Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Apply the compound interest formula for investments and loans.
• Calculate the present value of a future amount.
• Compare nominal and effective interest rates.
• Apply formulas for straight-line and reducing-balance depreciation.
• Solve real-world financial problems.

Lesson summary

Finance, growth, and decay are fundamental concepts in mathematics with widespread applications in our daily lives. Understanding these concepts allows us to make informed decisions about investments, loans, and understanding the implications of inflation. In the South African context, where economic realities often impact individuals and communities significantly, a solid grasp of these principles is crucial for financial literacy and responsible financial planning. For example, understanding compound interest can empower individuals to save effectively for retirement or education, while grasping the concept of depreciation can help businesses manage their assets wisely.

Lesson notes

2.1 Compound Interest Compound interest is the interest calculated on the initial principal, which also includes all of the accumulated interest from previous periods. This means that interest earns interest, leading to exponential growth.

Formula: A = P (1 + i)^n Where: A = Future value (accumulated amount) P = Principal amount (initial investment) i = Interest rate per compounding period (expressed as a decimal) n = Number of compounding periods Compounding Frequency: The frequency at which interest is compounded significantly affects the final amount.

Common compounding periods include: Annually: n = number of years, i = interest rate per year Semi-annually: n = number of years 2, i = interest rate per year / 2 Quarterly: n = number of years 4, i = interest rate per year / 4 Monthly: n = number of years 12, i = interest rate per year / 12 Example 1: Sarah invests R10,000 in a fixed deposit account that pays 8% interest per annum, compounded quarterly. Calculate the value of her investment after 5 years.

Solution: P = R10,000 i = 8% / 4 = 0.08 / 4 = 0.02 n = 5 4 = 20 A = 10000 (1 + 0.02)^20 A = 10000 (1.02)^20 A = 10000 * 1.485947 A = R14,859.47 Therefore, the value of Sarah's investment after 5 years is R14,859.

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7. Example 2: How long will it take for an investment of R5,000 to double if it earns 10% interest per annum, compounded monthly?

Solution: We want A = 2 P = 2 R5,000 = R10,000 P = R5,000 i = 10% / 12 = 0.10 / 12 = 0.008333 A = R10,000 10000 = 5000 (1 + 0.008333)^n 2 = (1.008333)^n Take the logarithm of both sides: ln(2) = n * ln(1.008333) n = ln(2) / ln(1.008333) n = 0.693147 / 0.0083 n ≈ 83.5 months Therefore, it will take approximately 84 months (7 years) for the investment to double. 2.2 Present Value Present value is the current worth of a future sum of money, given a specified rate of return. It's essentially the reverse of compound interest.

Formula: P = A / (1 + i)^n Where: P = Present value A = Future value i = Interest rate per compounding period n = Number of compounding periods Example 3: Thando wants to have R20,000 in 3 years. How much should she invest now if the interest rate is 9% per annum, compounded annually?

Solution: A = R20,000 i = 9% = 0.09 n = 3 P = 20000 / (1 + 0.09)^3 P = 20000 / (1.09)^3 P = 20000 / 1.295029 P = R15,443.53 Thando should invest R15,443.53 now. 2.3 Nominal and Effective Interest Rates Nominal Interest Rate: The stated annual interest rate.

Effective Interest Rate: The actual annual interest rate earned after taking into account the effects of compounding.

Formula: Effective Interest Rate = (1 + i/m)^m - 1 Where: i = Nominal interest rate m = Number of compounding periods per year Example 4: Which investment is better: 12% per annum compounded quarterly or 12.5% per annum compounded annually?

Solution: Option 1: Nominal = 12%, m = 4 Effective rate = (1 + 0.12/4)^4 - 1 = (1.03)^4 - 1 = 1.125509 - 1 = 0.125509 or 12.55% Option 2: Nominal = 12.5%, m = 1 Effective rate = (1 + 0.125/1)^1 - 1 = 1.125 - 1 = 0.125 or 12.5% Option 1 (12% compounded quarterly) is slightly better because its effective interest rate (12.55%) is higher than option 2 (12.5% compounded annually, with effective rate 12.5%). 2.4 Depreciation Depreciation is the decrease in value of an asset over time.

We will look at two methods: Straight-line and Reducing-balance. 2.4.1 Straight-Line Depreciation The asset depreciates by the same amount each year.

Formula: Depreciation amount per year = (Original Cost - Salvage Value) / Useful Life Book Value = Original Cost - (Depreciation amount per year Number of years) 2.4.2 Reducing-Balance Depreciation The asset depreciates by a fixed percentage of its book value each year.

Formula: Book Value = Original Cost (1 - i)^n where i = depreciation rate (expressed as a decimal) and n = number of years.

Example 5: A delivery truck costs R250,

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0. It has a salvage value of R50,000 after 5 years. (a) Calculate the annual depreciation using the straight-line method. (b) Calculate the book value after 3 years using the straight-line method. (c) Calculate the book value after 3 years using the reducing-balance method if the depreciation rate is 20% per annum.

Solution: (a)

Straight-Line Depreciation: Depreciation amount per year = (250000 - 50000) / 5 = 200000 / 5 = R40,000 (b)

Book Value after 3 years (Straight-Line): Book Value = 250000 - (40000 * 3) = 250000 - 120000 = R130,000 (c) Book Value after 3 years (Reducing-Balance): Book Value = 250000 (1 - 0.20)^3 = 250000 (0.8)^3 = 250000 * 0.512 = R128,000 2.5 Annuities (Introduction linked to Geometric Sequences and Series) An annuity is a series of equal payments made at regular intervals. We will focus on future value annuities. This topic links directly to Geometric Series. Each payment grows with compound interest until the final payment.