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Math Help - An Elementary School Maths Problem Sum

  1. #1
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    An Elementary School Maths Problem Sum

    Hi Guys,

    I am stuck with this elementary school maths problem sum which looks relatively easy but i just simply cannot solve it !

    Algebra method cannot be used. The school only allows me to use other methods other than algebra =(

    Here goes the question :

    John , Peter and Alex bought a watch. Alex paid 1/4 of what Peter and John Paid for. John Paid 1/5 of what Peter and Alex paid. If Peter paid $56 more than John, how much is the watch?

    Please help !

    My Brain cells are dying !!
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  2. #2
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    Re: An Elementary School Maths Problem Sum

    Hello, liukawa!

    Algebra method cannot be used. . What?
    The school only allows me to use other methods other than algebra.
    I'd like to see the school's solution!

    John , Peter and Alex bought a watch.
    Alex paid 1/4 of what Peter and John paid.
    John Paid 1/5 of what Peter and Alex paid.
    Peter paid $56 more than John.
    How much is the watch?

    Without algebra, trial-and-error is the only way.
    Why would ANYONE want us to do that?


    After several minutes (hours? days?) of scribbling,

    . . we might find: . \begin{Bmatrix} A&=&6 \\P&=&19 \\ J&=&5\end{Bmatrix}


    This satisfies the first two restrictions:

    . . A \:=\:\tfrac{1}{4}(P + J) \quad\Rightarrow\quad 6 \:=\:\tfrac{1}{4}(19+5)

    . . J \:=\:\tfrac{1}{5}(A + P) \quad\Rightarrow\quad 5 \:=\:\tfrac{1}{5}(6+19)


    It does not satisfy the third restriction: P is 56 more than J.

    In our solution, P is only 14 more than J.


    Maybe if we multiplied everything by 4, we might get the answer.

    So, we try: . \begin{Bmatrix}A \:=\:24 \\ P \:=\:76 \\ J \:=\:20 \end{Bmatrix}\;\hdots\;\text{ and it works!}


    I can't imagine ANYONE assigning a problem this complicated
    . . and expecting one to solve it by arithmetic? .guessing?
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  3. #3
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    Re: An Elementary School Maths Problem Sum

    Quote Originally Posted by liukawa View Post
    Alex paid 1/4 of what Peter and John Paid for. [1]
    John Paid 1/5 of what Peter and Alex paid. [2]
    Peter paid $56 more than John. [3]
    Soroban seems to be in an unusually grouchy mood! Wife had a headache?

    [1][3]: J + P + A = "60", 64, 68, 72 ........ 116, "120", 124 ....
    [2][3]: A + P + J = "60", 65, 70, 75 ........ 115, "120", 125
    60 too low (since P paid over 56), so watch = 120
    Last edited by Wilmer; July 17th 2012 at 07:00 AM.
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  4. #4
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    Re: An Elementary School Maths Problem Sum

    John , Peter and Alex bought a watch.
    Alex paid 1/4 of what Peter and John paid.
    John Paid 1/5 of what Peter and Alex paid.
    Peter paid $56 more than John.
    How much is the watch?
    So A paid 1/5 of total, J paid 1/6 of total and P paid 19/30 of total.

    Now set up a spreadsheet with total in the first column running upwards from $60 in steps of $1, and with columns for what A,J and P paid and a further column with P-J. This last column increases as you move down, look for where it comes to $56, and you have the answer. (Yes, it is $120)
    Last edited by zzephod; July 19th 2012 at 12:00 AM. Reason: improve syntax
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