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Math Help - Tough Algebraic question for a 10 year old

  1. #1
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    Tough Algebraic question for a 10 year old

    Here's the problem:

    If X horses can eat through Y bags of feed in Z days. How many days would U bags of feed last with V similar horses eating at the same rate?

    My answer:

    Let the number of days be a = U/V

    The rate of eating for each horse is Y/X per day

    V similar horses would consume V(Y/X) of bags per day

    Therefore it would take U/V(Y/X) days

    Now my teacher corrected this by putting (UX/VY)Z which I do not understand. Could someone explain this very clearly? With many thanks...
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  2. #2
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    Re: Tough Algebraic question for a 10 year old

    Quote Originally Posted by Natasha1 View Post
    Here's the problem:

    If X horses can eat through Y bags of feed in Z days. How many days would U bags of feed last with V similar horses eating at the same rate?

    My answer:

    Let the number of days be a = U/V
    This doesn't make sense. You can say "let the number of days be a" but you cannot say then "a= U/V". You are told that U is a number of bags of feed and V is a number of horses. "U/V" has units of "bags of feed per horse".

    The rate of eating for each horse is Y/X per day
    No, it isn't. You are told that "X horses can eat through Y bags of feed in Z days." That means that Y/X is the "number of bags of feed per horse" for Z days, not one day.

    [/quote]V similar horses would consume V(Y/X) of bags per day[/quote]
    Not "per day"- "per Z days".

    Therefore it would take U/V(Y/X) days

    Now my teacher corrected this by putting (UX/VY)Z which I do not understand. Could someone explain this very clearly? With many thanks...
    For exactly the reason I said. Since "X horses can eat through Y bags of feed in Z days." "X/Y" would be the amount of feed per horse for Z days, not for one day. V(Y/X) is the amount of feed for V horses for Z days, not for one day. The amount of feed for V horses per day would be [V(Y/X)]/Z= (VY)/(XZ).
    Thanks from topsquark
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  3. #3
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    Re: Tough Algebraic question for a 10 year old

    in such type of questions we form a statement with what we want to find appearing at the end of the statement. In this case we want to find the days for which the feed will last. So we write the statement.
    Y bags of feed are consumed by X horses in = Z days.
    [ now we work out by unitary method. i.e. 1 bag of feed will be consumed by the same number of horses in lesser number of days , thus ]
    1 bag of feed are consumed by X horses in = Z / Y days.
    Now U bag of feed will be consumed by the same number of horses in more number of days , thus
    U bag of feed are consumed by X horses in = {Z* U } / Y days.
    [ When number of horses is reduced to 1 the feed will last for more number of days, thus ]
    U bag of feed are consumed by 1 horses in = {Z* U } / {X * Y } days.
    [ and when number of horses is increased to V the feed will last for less number of days, thus ]
    U bag of feed are consumed by V horses in = {Z* U*V } / {X * Y } days.
    Thanks from Natasha1
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