# 1.3 Finding i and n

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## Finding $i$

By this stage in your studies of the mathematics of finance, you have always known what interest rate to use in the calculations, and how long the investment or loan will last. You have then either taken a known starting point and calculated a future value, or taken a known future value and calculated a present value.

But here are other questions you might ask:

1. I want to borrow R2 500 from my neighbour, who said I could pay back R3 000 in 8 months time. What interest is she charging me?
2. I will need R450 for some university textbooks in 1,5 years time. I currently have R400. What interest rate do I need to earn to meet this goal?

Each time that you see something different from what you have seen before, start off with the basic equation that you should recognise very well:

$A=P·{\left(1+i\right)}^{n}$

If this were an algebra problem, and you were told to “solve for $i$ ", you should be able to show that:

$\begin{array}{ccc}\hfill \frac{A}{P}& =& {\left(1+i\right)}^{n}\hfill \\ \hfill \left(1+i\right)& =& {\left(\frac{A}{P}\right)}^{1/n}\hfill \\ \hfill i& =& {\left(\frac{A}{P}\right)}^{1/n}-1\hfill \end{array}$

You do not need to memorise this equation, it is easy to derive any time you need it!

So let us look at the two examples mentioned above.

1. Check that you agree that $P$ =R2 500, $A$ =R3 000, $n$ =8/12=0,666667. This means that:
$\begin{array}{ccc}\hfill i& =& {\left(\frac{\mathrm{R}3\phantom{\rule{3.33333pt}{0ex}}000}{\mathrm{R}2\phantom{\rule{3.33333pt}{0ex}}500}\right)}^{1/0,666667}-1\hfill \\ & =& 31,45%\hfill \end{array}$
Ouch! That is not a very generous neighbour you have.
2. Check that $P$ =R400, $A$ =R450, $n$ =1,5
$\begin{array}{ccc}\hfill i& =& {\left(\frac{\mathrm{R}450}{\mathrm{R}400}\right)}^{1/1,5}-1\hfill \\ & =& 8,17%\hfill \end{array}$
This means that as long as you can find a bank which pays more than 8,17% interest, you should have the money you need!

Note that in both examples, we expressed $n$ as a number of years ( $\frac{8}{12}$ years, not 8 because that is the number of months) which means $i$ is the annual interest rate. Always keep this in mind - keep years with years to avoid making silly mistakes.

## Finding $i$

1. A machine costs R45 000 and has a scrap value of R9 000 after 10 years. Determine the annual rate of depreciation if it is calculated on the reducing balance method.
2. After 5 years an investment doubled in value. At what annual rate was interest compounded ?

## Finding $n$ - trial and error

By this stage you should be seeing a pattern. We have our standard formula, which has a number of variables:

$A=P·{\left(1+i\right)}^{n}$

We have solved for $A$ (in Grade 10), $P$ (in "Present Values or Future Values of an Investment or Loan" ) and $i$ (in "Finding i" ). This time we are going to solve for $n$ . In other words, if we know what the starting sum of money is and what it grows to, and if we know what interest rate applies - then we can work out how long the money needs to be invested for all those other numbers to tie up.

This section will calculate $n$ by trial and error and by using a calculator. The proper algebraic solution will be learnt in Grade 12.

Solving for $n$ , we can write:

$\begin{array}{ccc}\hfill A& =& P{\left(1+i\right)}^{n}\hfill \\ \hfill \frac{A}{P}& =& {\left(1+i\right)}^{n}\hfill \end{array}$

Now we have to examine the numbers involved to try to determine what a possible value of $n$ is. Refer to your Grade 10 notes for some ideas as to how to go about finding $n$ .

We invest R3 500 into a savings account which pays 7,5% compound interest for an unknown period of time, at the end of which our account is worth R4 044,69. How long did we invest the money?

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

We are required to find $n$ .

1. We know that:

$\begin{array}{ccc}\hfill A& =& P{\left(1+i\right)}^{n}\hfill \\ \hfill \frac{A}{P}& =& {\left(1+i\right)}^{n}\hfill \end{array}$
2. $\begin{array}{ccc}\hfill \frac{\mathrm{R}4\phantom{\rule{3.33333pt}{0ex}}044,69}{\mathrm{R}3\phantom{\rule{3.33333pt}{0ex}}500}& =& {\left(1+7,5%\right)}^{n}\hfill \\ \hfill 1,156& =& {\left(1,075\right)}^{n}\hfill \end{array}$

We now use our calculator and try a few values for $n$ .

 Possible $n$ $1,{075}^{n}$ 1,0 1,075 1,5 1,115 2,0 1,156 2,5 1,198

We see that $n$ is close to 2.

3. The R3 500 was invested for about 2 years.

## Finding $n$ - trial and error

1. A company buys two types of motor cars: The Acura costs R80 600 and the Brata R101 700 VAT included. The Acura depreciates at a rate, compounded annually, of 15,3% per year and the Brata at 19,7%, also compounded annually, per year. After how many years will the book value of the two models be the same ?
2. The fuel in the tank of a truck decreases every minute by 5,5% of the amount in the tank at that point in time. Calculate after how many minutes there will be less than $30l$ in the tank if it originally held $200l$ .

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Source:  OpenStax, Siyavula textbooks: grade 11 maths. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11243/1.3
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