Compound Interest Exam Question Walkthrough

Why Compound Interest Questions Matter

Compound interest is one of the most practical topics in IGCSE Maths. It appears on almost every Paper 2 and Paper 4, testing your ability to apply a formula in financial contexts including savings, loans, investments, and depreciation. Questions range from simple two-mark calculations to complex multi-step problems worth six or more marks.

The formula itself is straightforward, but the exam tests whether you can adapt it to different situations — including depreciation, finding the rate, and finding the number of years.

The Formula

A = P(1 + r/100)^n

Where:

A = the final amount
P = the principal (initial amount)
r = the annual interest rate (as a percentage)
n = the number of years

For depreciation (value decreasing), the formula becomes:

A = P(1 − r/100)^n

The only change is the minus sign inside the bracket.

Problem 1: Basic Compound Interest

Rajesh invests RM5000 at a compound interest rate of 3.5% per year. Find the value of his investment after 4 years.

Solution:

P = 5000, r = 3.5, n = 4
A = 5000(1 + 3.5/100)⁴
A = 5000(1.035)⁴
A = 5000 × 1.14752…
A = RM5737.62 (to 2 decimal places)

The compound interest earned is 5737.62 − 5000 = RM737.62.

Problem 2: Depreciation

A car is bought for RM85,000. It depreciates at 12% per year. Find its value after 3 years.

Solution:

P = 85000, r = 12, n = 3
A = 85000(1 − 12/100)³
A = 85000(0.88)³
A = 85000 × 0.681472
A = RM57,925.12 (to 2 decimal places)

The car has lost 85000 − 57925.12 = RM27,074.88 in value over three years.

Problem 3: Finding the Number of Years

This is the type that catches many students. You need to use logarithms or trial and improvement.

An investment of RM2000 is made at 5% compound interest per year. After how many complete years will the investment first exceed RM3000?

Setting up:

2000(1.05)^n > 3000
(1.05)^n > 1.5

Using trial and improvement:

n = 5: (1.05)⁵ = 1.2763 (not enough)
n = 8: (1.05)⁸ = 1.4775 (still not enough)
n = 9: (1.05)⁹ = 1.5513 (exceeds 1.5)

So the investment first exceeds RM3000 after 9 complete years.

Using logarithms (if you have studied them):

n > log(1.5) / log(1.05)
n > 0.17609 / 0.02119
n > 8.31

Since n must be a whole number of years and 8.31 rounds up, n = 9 years.

Problem 4: Finding the Rate

A painting was bought for RM1200 and sold 5 years later for RM1800. Assuming the value increased at a constant compound rate, find the annual rate of increase.

Setting up:

1800 = 1200(1 + r/100)⁵
1.5 = (1 + r/100)⁵

Solving:

Take the fifth root of both sides:

(1 + r/100) = 1.5^(1/5)
(1 + r/100) = 1.5^0.2
(1 + r/100) = 1.08447…
r/100 = 0.08447
r = 8.45% (to 3 significant figures)

On a calculator, you compute 1.5^(1/5) using the power button: 1.5, then the x^y button, then 0.2.

Problem 5: Compound Interest vs. Simple Interest

Sometimes the exam asks you to compare compound and simple interest. Here is how they differ:

Simple interest: Interest is calculated on the original principal only.

Simple interest = P × r × n / 100
Final amount = P + P × r × n / 100 = P(1 + rn/100)

Compound interest: Interest is calculated on the principal plus any previously earned interest.

Final amount = P(1 + r/100)^n

Example: RM10,000 invested for 3 years at 6%.

Simple interest: 10000 × 0.06 × 3 = RM1800. Total = RM11,800.

Compound interest: 10000(1.06)³ = 10000 × 1.191016 = RM11,910.16.

The difference is RM110.16 — compound interest earns more because you earn interest on your interest.

Handling Monthly or Quarterly Compounding

Some harder questions specify that interest is compounded monthly or quarterly rather than annually. The formula adjusts:

Monthly: A = P(1 + r/(100 × 12))^(12n)
Quarterly: A = P(1 + r/(100 × 4))^(4n)

For example, RM5000 at 6% compounded monthly for 2 years:

A = 5000(1 + 6/1200)^24
A = 5000(1.005)^24
A = 5000 × 1.12716
A = RM5635.80

Compare this with annual compounding: 5000(1.06)² = RM5618.00. Monthly compounding gives a slightly higher return.

Common Mistakes

Using simple interest instead of compound interest. Always check whether the question says “compound” or “simple.” If it says compound, use the power formula.
Forgetting to convert the percentage. r = 3.5 means you use 3.5/100 = 0.035 inside the bracket, giving (1.035).
Rounding too early. Keep the multiplier to at least 4 decimal places during calculations. Only round your final answer.
Getting the depreciation formula wrong. For depreciation, use (1 − r/100), not (1 + r/100). The value is decreasing.
Not answering “complete years” correctly. If the question asks for complete years, you must round up to the next whole number. 8.31 years means 9 complete years.
Confusing the final amount with the interest earned. The interest earned is A − P, not A.

Mark Scheme Insights

IGCSE examiners typically award marks as follows:

1 mark for identifying the correct formula
1 mark for correct substitution
1 mark for the final answer

For multi-step problems (finding years or rates), additional marks are given for the algebraic rearrangement and for the final interpretation (e.g., stating the answer as a whole number of years).

Practice Question

A tree is 1.5 metres tall. It grows at a compound rate of 8% per year. After how many complete years will the tree first be taller than 3 metres?

Set up the inequality 1.5(1.08)^n > 3 and solve using trial and improvement or logarithms.

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