Question

A student borrows $800 at 5.3% annual interest compounded semiannually. How much does he owe after 4 years?

Answer 1

`$186.19`

Explanation:

For this type of question, we would use Compound Interest which is gaining interest continuously over a certain amount of years, instead of working them all out separately. The formula for Compound Interest is:

`A=P(1+r// nxx100)^t`

Where `A` is the amount of interest, `P` is the original amount, `R` is the interest rate, `N` is how many times interested per year and `T` is the time.

Plugging in values:

`A=$800(1+5.3/200)^8`

We change this to `^8` as semiannually means twice a year, and as this is after `4` years, we double it to get `8.`

Plugging into calculator:

`=$986.1923212`

Rounding to 2 d.p:

`$986.19`

Minusing the before value from the after value:

`$986.19-800=$186.19`

Answer 2

Rounding to 2 decimal places gives `$186.18` being owed in interest

Explanation:

`color(blue)("Some thoughts")`

Note that 'semiannually' means twice per year. So each year has 2 calculation cycles

You have to 'split up' the annual percentage into a proportion that reflects the calculation cycle.

So the annual interest of `5.3%` becomes `5.3/2%` at each calculation cycle.

The general form for this context is

`color(white)("dddddddd")P(1+x/(2xx100))^(2n)`

where `x=5.3` and `n` is the count in years
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Answering the question")`

`P(1+x/(2xx100))^(2n)color(white)("ddd")->color(white)("ddd")$800(1+5.3/(2xx100))^(2xx4)`

`color(white)("dddddddddddddddddd")->color(white)("ddd")$800(205.3/200)^8 ~~986.1923...`

The amount owed is
`$986.1923...`
`ul($800.0000 larr" Subtract")`
`$186.1923...`

Rounding to 2 decimal places gives `$186.18` being owed in interest