Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

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Performance objectives

• Calculate the future value of an investment using the compound interest formula.
• Determine the total repayment amount for a loan.
• Compare different investment options.
• Analyze the impact of different loan terms.
• Apply compound interest knowledge to real-world scenarios.

Lesson summary

Finance is a crucial aspect of everyday life. Understanding compound interest, loans, and investments is not just a theoretical exercise; it's a practical skill that empowers you to make informed financial decisions. In South Africa, where many face financial challenges, this knowledge is even more critical. From saving for tertiary education (fees are high!) to buying a car or a home, or even starting a small business, understanding how money grows (or shrinks) over time is essential for securing your financial future and contributing to the economic well-being of your community.

Lesson notes

Compound Interest: Unlike simple interest, which is calculated only on the principal amount, compound interest is calculated on the principal amount and the accumulated interest from previous periods. This means your money grows exponentially over time. The more frequently the interest is compounded (e.g., monthly, daily), the faster your investment grows.

Formula: The formula for compound interest is: A = P (1 + i)^n Where: A = Future Value (the total amount after n periods) P = Principal Amount (the initial investment or loan amount) i = Interest Rate per compounding period (expressed as a decimal – divide the annual interest rate by the number of compounding periods per year) n = Number of compounding periods (total number of times interest is compounded – multiply the number of years by the number of compounding periods per year)

Example 1: Investing in a Savings Account Sipho invests R5,000 in a savings account that offers an annual interest rate of 8% compounded quarterly. He plans to leave the money in the account for 5 years. How much will he have at the end of the 5 years? P = R5,000 i = 8% per year / 4 quarters per year = 0.08 / 4 = 0.02 per quarter n = 5 years 4 quarters per year = 20 quarters A = 5000 (1 + 0.02)^20 A = 5000 (1.02)^20 A = 5000 * 1.485947 A = R7,429.74 Therefore, Sipho will have R7,429.74 at the end of 5 years.

Loans and Compound Interest: Loans operate on the same principle of compound interest, but from the lender's perspective. You borrow a principal amount, and you're charged interest on the outstanding balance. Understanding how interest accrues on loans is crucial for making informed borrowing decisions. The longer the loan term, the more interest you'll pay.

Example 2: Taking out a Loan for a Car Zanele wants to buy a used car for R80,

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0. She takes out a loan from a bank with an annual interest rate of 12% compounded monthly. She plans to repay the loan over 3 years. What will be her monthly payment? This requires the Loan Amortization formula, which is a bit more complex but essential for understanding loan repayments: M = P [ i(1 + i)^n ] / [ (1 + i)^n – 1] Where: M = Monthly Payment P = Principal Amount (R80,000) i = Monthly Interest Rate (12% per year / 12 months per year = 0.12 / 12 = 0.01) n = Number of Months (3 years 12 months per year = 36) M = 80000 [ 0.01(1 + 0.01)^36 ] / [ (1 + 0.01)^36 – 1] M = 80000 [ 0.01(1.01)^36 ] / [ (1.01)^36 – 1] M = 80000 [ 0.01(1.430769) ] / [ 1.430769 – 1] M = 80000 [ 0.01430769 ] / [ 0.430769] M = 80000 * 0.033219 M = R2,657.52 Zanele's monthly payment will be approximately R2,657.

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2. Over 3 years, she will pay a total of 36 * R2,657.52 = R95,670.

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2. The total interest paid is R95,670.72 - R80,000 = R15,670.72 Example 3: Comparing Investment Options Thabo wants to invest R10,000 for 2 years.

He has two options: Option A: An account offering 9% interest per year, compounded annually.

Option B: An account offering 8.5% interest per year, compounded monthly. Which option is better?

Option A: A = 10000 (1 + 0.09)^2 A = 10000 (1.09)^2 A = 10000 * 1.1881 A = R11,881 Option B: i = 0.085 / 12 = 0.0070833 n = 2 * 12 = 24 A = 10000 (1 + 0.0070833)^24 A = 10000 (1.0070833)^24 A = 10000 * 1.18448 A = R11,844.80 Although Option B has a lower annual interest rate, the monthly compounding results in a slightly lower final amount than Option

A. Thabo should choose Option

A. Guided Practice (With Solutions)

Question 1: Lerato invests R3,000 in a fixed deposit account that pays an annual interest rate of 7% compounded semi-annually (twice a year). How much will she have after 4 years?

Solution: P = R3,000 i = 7% per year / 2 = 0.07 / 2 = 0.035 per half-year n = 4 years 2 = 8 half-years A = 3000 (1 + 0.035)^8 A = 3000 (1.035)^8 A = 3000 * 1.316809 A = R3,950.43 Lerato will have R3,950.43 after 4 years. The key here is recognizing the compounding period is semi-annual, so you adjust both the interest rate and the number of periods accordingly.

Question 2: David takes out a personal loan of R15,000 with an annual interest rate of 15% compounded monthly. He agrees to repay the loan over 2 years. Calculate his monthly payment using the loan amortisation formula.

Solution: M = P [ i(1 + i)^n ] / [ (1 + i)^n – 1] P = R15,000 i = 15% per year / 12 = 0.15 / 12 = 0.0125 n = 2 years 12 = 24 M = 15000 [ 0.0125(1 + 0.0125)^24 ] / [ (1 + 0.0125)^24 – 1] M = 15000 [ 0.0125(1.0125)^24 ] / [ (1.0125)^24 – 1] M = 15000 [ 0.0125(1.347351) ] / [ 1.347351 – 1] M = 15000 [ 0.016842 ] / [ 0.347351] M = 15000 * 0.048486 M = R727.29 David's monthly payment will be R727.

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9. Remember to use the loan amortisation formula.

Question 3: Nomusa invests R8,000 in an account that offers an interest rate of 6% per year compounded quarterly. How long will it take for her investment to reach R10,000? (Round your answer to the nearest quarter).

Solution: A = P (1 + i)^n 10000 = 8000 (1 + 0.06/4)^n 10000/8000 = (1.015)^n 1.25 = (1.015)^n This requires logarithms to solve for n.