Lesson One level more complicated non-standard problems on loans

One level more complicated non-standard problems on loans

Problem 1

The purchase price in this problem is $1500 plus the loaned amount.


About the loan, we know that the payments are $265 at the end of each month 
for 4 years at the annual interest rate of 9% compounded quarterly.


So, it is a non-standard loan payment/compounding scheme, because the payment schedule 
does not coincide with the compounding schedule.


But it is equivalent to (it works as) a standard payment schedule with quarterly payments 
of  3*265 = 795 dollars at the end of each quarter compounded at effective rate r = 0.09/4 
at the end of each quarter.


For this standard schedule scheme, we can use the formula for periodic payments for the loan 
at given conditions. The formula is

    Q = ,    (1)


where L is the loan amount; r =  is the effective interest rate per quarter;
n is the number of payments (same as the number of quarters, n = 4*4 = 16); 
Q is the quarterly payment of $795.


Substitute these values into the formula and get an equation for quarterly payments

    795 = .    (2)


We can calculate the coefficient (the factor in the formula) separately

     = 0.07511663.


Now from (1) we find L

    L =  = 10583.54 dollars (rounded to closest cent).


Thus the purchase price for the car was  $10583.54 + $1500 = $12083.54.


It is the ANSWER to question (a).



The amount paid back for this loan was  16 payments by 795 dollars = 16*795 = 12720 dollars.

The interest paid for the loan is the difference $12720 - $10583.54 = $2136.46


It is the ANSWER to question (b).