Lesson

Some types of loans and investments attract *simple interest*. This is an amount calculated on the principle amount as a percentage, usually on an annual basis.

Suppose a car finance agreement includes a $9%$9% per annum simple interest charge on the initial amount borrowed over a period of $4$4 years. If the amount borrowed is $\$10000$$10000, how much interest is paid?

The interest in $1$1 year is $\$10000\times\frac{9}{100}=\$900$$10000×9100=$900. So, over the $4$4 year term of the loan the total interest paid is $\$900\times4$$900×4 = $\$3600$$3600.

At the end of the $4$4 years, the borrower would have repaid the $\$10000$$10000 and the $\$3600$$3600 interest. Often this total amount of $\$13600$$13600 would be spread over $48$48 equal monthly repayments during the term of the loan.

The amount of each monthly repayment, in this case, would be $\frac{\$13600}{48}=\$283.35$$1360048=$283.35, rounded to the nearest five cents.

Following the steps in Example 1, we see that the general formula for the amount of interest paid must be

$I=PRT$`I`=`P``R``T`

where $I$`I` is the amount of interest, $P$`P` is the principle, $R$`R` is the periodic interest rate and $T$`T` is the number of time periods.

Interest usually accrues at the beginning of each interest period. This could be at the beginning of each year or at the beginning of each month or even at the beginning of each day. So, if a graph is drawn that shows how much interest has been charged at any time during the term of the loan, it will be in the form of a series of steps.

The following graph illustrates a case where $\$400$$400 was charged each year over a $5$5 year period.

We might imagine the interest accruing monthly or daily or perhaps continuously. The corresponding graphs would have smaller and smaller steps until, in the continuous case, the graph would appear as a line.

The formula $I=PRT$`I`=`P``R``T` in the case where $T$`T` is allowed to vary continuously and the parameters $P$`P` and $R$`R` are constants has a straight line graph. Its gradient is the product $P\times R$`P`×`R`.

An amount of $\$4000$$4000 is to be invested at a simple interest rate of $6%$6% per annum.

How much interest is earned each year?

Which of the following shows the amount of the investment after $x$

`x`years?Loading Graph...$y=4000x$ `y`=4000`x`ALoading Graph...$y=4000x+240$ `y`=4000`x`+240BLoading Graph...$y=240x+4000$ `y`=240`x`+4000CLoading Graph...$y=240x$ `y`=240`x`DLoading Graph...$y=4000x$ `y`=4000`x`ALoading Graph...$y=4000x+240$ `y`=4000`x`+240BLoading Graph...$y=240x+4000$ `y`=240`x`+4000CLoading Graph...$y=240x$ `y`=240`x`DWhich feature of the graph represents the interest earned each year?

The value of $y$

`y`when $x=1$`x`=1.AThe gradient.

BThe $y$

`y`-intercept.CThe value of $y$

`y`when $x=1$`x`=1.AThe gradient.

BThe $y$

`y`-intercept.CHow would the graph change if the amount was invested at a rate of $3%$3% p.a. instead? Select all correct options.

The straight line graph would be flatter.

AThe straight line graph would increase more quickly.

BThe straight line graph would be steeper.

CThe graph would shift vertically downwards.

DThe straight line graph would increase more slowly.

EThe graph would shift vertically upwards.

FThe straight line graph would be flatter.

AThe straight line graph would increase more quickly.

BThe straight line graph would be steeper.

CThe graph would shift vertically downwards.

DThe straight line graph would increase more slowly.

EThe graph would shift vertically upwards.

F

This graph shows an amount of $\$17000$$17000 being deposited into two different banks.

Each bank offers simple interest at a different rate, and for different lengths of time.

Loading Graph...

Lisa wants her investment of $\$17000$$17000 to grow to $\$18000$$18000 by the time $8$8 years have passed.

Which investment option should she take?

Option A

AOption B

BOption A

AOption B

BHow many years longer will it take for $\$17000$$17000 to grow to $\$18000$$18000 under option B than under option A?

Apply everyday compounding rates

Apply numeric reasoning in solving problems