Interest, depreciation and growth · Commercial arithmetic

Simple interest

Interest is paid on the original sum (the principal $P$ ) only, the same amount every year:

I = \frac{PRT}{100}, \qquad \text{amount } A = P + I

where $R$ is the rate per cent per annum and $T$ is the time in years.

Simple interestThe same interest every year: I = PRT/100

Compound interest

Each year’s interest is added to the amount, and the next year’s interest is worked out on the new amount. Each year multiplies by $1 + \frac{r}{100}$ :

\begin{aligned} A &= P\left(1 + \frac{r}{100}\right)^n \\ \text{compound interest} &= A - P \end{aligned}

Compound interestEach year × (1 + r/100), so the steps grow

Simple and compound interestChange the principal, rate and time

5000simple interest, PRT/1006105.1compound: 10,000 × 1.1^5 − P16105.1compound amount

principal = 10,000rate = 10%years = 5

Simple interest adds 10% of the original 10,000 every year: a straight line, total 5000. Compound interest adds 10% of the amount so far, so each year's interest is bigger than the last: the curve pulls away, reaching 6105.1 in interest.

The straight line is simple interest; the curve is compound interest, which grows faster because the interest itself earns interest.

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q10 (a, b)

A man bought a house at $350,000.00. He paid $20\%$ of the cost from his own resources and the rest with a loan he took from the bank at $7\%$ simple interest per annum for 8 years. Calculate the:

total cost of the house to the man;

percentage increase in the cost of the house as a result of the loan;

The loan
He paid $20\%$ himself: $0.2 \times 350\,000 = 70\,000$ dollars. The loan was the other $280,000.
Think first. How much did he borrow?
The interest
$I = \frac{280\,000 \times 7 \times 8}{100} = 156\,800$ , so the interest is $156,800.
Think first. Simple interest: which formula?
(a) Total cost
$350\,000 + 156\,800 = 506\,800$ , so $506,800.
(b) Percentage increase
$\frac{156\,800}{350\,000} \times 100\% = 44.8\%$ .

Depreciation and growth

Depreciation is compound interest going down: each year the value falls by $r\%$ of its value at the start of that year, so it’s multiplied by $1 - \frac{r}{100}$ each year. Population growth works like compound interest going up.

DepreciationEach year × (1 − r/100), so the value falls by less each year

Worked example · WAEC 2022

WAEC 2022 · Paper 2 · Q7 (a)

A television set purchased for ₦65,000.00 depreciates by $14\%$ per year. Find the value of the television at the end of the third year.

The yearly multiplier
$1 - 0.14 = 0.86$ .
Think first. Losing 14% a year leaves what fraction each year?
Three years
$65\,000 \times 0.86^3 = 65\,000 \times 0.636056 \approx ₦41{,}343.64$ .

Your turn

WAEC 2021 · Paper 2 · Q13 (a)

(a)
On Sam's first birthday celebration, his grandfather deposited an amount of $1,000.00 in a bank compounded at $4\%$ interest annually. Find how much is in the account if Sam is 4 years old.

Worked solution (try it first)

(a)

From his first birthday to when he is 4 is 3 years.
Compound interest at $4\%$ : $1000 \times 1.04^3 = 1000 \times 1.124864$
$\approx 1124.86$ , so the account holds $1,124.86.

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