Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

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Performance objectives

• Calculate simple and compound interest.
• Apply formulas for present and future value.
• Calculate depreciation using two methods.
• Analyse the impact of inflation.
• Model population growth and decay.

Lesson summary

Finance, growth, and decay are fundamental concepts in mathematics that directly impact your lives. Understanding these principles allows you to make informed financial decisions, predict future trends, and appreciate the dynamics of population growth, depreciation of assets, and compound interest. This knowledge is vital for managing personal finances, understanding investments, and participating effectively in the South African economy. Consider the implications of compound interest when saving for tertiary education or buying a car – understanding it is crucial for making sound financial choices.

Lesson notes

2.1 Simple Interest Simple interest is calculated only on the principal amount. It is a straightforward method of calculating interest, but less common in long-term investments.

Formula: A = P(1 + in)

Where: A = Accumulated amount (final amount) P = Principal amount (initial amount) i = Interest rate (expressed as a decimal) n = Number of years Example 1: Sipho invests R5000 in a fixed deposit account that earns simple interest at a rate of 8% per annum. How much will he have after 3 years?

Solution: P = R5000 i = 8% = 0.08 n = 3 A = 5000(1 + 0.08 * 3) A = 5000(1 + 0.24) A = 5000(1.24) A = R6200 Sipho will have R6200 after 3 years. 2.2 Compound Interest Compound interest is calculated on the principal amount and also on the accumulated interest of previous periods. This leads to exponential growth. It's vital for long-term investments and loans.

Formula: A = P(1 + i)^n Where: A = Accumulated amount (final amount) P = Principal amount (initial amount) i = Interest rate per compounding period (expressed as a decimal) n = Number of compounding periods Important Notes: Compounding Frequency: The interest rate and number of periods must be adjusted if interest is compounded more than once a year.

Semi-annually (twice a year): Divide the annual interest rate by 2 and multiply the number of years by

2. Quarterly (four times a year): Divide the annual interest rate by 4 and multiply the number of years by

4. Monthly (12 times a year): Divide the annual interest rate by 12 and multiply the number of years by

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2. Example 2: Thandi invests R8000 in a savings account that earns compound interest at a rate of 10% per annum, compounded quarterly. How much will she have after 5 years?

Solution: P = R8000 i = 10%/4 = 0.025 n = 5 4 = 20 A = 8000(1 + 0.025)^20 A = 8000(1.025)^20 A = 8000(1.638616) A ≈ R13108.93 Thandi will have approximately R13108.93 after 5 years. 2.3 Present Value The present value (PV) is the current worth of a future sum of money or stream of cash flows given a specified rate of return. In essence, it's the inverse of compound interest.

Formula: P = A / (1 + i)^n Where: P = Present Value A = Accumulated amount (future value) i = Interest rate per compounding period (expressed as a decimal) n = Number of compounding periods Example 3: Zola wants to have R20,000 in 4 years. How much must he invest now in an account that earns compound interest at a rate of 9% per annum, compounded annually?

Solution: A = R20000 i = 9% = 0.09 n = 4 P = 20000 / (1 + 0.09)^4 P = 20000 / (1.09)^4 P = 20000 / 1.411582 P ≈ R14168.58 Zola needs to invest approximately R14168.58 now. 2.4 Depreciation Depreciation is the decrease in the value of an asset over time due to wear and tear, obsolescence, or other factors.

Two Common Methods: Straight-Line Depreciation: The asset depreciates by the same amount each year.

Formula: A = P(1 - in)* A = Accumulated value P = Initial Value i = Depreciation rate n = Number of years Reducing Balance Depreciation: The asset depreciates by a percentage of its current value each year.

Formula: A = P(1 - i)^n A = Accumulated value P = Initial Value i = Depreciation rate n = Number of years Example 4: A delivery van is purchased for R250,000. (a) Calculate its value after 5 years using straight-line depreciation at a rate of 15% per annum. (b) Calculate its value after 5 years using reducing balance depreciation at a rate of 15% per annum.

Solution: (a)

Straight-Line Depreciation: P = R250000 i = 15% = 0.15 n = 5 A = 250000(1 - 0.15 * 5) A = 250000(1 - 0.75) A = 250000(0.25) A = R62500 The van's value after 5 years using straight-line depreciation is R62500. (b)

Reducing Balance Depreciation: P = R250000 i = 15% = 0.15 n = 5 A = 250000(1 - 0.15)^5 A = 250000(0.85)^5 A = 250000(0.443705) A ≈ R110926.25 The van's value after 5 years using reducing balance depreciation is approximately R110926.

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5. Notice it's higher than with straight-line depreciation. 2.5 Inflation Inflation is the rate at which the general level of prices for goods and services is rising, and, subsequently, purchasing power is falling. It erodes the real value of money over time. The same amount of money will buy less in the future.

Effect on Investments: It’s vital to consider the real rate of return on investments, which is the nominal rate of return minus the inflation rate.

Example 5: You invest R10,000 in an account that earns 7% interest per year. The inflation rate is 4%. What is the real rate of return on your investment? What is the approximate real value of your investment after 1 year?

Solution: Nominal interest rate = 7% Inflation rate = 4% Real rate of return ≈ Nominal rate - Inflation rate = 7% - 4% = 3% Value after 1 year (Nominal): R10,000 * (1 + 0.07) = R10,700 Approximate Real value after 1 year: R10,000 * (1 + 0.03) = R10,

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0. The investment is nominally worth R10,700, but its purchasing power is only equivalent to about R10,300 in today’s terms due to inflation.