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Help: Loan Repayments. (1 Viewer)

Fortify

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Mr Noonan borrowed $30000 in 1979 to buy his house in Woolooware. He was charged 11.5% interest and the loan was to be repaid over 30 years by yearly payments.
a. How much is needed to be paid each year? Done - $3586.92
b. How much of the loan will still be owing at the end of the 18th year? Done - $22743.33
c. If he decided to make repayments at the rate of $4500 per year, over what period will the loan be repaid ?

I'm a bit lost and confused with C. Can anyone help?
 

addikaye03

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just quickly looking it i think it will require the use of the logarithmic principle:
a^k=kloga
Solve for the unknown basically, in this case the # of years.

If noone, does a worked solution by tonight ill do it for you.
 

Fortify

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I got to:

4500 = 30000*(1.115)^n / { [ (1.115^n) - 1) / (1.115 - 1) ] }

An = 30000*(1.115)^n - m*{ [ a * (r^n) ] / [ r - 1 ] }

Where:
m = 4500
r = 1.115
a = 1
 
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Manboobs

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did you mean
An = 30000*(1.115)^n - m*{ [ a * (r^n - 1) ] / [ r - 1 ] }

so
An = 30000*(1.115)^n -:pirate: 4500*(1.115^n - 1)/(1.115-1)

with An being the amount.

The question is asking you: Over what period will the loan be repaid?
So they want to know when the amount goes to zero.


With that, you can now reorganise it so that you'll have

number = 1.115^n. :)

Hope this helps. Post your solution when you get the answer. If you need the solution i'll reluctantly post it.

*Make sure you read what the question asks and get the correct year. Most exam questions ask in what year was the loan fully paid back, or what year was the final payment made, so you usually round up.
 

Fortify

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Thanks for the help guys. I got finally got the question after reading the Log Laws part in my textbook!
 

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