Math Problem Statement

You have a credit card that has a balance of $9750 at an APR of 18.99%. You plan to pay $500$ each month in an effort to clear the debt quickly. How many months will it take you to pay off the balance?

Solution

To calculate how many months it will take to pay off the credit card balance of $9,750 at an APR of 18.99%, with monthly payments of $500, we can use the following formula for paying off a loan with fixed payments:

M=log(1Prn)log(1+r)M = \frac{\log\left(1 - \frac{P \cdot r}{n}\right)}{\log\left(1 + r\right)}

Where:

  • MM is the number of months it will take to pay off the debt.
  • PP is the monthly payment ($500).
  • BB is the balance ($9,750).
  • rr is the monthly interest rate (APR divided by 12).
  • nn is the balance ($9,750).

The monthly interest rate rr is given by:

r=18.99%12=0.01899÷12=0.015825 per month.r = \frac{18.99\%}{12} = 0.01899 \div 12 = 0.015825 \text{ per month}.

Now we can use these values to calculate the number of months needed to pay off the debt. I'll perform the calculations.It will take approximately 24 months (or 2 years) to pay off the credit card balance of $9,750 with monthly payments of $500 at an APR of 18.99%.

Would you like more details or have any questions about this calculation?

Here are 5 related questions to expand on the concept:

  1. What happens if the APR increases or decreases by a few percentage points?
  2. How much total interest would you end up paying over the 24 months?
  3. How does making a higher monthly payment, like $600, affect the time to pay off the balance?
  4. Can you calculate the balance left after paying for 12 months?
  5. What are the advantages of paying more than the minimum required?

Tip: Paying more than the minimum monthly payment can significantly reduce the total interest paid over time.

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Math Problem Analysis

Mathematical Concepts

Algebra
Logarithms
Compound Interest

Formulas

M = log(1 - P * r / n) / log(1 + r)
r = APR / 12

Theorems

Compound interest calculation

Suitable Grade Level

College level or Grades 11-12