₹32,000. Remaining = 72,000 − 32,000 = ₹40,000.
Why this question: Finding rate from two consecutive year amounts — a direct ratio approach, no equation-solving needed.
Solving path: Growth factor = 3149.28 ÷ 2916 = 1.08. Rate = 8%.
Why this question: Partial repayment problem with 10% rate. Tests compound-first, subtract-second discipline.
Solving path: Amount = 80,000 × (1.1)² = 80,000 × 1.21 = ₹96,800. After paying ₹50,000: 96,800 − 50,000 = ₹46,800.
Subtracting payment before compounding in partial repayment problems. The loan grows first, then you pay. Never reduce the principal first and then compound on the reduced figure.
Using annual rate directly in half-yearly problems. If the question says "10% per annum, compounded half-yearly", the per-period rate is 5%, not 10%. Always adjust before plugging into the formula.
Confusing CI with Amount. The question asks for CI, you compute Amount and mark it. CI = Amount − Principal. Read the question once more before marking.
Not simplifying fractions before raising to powers. Trying to compute (0.9524)³ in your head is painful. Always convert to fraction form first — 100/105 = 20/21 — and then cube numerator and denominator separately.
Using the effective 2-year rate formula for non-annual compounding. The formula 2R + R²/100 works only for annual compounding with n = 2. For half-yearly, first adjust R and n, then apply the standard formula.
Rounding intermediate results too early. In problems like the half-yearly one (1.05³ = 1.157625), rounding to 1.158 can shift your answer by a rupee or two, which matters when options are close together.