# Maths puzzleWatch

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Thread starter 15 years ago
#21
(Original post by theone)
Can you confirm whether my solution is right?
It seems to make sense logically so I can't see why it wouldn't be right. But I don't have an answer so I can't tell you yes or no as such. Will let you know when model answers come out.

Ed
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15 years ago
#22
(Original post by ZJuwelH)
Truth be told, the question says: "A person takes out a mortgage of 150,000... how much does the annual repayment need to be if the person is to owe 125,000 five years from now?"

All the rest of it seems to be chatcrap if you read it like that so Sire is probably right.
Yeah, it didn't say anything about calculating the interest is all. That is what I was getting at. Thanks ZJuwelH mate. I do understand why the one feels I've missed it though.
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15 years ago
#23
(Original post by Tednol)
Regardless of the question being well or badly phrased, this is what it is trying to ask. Put simply, how much will you have to pay each year to turn £150k to £125k in 5 years, given that each year the amount you have left to payoff will increase by 1.05 times.

It's worth 15 marks, that is to say 15% of the exam. The answer is not going to be £5000 sadly.
Ah so it isn't monthly. *scratches that one out* Ok this is relatively simple. Assuming you're paying this off at the end of each year, after the interst has taken effect. You're starting total is now 157,500. Please tell me this is right before I continue.
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15 years ago
#24
i think i got it kinda right?
i got the constant repayments as 10411.84583(lol)

you start off with 150000
you pay in at beginning of year 10411.84583
at end of first year you pay in 10411.84583
interest 0.05 applies
total to pay now = 142963.5018

end of second year total to pay after interest and repayment.
139179.2388

end of 3rd year
1352505.7626

end of 4th year
131033.6126

end of 5th year
126652.8551

kind of near 125000
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15 years ago
#25
(Original post by calumc)
That's what I got aswell, £10411.85 / year.

Year 1 : (150000 - b)x1.05

And since b is the same every year:
Year 2 : (150000 - 2b)x(1.05^2)

And so on, up to year 5:

(150000 - 5b)x(1.05^5) which equals 125,000.

So b is £10411.85
This logic is flawed:

After one year we agree we have (150000-b)x(1.05) to pay off.

So after our second year we have payed off b more but gained 1.05 interest after paying off b. So the amount is ((150000-b)x(1.05) - b)x(1.05).
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15 years ago
#26
does K really matter, as we're only looking at the first 5 years?
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15 years ago
#27
(Original post by calumc)
I don't think it should, but a correct answer should come to zero after K years.
i know it should, but the percentage rate is only 5% for the first 5 years, so the annual repayments could change after that, so K is independent of the answer
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15 years ago
#28
I get b = £11,452

I got this from the following argument which having read the thread I think theone already stated. If A_n is the amount at the end of the nth year (where A_0 is 150,000) then:

A_n = (A_{n-1} - b)p, where p = 1 + I (in this case 1.05)

By replacing A_{n-1} with this formula and so on you can get the general formula:

A_n = (p^n)A_0 - b*( sum_r=1->n p^r )

When I substituted the values I got b = £11,452. Can anybody verify?
0
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