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    kofigc's Avatar
    kofigc Posts: 21, Reputation: 1
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    #1

    Oct 26, 2010, 02:49 PM
    Arithmetic and Geometrical Progression
    Hey everyone, looking for some more help / advice :)
    Questions first, my answers at the bottom :)

    Question One
    The value of a machine which costs £35000 depreciates by 10% per annum.

    A) What is its value after 10 years?

    The costs of repairs and maintenance is £2000 in the first year and increases by 10% per year.

    b) After how many years will it be cheaper to buy a new machine than to pay for the repairs and maintenance up till then?

    Question Two

    A man wishes to replay a debt of £3000 by paying a fixed sum of £60 per month to reduce the debt, together with a payment of 1% interest for that month indebtedness. Make a schedule of payment and find the total interest pain until the debt is completely discharged.

    Question Three

    A man deposits £1000 every year into a pension fund earning 6% interest per annum, compounded yearly. Deposits are made at the start of each year. How much will the fund be woth at the end of 15 years after the initial deposit?

    My Answers (50% marking)

    Q1 a- Calculated the long way of reducing the value by 10% for 10 years, value after 10 years = £12203.7454

    b - 2000 * 1.1^30 = £34898.80454
    2000 * 1.1^31 = £38388.68499
    so, it will be cheaper to buy a new machine after 30 years.

    Question Two - There are 50 "£60's" in £3000 so N= 50, D= 0.6, A= 0.6.

    Sn= N/2*[(2A+(n-1)*d]
    = 50/2*[2*0.6+(50-1)*0.6]
    = 25*[1.2+(49*0.6)]
    = 25*(1.2+29.4)
    = 765

    Question Three - Again calculated the long and painful way of adding 6% to the initial year and for each year thereafter adding £1000 then the 6%. Totaling a value of £24672.52808 after 15 years.
    ArcSine's Avatar
    ArcSine Posts: 969, Reputation: 106
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    #2

    Oct 27, 2010, 06:43 AM
    You're good on 1A ; (rounded).

    For 1B, I'm thinking that the question's language "up till then" indicates that they're looking for the sum of the increasing costs up to the point at which this sum (of a geometric sequence) first exceeds the 35K cost of a machine. In other words, when does (2,000 + 2,200 + 2,420 +... ) first crack through the 35K ceiling? Letting the "sum of a geo sequence" formula do the heavy lifting, we have that and .

    You answer on (2) is spot-on.

    You're also good on (3), but there's a formula that would've saved you a measure of pain. Google for "future value of an annuity due", and you'll find variants on the formula, all of which are equivalent.

    All in all, nice work!
    ArcSine's Avatar
    ArcSine Posts: 969, Reputation: 106
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    #3

    Oct 27, 2010, 06:48 AM
    Okay, where's the darn preview or edit feature? Those should read .
    kofigc's Avatar
    kofigc Posts: 21, Reputation: 1
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    #4

    Oct 28, 2010, 01:23 PM
    Comment on ArcSine's post
    Hi, thank you very much for the reply!
    I'm not sure if I'm just being slow here or not, but how can I then solve the issue in 1b? If I'm misreading your reply, I apologise and would be grateful if you point me toward the right direction :S
    ArcSine's Avatar
    ArcSine Posts: 969, Reputation: 106
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    #5

    Oct 28, 2010, 03:05 PM
    Not a problem, comrade... I was just saying that it appears to me the question wants to know the point at which the accumulated repair expenditures first exceeds the cost of a new machine. I then went on to note that (the sum of the first 10 years' repair & maint costs, using your notation for the partal sum of a sequence) was less than 35K, whereas the 11th year of expense puts you over the threshold in total.

    Hence my approach says that 11 years of repair & maint costs, starting at 2K and increasing by 10% per year, is in total greater than the cost of a new unit. Your approach to (1)(b) was to determine the year for which the R&M cost for that year alone would be > 35K. And frankly, the wording of the question has some amgibuity as to which answer it's really looking for. (So maybe you take both answers into class, and have the instructor clarify which?)

    Cheers!
    kofigc's Avatar
    kofigc Posts: 21, Reputation: 1
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    #6

    Oct 28, 2010, 03:34 PM
    Comment on ArcSine's post
    Ah, I understand now - apologies for my slowness lol! The wording is a bit of a mystery with most questions we get I'm afraid as English is not the lecturer's native language so it can get confusing.
    Thank you very much for your help again!
    kofigc's Avatar
    kofigc Posts: 21, Reputation: 1
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    #7

    Nov 3, 2010, 04:48 AM
    *Update*

    I reviewed my answer to (1B) and am now getting a higher score (75%) but am still frustrated that I am incapable of getting 100%.
    Using the same solutions as above for (1A), (2) and (3), I have tried using 31, 11 and 10 for 1B and all of these produce a 75% score.

    Does this indicate that this is the area I need correcting? Although the fact I am now scoring 25% more than previously would suggest I'm getting closer to the answer required?
    ArcSine's Avatar
    ArcSine Posts: 969, Reputation: 106
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    #8

    Nov 3, 2010, 05:48 AM
    I'm not familiar with the scoring system that's at work here, so I'm afraid I can't offer much help. Regarding (1b), the wording of the question leaves a lot to be desired. My only two suggestions: (1) Review your background study material carefully, trying to find a worked-out example that uses similar terminology, or seems to fit the scenario of question (1b). That might give you a better clue what the question is really asking. (2) Ask your instructor. The English language has a real knack for ambiguity sometimes, and the fact that things in the real world often have to be clarified, or re-stated, is proof positive of that. I'm guessing he'd be sympathetic to that.

    Sorry I couldn't help out more, but if neither 31 years (your approach) nor 11 years (mine) is correct, I'm unsure what (1b) is asking.

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