$5000 is invested at 8% interest compounded twice a year. How much… Step by Step

Step by step

The question

Question: $5000 is invested at 8% interest compounded twice a year.

How much is in the account after 3 years?

1 Read the numbers

P is the money put in: $5000.

P = $5000: the money put in

r is the rate.

8% is 8/100: 0.08.

r = 8% = 0.08

n is how many times a year the interest is added: twice a year, so n = 2.

n = 2: twice a year

t is the time in years: 3.

t = 3 years

The amount after t years is A = P times (1 + r/n) to the power nt.

Each time, the money is multiplied by 1 + r over n.

A = P(1 + rn)^nt

Careful!

The 8% is for a whole year.

Each time we add only r/n: 0.08 divided by 2.

Careful: not 5000 × 1.08⁶ ✗

2 Put them in

We put the numbers in.

A = 5000(1 + 0.082)^{2 × 3}

3 Work it out

Inside the bracket first: 0.08 divided by 2 is 0.04.

The power: 2 times 3 is 6.

0.08 ÷ 2 = 0.04

2 × 3 = 6

1 + 0.04 is 1.04.

So A is 5000 times 1.04 to the power 6.

A = 5000 × 1.04⁶

With a calculator, that power is about 1.26532.

1.04⁶ ≈ 1.26532

Now times 5000.

A ≈ 5000 × 1.26532

6326.5951...

Money is rounded to cents: about $6326.60.

A ≈ $6326.60

4 Check

Let's check by going back: 6326.60 divided by 1.26532 is about $5000, the money put in.

$6326.60 ÷ 1.26532 ≈ $5000

And it makes sense.

Simple interest would give $6200.

Interest on interest gives a bit more.

So the answer is right!

simple interest: $5000 + $1200 = $6200 < $6326.60

Let's look back.

First: read the numbers.

Then: put them in.

Then: work it out.

Last: check.

Answer