repayments (1 Viewer)

[Seraph]

hmm

A person invests $800 at the beginning of each year in a superannuation fund. Compound interest is
paid at 10% per annum on the investment. The first $800 is to be invested at the beginning of 1988
and the last is to be invested at the beginning of 2017. Calculate to the nearest dollar:
(i) the amount to which the 1988 investment will have grown by the beginning of 2018.
(ii) the amount to which the total investment will have grown by the beginning of 2018.

i get (i) pretty straight-forward superannuation question (although i didnt take note of 2018 at the start)

now i looked at the solution to (ii) , yea but i dotn really understand what this question is saying????
wth is the difference???? if its the total investment why is it just 800(1.01)^(30) ???

meh... i dislike the wording of these two questions !!!

 

[velox]

Hmmm, i dont think that the answer to the 2^ND question is right. Because if that was the answer it would'nt account for the $800 being deposited each year. Which book is it from?

 

[Brad]

If the last deposit doesnt accure interest can you just let A=1 in the geometric sum thingy?

 

[Binky]

Just a minute.. that's not loan repayments, that's investments.
loan repayments have M(bla bla bla) sorta thing

 

[lucyinthehole]

just a quick note on the whole "if it's invested at the beginning of each year, but they ask about the end of 40 years" or whatever, i read the markers notes for 2002 HSC on BOS, and they said that for a question like that, you still get full marks if you do everything correctly, but your "n" is off by one.

apparently maths marker want to give you marks.... go figure

i'm not saying you still shouldn't try to get the actual answr, just don't worry so much, and make sure you put down all working you can, cos you often can get heaps of marks just in the working, whether it's right or not

 

[Seraph]

Thats pretty odd

since IMO the hardest thing about time repayments superannuation etc is the years

(although now i understand it better )

 

[Seraph]

Hmmm, i dont think that the answer to the 2^ND question is right. Because if that was the answer it would'nt account for the $800 being deposited each year. Which book is it from?

HSC 1987

 

[velox]

lol ok, well that got me, i still dont know.

 

[velox]

Loan Repayment - Revision

A man borrows $5000 at an interest rate of 18% per annum where the interest is compounded monthly on the balance owing. The loan is to be repaid in equal monthly repayments. What should be the amount of each installment, if the loan is to be paid off in 4 years?

1. Let the amount of each monthly installment be $P
2.Let $A_N be the amount left owing after N months of paying it back

3. $A₁=5000 x 0.015 x 1 -P
= 5000(1+0.015) - P

4. $A₂ = [5000(1+0.015) - P] +5000(1+0.015) x 0.015 x 1 -P
= [5000(1+0.015) -P] [1+0.015] - P
= 5000(1+0.015) ² - p - [p(1+0.015)]
.
.
.
.
.
5. $A₄₈ = 5000(1+0.015) ⁴⁸ - p - p[(1+0.015)¹] + .............+(1+0.015)⁴⁷]

6. You want the total amount to equal 0, i.e he has paid off the loan. So you let that equal to zero

7. 0 = 5000(1+0.015) ⁴⁸ - p - p[(1+0.015)¹] + .............+(1+0.015)⁴⁷]

8. Rearrange the equation

9. 5000(1+0.015) ⁴⁸ - p = p[(1+0.015)¹] + .............+(1+0.015)⁴⁷]

10. Rearrange to make it P=

11. P = [5000(1+0.015)⁴⁸] / [1+(1+0.015)+.....+(1+0.015)⁴⁷]

12. Then use the Sum to N terms formula on the bottom and then divide to find P.


13. S<SUB>n</SUB> = (a(1-r<SUP>n</SUP>))/(1-r)

Voila!