Compound interest

1. Meaning and core model

Simple interest is calculated on the original principal throughout the stated duration. Under compound interest, interest is added to the balance at specified intervals, and subsequent interest is calculated on this enlarged balance. This addition is called capitalisation. The interval between successive capitalisations is the compounding period. Annual, half-yearly, quarterly and monthly compounding are common examination conventions.

Let P denote principal, A the final amount, r the annual percentage rate and n the number of years. For annual compounding at a constant rate, A = P(1 + r/100)^n, and CI = A − P. Write x = r/100 to simplify the expression to A = P(1 + x)^n. The formula assumes that interest remains invested and that there are no intermediate deposits or withdrawals.

For example, ₹10,000 invested at 10% per annum becomes ₹11,000 after one year and ₹12,100 after two years. The second year's interest is ₹1,100, not ₹1,000, because the first year's interest also earns interest. Thus, the final amount is ₹12,100 while the compound interest is ₹2,100. Confusing amount with interest is a frequent avoidable error.

• Identify principal, amount, rate, duration and compounding frequency before calculating.
• Convert percentage rates into decimal or fractional multipliers: 10% gives 1.10 or 11/10.
• Keep the rate period and the number of periods consistent.

2. Frequency, effective rates and fractional periods

If an annual nominal rate r is compounded m times a year, use A = P[1 + r/(100m)]^(mn). Half-yearly compounding divides the annual rate by two and doubles the number of years; quarterly compounding divides the rate by four and multiplies the years by four. At 12% nominal annual interest compounded half-yearly, ₹10,000 becomes ₹10,000 × 1.06 × 1.06 = ₹11,236 after one year.

The effective annual rate measures the actual percentage increase over one year. Its formula is {[1 + r/(100m)]^m − 1} × 100. Thus, 12% nominal interest compounded half-yearly produces an effective annual rate of 12.36%. For the same positive nominal annual rate, more frequent compounding produces a higher effective annual return. Do not divide an explicitly stated effective annual rate by the number of subperiods; the equivalent subperiod multiplier must instead be obtained by taking the appropriate root.

Fractional durations require attention to the wording. Some arithmetic problems specify annual compounding for complete years and simple interest for the remaining months. Under this convention, A = P(1 + x)^k(1 + fx), where k is the number of complete years and f is the remaining fraction of a year. For 18 months at 10%, this gives P × 1.10 × 1.05. A fractional-exponent model, P(1.10)^1.5, is different and should not be substituted without justification.

• For ₹10,000 at 12% compounded quarterly for one year, calculate ₹10,000 × (1.03)^4.
• Treat annual, nominal annual and effective annual rates as distinct descriptions.
• When the treatment of a broken period is unspecified, recognise the ambiguity rather than assuming all conventions give the same answer.

3. Simple-interest comparisons and successive changes

For the same principal and rate, simple and compound interest are equal after one annual period. Over two years, SI = 2Px and CI = P(2x + x²), so CI − SI = Px². If the difference is ₹100 at 10% per annum, P = 100/(0.10)² = ₹10,000. This shortcut applies to two annual periods at an unchanged rate, not automatically to every problem mentioning two years.

For three annual periods, CI − SI = P(3x² + x³). Instead of memorising many separate formulas, understand expansion of (1 + x)^n. The difference represents the additional return generated by interest earning further interest. Under constant annual compounding, consecutive yearly interest amounts also form a geometric progression: each year's interest is the preceding year's interest multiplied by 1 + x.

When rates vary, multiply the relevant factors. A 10% increase followed by a 20% increase gives 1.10 × 1.20 = 1.32, or 32% overall growth. For signed changes a% and b%, the combined percentage change is a + b + ab/100. A 20% rise followed by a 20% fall gives 1.20 × 0.80 = 0.96, a net decline of 4%, not zero.

Depreciation uses the same structure with a negative rate: A = P(1 − d/100)^n. A machine worth ₹50,000 depreciating by 10% annually is worth ₹40,500 after two years. Population growth, price indices and investment returns can similarly be modelled using successive multipliers, provided their assumed rates and periods are clearly defined.

• Unchanged positive percentage growth produces increasing absolute annual gains.
• Unchanged percentage depreciation produces decreasing absolute annual losses.
• Without intermediate cash flows, reversing the order of percentage multipliers does not change the final amount.

4. Reverse problems, instalments and cash-flow timing

To recover the principal, divide the final amount by the accumulated multiplier: P = A/(1 + x)^n. If ₹12,100 is obtained after two years at 10%, the principal is ₹10,000. If the amount and principal are known for two years, take the square root of A/P to identify the annual multiplier. For example, A/P = 1.44 implies a multiplier of 1.20 and an annual rate of 20%.

Doubling questions often require reasoning rather than finding the rate. If money doubles in t years under an unchanged compound-growth model, it becomes four times in 2t years and eight times in 3t years. It does not become three times in 2t years. The Rule of 72 estimates doubling time as 72 divided by the annual percentage rate, but it is only an approximation and should not replace exact reasoning where options are close.

For loans repaid in instalments, each payment changes the outstanding balance. If a loan P is repaid through two equal end-of-year instalments X at annual rate x, valuation at the end of year two gives P(1 + x)² = X(1 + x) + X. The first payment is carried forward for one year, while the second is already at the comparison date.

For a ₹21,000 loan at 10%, the equation is ₹25,410 = 2.1X, giving X = ₹12,100. After the first year's interest and repayment, the balance is ₹23,100 − ₹12,100 = ₹11,000. After the second year's interest, ₹12,100 is due and the second payment clears it. Simply dividing the no-repayment maturity amount into two equal parts would ignore the first instalment's reduction of the debt.

• Draw a short time line for deposits, withdrawals and repayments.
• Move all cash flows to a common valuation date before adding or comparing them.
• Beginning-of-period and end-of-period payments produce different answers.

5. CSAT calculation strategy and error control

Translate the wording into multipliers before doing arithmetic. Useful fractions include 12.5% = 1/8, 20% = 1/5 and 25% = 1/4. Thus, two years at 12.5% gives an amount multiplier of (9/8)² = 81/64. If the principal is ₹6,400, the amount is immediately ₹8,100 and the compound interest is ₹1,700. Fractional methods are particularly efficient when the principal cancels the denominator.

Use ratios when the question asks only for a rate, a relative amount or a doubling relationship. Avoid introducing an arbitrary principal unless it simplifies the calculation. Preserve exact fractions until the final step, and use answer options for elimination when justified. With a positive rate and more than one annual period, compound interest must exceed simple interest under otherwise identical assumptions.

Before marking an answer, check whether the question seeks principal, interest, amount, rate or difference. Then verify units, payment timing and compounding frequency. Real financial products may involve changing rates, tax, fees or specific crediting rules; CSAT problems normally abstract from these unless stated. Solve the stated mathematical model, not an assumed banking practice.

• Use estimation to test plausibility, not to conceal a mismatch between periods.
• Do not add annual percentage rates when growth is compounded.
• Leave a calculation-heavy question for later if a clean setup does not emerge quickly.

Previous year questions

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Practice questions