Question

A person invest $3000 in a bank the bank pays 6.75% interest compounded semi annually to the nearest 10th of a year how long must a person leave the money in the bank until it reaches $6700

Answer 1

Answer:

It would take 12.1 years.

Step-by-step explanation:

We are given with $3000 invested in bank pays 6.75% interest compounded semi annually to the nearest 10th. We are asked to find the number of years the bank reaches $6700.

Let's use the compound interest formula.

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

We know A=6700

P=3000

r =0.0675

n=2(because compounded semiannually)

t =?( we need to find it)

Plug in the known values into the formula.

$6700=3000(1+\frac{0.0675}{2}) ^{2*t}$

Simplify and solve for 't'

Divide both sides by 3000

$2.23333= (1.03375)^{2t}$

Take log on both sides

$log(2.23333) = 2t log(1.03375)$

Divide both sides by log(1.03375)

24.2067=2t

Divide both sides by 2.

t=12.10...

To round to nearest 10th, we get t=12.1 years.