Solving the Mystery: Identifying the Faulty Bag of Balls

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The discussion revolves around a problem involving eight bags, each containing 10 balls, where one bag has balls weighing 2 kg and the others weigh 1 kg. The challenge is to identify the bag with the heavier balls using a single weighing on a scale. Initial suggestions highlight that weighing two bags can quickly reveal the faulty one, but this is deemed too simple. The correct approach involves taking a different number of balls from each bag: one from bag 1, two from bag 2, and so on, up to eight from bag 8. By weighing all these balls together and subtracting 36 from the total weight, the result indicates which bag contains the 2 kg balls. The conversation emphasizes the importance of accurately interpreting the problem's parameters, particularly the use of a scale rather than a balance.
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try this,
there are eight bags. each bag contains 10 balls. one of those bags contains balls weighing 2 kg(all balls weigh the same). All the other balls weigh 1 kg. you have a weighing machine (not a balance). in one weighing u have to find out which bag contains the faulty balls.
 
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I wouldn't be surprised if there are a lot of threads with problems like this one. This version is far too easy though. Just weigh any two bags. If they balance, then the third bag contains the faulty balls. If they don't balance, the heavier of the two contains them.
 
AKG said:
I wouldn't be surprised if there are a lot of threads with problems like this one. This version is far too easy though. Just weigh any two bags. If they balance, then the third bag contains the faulty balls. If they don't balance, the heavier of the two contains them.

Nope, this is the one where you have a scale - rather than a balance so you weigh a different number of balls from each bag. (This is typically referrred to as the counterfit coin problem.)

Vikasj007 graciously posted a more interesting version which involved 81 bags:
https://www.physicsforums.com/showthread.php?t=33784
 
Oops. He even said "not a balance". Still, this one is too easy. Weigh two balls from bag A together with one ball from bag B. If the total weight is 3kg, then it's bag C, if the weight is 4 kg, then it's bag B, and if it's 5 kg then it's bag A.
 
AKG said:
Oops. He even said "not a balance". Still, this one is too easy. Weigh two balls from bag A together with one ball from bag B. If the total weight is 3kg, then it's bag C, if the weight is 4 kg, then it's bag B, and if it's 5 kg then it's bag A.


How do you keep missing that he said 8 Bags? Its the first line for crying out loud... It still may be easy but at least read what he wrote...

geniusprahar_21 said:
try this,
there are eight bags. each bag contains 10 balls. one of those bags contains balls weighing 2 kg(all balls weigh the same). All the other balls weigh 1 kg. you have a weighing machine (not a balance). in one weighing u have to find out which bag contains the faulty balls.

In white for anyone who doesent know the answer: Number the bags 1-8. Take 1 ball from bag 1, 2 from bag 2, exc.. Weigh the lot of them and subtract 34 from the answer... That is the bag number with the 2kg balls.

EDIT: Thanks Everneo! I did mean 36 :rolleyes:
 
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did you mean 'subtract 36' ? [/color]
 
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