Question

Xavier deposits $6 daily into an interest bearing account to save for renovations. The account earns 4.57% which compounds annually. What is the present value of the investment if Xavier renovates in five years?

Answer 1

To solve this, we are going to use the present value formula: $PV=P[ \frac{1-(1+ \frac{r}{n})^{-kt} }{ \frac{r}{n} } ]$
where
$PV$ is the present value 
$P$ is the periodic payment
$n$ is the number of times the interest is compounded per year $k$ is the number of payments per year $t$ is the number of years 
We know for our problem that Xavier deposits $6 daily into an interest bearing account, so $P=6$ and $k=365$. To convert the interest rate to decimal form, we are going to divide the rate by 100%
$r= \frac{4.57}{100} =0.0457$
Since the interest is compounded annually, it is compounded 1 time per year; therefore, $n=1$. We also know that Xavier renovates in five years, so 
Lets replace the values in our formula:
$PV=P[ \frac{1-(1+ \frac{r}{n})^{-kt} }{ \frac{r}{n} } ]$
$PV=6[ \frac{1-(1+ \frac{0.0457}{1})^{-(365)(5)} }{ \frac{0.0457}{1} } ]$
$PV=131.29$

We can conclude that the present value of Xavier's investment is $131.29