Simple Interest - Waokmath

Lesson Objectives Define simple interest and its related terms (principal, rate, time). Derive and use the formula for simple interest: [latex]I = frac{P times R times T}{100}[/latex] 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: [latex]I = frac{P times R times T}{100}[/latex] Where: [latex]I[/latex] = Interest earned or paid [latex]P[/latex] = Principal amount (initial amount) [latex]R[/latex] = Rate of interest per annum [latex]T[/latex] = Time (in years) The total amount after interest is: [latex]A = P + I[/latex] Worked Examples Example 1 (Basic): Find the simple interest on [latex]P = 2000[/latex], [latex]R = 5%[/latex], [latex]T = 3[/latex] years. [latex]I = frac{2000 times 5 times 3}{100} = frac{30000}{100} = 300[/latex] Example 2 (Finding Rate): If [latex]I = 400[/latex], [latex]P = 1000[/latex], and [latex]T = 2[/latex] years, find [latex]R[/latex]. [latex]R = frac{100 times I}{P times T} = frac{100 times 400}{1000 times 2} = frac{40000}{2000} = 20%[/latex] Example 3 (Finding Time): If [latex]P = 1500[/latex], [latex]R = 8%[/latex], [latex]I = 360[/latex], find [latex]T[/latex]. [latex]T = frac{100 times I}{P times R} = frac{100 times 360}{1500 times 8} = frac{36000}{12000} = 3[/latex] years Example 4 (WAEC-style): Calculate the total amount to be paid after borrowing [latex]5000[/latex] at [latex]6%[/latex] per annum for [latex]4[/latex] years. [latex]I = frac{5000 times 6 times 4}{100} = 1200[/latex], so [latex]A = 5000 + 1200 = 6200[/latex] Example 5 (Challenging): A sum amounts to [latex]8800[/latex] in 4 years at 10% per annum. Find the principal. [latex]A = 8800[/latex], [latex]T = 4[/latex], [latex]R = 10%[/latex] Let [latex]P[/latex] be the principal. Then: [latex]I = A - P = 8800 - P[/latex] Substituting: [latex]8800 - P = frac{P times 10 times 4}{100} = frac{40P}{100} = 0.4P[/latex] [latex]8800 - P = 0.4P Rightarrow 8800 = 1.4P Rightarrow P = frac{8800}{1.4} = 6285.71[/latex] Exercises Calculate the simple interest on [latex]N2500[/latex] for 3 years at 6% per annum. If [latex]P = 3000[/latex], [latex]I = 450[/latex], [latex]T = 3[/latex] years, find the rate. How long will it take [latex]N4000[/latex] to earn [latex]N640[/latex] at 8% per annum? [WAEC] Find the amount after 5 years on [latex]N12000[/latex] invested at 10% simple interest. (Past Question) [NECO] A loan of [latex]N7000[/latex] was taken at 5% for 4 years. Find the total repayment. (Past Question) [JAMB] A sum of [latex]N5000[/latex] becomes [latex]N6200[/latex] in 4 years. Find the rate. (Past Question) A man invests [latex]N10000[/latex] at 12% per annum. Find the interest earned in 1.5 years. If [latex]I = N240[/latex], [latex]R = 4%[/latex], [latex]P = N1500[/latex], find [latex]T[/latex]. Calculate the interest earned on [latex]N1800[/latex] for 9 months at 10% per annum. [NABTEC] What principal will yield [latex]N540[/latex] in 3 years at 6% simple interest? (Past Question) 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! Clip It! Share your ANSWER in the Chat. Indicate TITLE e.g Linear Equation 1. .....2. e.t.c