Question

How much money will you need to invest initially to have $750.00 in 10 years and 8 months if the money is compounded daily at an annual rate of 2 1/2%?A.$585.90B. $574.45C. $576.33D. $574.60

Answer 1

Given:

The amount after 10 years and 8 months, A=$750.00.

The rate of interest, r =2 1/2 %.

The period of time, t =10 years and 8 months.

The interest is compounded daily

Required:

We need to find the intial investment amount.

Explanation:

Conver the period of time to years.

$1\text{ year =12 months.}$$\frac{8}{12}\text{ year =8 months.}$

$10\text{ +}\frac{8}{12}\text{ years =10 years and 8 months.}$

$10\text{ +}\frac{8}{12}\text{ years =10 years and 8 months.}$

$10\text{ }\times\frac{12}{12}\text{+}\frac{8}{12}\text{ years =10 years and 8 months.}$

$\frac{120}{12}\text{+}\frac{8}{12}\text{ years =10 years and 8 months.}$

$\frac{120+8}{12}\text{ years =10 years and 8 months.}$

$\frac{128}{12}\text{ years =10 years and 8 months.}$

We get t =128/12.

The annual interest rate is

$r=2.5\text{ \%.}$$r=0.025.$

The number of days in a year = 365 days.

The money is compounded daily, n=365.

Consider the formula to find the amount in compound interest.

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

Substitute A =750, r=0.025, n=365 and t =128/12 in the formula.

$750=P(1+\frac{0.025}{365})^{365\times\frac{128}{12}}$

$750=P(1+\frac{0.025}{365})^{3893.333}$

$P=\frac{750}{(1+\frac{0.025}{365})^{3893.333}}$$P=574.4516$

Final answer:

The initial amount is $ 574.45.