Question

John deposited \$10,000 to open a new savings account that earned 4 percent annual interest, compounded quarterly. If there were no other transactions in the account, what was the amount of money in John's account 6 months after the account was opened?

• A. \$10,100
• B. \$10,101
• C. \$10,200
• D. \$10,201
• E. \$10,400

Hint:

For simple cases of compound interest (like two periods), you can often calculate it step-by-step without the full formula. Period 1 (3 months): Interest = 1% of \$10,000 = \$100. New balance = \$10,100. Period 2 (6 months): Interest = 1% of \$10,100 = \$101. New balance = \$10,100 + \$101 = \$10,201.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This is a compound interest problem. We need to calculate the future value of an investment with interest compounded more than once per year.
Step 2: Key Formula or Approach:
The formula for compound interest is: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] Where:

A = the future value of the investment/loan, including interest
P = the principal amount (\$10,000)
r = the annual interest rate (4% or 0.04)
n = the number of times that interest is compounded per year (quarterly = 4)
t = the time the money is invested for in years (6 months = 0.5 years)
Step 3: Detailed Explanation:
First, identify all the variables from the problem:

P = 10,000
r = 0.04
n = 4
t = 0.5
The interest rate per period is \(\frac{r}{n} = \frac{0.04}{4} = 0.01\).
The total number of compounding periods is \(nt = 4 \times 0.5 = 2\).
Now, substitute these values into the formula: \[ A = 10000 \left(1 + 0.01\right)^2 \] \[ A = 10000 (1.01)^2 \] Calculate \((1.01)^2\): \[ 1.01 \times 1.01 = 1.0201 \] Finally, calculate the total amount A: \[ A = 10000 \times 1.0201 = 10201 \] Step 4: Final Answer:
The amount of money in the account after 6 months is \$10,201.