Simple interest: WASSCE Mathematics

Aminata deposits $5000$ NLe in a savings account at $8\%$ per annum. How much interest does she earn in $3$ years? This is a simple interest calculation — one of the most common WAEC financial questions.

Simple interest is paid only on the original sum (the principal), not on accumulated interest. WAEC tests it heavily in objective questions and word problems involving loans, deposits, and investments.

This lesson covers the simple interest formula, finding any of the four variables (principal, rate, time, interest), and word problems applying the formula.

Definition

Simple interest formula:

I = \dfrac{PRT}{100}

where:

$I$ = interest earned (or paid).
$P$ = principal (original sum invested or borrowed).
$R$ = annual interest rate (as a percentage).
$T$ = time in years.

Total amount (Principal + Interest):

A = P + I = P\left(1 + \dfrac{RT}{100}\right)

\begin{aligned}A &= P + I \\ &= P\left(1 + \dfrac{RT}{100}\right)\end{aligned}

In simple interest, the interest each year is the same: $\dfrac{PR}{100}$ .

Learning objectives

By the end of this lesson you should be able to:

use the simple interest formula $I = \dfrac{PRT}{100}$
find the interest, principal, rate or time when the other three are known
find the total amount $A = P + I$
solve simple interest problems in real contexts

Finding interest given P, R, T

Example 1

Find the simple interest on $5000$ NLe at $8\%$ per annum for $3$ years.

$I = \dfrac{5000 \times 8 \times 3}{100} = \dfrac{120000}{100} = 1200$ NLe.

Example 2

Aminata invests $20000$ NLe at $5\%$ per year for $4$ years. Find the simple interest and the total amount.

$I = \dfrac{20000 \times 5 \times 4}{100} = 4000$ NLe. Total $A = 20000 + 4000 = 24000$ NLe.

Example 3

Calculate the simple interest on $15000$ NLe at $7.5\%$ per annum for $5$ years.

$I = \dfrac{15000 \times 7.5 \times 5}{100} = \dfrac{562500}{100} = 5625$ NLe.

Example 4

A man borrowed $8000$ NLe at $12\%$ per year for $2.5$ years. Find interest.

$I = \dfrac{8000 \times 12 \times 2.5}{100} = \dfrac{240000}{100} = 2400$ NLe.

Example 5 — Time given in months

Find the interest on $3000$ NLe at $8\%$ for $6$ months.

Convert $6$ months $= 0.5$ years. $I = \dfrac{3000 \times 8 \times 0.5}{100} = \dfrac{12000}{100} = 120$ NLe.

Example 6

Find the interest on $2400$ NLe at $5\%$ for $9$ months.

$9$ months $= 9/12 = 0.75$ years. $I = \dfrac{2400 \times 5 \times 0.75}{100} = 90$ NLe.

Finding principal given I, R, T

Rearrange: $P = \dfrac{100I}{RT}$ .

Example 7

Find the principal that produces $300$ NLe interest at $5\%$ for $2$ years.

$P = \dfrac{100 \times 300}{5 \times 2} = \dfrac{30000}{10} = 3000$ NLe.

Example 8

What principal at $8\%$ per year for $5$ years gives $4800$ NLe interest?

$P = \dfrac{100 \times 4800}{8 \times 5} = \dfrac{480000}{40} = 12000$ NLe.

Example 9

A man earned $2000$ NLe interest in $3$ years at $7.5\%$ . Find his principal.

$P = \dfrac{100 \times 2000}{7.5 \times 3} = \dfrac{200000}{22.5} \approx 8888.89$ NLe.

Finding rate given P, I, T

Rearrange: $R = \dfrac{100I}{PT}$ .

Example 10

$2000$ NLe earned $300$ NLe in $2$ years. Find the rate.

$R = \dfrac{100 \times 300}{2000 \times 2} = \dfrac{30000}{4000} = 7.5\%$ .

Example 11

At what rate must $5000$ NLe be invested to earn $750$ NLe in $3$ years?

$R = \dfrac{100 \times 750}{5000 \times 3} = \dfrac{75000}{15000} = 5\%$ .

Example 12

An investment of $10000$ NLe earned $2400$ NLe in $4$ years. Find the rate.

$R = \dfrac{100 \times 2400}{10000 \times 4} = \dfrac{240000}{40000} = 6\%$ .

Finding time given P, R, I

Rearrange: $T = \dfrac{100I}{PR}$ .

Example 13

At $6\%$ per year, how long for $5000$ NLe to earn $1500$ NLe?

$T = \dfrac{100 \times 1500}{5000 \times 6} = \dfrac{150000}{30000} = 5$ years.

Example 14

A $12000$ NLe loan at $7\%$ accrued $2520$ NLe interest. How long?

$T = \dfrac{100 \times 2520}{12000 \times 7} = \dfrac{252000}{84000} = 3$ years.

Example 15

How long must $2500$ NLe be invested at $10\%$ to earn $625$ NLe?

$T = \dfrac{100 \times 625}{2500 \times 10} = \dfrac{62500}{25000} = 2.5$ years.

Worked example

Apply the simple interest formula.

(a) Find SI on $1500$ at $6\%$ for $4$ years. $I = 1500 \times 6 \times 4 / 100 = 360$ NLe.

(b) Find SI on $8000$ at $5.5\%$ for $3$ years. $I = 8000 \times 5.5 \times 3/100 = 1320$ NLe.

(c) Find amount on $3000$ at $4\%$ for $5$ years. SI = $3000 \times 4 \times 5 / 100 = 600$ . Amount = $3600$ NLe.

(d) $2000$ NLe earns $150$ NLe in $2.5$ years. Rate? $R = 100 \times 150 / (2000 \times 2.5) = 15000/5000 = 3\%$ .

(e) At $8\%$ , how long for $6000$ NLe to earn $1440$ ? $T = 100 \times 1440 / (6000 \times 8) = 144000/48000 = 3$ years.

(f) Find P given $I = 450$ , $R = 5\%$ , $T = 3$ . $P = 100 \times 450 / (5 \times 3) = 45000/15 = 3000$ NLe.

(g) Find SI on $4000$ at $10\%$ for $9$ months. $T = 0.75$ . $I = 4000 \times 10 \times 0.75/100 = 300$ NLe.

Word problems with simple interest

Example 16

Mohamed deposits $20000$ NLe in a savings account paying $4.5\%$ simple interest. How much interest after $6$ months? Total balance?

$T = 0.5$ years. $I = 20000 \times 4.5 \times 0.5/100 = 450$ NLe. Balance: $20000 + 450 = 20450$ NLe.

Example 17

Foday takes a $50000$ NLe loan at $12\%$ simple interest for $4$ years. Total amount owed at end?

$I = 50000 \times 12 \times 4 / 100 = 24000$ NLe. Total: $50000 + 24000 = 74000$ NLe.

Example 18

A school treasurer invests funds in two accounts. Account A pays $5\%$ on $20000$ NLe. Account B pays $7.5\%$ on $30000$ NLe. After $3$ years, total interest from both?

$I_A = 20000 \times 5 \times 3 / 100 = 3000$ . $I_B = 30000 \times 7.5 \times 3 / 100 = 6750$ . Total: $9750$ NLe.

Example 19

A businessman wants to triple his $5000$ NLe investment via simple interest at $10\%$ per year. How long?

Final amount = $15000$ . Interest needed = $10000$ . $T = 100 \times 10000 / (5000 \times 10) = 100000/50000 = 20$ years.

Example 20

A loan was repaid with $7200$ NLe (principal + interest) after $4$ years at $7.5\%$ . Find the original loan.

Amount = $P(1 + RT/100) = P(1 + 7.5 \times 4/100) = P(1 + 0.3) = 1.3 P$ . $1.3 P = 7200$ . $P = 7200/1.3 \approx 5538.46$ NLe.

Example 21

Mohamed lent $8000$ NLe to a friend at $5\%$ for $3$ years, but the friend repaid only the principal after $2$ years. How much interest did Mohamed lose by not getting it back at the agreed time?

At $3$ years: I_planned = $8000 \times 5 \times 3/100 = 1200$ NLe. At $2$ years (only): I_actual = $8000 \times 5 \times 2/100 = 800$ NLe.

If Mohamed considers the planned $3$ -year interest as what he should have received, then lost interest = $1200 - 800 = 400$ NLe.

But typically, if repaid early at $2$ years, Mohamed receives $800$ in interest (since simple interest is for $2$ years actually elapsed). The "loss" framing here depends on interpretation.

Example 22

Two investments of $2000$ NLe each at $6\%$ for different periods: $2$ years and $5$ years. Total interest from both?

First: $2000 \times 6 \times 2/100 = 240$ . Second: $2000 \times 6 \times 5/100 = 600$ . Total: $840$ NLe.

Two investments — same total, different durations or rates

Example 23

A man invests $10000$ NLe in two parts: part at $5\%$ and part at $8\%$ . After $3$ years, total interest is $1800$ NLe. How much was invested at each rate?

Let $x$ = amount at $5\%$ . Then $(10000 - x) =$ amount at $8\%$ .

Total interest: $\dfrac{x \cdot 5 \cdot 3}{100} + \dfrac{(10000 - x) \cdot 8 \cdot 3}{100} = 1800$ .

Simplify: $0.15 x + 0.24(10000 - x) = 1800$ . $0.15 x + 2400 - 0.24 x = 1800$ . $-0.09 x = -600$ . $x = 6666.67$ NLe at $5\%$ . $10000 - 6666.67 = 3333.33$ NLe at $8\%$ .

Note: not clean numbers. Adjust to cleaner case.

Alternative cleaner problem: amount at $5\%$ is $4000$ , amount at $8\%$ is $6000$ . Total interest in $3$ years: $4000 \times 5 \times 3/100 + 6000 \times 8 \times 3/100 = 600 + 1440 = 2040$ NLe.

Worked example

Simple interest word problems.

(a) Aminata deposits $15000$ NLe at $4\%$ for $5$ years. Total amount? $I = 15000 \times 4 \times 5/100 = 3000$ . Total: $18000$ NLe.

(b) Mohamed needs $2000$ NLe in $2$ years from an account paying $5\%$ simple interest. How much must he deposit now? $2000 = P (1 + 5 \times 2/100) = 1.1 P$ . $P = 2000/1.1 \approx 1818.18$ NLe.

(c) Bought a TV on a $24000$ NLe loan at $9\%$ . After $2$ years, total to repay? $I = 24000 \times 9 \times 2/100 = 4320$ . Total: $28320$ NLe.

(d) A man invested $5000$ NLe and after $3$ years had $6500$ NLe. Annual rate? Interest = $1500$ . $R = 100 \times 1500 / (5000 \times 3) = 10\%$ .

(e) An investment doubles in $20$ years at simple interest. Find the rate. Let $P$ be principal. After $20$ years, amount $= 2P$ , so $I = P$ . $R = 100 \times P / (P \times 20) = 5\%$ .

(f) At $6\%$ , how long for an amount to double? Let $P$ be principal. Want $I = P$ . $T = 100 \times P / (P \times 6) = 100/6 \approx 16.67$ years.

Simple interest in instalments

Sometimes payments are made in regular instalments over the loan period.

Example 24

A man takes a $12000$ NLe loan at $8\%$ simple interest for $3$ years, repayable in equal annual instalments. Find each instalment.

Total owed: $12000 + (12000 \times 8 \times 3/100) = 12000 + 2880 = 14880$ NLe. Each instalment: $14880/3 = 4960$ NLe per year.

Example 25

Aminata borrows $4000$ NLe at $5\%$ for $2$ years. She repays equal quarterly instalments. Find each instalment.

Total owed: $4000 + (4000 \times 5 \times 2/100) = 4400$ . Quarters in $2$ years: $8$ . Each instalment: $4400/8 = 550$ NLe.

In Sierra Leone

A Sierra Leonean small businesswoman, Foday's mother, takes a $30000$ NLe loan from her village savings group at $6\%$ simple interest for $2$ years. She'll repay in $24$ equal monthly instalments.

Total amount owed: $I = 30000 \times 6 \times 2/100 = 3600$ . Total = $33600$ NLe.

Each monthly instalment: $33600/24 = 1400$ NLe.

Verification: $24 \times 1400 = 33600$ ✓ Total. Of that, $30000$ is principal repayment and $3600$ is interest paid over $2$ years.

This kind of loan-repayment calculation is exactly what WAEC tests — combining the simple interest formula with the schedule of payments.

Common mistake

Common errors in simple interest:

1. Forgetting to divide by $100$ .

The formula has $\div 100$ because $R$ is a percentage. $\dfrac{PRT}{100}$ , not $PRT$ .

2. Time not in years.

$6$ months $= 0.5$ years. $9$ months $= 0.75$ years. $T$ must be in years for $R$ expressed as % per annum.

3. Confusing interest with total amount.

$I$ is just the interest; $A = P + I$ is the total amount.

4. Wrong direction in rearrangement.

Memorise $I = PRT/100$ , then rearrange systematically for $P$ , $R$ , $T$ .

5. Simple vs compound confusion.

Simple: interest only on principal. Compound: interest on accumulated total.

6. Sign or arithmetic errors.

Always re-check the final answer with the original numbers.

WASSCE examination focus

WAEC tests this

Simple interest WAEC questions:

Find interest. Given P, R, T.
Find principal. Given I, R, T.
Find rate. Given P, I, T.
Find time. Given P, R, I.
Find total amount. A = P + I.
Repayment calculations. Total/installments.
Comparison. Two investments at different rates.
Reverse from final amount. Find P given A, R, T.

Apply the formula directly; rearrange for the unknown variable.

Practice problems

1. Find SI on $3000$ NLe at $6\%$ for $4$ years.

2. Find SI on $4500$ NLe at $5\%$ for $3$ years.

3. Find SI on $2000$ NLe at $7.5\%$ for $2.5$ years.

4. Find SI on $1200$ NLe at $8\%$ for $9$ months.

5. Find the principal yielding $480$ NLe in $4$ years at $6\%$ .

6. Find the principal yielding $750$ NLe in $5$ years at $5\%$ .

7. At what rate must $5000$ NLe be invested to earn $1200$ NLe in $3$ years?

8. $10000$ NLe earned $1800$ NLe in $4$ years. Find the rate.

9. At $4\%$ , how long for $3000$ NLe to earn $480$ NLe?

10. At $7\%$ , how long for $2000$ NLe to grow to $2800$ NLe?

11. A man invested $7500$ NLe. After $5$ years, his total is $9750$ NLe. Find the rate.

12. A woman wants to triple her $3000$ NLe at $10\%$ simple interest. How long?

13. Aminata took a $15000$ NLe loan at $9\%$ simple interest for $3$ years. Repay in equal monthly instalments. Each instalment?

14. A bank pays $4\%$ per year. Foday deposits $2000$ NLe and after $T$ years has $2360$ NLe. Find $T$ .

15. An investor splits $10000$ NLe: $X$ NLe at $5\%$ and $(10000 - X)$ NLe at $8\%$ . After $4$ years, total interest is $2640$ NLe. Find $X$ .

Worked solutions

1. $3000 \times 6 \times 4/100 = 720$ NLe.

2. $4500 \times 5 \times 3/100 = 675$ NLe.

3. $2000 \times 7.5 \times 2.5/100 = 375$ NLe.

4. $T = 0.75$ . $1200 \times 8 \times 0.75/100 = 72$ NLe.

5. $P = 100 \times 480 / (6 \times 4) = 2000$ NLe.

6. $P = 100 \times 750 / (5 \times 5) = 3000$ NLe.

7. $R = 100 \times 1200 / (5000 \times 3) = 8\%$ .

8. $R = 100 \times 1800 / (10000 \times 4) = 4.5\%$ .

9. $T = 100 \times 480 / (3000 \times 4) = 4$ years.

10. Interest $= 800$ . $T = 100 \times 800 / (2000 \times 7) = 80000/14000 \approx 5.71$ years.

11. Interest = $2250$ . $R = 100 \times 2250/(7500 \times 5) = 225000/37500 = 6\%$ .

12. Final amount $= 9000$ . Interest = $6000$ . $T = 100 \times 6000/(3000 \times 10) = 20$ years.

13. $I = 15000 \times 9 \times 3/100 = 4050$ . Total = $19050$ . Months = $36$ . Each = $529.17$ NLe.

14. Interest = $360$ . $T = 100 \times 360/(2000 \times 4) = 4.5$ years.

15. Set up: $\dfrac{X \cdot 5 \cdot 4}{100} + \dfrac{(10000 - X) \cdot 8 \cdot 4}{100} = 2640$ . $0.2X + 0.32(10000 - X) = 2640$ . $0.2X + 3200 - 0.32X = 2640$ . $-0.12X = -560$ . $X = 4666.67$ NLe. Not integer — adjust problem in WAEC.

What's next

In the next lesson we cover compound interest — where interest is added to the principal each period, so subsequent interest is calculated on the new total. This grows faster than simple interest.

Simple interest