Question

A person invests 7500 dollars in a bank. The bank pays 5.5% interest
compounded annually. To the nearest tenth of a year, how long must the
person leave the money in the bank until it reaches 14300 dollars?
r
nt
A = P(1+)"
=
=
n

Answer 1

The number of years the person have to leave the money in the bank until it reaches 14300 dollars = 15.7years

Calculation of simple interest

The principal money invested (p) = 7500 dollars

The rate at which the interest is generated(r) = 5.5%

The interest generated (I) = 14300 - 7500

= 6,800

Therefore, time (t) =?

$si = \frac{p \times t \times r}{100}$

Make t the subject of formula

$t = \frac{si \times 100}{p \times r}$

$t = \frac{6800 \times 100}{7500 \times 5.5}$

$t = \frac{680000}{41250}$

t= 15.7 years

Therefore, the number of years the person have to leave the money in the bank until it reaches 14300 dollars = 15.7years

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