Question

Persevere with Problems Suppose Serena invests $2,500 for 3 years and 6 months and earns $328.13. What was the rate of interest as a percentage? Round your answer to the nearest hundredth of a percent.

Answer 1

To find the rate of interest earned on Serena's investment, we can use the formula for simple interest:

\[
I = P \times r \times t
\]

where:
- \( I \) is the interest earned,
- \( P \) is the principal amount (initial investment),
- \( r \) is the rate of interest (as a decimal),
- \( t \) is the time the money is invested (in years).

From the problem, we have:
- \( I = 328.13 \),
- \( P = 2500 \),
- \( t = 3 \text{ years} + 6 \text{ months} = 3.5 \text{ years} \) (since 6 months is half a year).

Now, we can rearrange the formula to solve for \( r \):

\[
r = \frac{I}{P \times t}
\]

Substituting the values we have:

\[
r = \frac{328.13}{2500 \times 3.5}
\]

Calculating the denominator:

\[
2500 \times 3.5 = 8750
\]

Now substitute this value back into the formula:

\[
r = \frac{328.13}{8750}
\]

Calculating the value of \( r \):

\[
r \approx 0.0375
\]

To convert this into a percentage, we multiply by 100:

\[
r \approx 0.0375 \times 100 \approx 3.75\%
\]

Therefore, the rate of interest as a percentage, rounded to the nearest hundredth of a percent, is:

\[
\boxed{3.75\%}
\]