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

You need to deposit approximately \(\$3855.40\) into the bank account to have \(\$10,000\) in ten years with a \(10\%\) interest rate compounded annually.

Step-by-step solution

Convert interest rate to decimal form

In the given problem, the annual interest rate is \(10\%\). To convert it into a decimal, divide it by 100: \[ r = 10/100 = 0.1 \] So, the interest rate as a decimal is 0.1.

Rearrange the compound interest formula to solve for P

We need to find the initial deposit (P) using the compound interest formula: \[ A = P(1 + r/n)^{nt} \] Since we are given that interest is compounded annually, n = 1. We need to rearrange the formula to solve for P: \[ P = \frac{A}{(1 + r/n)^{nt}} \] Now we have the formula to find the initial deposit amount.

Plug in the given values and solve for P

In the problem, we are given: - A = $10,000 - r = 0.1 (converted from \(10\%\)) - n = 1 (compounded annually) - t = 10 years Plug in these values into the formula: \[ P = \frac{10,000}{(1 + 0.1/1)^{1 \cdot 10}} \] Simplify the formula: \[ P = \frac{10,000}{(1 + 0.1)^{10}} \] Calculate the result: \[ P \approx 3855.40 \] Therefore, you need to deposit approximately \(\$3855.40\) into the bank account to have \(\$10,000\) in ten years with a \(10\%\) interest rate compounded annually.