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Introduction

In earlier grades simple interest and compound interest were studied, together with the concept of depreciation. Nominal and effective interest rates were also described. Since this chapter expands on earlier work, it would be best if you revised the work done in Grades 10 and 11.

If you master the techniques in this chapter, when you start working and earning you will be able to apply the techniques in this chapter to critically assess how to invest your money. And when you are looking at applying for a bond from a bank to buy a home, you will confidently be able to get out the calculator and work out with amazement how much you could actually save by making additional repayments. Indeed, this chapter will provide you with the fundamental concepts you will need to confidently manage your finances and with some successful investing, sit back on your yacht and enjoy the millionaire lifestyle.

Finding the length of the investment or loan

In Grade 11, we used the Compound Interest formula A = P ( 1 + i ) n to determine the term of the investment or loan, by trial and error. Remember that P is the initial amount, A is the current amount, i is the interest rate and n is the number of time units (number of months or years). So if we invest an amount and know what the interest rate is, then we can work out how long it will take for the money to grow to the required amount.

Now that you have learnt about logarithms, you are ready to work out the proper algebraic solution. If you need to remind yourself how logarithms work, go to Chapter  [link] (on page  [link] ).

The basic finance equation is:

A = P · ( 1 + i ) n

If you don't know what A , P , i and n represent, then you should definitely revise the work from Grade 10 and 11.

Solving for n :

A = P ( 1 + i ) n ( 1 + i ) n = ( A / P ) log ( ( 1 + i ) n ) = log ( A / P ) n log ( 1 + i ) = log ( A / P ) n = log ( A / P ) / log ( 1 + i )

Remember, you do not have to memorise this formula. It is very easy to derive any time you need it. It is simply a matter of writing down what you have, deciding what you need, and solving for that variable.

Suppose we invested R3 500 into a savings account which pays 7,5% compound interest. After an unknown period of time our account is worth R4 044,69. For how long did we invest the money? How does this compare with the trial and error answer from Chapters  [link] .

    • P =R3 500
    • i =7,5%
    • A =R4 044,69

    We are required to find n .

  1. We know that:

    A = P ( 1 + i ) n ( 1 + i ) n = ( A / P ) log ( ( 1 + i ) n ) = log ( A / P ) n log ( 1 + i ) = log ( A / P ) n = log ( A / P ) / log ( 1 + i )
  2. n = log ( A / P ) / log ( 1 + i ) = log ( 4 044 , 69 3 500 ) log ( 1 + 0 . 075 ) Remember that : 7 . 5 % = 7 . 5 100 = 0 . 075 = 2 . 0
  3. The R3 500 was invested for 2 years.

A series of payments

i

i

By this stage, you know how to do calculations such as `If I want R1 000 in 3 years' time, how much do I need to invest now at 10% ?'

What if we extend this as follows: `If I want to draw R1 000 next year, R1 000 the next year and R1 000 after three years ... how much do I need to initially put into a bank account earning 10% p.a. to be able to afford to be able to do this?'

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Source:  OpenStax, Siyavula textbooks: grade 12 maths. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11242/1.2
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