Question

How much money do you have to put into a bank account that pays \(10 \%\) interest compounded annually to have \(\$ 10,000\) in ten years?

Short answer

To have \(\$ 10,000\) in ten years with a 10% annual interest rate compounded annually, you need to initially deposit approximately \(\$ 3855.43\) into the bank account.

Step-by-step solution

Understand the compound interest formula

The compound interest formula is given by: \(A = P (1 + \frac{r}{n})^{nt}\), where: A - the final balance after t years P - the initial principal (amount of money deposited) r - the annual interest rate in decimal n - the number of times interest is compounded per year t - number of years In this exercise, we have: A = $10,000 r = 10% = 0.1 (converted to decimal) n = 1 (as the interest is compounded annually) t = 10 years Our goal is to find the initial principal (P) that satisfies this equation.

Rearrange the formula to solve for P

We need to find the value of P, so we will rearrange the formula as: \(P = \frac{A}{(1 + \frac{r}{n})^{nt}}\) Now, we will plug in the values we have into the formula.

Substitute the values and solve

Substitute the values into the formula to find P: \(P = \frac{10000}{(1 + \frac{0.1}{1})^{1*10}}\) Simplify the expression: \(P = \frac{10000}{(1 + 0.1)^{10}}\) \(P = \frac{10000}{(1.1)^{10}}\) Now, calculate the value of P: \(P ≈ 3855.43\)

Interpret the result

The result shows that an initial deposit of about \( \ \$3855.43\) is needed to have \(\$10,000\) in ten years with a 10% interest rate compounded annually.