Question

Jerod hopes to earn $1200 in interest in 4.9 years time from $24,000 that he has available to invest. To decide if it's feasible to do this by investing In an account that compounds monthly, he needs to determine the annual interest rate such an account would have to offer for him to meet his goal. What would the annual rate of interest have to be? Round to two decimal places

Answer 1

Answer:

The annual interest rate would have to be of 0.1%.

Step-by-step explanation:

Compound interest:

The compound interest formula is given by:

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

Where A(t) is the amount of money after t years, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per year and t is the time in years for which the money is invested or borrowed.

Jerod hopes to earn $1200 in interest in 4.9 years time from $24,000 that he has available to invest.

This means that:

$A(4.9) = 1200 + 24000 = 25200$

$t = 4.9$

$P = 24000$

Compounded monthly:

This means that $n = 12$

What would the annual rate of interest have to be?

We have to solve for r, so:

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

$25200 = 24000(1 + \frac{r}{12})^{12*4.9}$

$(1 + \frac{r}{12})^{12*4.9} = \frac{25200}{24000}$

$(1 + \frac{r}{12})^{58.8} = 1.05$

$\sqrt[58.8]{(1 + \frac{r}{12})^{58.8}} = \sqrt[58.8]{1.05}$

$1 + \frac{r}{12} = (1.05)^{\frac{1}{58.8}}$

$1 + \frac{r}{12} = 1.00083$

$\frac{r}{12} = 0.00083$

$r = 12*0.00083$

$r = 0.001$

The annual interest rate would have to be of 0.1%.