Answer
The first investment would earn $\$649,083$ more than the second investment.
Work Step by Step
This is the formula we use when we make calculations with compound interest:
$A = P~(1+\frac{r}{n})^{nt}$
$A$ is the final amount in the account
$P$ is the principal (the amount of money invested)
$r$ is the interest rate
$n$ is the number of times per year the interest is compounded
$t$ is the number of years
We can find the total amount in the account $A_1$ after 30 years when we invest at a rate of 10% compounded annually.
$A = P~(1+\frac{r}{n})^{nt}$
$A_1 = (\$50,000)~(1+\frac{0.10}{1})^{(1)(30)}$
$A_1 = \$872,470.11$
After 30 years, there will be $\$872,470.11$ in the account.
We can find the total amount in the account $A_2$ after 30 years when we invest at a rate of 5% compounded monthly.
$A = P~(1+\frac{r}{n})^{nt}$
$A_2 = (\$50,000)~(1+\frac{0.05}{12})^{(12)(30)}$
$A_2 = \$223,387.22$
After 30 years, there will be $\$223,387.22$ in the account.
We can find the difference between the first investment and the second investment.
$A_1-A_2 = \$872,470.11-\$223,387.22$
$A_1-A_2 = \$649,083$
The first investment would earn $\$649,083$ more than the second investment.
