Questions: Omar has just received a check for 32,595. This is a return from an investment that he made 18 years ago. He was told that the return was the equivalent of 11% per year. How much was his original investment? 2,954.70. 3,566.90. 5,760.98. 4,981.24.

Solution Steps

To find Omar's original investment, we need to use the formula for compound interest. The formula is:

\[ A = P (1 + r/n)^{nt} \]

where:

• \( A \) is the amount of money accumulated after n years, including interest.
• \( P \) is the principal amount (the initial amount of money).
• \( r \) is the annual interest rate (decimal).
• \( n \) is the number of times that interest is compounded per year.
• \( t \) is the time the money is invested for in years.

Given:

• \( A = 32595 \)
• \( r = 0.11 \)
• \( t = 18 \)
• \( n = 1 \) (since the interest is compounded annually)

We need to solve for \( P \).

1. Rearrange the compound interest formula to solve for \( P \).
2. Substitute the given values into the formula.
3. Calculate the value of \( P \) using Python.

We start with the compound interest formula:

\[ A = P (1 + r/n)^{nt} \]

To find the original investment \( P \), we rearrange the formula:

\[ P = \frac{A}{(1 + r/n)^{nt}} \]

We substitute the known values into the rearranged formula:

• \( A = 32595 \)
• \( r = 0.11 \)
• \( t = 18 \)
• \( n = 1 \)

This gives us:

\[ P = \frac{32595}{(1 + 0.11/1)^{1 \cdot 18}} = \frac{32595}{(1 + 0.11)^{18}} = \frac{32595}{(1.11)^{18}} \]

Now we calculate \( (1.11)^{18} \):

\[ (1.11)^{18} \approx 5.0802 \]

Then we compute \( P \):

\[ P \approx \frac{32595}{5.0802} \approx 4981.2389 \]

Final Answer