An investment of $1800 earned $60.75 at a simple-interest rate of 4.5% per year. For how long was the investment held?

To determine the time for which the investment was held, we can use the simple interest formula:

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

Where:
- $ I $ is the interest earned.
- $ P $ is the principal amount (the initial amount of money).
- $ r $ is the annual interest rate (decimal).
- $ t $ is the time the money is invested for in years.

Given:
- $ I = 60.75 $
- $ P = 1800 $
- $ r = 4.5\% = 0.045 $

We need to solve for $ t $. Rearranging the formula to isolate $ t $:

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

Plugging in the given values:

\[ t = \frac{60.75}{1800 \times 0.045} \]
\[ t = \frac{60.75}{81} \]
\[ t = 0.75 \]

So, the investment was held for $ 0.75 $ years, which is equivalent to 9 months.

Would you like more details or have any questions?

Here are 5 more questions you can ask:

1. How do you convert years into months when dealing with interest calculations?
2. What is the difference between simple interest and compound interest?
3. How would the calculation change if the interest rate was compounded monthly?
4. How do you find the principal amount if the interest earned, rate, and time are known?
5. Can you explain the impact of increasing the interest rate on the amount of interest earned?

**Tip:** Always ensure that the interest rate and time period are in the same units (e.g., both in years) when using the simple interest formula.

Learn how to calculate the time an investment was held using the simple interest formula. This problem involves an investment of $1800 earning $60.75 at an annual interest rate of 4.5%. Suitable for Grades 7-9.

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