Lesson Notes By Weeks and Term v5 - Grade 12
Finance, growth and decay – Week 7 focus
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Finance, growth and decay – Week 7 focus
Download the Lessonotes Mobile South Africa app for faster lesson access on Android and iPhone.
Subject: Mathematics
Class: Grade 12
Term: 1st Term
Week: 7
Theme: General lesson support
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Finance, growth, and decay form a cornerstone of mathematical literacy, especially for South African learners preparing to navigate the economic realities of our country. Understanding these concepts empowers you to make informed decisions about investments, loans, savings, and other crucial financial matters. Without this understanding, you are vulnerable to poor financial choices, which can have long-term consequences. Imagine wanting to buy a car or a house; knowing how interest works can save you thousands of Rands. Moreover, these concepts extend beyond personal finance.
2.1 Compound Interest: Compound interest is interest earned not only on the principal amount but also on any accumulated interest. This is a powerful tool for wealth creation and should be thoroughly understood.
The formula for compound interest is: A = P(1 + i)^n Where: A = Accumulated amount (future value) P = Principal amount (present value) i = Interest rate per compounding period (expressed as a decimal) n = Number of compounding periods Example 1: Sarah invests R5,000 in a savings account that pays 8% interest per year, compounded annually. How much will she have after 5 years?
Solution: P = R5,000 i = 0.08 (8% expressed as a decimal) n = 5 A = 5000(1 + 0.08)^5 A = 5000(1.08)^5 A = R7,346.64 Therefore, Sarah will have R7,346.64 after 5 years.
Example 2: Sipho wants to have R10,000 in 3 years. How much should he invest now in an account that pays 10% interest per year, compounded quarterly?
Solution: A = R10,000 i = 0.10/4 = 0.025 (10% divided by 4 quarters in a year) n = 3 4 = 12 (3 years multiplied by 4 quarters per year) 10000 = P(1 + 0.025)^12 10000 = P(1.025)^12 10000 = P(1.344888824) P = 10000 / 1.344888824 P = R7,435.56 Sipho needs to invest R7,435.56 now.
Why it works: Compounding annually means the interest is calculated and added to the principal once a year. Compounding quarterly means it is calculated and added four times a year. This frequency impacts the total interest earned because you earn interest on interest more frequently. 2.2 Nominal and Effective Interest Rates: The nominal interest rate is the stated annual interest rate. The effective interest rate is the actual annual interest rate taking into account the effect of compounding more frequently than once a year. The effective interest rate is always higher than the nominal interest rate when compounding occurs more than annually. The formula for effective interest rate is: Effective rate = (1 + i/n)^n - 1 Where: i = Nominal interest rate (expressed as a decimal) n = Number of compounding periods per year Example 3: A bank offers a nominal interest rate of 12% per year, compounded monthly. What is the effective annual interest rate?
Solution: i = 0.12 n = 12 Effective rate = (1 + 0.12/12)^12 - 1 Effective rate = (1 + 0.01)^12 - 1 Effective rate = (1.01)^12 - 1 Effective rate = 1.12682503 - 1 Effective rate = 0.12682503 Effective rate = 12.68% The effective annual interest rate is 12.68%. This is what you actually earn in a year due to monthly compounding. 2.3 Depreciation: Depreciation is the decrease in the value of an asset over time.
There are two main methods: straight-line depreciation and reducing-balance depreciation.
Straight-Line Depreciation: The asset depreciates by the same amount each year.
Formula: A = P(1 - ni)
Where: A = Book value of the asset after n years P = Original cost of the asset n = Number of years i = Rate of depreciation per year (expressed as a decimal)
Reducing-Balance Depreciation: The asset depreciates by a percentage of its current value each year. This means the depreciation amount decreases over time.
Formula: A = P(1 - i)^n Where: A = Book value of the asset after n years P = Original cost of the asset n = Number of years i = Rate of depreciation per year (expressed as a decimal)
Example 4: A delivery van costs R200,
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0. It depreciates using the straight-line method at a rate of 15% per year. What is its value after 4 years?
Solution: P = R200,000 n = 4 i = 0.15 A = 200000(1 - 4 0.15) A = 200000(1 - 0.6) A = 200000(0.4) A = R80,000 The value of the van after 4 years is R80,
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0. Example 5: A computer costs R15,000 and depreciates using the reducing-balance method at a rate of 20% per year. What is its value after 3 years?
Solution: P = R15,000 n = 3 i = 0.20 A = 15000(1 - 0.20)^3 A = 15000(0.8)^3 A = 15000(0.512) A = R7,680 The value of the computer after 3 years is R7,
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0. Why different methods? Straight-line depreciation is simple, but reducing-balance is often more realistic, as assets tend to lose value faster when they are newer. 2.4 Annuities: An annuity is a series of equal payments made at regular intervals.
There are two main types: future value annuities and present value annuities.
Future Value Annuity: The accumulated value of a series of payments made over time, earning interest. This is what you'd use to calculate how much you'll have saved if you contribute a certain amount each month.
Formula: FV = x[((1 + i)^n - 1)/i] Where: FV = Future Value of the annuity x = Periodic payment i = Interest rate per period n = Number of periods Present Value Annuity: The current value of a series of future payments. This is what you'd use to calculate how much you can borrow if you can afford to pay a certain amount each month.
Formula: PV = x[ (1 - (1 + i)^-n) / i] Where: PV = Present Value of the annuity x = Periodic payment i = Interest rate per period n = Number of periods Example 6: Thando deposits R500 per month into a savings account that pays 6% interest per year, compounded monthly.