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Superannuation Q (1 Viewer)

Aysce

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Oh hai guys.

Im stuck on this Q and I don't recognise what im doing wrong.

Q. A school invests $5000 at the end of each school year at 6% p.a towards a new library. How much will the school have after 10 years?

I used A=P(1+r)^n, where r = 0.06, P = 5000 and n=10. Then I did n=9,8,7.. like you usually would, although obviously its wrong.

Don't judge me :cry:
 

Demento1

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Oh hai guys.

Im stuck on this Q and I don't recognise what im doing wrong.

Q. A school invests $5000 at the end of each school year at 6% p.a towards a new library. How much will the school have after 10 years?

I used A=P(1+r)^n, where r = 0.06, P = 5000 and n=10. Then I did n=9,8,7.. like you usually would, although obviously its wrong.

Don't judge me :cry:
Could I please know what your answer was and what the actual answer to the question is?
 

nightweaver066

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You included another years worth of investment. It's best to set out your working like below so you don't make any mistakes (don't remember formulae!)















 
Last edited:

Carrotsticks

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I agree with nightweaver066.

The best way to set our your answer is a nice clear and logical manner such that you can easily tell where you made a mistake (if you make one!).

It also prevents silly mistakes because you are not getting lost in any sort of 'crazy working out'.
 

Demento1

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You included another years worth of investment. It's best to set out your working like below so you don't make any mistakes (don't remember formulae!)















Good answer, though in my opinion you should say something like













This is just so that he knows he has to find 1.06 percent of the previous year's amount and that he knows where that 5000(1.06) + 5000 comes from.

Edit: I originally didn't know where the numbers came from until I was told they came from the previous. As nightweaver has correctly said, don't memorise the formula at all.
 

Timske

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Oh hai guys.

Im stuck on this Q and I don't recognise what im doing wrong.

Q. A school invests $5000 at the end of each school year at 6% p.a towards a new library. How much will the school have after 10 years?

I used A=P(1+r)^n, where r = 0.06, P = 5000 and n=10. Then I did n=9,8,7.. like you usually would, although obviously its wrong.

Don't judge me :cry:
A=P(1+r)^n

It's A = P(1 + r/100)^n

1 + r/100 = 1 + 6/100 = 1.06
 

Timske

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You included another years worth of investment. It's best to set out your working like below so you don't make any mistakes (don't remember formulae!)















Isn't a = 1.06?

Let An be the amount which the nth investment of $3000 accumulates.

First $5000 is invested at 6% p.a. for 10 years.
:. A1 = 5000(1.06)^10
Second $5000 is invested at 6% p.a. for 9 years.
:. A2 = 5000(1.06)^9
Third $5000 is invested at 6% p.a. for 8 years.
:. A3 = 5000(1.06)^8
Final $5000 is invested at 6% p.a. for 1 year.
:. A10 = 5000(1.06)^1
:. Total amount = A1 + A2 + A3 + ... + A10

= 5000(1.06)^10 + 5000(1.06)^9 + 5000(1.06)^8 + ... + 5000(1.06)^1

= 5000[(1.06)^10 + (1.06)^9 + (1.06)^8 + ... + (1.06)^1]

= 5000[(1.06)^1 + ... + (1.06)^8 + (1.06)^9 + (1.06)^10] The part in bold is a sum of a GP hence you apply the formula,
[a(r^n - 1)/(r - 1)]

Where: a = 1.06, r = 1.06, n = 10
 

nightweaver066

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Timske, the question says that the $5000 is invested at the end of each year.

So after 10 years (at the end of the 10th year), the school would have invested $5000 but this wouldn't gain any interest.

Demento1, that's a nicer way to write it when explaining it but i thought the OP would understand it. I wouldn't waste my time on that line in an exam situation :p
 

Aysce

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You included another years worth of investment. It's best to set out your working like below so you don't make any mistakes (don't remember formulae!)















Alright, I get where I went wrong, cheers mate.
 

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