Question

Desmond deposits $50 into a savings account every month at an annual interest rate of

Answer 1

Given:

Desmond deposits $ 50 monthly.

Yearly he deposits = $50×12 = $ 600

Rate of interest compounded monthly = 4.7%

To find the amount he will receive after 10 years and the rate of change the value of his account after 10 years.

Formula

$A = P(1+\frac{r}{(n)(100)} )^{nt}$

where,

A be the final amount

P be the principal

r be the rate of interest

t be the time and

n be the number of times the interest is compounded.

Now,

Taking,

P = 600, r = 4.7, n = 12, t = 10 we get,

$A = 600(1+\frac{4.7}{(12)(100)}) ^{(12)(10)}$

or, $A = 600(1.0039)^{120}$

or, $A = 959.1$

Now,

At starting he has $ 600

At the end of 10 years he will be having $ 959.1

So,

The amount of change in his account = $ (959.1-600) = $ 359.1

Therefore the rate of change = $(\frac{359.1}{600} )(100) %$

= 59.85%

Hence,

a) His account will contain $ 959.1 after 10 years.

b) The rate of change in his account is 59.85% after 10 years.