superannuation help please :S (1 Viewer)

[bachviete]

I cant seam to understand how to solve these types of superannuation problems, i would really appreciate if you could show me the complete way please =) 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? Thankyou

 

[biopia]

This question is actually quite easy =] It shouldn't be too hard to understand lol.
When I see these kinds of questions, I like to create a formula for how much money will be the account after n years. To do this, you need to assess the information given.
Let A1 be the amount of money in the account after 1 year. As the question says "at the end of every year" A1 = 5000
A2 (The amount after the 2nd year) will have the interest applied to it so:
A2 = A1(1 + 6/100) + 5000 (The extra 5000 is because they've deposited another 5000 at the end of the second year.
A2 = 5000(1 + 6/100) + 5000
Similarly for A3:
A3 = A2(1 + 6/100) + 5000
Expanding gives:
A3 = 5000(1 + 6/100)^2 + 5000(1 + 6/100) + 5000
Hopefully by now, a pattern should start to emerge. It looks as if the first term in the sum is always one less power than the year, and the next term is one less than that, and the next one is one less than that until you get to a power of 1.
So:
An = 5000(1 + 6/100)^(n - 1) + 5000(1 + 6/100)^(n - 2) ... + 5000^1
If you take the common factor of 5000 out, you get
An = 5000[1 + 6/100)^(n - 1) + (1 + 6/100)^(n - 2) + ... + 1]
If you reverse the order of the terms in the brackets, to make the 1 first, hopefully you should notice the sum is taking on the form of a geometric series where a=1 r=(1 + 6/100). First though, substitute for n=10.
A10 = 5000[1 + (1 + 6/100) + ... + (1 + 6/100)^9]
Notice how the brackets have 10 terms even though the highest power is 9.
Use the geometric sum formula where r>1:
A10 = 5000[{1(1 + 6/100)^10 - 1}/{(1 + 6/100) - 1}]
And if all is done correctly, you should get an answer of $65903.97 which makes sense, because if 5000 dollars is donated for 10 years, the answer has to be at least 50000.

I hope I haven't made any mistakes lol. Still, the process is the same every time.
Good luck for Tuesday!

 

[bachviete]

(

This question is actually quite easy =] It shouldn't be too hard to understand lol.
When I see these kinds of questions, I like to create a formula for how much money will be the account after n years. To do this, you need to assess the information given.
Let A1 be the amount of money in the account after 1 year. As the question says "at the end of every year" A1 = 5000
A2 (The amount after the 2nd year) will have the interest applied to it so:
A2 = A1(1 + 6/100) + 5000 (The extra 5000 is because they've deposited another 5000 at the end of the second year.
A2 = 5000(1 + 6/100) + 5000
Similarly for A3:
A3 = A2(1 + 6/100) + 5000
Expanding gives:
A3 = 5000(1 + 6/100)^2 + 5000(1 + 6/100) + 5000
Hopefully by now, a pattern should start to emerge. It looks as if the first term in the sum is always one less power than the year, and the next term is one less than that, and the next one is one less than that until you get to a power of 1.
So:
An = 5000(1 + 6/100)^(n - 1) + 5000(1 + 6/100)^(n - 2) ... + 5000^1
If you take the common factor of 5000 out, you get
An = 5000[1 + 6/100)^(n - 1) + (1 + 6/100)^(n - 2) + ... + 1]
If you reverse the order of the terms in the brackets, to make the 1 first, hopefully you should notice the sum is taking on the form of a geometric series where a=1 r=(1 + 6/100). First though, substitute for n=10.
A10 = 5000[1 + (1 + 6/100) + ... + (1 + 6/100)^9]
Notice how the brackets have 10 terms even though the highest power is 9.
Use the geometric sum formula where r>1:
A10 = 5000[{1(1 + 6/100)^10 - 1}/{(1 + 6/100) - 1}]
And if all is done correctly, you should get an answer of $65903.97 which makes sense, because if 5000 dollars is donated for 10 years, the answer has to be at least 50000.

I hope I haven't made any mistakes lol. Still, the process is the same every time.
Good luck for Tuesday!

I think i see where i went wrong. When it says at the end of every year i think of it as that they put the money at the start, and hence i get the first term of the GP as 5000 x 1.06 instead of 5000. Thankyou and good luck with your HSC exams too =)

 

[tonyharrison]

Re: (

I think i see where i went wrong. When it says at the end of every year i think of it as that they put the money at the start, and hence i get the first term of the GP as 5000 x 1.06 instead of 5000. Thankyou and good luck with your HSC exams too =)

Yeah i did the same things as well. Except, annoyingly, my question didn't specify when the interest was compounded...

Does that mean that for every superannuation question, (A1 = principle amount) and (A2 = principle amount x interest + investment) regardless of whether the question specifies when the interest is compounded?