To determine how much you need to deposit now in order to have a balance of Br. 10,000 on December 31, 2003, with an interest rate of 8% compounded annually, 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 initial amount of money).
- \( r \) is the annual interest rate (decimal).
- \( n \) is the number of years the money is invested or borrowed.
In this scenario:
- \( A = 10,000 \) Br (the future value)
- \( r = 0.08 \) (8% annual interest rate)
- The time period is from January 1, 1999, to December 31, 2003, which is 5 years (1999, 2000, 2001, 2002, 2003).
First, we need to rearrange the formula to solve for \( P \):
\[ P = \frac{A}{(1 + r)^n} \]
Substituting the known values into the formula:
\[ P = \frac{10,000}{(1 + 0.08)^5} \]
Calculating \( (1 + 0.08)^5 \):
\[ (1.08)^5 \approx 1.4693 \]
Now we calculate \( P \):
\[ P = \frac{10,000}{1.4693} \approx 6,805.59 \]
Thus, you must deposit approximately Br. 6,805.59 on January 1, 1999, to have a balance of Br. 10,000 on December 31, 2003, with an interest rate of 8% compounded annually.
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