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

2 Answers
Feb 28, 2018

$186.19$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)^tA=P(1+r/n×100)t

Where AA is the amount of interest, PP is the original amount, RR is the interest rate, NN is how many times interested per year and TT is the time.

Plugging in values:

A=$800(1+5.3/200)^8A=$800(1+5.3200)8

We change this to ^8^8 as semiannually means twice a year, and as this is after 44 years, we double it to get 8.8.

Plugging into calculator:

=$986.1923212=$986.1923212

Rounding to 2 d.p:

$986.19$986.19

Minusing the before value from the after value:

$986.19-800=$186.19$986.19800=$186.19

Feb 28, 2018

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

Explanation:

color(blue)("Some thoughts")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%5.3% becomes 5.3/2%5.32% at each calculation cycle.

The general form for this context is

color(white)("dddddddd")P(1+x/(2xx100))^(2n)ddddddddP(1+x2×100)2n

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

P(1+x/(2xx100))^(2n)color(white)("ddd")->color(white)("ddd")$800(1+5.3/(2xx100))^(2xx4)P(1+x2×100)2ndddddd$800(1+5.32×100)2×4

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