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Math Help - Another word problem

  1. #1
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    Another word problem

    hi everyone,

    Some of the balls in bag A are divided into two bags B and C. If five are put in bag B and a third of the remainder are put in bag C, twelve remain.
    Find the number of balls originally in bag A.

    first attempt was

    5+\frac{x}{3}=12

    which is wrong, then

    5+\frac{1}{3}(x-5)=12

    which is still wrong

    the answer is 23. but can someone please show me. thanks
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Lets say we have x balls in bag A

    Take 5 balls to bag B


    remainder = (x-5)

    Balls taken to Bag C is

    \frac{1}{3}(x-5)


    Now remainder = (x-5) - (x-5)/3


    This remainder is = 12

    2(x-5)/3 = 12

    x =36/2 + 5

    x = 18+5

    x= 23

    ------------------------
    Adarsh
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  3. #3
    Member
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    adarsh, you did it again...
    thanks.
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