superannuation question (1 Viewer)

[kloudsurfer]

Hey,
Sorry to be a pain, but im stuck on these questions. I know how to do them, but i keep getting the answers wrong! Im not sure if its something I am doing or its the textbook (which has been known to be wrong!) and if it is me I dont know what it is!

Would someone be able to help me out?

1) 1) Micheal woks for a company that puts aside $1200 at the begining of each yr for his superannuation. If the money earns interest at the rate of 15% pa, find the amount of superannuation avaliable to micheal at the end of 30 yrs. (SOLVED ignore this question)
I got $749935.38, but the answer in the textbook says $599948.30

2) A sum of 1500 is invested at the end of each yr for 15 yrs at 14% p.a.. Find the amount of superannuation avaliable at the end of the 15 yrs.
I got $64263.62 but the texbook answer is $65763.62 (which is strangely close...)

Could someone please help me out here? Are my answers wrong?
Thanks in advance.

 

[zingerburger]

Shouldn't this be in the General Maths sub-forum?

Well, anyway isn't there a formula for this? I think I remember there being one when we did Consumer Arithemetic in Year 10... Help me so I can help you. Could you show me your calculations please?

 

[kloudsurfer]

Shouldn't this be in the General Maths sub-forum?

Well, anyway isn't there a formula for this? I think I remember there being one when we did Consumer Arithemetic in Year 10... Help me so I can help you. Could you show me your calculations please?

No its part of the series topic in maths because you have to use the compound interest formula but then you have to arrange them in a series and then you have to find the sum of the series...

Hang on ill post my working out to show you what i mean (my method is right i think, i might have made a little error or something)...wheres my maths book?

 

[zingerburger]

Hmm, I don't think I've done that topic yet... But I'll try.

 

[kloudsurfer]

In typing up my working for the first question i just found where i made an error! Ive got the answer to that one, thanks anyway!
But im still stuck on the second one...:

2) A sum of 1500 is invested at the end of each yr for 15 yrs at 14% p.a.. Find the amount of superannuation avaliable at the end of the 15 yrs.

Heres my working:

1st yr A= 1500(1+0.14)^14 (only 14 yrs because no interest is earned either the first or last yr ive forgotten)
2nd yr A= 1500 (1+0.14)^13

and so on...

final yr A = 1500 (1+0.14)^1

So then you get the series...
1500[1.14+1.14^2+...1.14^14)

find the sum of the series using Sn=(a(r^n-1))/(r-1)
a=1.14, r=1.14, n=14

Sn=(1.14((1.14^14)-1))/(1.14-1)
=$64263.62

But the answer says its $65763.62

 

[pLuvia]

You need to form a GP for these type of questions so first one it would look something like this;
1st year=(1200)1.15
2nd year=(1200(1.15))+1200)1.15
=1200(1.15)²+1200(1.15)
nth year=1200(1.15)ⁿ+1200(1.15)^n-1....+(1.15)

This is a GP as I said, nth year=1200*1.15*(1-1.15³⁰)/(1-1.15)=599948.30

2nd one
1st year=1500
2nd year=1500(1.14)
nth year=1500(1.14)ⁿ+1500(1.14)^n-1+...+1500

Similarly the GP is 1500(1-1.14¹⁵)/(1-1.14)=65763.62

kloudsurfer: Even if there isn't any interest added to the 1st year there will still be that 1500 amount that the person deposited

 

[kloudsurfer]

You need to form a GP for these type of questions so first one it would look something like this;
1st year=(1200)1.15
2nd year=(1200(1.15))+1200)1.15
=1200(1.15)²+1200(1.15)
nth year=1200(1.15)ⁿ+1200(1.15)^n-1....+(1.15)

This is a GP as I said, nth year=1200*1.15*(1-1.15³⁰)/(1-1.15)=599948.30

2nd one
1st year=1500
2nd year=1500(1.14)
nth year=1500(1.14)ⁿ+1500(1.14)^n-1+...+1500

Similarly the GP is 1500(1-1.14¹⁵)/(1-1.14)=65763.62

kloudsurfer: Even if there isn't any interest added to the 1st year there will still be that 1500 amount that the person deposited

Sorry! I found out I just made a silly mistake in that first one! Ive fixed it now. Thanks anyway!

Hmmm, as for the method, I used a different one, but its the same idea and i still got the same answer


Anyone wanna have a crack at question two? Or to read my working out and tell me where i went wrong?

Thanks in advance!

 

[zingerburger]

Glad to be of help without even trying.

Nah, sorry. We haven't done the topic yet and I really have no idea. But just in case you haven't noticed, your answer is exactly $1500 off from the "correct" answer. Would it have anything to do with the fact that $1500 is deposited each year? You might have forgotten to add that $1500.

 

[kloudsurfer]

kloudsurfer: Even if there isn't any interest added to the 1st year there will still be that 1500 amount that the person deposited

Ohhhhhh!!!!!!
Thanks, totally didnt think of that.
It makes sense now (even though i do not understand your method...wats a GP out of curiousity?)

Thanks heaps! And thanks zingerburger!

 

[SoulSearcher]

Ohhhhhh!!!!!!
Thanks, totally didnt think of that.
It makes sense now (even though i do not understand your method...wats a GP out of curiousity?)

Thanks heaps! And thanks zingerburger!

Geometric Progression

 

[kloudsurfer]

Geometric Progression

Right...am i supposed to know what that is at this point in 2U?

 

[SoulSearcher]

Right...am i supposed to know what that is at this point in 2U?

Basically, it's a geometric series

 

[kloudsurfer]

Basically, it's a geometric series

Eh good enough.

Anyway, Ive figured it all out. Thanks for your help guys!

 

[Forbidden.]

You need to form a GP for these type of questions so first one it would look something like this;
1st year=(1200)1.15
2nd year=(1200(1.15))+1200)1.15
=1200(1.15)²+1200(1.15)
nth year=1200(1.15)ⁿ+1200(1.15)^n-1....+(1.15)

This is a GP as I said, nth year=1200*1.15*(1-1.15³⁰)/(1-1.15)=599948.30

2nd one
1st year=1500
2nd year=1500(1.14)
nth year=1500(1.14)ⁿ+1500(1.14)^n-1+...+1500

Similarly the GP is 1500(1-1.14¹⁵)/(1-1.14)=65763.62

kloudsurfer: Even if there isn't any interest added to the 1st year there will still be that 1500 amount that the person deposited

That line I've marked in bold, does it even correspond to the formula S_n = a(rⁿ-1)/r-1 for |r| > 1 ?

 

[pLuvia]

That line I've marked in bold, does it even correspond to the formula S_n = a(rⁿ-1)/r-1 for |r| > 1 ?

That's only if the GP converges which it surely does not