To determine how much Bruce will owe after 5 years, we use the compound interest formula:
[tex]\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\][/tex]
where:
- [tex]\( P \)[/tex] is the principal amount (the initial amount of money),
- [tex]\( r \)[/tex] is the annual interest rate (in decimal form),
- [tex]\( n \)[/tex] is the number of times interest is compounded per year,
- [tex]\( t \)[/tex] is the time the money is invested for in years,
- [tex]\( A \)[/tex] is the amount of money accumulated after n years, including interest.
From the question:
- The principal ([tex]\( P \)[/tex]) is [tex]\( \\$1,000 \)[/tex].
- The annual interest rate ([tex]\( r \)[/tex]) is [tex]\( 10\% \)[/tex] or [tex]\( 0.10 \)[/tex] in decimal form.
- The loan compounds once per year, so [tex]\( n = 1 \)[/tex].
- Bruce waits for 5 years to begin paying back his loan, so [tex]\( t = 5 \)[/tex] years.
Substitute these values into the formula:
[tex]\[
A = 1000 \left(1 + \frac{0.10}{1}\right)^{1 \cdot 5}
\][/tex]
Simplify inside the parentheses first:
[tex]\[
A = 1000 \left(1 + 0.10\right)^5
\][/tex]
[tex]\[
A = 1000 \left(1.10\right)^5
\][/tex]
Now calculate [tex]\( 1.10^5 \)[/tex]:
[tex]\[
1.10^5 \approx 1.61051
\][/tex]
So,
[tex]\[
A = 1000 \cdot 1.61051
\][/tex]
[tex]\[
A \approx 1610.51
\][/tex]
Therefore, Bruce will owe approximately [tex]\(\$ 1610.51\)[/tex] after 5 years. This matches the third option provided:
[tex]\[
\boxed{1,610.51}
\][/tex]
