Help with math question (1 Viewer)

[gosfa747]

I'm doing a couple of question and i seem to get it wrong every time I'm doing it. I don't know why but can someone please answer this question for me

Ooh BTW this is superannuation in series

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

The second question is an employee contributes $200 into superannuation fund at the end of each year. If the interest rate on this fund is 11.5% p.a.., how much will the employee have at the end of 20 years


I would really appreciate it if someone could answer it

 

[bluecrisps]

Dunno if question one is a GP or AP and not sure if the school invests 5000$ then gets 6% interest and also adds another extra 5000 per year. or it could just say that the school got $5000 and is getting 6% interest, i will assume the latter cos it's easier for me to do (even though I think it's probably the first case). [what i just said was it could either be a summing of sequences question i.e. adding up all the terms or just a normal a.p. or g.p. question]

If it's A.P. < simple interest > then use Tn= a + (n-1)d
the common difference, d, would 5000 * 6% = $300.00
a is the starting term which would be $5000.00
Tn = 5000 + (n-1) * 300
T10 = 5000 + (10-1) * 300
T10 = $7700.00 --- --- --- so the school would have $7700.00 after 10 years

If it was a G.P. meaning compound interest
Tn = ar^(n-1)
a = $5000.00
So the ratio in this case would be 1.06
Tn = 5000* (1.06)^(n-1)
T10 = 5000 * (1.06)^9
= 8447.394795 = 8447.39 (nearest cent) would be saved after 10 years

Q2. try it out
If it's simple interest use Tn = a + (n-1)d
If it's compound interest use Tn = a r^(n-1)
[r is the common ratio, in this case it would be 1.115 (i coverted the percentage into decimal point)]
a = 200

lol im probably wrong anyways so wait for someone else to answers

 

[bluecrisps]

ok now that i think about it im probably wrong cos for super you put in money each year AND get interest so it's a summing of sequences........... sigh

 

[Menomaths]

((1.06¹⁰-1)/.06)*5000=$65903.97

200((1.115²⁰-1)/.115)=$13601.02

 

[Zeref]

((1.06¹⁰-1)/.06)*5000=$65903.97

wouldnt ((1.06¹⁰-1)/.06) be multiplied by 5000(1.06)?

200((1.115²⁰-1)/.115)=$13601.02

again, wouldn't ((1.115²⁰-1)/.115) be multiplied by 200(1.115)?

 

[Menomaths]

((1.06¹⁰-1)/.06)*5000=$65903.97

200((1.115²⁰-1)/.115)=$13601.02

wouldnt ((1.06¹⁰-1)/.06) be multiplied by 5000(1.06)?




again, wouldn't ((1.115²⁰-1)/.115) be multiplied by 200(1.115)?

No, because the question says 'end' which means he put the money at the end of the year therefore he does not earn interest, thus a = 500 (1)
since the sequence is 500 (1+1.06+1.06^2...)

 

[Zeref]

No, because the question says 'end' which means he put the money at the end of the year therefore he does not earn interest, thus a = 500 (1)
since the sequence is 500 (1+1.06+1.06^2...)

ahhh i see