Superannuation question (1 Viewer)

[sinophile]

Ive tried this question seven times and they're all wrong, jesus christ someone please go through the working out as if you were explaining to an idiot please.

John puts $2000 into a superannuation account on his 40th birthday. He continues to do this on his birthday up to and including his 60th birthday. The interest he earns is 10% per annum compunded yearly.

On his 61st birthday he deposits the accumulated amount into an account which earns 8% per annum compunded yearly.

He will colect his accumulated amount on his 65th birthday.

i) How much does the first $2000 accumulate to when John celebrates his 61st birthday?
ii) How much will John collect on his 65th birthday?

 

[addikaye03]

Ive tried this question seven times and they're all wrong, jesus christ someone please go through the working out as if you were explaining to an idiot please.

John puts $2000 into a superannuation account on his 40th birthday. He continues to do this on his birthday up to and including his 60th birthday. The interest he earns is 10% per annum compunded yearly.

On his 61st birthday he deposits the accumulated amount into an account which earns 8% per annum compunded yearly.

He will colect his accumulated amount on his 65th birthday.

i) How much does the first $2000 accumulate to when John celebrates his 61st birthday?
ii) How much will John collect on his 65th birthday?

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

i) On John's 40th birthday he puts $2000, the interest is COMPOUND annually.
Since this first amount stays in the bank for 20 years:

A1=2000(1.1)^21.

ii) John puts the same amount in the second year ( $2000) but since he pulls it all out after his 60th birthday this 2nd $2000 dollars will only be in the bank for 19 years.

A2=2000(1.1)^20

Same idea for 3rd year, this amount would be in the bank for 18 years

A3=2000(1.1)^19

Since 21 years is the nth term

A21(TOTAL)=[2000(1.1)^21]+[2000(1.1)^20]+[2000(1.1)^19+...+[2000(1.1)]

Notice this is an Geometric series: a=2000(1.1) r=(1.1)

Sn=a(r^n-1)/r-1 therefore 2000(1.1)[(1.1)^21-1]/0.1

S20=$140805.5 ( near dollar)

So then he put this money into an account which has 8% interest, calculated annually for 4 years ( Since he collects on his 61st birthday, then theres his 62,63,64,65 birthday till he withdraws)

A=140805.5(1.08)^4

A=$191564.33#

Sorry about that i misread and was a term out lol there ya go

 

[sinophile]

The answer are i)$14800.50 and ii)$191564.33

Either the answer is wrong, or you missed out on something.

 

[lyounamu]

The answer are i)$14800.50 and ii)$191564.33

Either the answer is wrong, or you missed out on something.

i)

The answer should be

140805.4988...

not 14800.50 from what I see here because that's too little. Think about it, you put 2000 every year for 20 years and it is compounded yearly. You gotta have much more than that.

My working out is quite similar to addikaye's.

But it's like this:

1st: 2000 x 1.1
2nd : 2000 x 1.1^2 + 2000 x 1.1
21st: 2000 x 1.1^21 + 2000 x 1.1^20 +....+ 2000 x 1.1 = (2000(1.1)(1.1^21-1))/0.1 = 140805.498....

ii)

140805.498... x 1.08^4 (because it is compunded for 4 years) = 191564.3266...

 

[addikaye03]

i)

The answer should be

140805.4988...

not 14800.50 from what I see here because that's too little. Think about it, you put 2000 every year for 20 years and it is compounded yearly. You gotta have much more than that.

My working out is quite similar to addikaye's.

But it's like this:

1st: 2000 x 1.1
2nd : 2000 x 1.1^2 + 2000 x 1.1
21st: 2000 x 1.1^21 + 2000 x 1.1^20 +....+ 2000 x 1.1 = (2000(1.1)(1.1^21-1))/0.1 = 140805.498....

ii)

140805.498... x 1.08^4 (because it is compunded for 4 years) = 191564.3266...

thats what i thought, 14800.50 is def not right. I mean if u put $2000 into a bank account that had 0% interest for 20 years, it would still be $40000 lol John is either dipping into his super illegally, or his interest is negative.

 

[lyounamu]

thats what i thought, 14800.50 is def not right. I mean if u put $2000 into a bank account that had 0% interest for 20 years, it would still be $40000 lol John is either dipping into his super illegally, or his interest is negative.

lol

We should have such scenarios in HSC maths.

 

[addikaye03]

lol

We should have such scenarios in HSC maths.

who knows how low the banks go these days, and im pretty sure John should start putting like $1000/week away for his upcoming court order lol

 

[sinophile]

Can someone pelase explain what im doing wrong? I keep getting $2000 less than you guys.

My working out is:

A1= 2000x1.1
A2= 2000x1.1^2 + 2000
A3= 2000x1.1^3 + 2000x1.1 + 2000
factorise out the 2000, leave the first 'chunk' out:
A3= 2000x1.1^3 2000(1+1.1)

etc. etc. We can see a pattern, make an expression for An and use the sum of GP

A21= 2000x1.1[ (1.1^(21)-1)/0.1]
=$140805.50

 

[lyounamu]

Can someone pelase explain what im doing wrong? I keep getting $2000 less than you guys.

My working out is:

A1= 2000x1.1
A2= 2000x1.1^2 + 2000
A3= 2000x1.1^3 + 2000x1.1 + 2000
factorise out the 2000, leave the first 'chunk' out:
A3= 2000x1.1^3 2000(1+1.1)

etc. etc. We can see a pattern, make an expression for An and use the sum of GP

A21= 2000x1.1[ (1.1^(21)-1)/0.1]
=$140805.50

uh...you are actually right though.

 

[sinophile]

OH SHIT sorry I mistyped the last step

A21= (2000X1.1^(21))x2000[(1.1^(21)-1)/0.1]
=$142805.50

what

 

[lyounamu]

OH SHIT sorry I mistyped the last step

A21= (2000X1.1^(21))x2000[(1.1^(21)-1)/0.1]
=$142805.50

what

It should be
A21= 2000x1.1[(1.1^(21)-1)/0.1]
=$140805.50

 

[sinophile]

Why? I dont get it.

 

[lyounamu]

Why? I dont get it.

Your series is incorrect.

It should be like:

A1 = 2000 x 1.1
A2 = 2000 x 1.1^2 + 2000 x 1.1

Then you will understand how I got that out.

 

[addikaye03]

Do it the way i done it, its more logical if u do it the reverse way i think. I.e the first 2000 will sit in the bank for 21 years, second 2000 will sit in the bank for 20 years.. forming a geometric series etc etc