Lesson Notes By Weeks and Term v5 - Grade 11
Finance, growth and decay – Week 8 focus
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Finance, growth and decay – Week 8 focus
Download the Lessonotes Mobile South Africa app for faster lesson access on Android and iPhone.
Subject: Mathematics
Class: Grade 11
Term: 3rd Term
Week: 8
Theme: General lesson support
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This week, we delve into the fascinating world of Finance, Growth, and Decay. This topic is crucial because it provides you with the tools to understand how money grows (or shrinks!) over time, and how to make informed financial decisions that will affect your future. Understanding compound interest, inflation, and depreciation are essential skills for managing personal finances, understanding investments, and even analyzing economic trends in South Africa. Consider how understanding these concepts can help you make informed decisions about saving for university, buying a car, or even starting a small business one day.
2.1 Compound Interest: Compound interest is the interest earned on both the principal amount and the accumulated interest. This "interest on interest" effect makes compound interest a powerful tool for wealth creation.
The formula for compound interest is: `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, e.g., 10% = 0.10) `n` = Number of compounding periods Important Considerations: Compounding Period: The frequency with which interest is calculated and added to the principal.
Common compounding periods are: Annually (once per year) Semi-annually (twice per year) Quarterly (four times per year) Monthly (12 times per year) Daily (365 times per year)
Adjusting 'i' and 'n': If the interest is compounded more frequently than annually, you need to adjust the interest rate (`i`) and the number of periods (`n`). Divide the annual interest rate by the number of compounding periods per year, and multiply the number of years by the number of compounding periods per year.
Example 1: Sarah invests R5,000 in a savings account that pays 8% interest compounded annually. How much will she have after 5 years? P = R5,000 i = 0.08 n = 5 A = 5000(1 + 0.08)^5 A = 5000(1.08)^5 A = R7,346.64 (rounded to the nearest cent) Therefore, Sarah will have R7,346.64 after 5 years.
Example 2: David invests R10,000 in an account that pays 12% interest compounded monthly. How much will he have after 3 years? P = R10,000 i = 0.12 / 12 = 0.01 (monthly interest rate) n = 3 12 = 36 (number of months) A = 10000(1 + 0.01)^36 A = 10000(1.01)^36 A = R14,307.69 (rounded to the nearest cent) Therefore, David will have R14,307.69 after 3 years. 2.2 Present Value: Present value is the current worth of a future sum of money, given a specified rate of return or discount rate.
It essentially answers the question: "How much money do I need to invest today to have a certain amount in the future?" The formula for present value is: `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 4 years. How much does she need to invest today if the interest rate is 10% compounded annually? A = R20,000 i = 0.10 n = 4 P = 20000 / (1 + 0.10)^4 P = 20000 / (1.10)^4 P = R13,660.28 (rounded to the nearest cent) Therefore, Thando needs to invest R13,660.28 today. 2.3 Nominal and Effective Interest Rates: Nominal Interest Rate: The stated annual interest rate. This is the rate usually quoted by banks and financial institutions.
Effective Interest Rate: The actual annual interest rate earned after taking into account the effects of compounding. It represents the true cost of borrowing or the true return on an investment. The formula to convert from nominal to effective interest rate is: `Effective Rate = (1 + i/n)^n - 1` Where: `i` = Nominal interest rate (as a decimal) `n` = Number of compounding periods per year Example 4: A bank offers a nominal interest rate of 12% compounded quarterly. What is the effective interest rate? i = 0.12 n = 4 Effective Rate = (1 + 0.12/4)^4 - 1 Effective Rate = (1 + 0.03)^4 - 1 Effective Rate = (1.03)^4 - 1 Effective Rate = 1.1255 - 1 Effective Rate = 0.1255 or 12.55% Therefore, the effective interest rate is 12.55%. This means that even though the nominal rate is 12%, the actual annual return is higher due to compounding. 2.4 Exponential Growth and Decay: Exponential growth and decay describe processes where a quantity increases or decreases at a rate proportional to its current value.
Exponential Growth: The quantity increases over time. Examples include population growth and the growth of an investment.
Exponential Decay: The quantity decreases over time. Examples include the depreciation of a car and the decay of a radioactive substance. The general formula for exponential growth and decay is: `A = A₀(1 + r)^t` (for growth) `A = A₀(1 - r)^t` (for decay)
Where: `A` = Final amount `A₀` = Initial amount `r` = Growth/decay rate (as a decimal) `t` = Time period Example 5: Population Growth The population of a town is currently 50,000 and is growing at a rate of 3% per year. What will the population be in 10 years? A₀ = 50,000 r = 0.03 t = 10 A = 50000(1 + 0.03)^10 A = 50000(1.03)^10 A ≈ 67,195.81 Therefore, the population will be approximately 67,196 in 10 years (rounding to the nearest whole number).
Example 6: Depreciation A car costs R200,000 new and depreciates at a rate of 15% per year. What will its value be after 4 years? A₀ = R200,000 r = 0.15 t = 4 A = 200000(1 - 0.15)^4 A = 200000(0.85)^4 A = R104,401.25 Therefore, the car's value will be R104,401.25 after 4 years. Guided Practice (With Solutions)
Question 1: Sipho invests R8,000 in a fixed deposit account that pays 9% interest compounded semi-annually. How much will he have after 6 years?