SOLUTION: For how many years will Prasad make payments on the $28,000 he borrowed to start his machine shop if he makes payments of $3400 at the end of every three months and interest is 8.0

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: For how many years will Prasad make payments on the $28,000 he borrowed to start his machine shop if he makes payments of $3400 at the end of every three months and interest is 8.0      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 1207428: For how many years will Prasad make payments on the $28,000 he borrowed to start his machine shop if he makes payments of $3400 at the end of every three months and interest is 8.08% compounded semi-annually?
Answer by ikleyn(52794) About Me  (Show Source):
You can put this solution on YOUR website!
.
For how many years will Prasad make payments on the $28,000 he borrowed to start his machine shop
if he makes payments of $3400 at the end of every three months and interest is 8.08% compounded semi-annually?
~~~~~~~~~~~~~~~~~~~~~~~


        The scheme of payments/compounding is non-standard and it is not totally clear how to it really works.

        So,  in my solution I will make some assumptions,  but I am not sure if they are adequate.

        In any case,  I will try.


It is equivalent to semi-annual payments on the loan with semi-annual payments of 2*3400 = 6800 dollars,
while the loan is compounded semi-annually.

 
So, use the formula for semi-annual payments for a loan

    M = P%2A%28r%2F%281-%281%2Br%29%5E%28-n%29%29%29


where P is the loan amount; r = 0.0808%2F2 is the effective interest rate semi-annually;
n is the number of payments (same as the number of semi-annual periods); 
M is the semi-annual payment of $6800.


In this problem  M = 6800;  P = $28000;  r = 0.0808%2F2.


Substitute these values into the formula and get for semi-annual payments

   6800 = 28000%2A%28%28%280.0808%2F2%29%29%2F%281-%281%2B0.0808%2F2%29%5E%28-n%29%29%29.


We should find "n" from this equation.

Divide both sides by 28000.  You will get

    6800%2F28000 = %28%280.0808%2F2%29%29%2F%281-%281%2B0.0808%2F2%29%5E%28-n%29%29,

or

    0.242857143 = %28%280.0808%2F2%29%29%2F%281-%281%2B0.0808%2F2%29%5E%28-n%29%29.


It implies, step by step

    0.242857143%2A%282%2F0.0808%29 = 1%2F%281-%281%2B0.0808%2F2%29%5E%28-n%29%29,

    6.011315417 = 1%2F%281-%281%2B0.0808%2F2%29%5E%28-n%29%29,

    1%2F6.011315417 = 1-%281%2B0.0808%2F2%29%5E%28-n%29,

    0.166352941 = 1-%281%2B0.0808%2F2%29%5E%28-n%29,
    
    %281%2B0.0808%2F2%29%5E%28-n%29 = 1 - 0.166352941,

    %281%2B0.0808%2F2%29%5E%28-n%29 = 0.833647059,

    %281%2B0.0808%2F2%29%5En = 1%2F0.833647059,

    %281%2B0.0808%2F2%29%5En = 1.199548405.


Take logarithm base 10

    n*log(1+0.0808/2) = ln(1.199548405)

and find "n"

    n = log%28%281.199548405%29%29%2Flog%28%281%2B0.0808%2F2%29%29 = 4.59.


From these calculations, I make the conclusion that input data in the problem are incorrect,
since they lead to the non-integer number of semi-annual payments.