prinipal is equal to 7000
interest rate = 6% per year compounded annually.
he pays off the loan with 1300 deposits at the end of each year.
here is what's happening.
all values are rounded to the nearest penny.
time period 0 is the start of the loan.
time period 1 is the end of the first year.
time period 2 is the end of the second year.
etc.
time period 0.
loan starts.
remaining balance is 7000
time period 1.
interest charged on 7000 = 6% of 7000 = 420
remaining balance becomes 7420
payment of 1300 is posted.
remaining balance becomes 7420 minus 1300 = 6120.
time period 2.
interest charged on 6120 = 6% of 6120 = 367.2
remaining balance becomes 6487.2
payment of 1300 is posted.
remaining balance becomes 6487.2 minus 1300 = 5187.20.
time period 3.
interest charged on 5187.2 = 6% of 5187.2 = 311.23.
remaining balance becomes 5498.43.
payment of 1300 is posted.
remaining balance becomes 5498.43. minus 1300 = 4198.43.
your solution is that he still owed 4198.43. at the end of the third year.
shown below is the loan carried to the end.
1300 is paid at the end of each month except for the last month.
in the last month, a payment of $913.42 is all that is required to close out the loan.
rembal int pmt
0 $7,000.00 $0.00
1 $6,120.00 $420.00 $1,300.00
2 $5,187.20 $367.20 $1,300.00
3 $4,198.43 $311.23 $1,300.00
4 $3,150.34 $251.91 $1,300.00
5 $2,039.36 $189.02 $1,300.00
6 $861.72 $122.36 $1,300.00
7 $0.00 $51.70 $913.42