# Cambridge Chat (previously New Cambridge Students Entry 2004)

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15 years ago
#4241
OK....so i assume that twice a year, 5% of the amount in the bank is calculated and then added to your total.

Thus after half a year, you have: 5,000,000*1.05. After 1 year you have: 5,000,000*1.05*1.05

But then you take out your set amount, X, so you have:

5,000,000*1.05^2 - X

so we have that going on for 50 years:

(5,000,000*1.05^2 - x)*1.05^2 - x)...and so on
we can make that simpler:
((5512500 - x)*1.1025 -x) and so on for 50 years (that is 2 years there)

if we multiply that out we will have:
5512500*1.1025 - 1.1025x - x
and so on

now the key point is that the term which doesnt involve X at all at the start, we can work out what that is after 50 years, it's just 5,000,000 * 1.1025^50
the first x term will be 1.1025^49.x (since it doesnt apply to the first year)
the second x term will be 1.1025^48.x
and so on all the way down to a final "x" (which is your last withdrawel)

so we have the equation: 5,000,000 * 1.1025^50 - (1.1025^49 + 1.1025^48 + 1.1025^47 + ..... + 1.1025 + 1)x = 0
so all you gotta do is figure out what those values are which satisfy are to work out x.

this is me working it out as i go along, please point out any mistakes I have made...I will try and come up with a definitive answer in the next hour.
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15 years ago
#4242
(Original post by Willa)
OK....so i assume that twice a year, 5% of the amount in the bank is calculated and then added to your total.

Thus after half a year, you have: 5,000,000*1.05. After 1 year you have: 5,000,000*1.05*1.05
No, it's 5% every year...it may be paid in in two installments, but you still only have 1.05*original amount at the end of the year, not 1.05*1.05*original amount.

I'll have a look later if someone else hasn't answered the question already.
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15 years ago
#4243
agh got it sorted just writing up
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15 years ago
#4244
all this maths talk
*cries*
oh how I'm dreading opening that maths workbook...
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15 years ago
#4245
(Original post by Willa)
so we have the equation: 5,000,000 * 1.1025^50 - (1.1025^49 + 1.1025^48 + 1.1025^47 + ..... + 1.1025 + 1)x = 0
so all you gotta do is figure out what those values are which satisfy are to work out x.
Yep I got this after I sorted out my signs, then summed the series as a GP,

a = 1
r = 1.05
sum to 49 = (1.05^49 - 1)/0.05 = 198.4266625857843054257412104992 6

x = 5*10^6 * 1.05^50 / 198.4266625857843054257412104992 6

= 0.288 * 10^6

It seems like rather a high number, perhaps write a program to iterate the model and see if it works?

£288958 - does that sound reasonable? You could take £300 000 out per year? =| I must have made a mistake
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15 years ago
#4246
(Original post by Fluffstar)
all this maths talk
*cries*
oh how I'm dreading opening that maths workbook...
ditto. I always think with the maths problems posted on UKL: "I'm sure I could have a stab at that, but I really can't be bothered!"

I've had a quick look at the maths workbook and it doesn't look too hideous though, and there are notes and answers provided, so I'm hoping I can teach myself any gaps.
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15 years ago
#4247
Guhh, I'm really confused now about what the question actually means, but I'm sure you guys are right. But um...I used an algorithm to remove £300,000 at the beginning of every year and add 5% to whatever was in the account every 6 months. Assuming it's not faulty, the amount of money in the bank account actually grows every year, so it can't be zero by the 50th year. If anything, he needs to take out more than £300k a year.

(Original post by Fluffstar)
all this maths talk
*cries*
oh how I'm dreading opening that maths workbook...
Oh well, it will be over in a year.
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15 years ago
#4248
(Original post by fishpaste)

£288958 - does that sound reasonable? You could take £300 000 out per year? =| I must have made a mistake

Hm, I wrote something to test it, and subtracting £277k has around the desired effect.
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15 years ago
#4249
(Original post by Squishy)
Guhh, I'm really confused now about what the question actually means, but I'm sure you guys are right. But um...I used an algorithm to remove £300,000 at the beginning of every year and add 5% to whatever was in the account every 6 months. Assuming it's not faulty, the amount of money in the bank account actually grows every year, so it can't be zero by the 50th year. If anything, he needs to take out more than £300k a year.

Oh well, it will be over in a year.
Well yes it adds money, but you end up removing all the interest and 'a bit more' when you subtract the money.

Like the first interest on 5 million will be 250k, then you remove 277k.

Sorry, just read your post properly, I'm going on the assumption here that it's 5% per year. I would have thought it was 1.05*1.05 but someone said it wasn't?
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15 years ago
#4250
(Original post by figgetyfig)
ditto. I always think with the maths problems posted on UKL: "I'm sure I could have a stab at that, but I really can't be bothered!"
i know what you mean. i hardly ever even bothered to answer physics and chemistry questions when i was revising!

whats been happening, btw? i haven't been on for a few days. have i missed anything, or has it been, the usual drivel?
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15 years ago
#4251
If you went with 1.05^2
then

x(1.1025^49 + 1.1024^48 + ... + 1.1025) = 1.1025^50 * 5 * 10^6

summing the GP again

S[49] = 1.1025(1.1025^49 - 1) / 0.1025 = 1272.183003378570292932658849694 8

so x = 516833
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15 years ago
#4252
well at least i was write in principle, i just didnt totally understand the question!

the two installments thing is pretty redundant then huh?
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15 years ago
#4253
(Original post by fishpaste)
Well yes it adds money, but you end up removing all the interest and 'a bit more' when you subtract the money.

Like the first interest on 5 million will be 250k, then you remove 277k.

Sorry, just read your post properly, I'm going on the assumption here that it's 5% per year. I would have thought it was 1.05*1.05 but someone said it wasn't?
Yeah, I wasn't sure if it was that. But if it is 1.05*1.05, then at least (1.05*1.05*5E6)-5E6 = £512,500 must be removed per year, or else his money will grow, despite money being taken away from it. If he takes away exactly £512,500, his money will continuously cycle and he will always have £5E6 in his account at the beginning of every year.
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15 years ago
#4254
(Original post by Squishy)
Yeah, I wasn't sure if it was that. But if it is 1.05*1.05, then at least (1.05*1.05*5E6)-5E6 = £512,500 must be removed per year, or else his money will grow, despite money being taken away from it. If he takes away exactly £512,500, his money will continuously cycle and he will always have £5E6 in his account at the beginning of every year.

yep! my magic number is: 516833
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15 years ago
#4255
i'm not even trying to keep up with all this maths... my brain hurts!
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15 years ago
#4256
i enjoyed it, was a chance to try and warmup again....i feel completely mentally out of shape...should have seen that a Gp would sum those values, grr! *kicks self*
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15 years ago
#4257
(Original post by Willa)
i enjoyed it, was a chance to try and warmup again....i feel completely mentally out of shape...should have seen that a Gp would sum those values, grr! *kicks self*
yeah I enjoyed that. at least you got your equation right in the first place, my RHS had really messed up positives/negatives.
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15 years ago
#4258
(Original post by fishpaste)
yep! my magic number is: 516833
Ahh, cheeky edit. You posted 569808 before, didn't you? That is too big...I think you run into negative figures 24 years after you start.

516833 sounds about right...I can't be bothered to check though, I deleted my program. But by the Willa interest interpretation, I'll go with that as well.
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15 years ago
#4259
(Original post by scarlet ibis)
i'm not even trying to keep up with all this maths... my brain hurts!
Stare long and hard at a brick Amy.
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15 years ago
#4260
nooooooooooooooooooooooooooooooo not MATHS
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