Simple Interest

Lesson Objectives

Define simple interest and its related terms (principal, rate, time).
Derive and use the formula for simple interest: $I = \frac{P \times R \times T}{100}$
Solve problems involving calculation of interest, amount, rate, or time.
Apply simple interest concepts to real-life financial situations.
Identify the difference between simple and compound interest.

Lesson Introduction

Have you ever borrowed or saved money? Banks and lenders often use the concept of interest to determine how much is paid for using money over time. Simple interest is calculated only on the principal amount. In this lesson, you'll learn how to calculate simple interest and solve problems related to borrowing and saving money.

Core Lesson Content

Simple Interest is calculated using the formula:
$I = \frac{P \times R \times T}{100}$
Where:
$I$ = Interest earned or paid
$P$ = Principal amount (initial amount)
$R$ = Rate of interest per annum
$T$ = Time (in years)

The total amount after interest is:
$A = P + I$

Worked Examples

Example 1 (Basic):
Find the simple interest on

P = 2000

R = 5\%

T = 3

years.

I = \frac{2000 \times 5 \times 3}{100} = \frac{30000}{100} = 300

Example 2 (Finding Rate):
If

I = 400

P = 1000

, and

T = 2

years, find

R

R = \frac{100 \times I}{P \times T} = \frac{100 \times 400}{1000 \times 2} = \frac{40000}{2000} = 20\%

Example 3 (Finding Time):
If

P = 1500

R = 8\%

I = 360

, find

T

T = \frac{100 \times I}{P \times R} = \frac{100 \times 360}{1500 \times 8} = \frac{36000}{12000} = 3

years

Example 4 (WAEC-style):
Calculate the total amount to be paid after borrowing

5000

6\%

per annum for

4

years.

I = \frac{5000 \times 6 \times 4}{100} = 1200

, so

A = 5000 + 1200 = 6200

Example 5 (Challenging):
A sum amounts to

8800

in 4 years at 10% per annum. Find the principal.

A = 8800

T = 4

R = 10\%

Let

P

be the principal. Then:

I = A - P = 8800 - P

Substituting:

8800 - P = \frac{P \times 10 \times 4}{100} = \frac{40P}{100} = 0.4P

8800 - P = 0.4P \Rightarrow 8800 = 1.4P \Rightarrow P = \frac{8800}{1.4} = 6285.71

Exercises

Calculate the simple interest on $N2500$ for 3 years at 6% per annum.
If $P = 3000$ , $I = 450$ , $T = 3$ years, find the rate.
How long will it take $N4000$ to earn $N640$ at 8% per annum?
[WAEC] Find the amount after 5 years on $N12000$ invested at 10% simple interest. (Past Question)
[NECO] A loan of $N7000$ was taken at 5% for 4 years. Find the total repayment. (Past Question)
[JAMB] A sum of $N5000$ becomes $N6200$ in 4 years. Find the rate. (Past Question)
A man invests $N10000$ at 12% per annum. Find the interest earned in 1.5 years.
If $I = N240$ , $R = 4\%$ , $P = N1500$ , find $T$ .
Calculate the interest earned on $N1800$ for 9 months at 10% per annum.
What principal will yield $N540$ in 3 years at 6% simple interest?

Conclusion / Recap

In this lesson, we explored how to calculate simple interest and solve problems involving rate, time, and amount. These concepts are widely used in real-life financial planning and transactions.
Next up: Compound Interest – Understanding interest on interest!

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