Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 8 - Personal Finance - 8.4 Compound Interest - Exercise Set 8.4 - Page 521: 35

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.
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