1. You owe $2,348.62 on a credit card with an 8.75% APR. You pay $300.00 toward the card at the beginning of the month and another $300.00 in a savings account at a 3.25% APR. What is the difference in interest accrued if you had paid $600.00 toward the card instead of $300.00? 2. Using the information from Problem 1, how many months will it take you to pay off your debt if you pay at least $600.00 at the beginning of each month? Include the last month even if the payment is less than $600.00.I don't understand these questions, please explain them to me?

Part 1:

After payment of $300, remaining balance = $2,348.62 - $300 = $2,048.62.
Interest accrued is given by:

[tex]I=Prt \\ \\ =2,048.62\times0.0875\times \frac{1}{12} \\ \\ =\$14.94[/tex]

Had it been $600 was paid, remaining balance = $2,348.62 - $600 = $1748.62. Interest accrued is given by:

[tex]I=1,748.62\times0.0875\times \frac{1}{12} \\ \\ =$12.75[/tex]

Difference in interest accrued = $14.94 - $12.75 = $2.19


Part 2:

The present value of an annuity is given by:

[tex]PV= \frac{P\left[1-\left(1+ \frac{r}{12} \right)^{-12n}\right]}{ \frac{r}{12} }[/tex]

Where PV is the amount to be repaid, P is the equal monthly payment, r is the annual interest rate and n is the number of years.

Thus,

[tex]2348.62= \frac{600\left[1-\left(1+ \frac{0.0875}{12}\right)^{-12n}\right]}{\frac{0.0875}{12}} \\ \\ \Rightarrow 1-(1+0.007292)^{-12n}= \frac{2348.62\times0.0875}{12\times600} =0.028542 \\ \\ \Rightarrow(1.007292)^{-12n}=1-0.028542=0.971458 \\ \\ \Rightarrow \log(1.007292)^{-12n}=\log0.971458 \\ \\ \Rightarrow-12n\log1.007292=\log0.971458 \\ \\ \Rightarrow-12n= \frac{\log0.971458}{\log1.007292} =-3.985559 \\ \\ \Rightarrow n= \frac{-3.985559}{-12} =0.332130[/tex]

Therefore, the number of months it will take to pay of the debt is 3.99 months which is approximately 4 months.