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Math Help - Simple average over time help please

  1. #1
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    Simple average over time help please

    I need to work out the average length of time that a Farmer keeps a tonne of potatoes in his store. I have been given the following monthly stock figures and need to work this out. probably very simple but having a mind blank - help please.

    Month Tonnes in Store

    Oct 2110
    Nov 2010
    Dec 1910
    Jan 1710
    Feb 1610
    Mar 1185
    Apr 885
    May 445
    Jun 0
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  2. #2
    MHF Contributor Unknown008's Avatar
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    How many tonnes are there in all?

    How much time are there in all?

    From this, can you find out how much time one tonne is kept in store using proportions.

    X tonnes --> Y days
    1 tonne --> ? days
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  3. #3
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    Confused

    I am confused. There are 2110 tonnes for 30 days, 2010 for 30 days, 1910 for 30 days, 1710 for 30 days etc...??
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  4. #4
    MHF Contributor Unknown008's Avatar
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    To, take the total of all of them.

    October through June, you have (31 + 30 + 31 + 31 + 28 + 31 + 30 + 31 + 30) = 273 days.
    In all, you have (2110 + 2010 + 1910 + 1710 + 1610 + 1185 + 885 + 445 + 0) = 11865 tonnes.

    11865 tonnes --> 273 days.
    1 tonne --> 273/11865 = 0.023 days = 33 minutes
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  5. #5
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    Perhaps I have not explained myself properly. The farmer only sells 100 tonnes of his total 2110 in the first month (30days). This means the minimum length of time a tonne is in store is 30 days, the maximum being the full 8 months. However, as the quantities in store to not reduce at the same rate over time (he sells less at the start and more of them in the later months), it is not as simple as just averaging the time.

    I hope this explains my problem better.
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  6. #6
    MHF Contributor Unknown008's Avatar
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    We can make a table for that then?

    \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline<br />
Oct & Nov & Dec &Jan &Feb &Mar &Apr &May &Jun \\ \hline<br />
2110 & 2010 & 1910 &1710 &1610 &1185 &885 &445 &0 \\ \hline<br />
100 & 100 & 200 &100 &425 &300&440 &445&0 \\ \hline \end{array}

    Then, from that we get:
    100 tonnes kept for 31 days
    100 tonnes kept for 61 days
    200 tonnes kept for 92 days
    100 tonnes kept for 123 days
    425 tonnes kept for 151 days
    300 tonnes kept for 182 days
    440 tonnes kept for 212 days
    445 tonnes kept for 243 days

    This gives an average of 2110/883 = 2.39 days
    Last edited by Unknown008; October 29th 2010 at 11:49 PM.
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  7. #7
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    This can't possible be 2.39 days. No potatoes are sold for the first 30 days on over half remain in store at the end of 5 months! There are no potatoes in store for as little as 2.39 days so this can't be the average!
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  8. #8
    MHF Contributor Unknown008's Avatar
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    Oops, I mean... this is the average amount of tonnes removed daily.

    It should be the other way round, that is: 0.418 days = 10 hours.

    So, if each 10 hours you remove 1 tonne of potatoes, by the end of 8 months, there will be none left.

    Either that, or I'm not understanding the problem properly...

    EDIT: Or maybe it's simply 273 / 2110 = 0.129 days = 3.1 hours

    So, if you remove 1 tonne every hour, by 8 months, there will be none left.
    Last edited by Unknown008; October 30th 2010 at 12:39 AM.
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