Question: How much must be deposited today into the following account in order to have $30,000 in 6 years for a down payment on a house? Assume no additional deposits are made.
An account with annual compounding and an APR of 8%
$[ ]$ should be deposited today.
(Do not round until the final answer. Then round to the nearest cent as needed.)
1. **State the problem:** We want to find the present value (the amount to deposit today) that will grow to $30,000 in 6 years with an annual interest rate of 8% compounded once per year.
2. **Formula used:** The formula for compound interest to find the future value is:
$$ A = P \left(1 + \frac{r}{n}\right)^{nt} $$
where:
- $A$ is the amount of money accumulated after $t$ years, including interest.
- $P$ is the principal (the initial deposit).
- $r$ is the annual interest rate (decimal).
- $n$ is the number of times interest is compounded per year.
- $t$ is the number of years.
Since we want to find $P$, rearrange the formula:
$$ P = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}} $$
3. **Given values:**
- $A = 30000$
- $r = 0.08$
- $n = 1$ (annual compounding)
- $t = 6$
4. **Substitute values:**
$$ P = \frac{30000}{\left(1 + \frac{0.08}{1}\right)^{1 \times 6}} = \frac{30000}{(1.08)^6} $$
5. **Calculate the denominator:**
$$ (1.08)^6 = 1.586874322 $$
6. **Calculate $P$:**
$$ P = \frac{30000}{1.586874322} $$
7. **Simplify with cancellation:**
$$ P = 30000 \times \frac{1}{1.586874322} $$
8. **Final calculation:**
$$ P \approx 30000 \times 0.629899 = 18896.97 $$
**Answer:**
You must deposit approximately **18896.97** today to have $30,000 in 6 years with 8% annual compounding interest.