Question

One time deposit investment account
Account paid 3% interest compounded annually
No additional deposits or withdrawals were made
At the end of 2 yrs the account had a balance of $2,652.25
How much was customers one time deposit into investment account

Answer 1

To find the original amount deposited into the investment account (the principal), we can use the formula for compound interest:

\[
A = P(1 + r)^n
\]

Where:
- \( A \) is the amount of money accumulated after n years, including interest.
- \( P \) is the principal amount (the original amount of money).
- \( r \) is the annual interest rate (decimal).
- \( n \) is the number of years the money is invested or borrowed.

In this case:
- \( A = 2,652.25 \)
- \( r = 0.03 \) (3%)
- \( n = 2 \)

We need to find \( P \). Rearranging the formula to solve for \( P \) gives us:

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

Now, plug in the values:

\[
P = \frac{2,652.25}{(1 + 0.03)^2}
\]
\[
P = \frac{2,652.25}{(1.03)^2}
\]
\[
P = \frac{2,652.25}{1.0609}
\]
\[
P \approx 2,500.00
\]

Therefore, the customer's one-time deposit into the investment account was approximately **$2,500.00**.