Compound Interest: Don't Just Multiply by the Number of Years

The question

Mei Ling invests RM 8000 in a savings account. The account pays compound interest at a rate of $3.5\%$ per year.

(a) Calculate the amount in the account at the end of 4 years. Give your answer correct to the nearest sen. [2 marks]

(b) Calculate the total interest Mei Ling earns in these 4 years. [1 mark]

The common wrong answer

Many students work out one year’s interest and then multiply it by the number of years:

3.5\% \text{ of } 8000 = 0.035 \times 8000 = 280

280 \times 4 = 1120

8000 + 1120 = 9120

So they write RM 9120 for (a) and RM 1120 for (b).

For part (c) they need the amount to grow by $10\,000 - 8000 = 2000$ , so they divide by 280:

\frac{2000}{280} \approx 7.14

and round up to 8 years.

This is simple interest. It assumes the interest every year is RM 280, worked out on the original RM 8000. But with compound interest, each year’s interest is added to the account, and the next year’s $3.5\%$ is worked out on the new, larger amount. The interest grows every year, so the simple-interest working gives too little money in (a) and (b), and too many years in (c). Cambridge examiner reports describe exactly this: candidates using the simple-interest method, or simple depreciation, when the question is about a compound process.

The correct working

(a) Adding $3.5\%$ to an amount is the same as multiplying it by

1 + \frac{3.5}{100} = 1.035

Each year the amount is multiplied by $1.035$ again, so after 4 years:

\begin{aligned} \text{Amount} &= 8000 \times 1.035^4 \\ &= 8000 \times 1.147\,523\ldots \\ &= 9180.184\ldots \end{aligned}

So the amount after 4 years is RM 9180.18.

(b) The interest is the amount minus the money originally invested:

9180.18 - 8000 = 1180.18

So the total interest is RM 1180.18.

(c) Keep multiplying by $1.035$ and check each year against RM 10 000:

\begin{aligned} \text{After 6 years: } 8000 \times 1.035^6 &= 9834.04\ldots \quad (\text{less than } 10\,000) \\ \text{After 7 years: } 8000 \times 1.035^7 &= 10\,178.23\ldots \quad (\text{more than } 10\,000) \end{aligned}

So the amount first goes over RM 10 000 after 7 years.

Show both lines in the exam: the year just below RM 10 000 and the year just above it. That proves 7 is the smallest number of years.

Why it works

With simple interest the account grows by the same amount, RM 280, every year. With compound interest it grows by the same percentage every year, and $3.5\%$ of a bigger number is a bigger number. In year 1 the interest is RM 280, but in year 2 it is $3.5\%$ of RM 8280, which is RM 289.80, and by year 7 it is over RM 340.

Graph of the amount in the account against number of complete years from 0 to 8. The compound interest line starts at RM 8000 and curves upwards, passing RM 10 000 between years 6 and 7, reaching RM 10 178.23 at year 7. The simple interest line is straight, reaching only RM 9960 at year 7. — Compound interest pulls away from simple interest, crossing RM 10 000 a year earlier.

The gap is small at first, only RM 60.18 after 4 years. That is exactly why the wrong method is tempting: the answer looks reasonable. But in part (c) the difference decides the answer, because after 7 years the simple-interest method gives RM 9960, just short of RM 10 000, while the real account has RM 10 178.23.

Writing the multiplier, $1.035$ , also gives you a quick general rule: after $n$ years the amount is $8000 \times 1.035^n$ . You don’t need to work through the years one at a time, and there is no formula sheet, so it’s worth knowing how to build this yourself.

How to spot it next time

Read the word in front of “interest”. Compound means the multiplier is raised to a power: $\text{amount} \times \left(1 + \dfrac{r}{100}\right)^n$ . Only simple interest is the same every year.
Check what is being asked for: the amount in the account (interest included) or just the interest. Subtract the original sum only for the interest.
For “how many years”, test whole numbers of years and write down the year either side of the target value.