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Math Help - Pencil Mania

  1. #1
    Junior Member MoniMini's Avatar
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    Smile Pencil Mania

    I know the answer of this sum but would be glad to get an explanation to how this is done:

    Box P has 5 more pencils than Box Q. How many pencils must be moved from Box P to Box Q, such that Box P has 9 fewer pencils than Box Q now?

    a) 4
    b) 5
    c) 6
    d) 7
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  2. #2
    Master Of Puppets
    pickslides's Avatar
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    \frac{5+9}{2}
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  3. #3
    Super Member

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    Hello, MoniMini!

    Box P has 5 more pencils than Box Q.
    How many pencils must be moved from Box P to Box Q,
    so that Box P has 9 fewer pencils than Box Q now?

    . . a) 4 . . b) 5 . . c) 6 . . d) 7

    Let \,q = number of pencils in Box Q.
    Then there are q + 5 pencils in Box P.

    Suppose \,x pencils are moved from Box P to Box Q.
    . . Then Box P will have: q + 5 - x pencils. .[1]
    . . And Box Q will have: q + x pencils. .[2]


    We want [1] to be 9 less than [2].

    . . . .  q + 5 - x \;=\;(q + x) - 9


    Got it?
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