IGCSE · Practice questions · Simple and compound interest

Add it, or multiply it.

Six original Cambridge-style questions on simple and compound interest, including comparisons and reverse problems.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Simple adds, compound multiplies.

Simple: $I = \dfrac{PRT}{100}$, the same every year.
Compound: amount $= P \times \left(1 + \dfrac{r}{100}\right)^{n}$.
Interest or amount? Subtract $P$ if only the interest is asked for.
Number of years: try values of $n$ until the target is passed.
Round at the end to the nearest cent.

Recall

[2 marks]

Calculate the simple interest on $\$1200$ invested for 5 years at 3% per year.

$I = \dfrac{1200 \times 3 \times 5}{100} = \$180$ ✓

Calculation

[3 marks]

$\$1200$ is invested for 5 years at 3% per year compound interest. Calculate the interest earned.

Amount $= 1200 \times 1.03^{5} = 1391.128...$ ✓
Interest $= 1391.13 - 1200 = \$191.13$ ✓

Calculation

[2 marks]

$\$2500$ is invested at 4% per year simple interest. How many years will it take to earn $\$500$ in interest?

Interest per year $= 0.04 \times 2500 = \$100$. ✓
$500 \div 100 = 5$ years. ✓

Analysis

[3 marks]

Bank A pays 4.5% per year simple interest. Bank B pays 4.2% per year compound interest. Amir invests $\$6000$ for 6 years.

Which bank pays more interest, and by how much?

Bank A: $6000 \times 0.045 \times 6 = \$1620$. ✓
Bank B: $6000 \times 1.042^{6} - 6000 = \$1679.94$. ✓
Bank B pays $\$59.94$ more. ✓

Analysis

[3 marks]

(Extended) $\$2000$ is invested at $r\%$ per year compound interest. After 2 years it is worth $\$2205$. Find $r$.

$2000 \times m^{2} = 2205$, so $m^{2} = 1.1025$. ✓
$m = \sqrt{1.1025} = 1.05$. ✓
$r = 5$ ✓

Analysis

[3 marks]

Nadia invests $\$5000$ at 3.5% per year compound interest. Find the number of complete years until her investment is first worth more than $\$6000$.

After 5 years: $5000 \times 1.035^{5} = \$5938.43$, not enough. ✓
After 6 years: $5000 \times 1.035^{6} = \$6146.28$, more than $\$6000$. ✓
6 years. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.