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Weighing bags

  1. Aug 9, 2007 #1
    Interesting problem

    Five bags each contain 10 balls. The first bag has yellow balls, the second blue balls, the third red balls, the fourth green balls and the fifth orange balls. All the balls in four of these bags (we don't know which ones) weigh 20 units each and the the balls in the remaining bag weigh 18 units. If we are to determine the bag containing the lighter balls by method of weighing on a single plate scale, what is the minimum number of weighings needed? Describe the procedure.
    Last edited: Aug 9, 2007
  2. jcsd
  3. Aug 9, 2007 #2


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    take 1 ball from the first bag , 2 from the second, 3 from the third, get the pattern?

    Weigh the lot of them, bag number = (300 - weight)/2
  4. Aug 10, 2007 #3
    This problem is well known I assume (not to take away any credit from you, Integral). I've came with this variation: if we're only allowed to take out 1 ball from each bag, what would be the minimum number of weighings?
  5. Aug 10, 2007 #4

    Then you have 5 balls (A,B,C,D,E) and one of them is different.
    A binary search will find the ball in 3 weighings (maximum).
    But sometimes, just 1 weighing will be enough.
    So, the minimum is 1 weighing.

    Weigh ABCD . If ABCD=80, then "E" is the ball. If ABCD=78, then weigh AB.
    If AB=40, then weigh C. If C=20 , D is the ball (and if C=18...)
    If AB=38, then weigh A. If A=20, B is the ball (and if A=18...)

  6. Aug 10, 2007 #5
    Good, but I didn't say that you can't add a ball to bag.
  7. Aug 11, 2007 #6
    Well,in this case, 2 weighings will be the minimum necessary to ever determine the bag.
    Put 1 ball "A" into the bag "B" , and 1 ball "C" into the bag "D".
    Weigh the bags "B" + "D".
    If B+D=440 then "E" is the bag. If not, weigh bag "B" .
    Now, if B+D=438 and B=220 , then "C" is the bag.
    If B+D=420 and B=220 , then "D" is the bag.

  8. Aug 11, 2007 #7


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