Mathematics Loan Question (1 Viewer)

[SharkeyBoy]

Can you guys help? Thanks

Ari takes out a loan of $360 000. The loan is to be repaid in equal monthly repayments, $M, at the end of each month, over 25 years (300 months). Reducible interest is charged at 6% per annum, calculated monthly.
Let $A be the amount owing after the nth repayment.
(i) Write down an expression for the amount owing after two months, $A,
(ii) Show that the monthly repayment is approximately $2319.50.
(iii) After how many months will the amount owing, $An, become less than $180 000?

 

[stampede]

Why don't you first show us your own attempt and we'll see where we can help you.

 

[SharkeyBoy]

Why don't you first show us your own attempt and we'll see where we can help you.

Alright, fair enough:
(i) A = 360000(1+ 0.06/12)^2
(ii) ((1 + 0.06/12)^300 - 1) / (1+ 0.06/12 - 1) = 692.99
M = 360000 (1+0.06/12)^300 / 692.99
= 2319.498 = 2319.50
(iii)
180000 = 360000 - 2319.50M
M = 77.6 months = 78months

I'm pretty sure (iii) is wrong, but I just wanted to see how you guys work it out, because I don't fully get how to retrieve monthly repayments, it's just the way my textbook does it and it seems like too much to remember - is there any other way? Btw, I don't have the answers to the questions

 

[Drongoski]

Your (i) is already wrong; so, how did you get formula in (ii) from (i)?



Let $A(n) = amount still owing after nth payment.

A(1) = 360000 x 1.005 - M

A(2) = [360000 x 1.005 - M] x 1.005 - M = 36000 x 1.005^2 -M(1 + 1.005) = amount still owing after 2 months

A(3) = [360000 x 1.005^2 - M(1+1.005)]x1.005 - M = 360000 x 1.005^3 - M(1+1.005+1.005^2)

In this way, after n months, amount still owing:

A(n) = 360000 x 1.005^n - M(1+1.005+1.005^2 + . . . + 1.005^(n-1)) = 360000 x 1.005^n - M[1(1.005^n - 1)/(1.005-1)]

.: A(300) = 360000 x 1.005^300 - M(1.005^300 - 1)x 200 = 0

Solving: M = (360000/200)x(1.005^300/(1.005^300 - 1)) = 2319.4850...

(iii) solve for n: 360000x1.005^300 - 200M(1.005^300 - 1) < 180000

200M = 463897.0091 (approx: 463897)

So solve: (360000 - 463897)x1.005^n + 463897 < 180000

i.e. 103897x1.005^n > (463897 - 180000) = 283 897

i.e.1.005^n > 283897/103897 = 2.732485..

i.e. n > log 2.732 .../log 1.005 = 201.544 ...

So, if workings correct: answer should be 202 months.

 

[SharkeyBoy]

Your (i) is already wrong; so, how did you get formula in (ii) from (i)?



Let $A(n) = amount still owing after nth payment.

A(1) = 360000 x 1.005 - M

A(2) = [360000 x 1.005 - M] x 1.005 - M = 36000 x 1.005^2 -M(1 + 1.005) = amount still owing after 2 months

A(3) = [360000 x 1.005^2 - M(1+1.005)]x1.005 - M = 360000 x 1.005^3 - M(1+1.005+1.005^2)

In this way, after n months, amount still owing:

A(n) = 360000 x 1.005^n - M(1+1.005+1.005^2 + . . . + 1.005^(n-1)) = 360000 x 1.005^n - M[1(1.005^n - 1)/(1.005-1)]

.: A(300) = 360000 x 1.005^300 - M(1.005^300 - 1)x 200 = 0

Solving: M = (360000/200)x(1.005^300/(1.005^300 - 1)) = 2319.4850...

(iii) solve for n: 360000x1.005^300 - 200M(1.005^300 - 1) < 180000

200M = 463897.0091 (approx: 463897)

So solve: (360000 - 463897)x1.005^n + 463897 < 180000

i.e. 103897x1.005^n > (463897 - 180000) = 283 897

i.e.1.005^n > 283897/103897 = 2.732485..

i.e. n > log 2.732 .../log 1.005 = 201.544 ...

So, if workings correct: answer should be 202 months.

Thanks so much!
Like I said, I was looking at different examples from my textbook to see how to answer this question, so that's why I have retrieved different formulas...

 

[Yeeeah]

WHERE DID YOU GET 200 from in III

 

[Drongoski]

1/(1.005-1) = 200