Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

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Performance objectives

• Calculate simple and compound interest.
• Find the future value of investments or loans.
• Understand different interest rates and compounding periods.
• Calculate depreciation using two different methods.
• Apply these concepts to solve real-world problems.

Lesson summary

Finance, growth, and decay are fundamental concepts in mathematics with widespread applications in personal finance, business, and science. Understanding these concepts equips you with the tools to make informed decisions about savings, investments, loans, and understanding how populations change over time. In a South African context, with socio-economic challenges and opportunities, mastering these skills empowers you to navigate the financial landscape, plan for your future, and contribute to the economic growth of our nation.

Lesson notes

Simple Interest Simple interest is calculated only on the principal amount (initial investment or loan amount). It's a straightforward way to calculate interest, but less common in many real-world financial products, except for very short-term loans.

Formula: `A = P(1 + in)` Where: `A` = Accumulated amount (final amount including principal and interest) `P` = Principal amount (initial investment or loan amount) `i` = Interest rate per period (expressed as a decimal) `n` = Number of periods Example 1: Sipho invests R5,000 in a savings account that pays simple interest at a rate of 8% per annum. How much will he have after 3 years?

Solution: P = R5,000 i = 8% = 0.08 n = 3 years A = 5000(1 + (0.08 * 3)) A = 5000(1 + 0.24) A = 5000(1.24) A = R6,200 Sipho will have R6,200 after 3 years. Compound Interest Compound interest is calculated on the principal amount and on the accumulated interest from previous periods. This means that you earn interest on your interest, leading to faster growth over time. This is the standard form of interest calculation for most investments.

Formula: `A = P(1 + i)^n` Where: `A` = Accumulated amount `P` = Principal amount `i` = Interest rate per period (expressed as a decimal) `n` = Number of periods Important

Note: If the interest is compounded more than once a year (e.g., monthly, quarterly), you need to adjust the interest rate (`i`) and the number of periods (`n`) accordingly. New `i` = Annual interest rate / Number of compounding periods per year New `n` = Number of years Number of compounding periods per year Example 2: Thandi invests R10,000 in a fixed deposit account that pays compound interest at a rate of 10% per annum, compounded monthly. How much will she have after 5 years?

Solution: P = R10,000 Annual interest rate = 10% = 0.10 Number of compounding periods per year = 12 (monthly) Number of years = 5 New `i` = 0.10 / 12 = 0.008333 (approximately) New `n` = 5 12 = 60 A = 10000(1 + 0.008333)^60 A = 10000(1.008333)^60 A = 10000(1.645309) A = R16,453.09 Thandi will have approximately R16,453.09 after 5 years. Depreciation Depreciation is the decrease in the value of an asset over time due to wear and tear, obsolescence, or other factors. Businesses need to account for depreciation when calculating their taxes and the true value of their assets. Straight-Line Depreciation (also called Fixed Installment Method) The asset depreciates by the same amount each year.

Formula: Depreciation per year = (Original cost - Salvage value) / Useful life Book Value = Original Cost - Accumulated Depreciation Where: Salvage value is the estimated value of the asset at the end of its useful life. Useful life is the estimated number of years the asset will be used. Book value is the value of the asset after depreciation.

Example 3: A small business in Durban buys a delivery van for R250,

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0. They estimate that the van will have a useful life of 5 years and a salvage value of R50,

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0. Calculate the annual depreciation and the book value after 3 years using the straight-line method.

Solution: Original cost = R250,000 Salvage value = R50,000 Useful life = 5 years Depreciation per year = (250000 - 50000) / 5 Depreciation per year = 200000 / 5 Depreciation per year = R40,000 Accumulated depreciation after 3 years = 40000 * 3 = R120,000 Book value after 3 years = 250000 - 120000 = R130,000 Reducing-Balance Depreciation (also called Diminishing Value Method) The asset depreciates by a fixed percentage each year. This method results in higher depreciation expense in the earlier years of the asset's life.

Formula: Book Value end of year = Original Cost (1 - i)^n Where: i = depreciation rate. n = number of years. To find the depreciation rate `i`: i = 1 - (Salvage Value / Original Cost)^(1/n)

Example 4: A mining company in Rustenburg buys a piece of equipment for R1,000,

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0. They estimate that it will have a useful life of 8 years and a salvage value of R100,

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0. Calculate the depreciation rate and the book value after 4 years using the reducing-balance method.

Solution: Original cost = R1,000,000 Salvage value = R100,000 Useful life = 8 years i = 1 - (100000 / 1000000)^(1/8) i = 1 - (0.1)^(0.125) i = 1 - 0.794328 i = 0.205672 (approximately 20.57%) Book Value after 4 years = 1000000(1 - 0.205672)^4 Book Value after 4 years = 1000000(0.794328)^4 Book Value after 4 years = 1000000(0.399715) Book Value after 4 years = R399,715 (approximately) Guided Practice (With Solutions)

Question 1: Nandi invests R2,000 in a unit trust that offers a simple interest rate of 9% per annum. How much interest will she earn after 5 years?

Solution: P = R2,000 i = 9% = 0.09 n = 5 years A = P(1 + in) = 2000(1 + (0.09 * 5)) = 2000(1 + 0.45) = 2000(1.45) = R2,900 Interest Earned = A - P = 2900 - 2000 = R900

Commentary: This question directly applies the simple interest formula. Remember to subtract the principal to find the interest earned.

Question 2: A farmer takes out a loan of R50,000 to buy equipment.