3.1 Compound interest (Page 2/2)

Siyavula textbooks: grade 10 Page 2 / 2

\begin{matrix} Closing Balance after 2 years & = & [P (1 + i)] \times (1 + i) \\ = & P {(1 + i)}^{2} \end{matrix}

And if we take that money out, then invest it for another year, the balance becomes:

\begin{matrix} Closing Balance after 3 years & = & [P {(1 + i)}^{2}] \times (1 + i) \\ = & P {(1 + i)}^{3} \end{matrix}

We can see that the power of the term $(1 + i)$ is the same as the number of years. Therefore,

Closing Balance after n years = P {(1 + i)}^{n}

Fractions add up to the whole

It is easy to show that this formula works even when $n$ is a fraction of a year. For example, let us invest the money for 1 month, then for 4 months, then for 7 months.

\begin{matrix} Closing Balance after 1 month & = & P {(1 + i)}^{\frac{1}{12}} \\ Closing Balance after 5 months & = & Closing Balance after 1 month invested for 4 months more \\ = & [P {(1 + i)}^{\frac{1}{12}}] {(1 + i)}^{\frac{4}{12}} \\ = & P {(1 + i)}^{\frac{1}{12} + \frac{4}{12}} \\ = & P {(1 + i)}^{\frac{5}{12}} \\ Closing Balance after 12 months & = & Closing Balance after 5 months invested for 7 months more \\ = & [P {(1 + i)}^{\frac{5}{12}}] {(1 + i)}^{\frac{7}{12}} \\ = & P {(1 + i)}^{\frac{5}{12} + \frac{7}{12}} \\ = & P {(1 + i)}^{\frac{12}{12}} \\ = & P {(1 + i)}^{1} \end{matrix}

which is the same as investing the money for a year.

Look carefully at the long equation above. It is not as complicated as it looks! All we are doing is taking the opening amount ( $P$ ), then adding interest for just 1 month. Then we are taking that new balance and adding interest for a further 4 months, and then finally we are taking the new balance after a total of 5 months, and adding interest for 7 more months. Take a look again, and check how easy it really is.

Does the final formula look familiar? Correct - it is the same result as you would get for simply investing $P$ for one full year. This is exactly what we would expect, because:

1 month + 4 months + 7 months = 12 months, which is a year.

Can you see that? Do not move on until you have understood this point.

The power of compound interest

To see how important this “interest on interest" is, we shall compare the difference in closing balances for money earning simple interest and money earning compound interest. Consider an amount of R10 000 that you have to invest for 10 years, and assume we can earn interest of 9%. How much would that be worth after 10 years?

The closing balance for the money earning simple interest is:

\begin{matrix} A & = & P (1 + i \cdot n) \\ = & R 10 000 (1 + 9 % \times 10) \\ = & R 19 000 \end{matrix}

The closing balance for the money earning compound interest is:

\begin{matrix} A & = & P {(1 + i)}^{n} \\ = & R 10 000 {(1 + 9 %)}^{10} \\ = & R 23 673, 64 \end{matrix}

So next time someone talks about the “magic of compound interest", not only will you know what they mean - but you will be able to prove it mathematically yourself!

Again, keep in mind that this is good news and bad news. When you are earning interest on money you have invested, compound interest helps that amount to increase exponentially. But if you have borrowed money, the build up of the amount you owe will grow exponentially too.

Mr Lowe wants to take out a loan of R 350 000. He does not want to pay back more than R625 000 altogether on the loan. If theinterest rate he is offered is 13%, over what period should he take the loan.

Determine what has been provided and what is required
- opening balance, $P = R 350 000$
- closing balance, $A = R 625 000$
- interest rate, $i = 13 % per year$
We are required to find the time period( $n$ ).
Determine how to approach the problem
We know from [link] that:

$A = P {(1 + i)}^{n}$

We need to find $n$ .

Therefore we convert the formula to:

$\frac{A}{P} = {(1 + i)}^{n}$

and then find $n$ by trial and error.
Solve the problem
$\begin{matrix} \frac{A}{P} & = & {(1 + i)}^{n} \\ \frac{625000}{350000} & = & {(1 + 0, 13)}^{n} \\ 1, 785 . . . & = & {(1, 13)}^{n} \\ Try n & = & 3 : {(1, 13)}^{3} = 1, 44 . . . \\ Try n & = & 4 : {(1, 13)}^{4} = 1, 63 . . . \\ Try n & = & 5 : {(1, 13)}^{5} = 1, 84 . . . \end{matrix}$
Write the final answer Mr Lowe should take the loan over four years (If he took the loan over five years, he would end up paying more than he wants to.)

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Other applications of compound growth

The following two examples show how we can take the formula for compound interest and apply it to real life problems involving compound growth or compound decrease.

South Africa's population is increasing by 2,5% per year. If the current population is 43 million, how many more people will there be in South Africa in two years' time ?

Determine what has been provided and what is required
- initial value (opening balance), $P = 43 000 000$
- period of time, $n = 2 year$
- rate of increase, $i = 2, 5 % per year$
We are required to find the final value (closing balance $A$ ).
Determine how to approach the problem
We know from [link] that:

$A = P {(1 + i)}^{n}$
Solve the problem
$\begin{matrix} A & = & P {(1 + i)}^{n} \\ = & 43 000 000 {(1 + 0, 025)}^{2} \\ = & 45 176 875 \end{matrix}$
Write the final answer
There will be $45 176 875 - 43 000 000 = 2 176 875$ more people in 2 years' time

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A swimming pool is being treated for a build-up of algae. Initially, $50 m^{2}$ of the pool is covered by algae. With each day of treatment, the algae reduces by 5%. What area is covered by algae after 30 days of treatment ?

Determine what has been provided and what is required
- Starting amount (opening balance), $P = 50 m^{2}$
- period of time, $n = 30 days$
- rate of decrease, $i = 5 % per day$
We are required to find the final area covered by algae (closing balance $A$ ).
Determine how to approach the problem
We know from [link] that:

$A = P {(1 + i)}^{n}$

But this is compound decrease so we can use the formula:

$A = P {(1 - i)}^{n}$
Solve the problem
$\begin{matrix} A & = & P {(1 - i)}^{n} \\ = & 50 {(1 - 0, 05)}^{30} \\ = & 10, 73 m^{2} \end{matrix}$
Write the final answer
Therefore the area still covered with algae is $10, 73 m^{2}$

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Compound interest

An amount of R3 500 is invested in a savings account which pays compound interest at a rate of 7,5% per annum. Calculate the balance accumulated by the end of 2 years.
If the average rate of inflation for the past few years was 7,3% and your water and electricity account is R 1 425 on average, what would you expect to pay in 6 years time ?
Shrek wants to invest some money at 11% per annum compound interest. How much money (to the nearest rand) should he invest if he wants to reach a sum of R 100 000 in five year's time ?

The next section on exchange rates is included for completeness. However, you should know about fluctuating exchange rates and the impact that this has on imports and exports. Fluctuating exchange rates lead to things like increases in the cost of petrol. You can read more about this in Fluctuating exchange rates .

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Source: OpenStax, Siyavula textbooks: grade 10 maths [caps]. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11306/1.4

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