To determine the initial investment \( C \) that will grow to $30,000 by the child's 10th birthday with an annual compounded interest rate of 7%, we use the formula for compound interest:
\[ A = P (1 + r/n)^{nt} \]
Where:
- \( A \) is the amount of money accumulated after n years, including interest.
- \( P \) is the principal amount (the initial investment).
- \( r \) 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.
In this scenario:
- \( A = 30,000 \)
- \( r = 0.07 \)
- \( n = 1 \) (since the interest is compounded annually)
- \( t = 10 \)
The formula simplifies to:
\[ 30,000 = C (1 + 0.07/1)^{1 \cdot 10} \]
\[ 30,000 = C (1.07)^{10} \]
Now calculate \( (1.07)^{10} \):
\[ (1.07)^{10} \approx 1.967151 \]
Next, solve for \( C \):
\[ 30,000 = C \cdot 1.967151 \]
\[ C = \frac{30,000}{1.967151} \]
\[ C \approx 15,247 \]
Rounding to the nearest dollar, the initial investment should be:
\[ C \approx 15,250 \]
Thus, the correct initial investment is approximately \( \boxed{15,250} \).
At the birth of a baby, a couple decides to make an initial investment of C at the rate of 7% compounded annually so that the amount will grow to $30,000 by her 10th birthday. What should their initial investment be? Round to the nearest dollars.
$10,000
$15,438
$15,250
$20,228
1 answer