Question

question 17
tony is offering two repayment plans to phil for a long - overdue loan. offer 1 is to receive a visit from an enforcer and the debt is due in full at once. offer 2 is to pay back $3,900 per year at a 20% interest rate until phil pays off the loan principal. phil owes tony $15,000. how many years will it take phil to pay off the loan if he takes offer 2? round your answer to the nearest year.
2 points
integer, decimal, or e notation allowed

Explanation:

Step1: Set up the loan - repayment formula

We use the simple - loan repayment formula for a fixed annual payment. Let $P$ be the principal amount of the loan, $A$ be the annual payment, $r$ be the interest rate, and $n$ be the number of years. The principal amount $P = 15000$, and the annual payment $A=3900$, and the interest rate $r = 0.2$. Each year, the interest on the remaining balance is charged, and then the payment is made.

We can also use the following approach. Let $B_0$ be the initial balance ($B_0 = 15000$), and $B_n$ be the balance after $n$ years. The balance after the first year $B_1=(1 + r)B_0−A$, after the second year $B_2=(1 + r)B_1−A$, and in general, $B_n=(1 + r)B_{n - 1}-A$.

We can solve this problem by considering the present - value of an ordinary annuity formula in a non - present - value context. We want to find $n$ such that the sum of the payments (with interest) pays off the loan.

We can also use a brute - force approach by calculating the remaining balance each year.

Step2: Calculate the remaining balance each year

Year 1:
The interest on the initial loan of $P = 15000$ at a 20% interest rate is $I_1=0.2\times15000 = 3000$. The amount owed after interest is $15000 + 3000=18000$. After making a payment of $A = 3900$, the remaining balance $B_1=18000 - 3900=14100$.

Year 2:
The interest on $B_1$ is $I_2=0.2\times14100 = 2820$. The amount owed after interest is $14100+2820 = 16920$. After making a payment of $3900$, the remaining balance $B_2=16920 - 3900 = 13020$.

Year 3:
The interest on $B_2$ is $I_3=0.2\times13020=2604$. The amount owed after interest is $13020 + 2604=15624$. After making a payment of $3900$, the remaining balance $B_3=15624 - 3900 = 11724$.

Year 4:
The interest on $B_3$ is $I_4=0.2\times11724 = 2344.8$. The amount owed after interest is $11724+2344.8 = 14068.8$. After making a payment of $3900$, the remaining balance $B_4=14068.8 - 3900 = 10168.8$.

Year 5:
The interest on $B_4$ is $I_5=0.2\times10168.8 = 2033.76$. The amount owed after interest is $10168.8+2033.76 = 12202.56$. After making a payment of $3900$, the remaining balance $B_5=12202.56 - 3900 = 8302.56$.

Year 6:
The interest on $B_5$ is $I_6=0.2\times8302.56 = 1660.512$. The amount owed after interest is $8302.56+1660.512 = 9963.072$. After making a payment of $3900$, the remaining balance $B_6=9963.072 - 3900 = 6063.072$.

Year 7:
The interest on $B_6$ is $I_7=0.2\times6063.072 = 1212.6144$. The amount owed after interest is $6063.072+1212.6144 = 7275.6864$. After making a payment of $3900$, the remaining balance $B_7=7275.6864 - 3900 = 3375.6864$.

Year 8:
The interest on $B_7$ is $I_8=0.2\times3375.6864 = 675.13728$. The amount owed after interest is $3375.6864+675.13728 = 4050.82368$. After making a payment of $3900$, the remaining balance $B_8=4050.82368 - 3900 = 150.82368$.

Year 9:
The interest on $B_8$ is $I_9=0.2\times150.82368 = 30.164736$. The amount owed after interest is $150.82368+30.164736 = 180.988416$. After making a payment of $180.99$ (rounding up), the loan is paid off.

Answer:

9