This puzzle asked in software AG Bangalore
Q. There are 10 boxes each one is numbered as 1 ,2 ...10; each have 10 ball of equal weight . one of them box contains the defective balls . how can you find out the box which contains the defective balls .you have to weight only once...

• since they have mentioned that each ball of equal size,
  1) let us assume that each ball in a box weighs 1 gram.
  2)now,we will take 1 ball from the first box,2 ball from 2nd box,3 from 3rd and so on upto 10 boxes...
  3)now weigh these all balls at a time in weighing machine....
  4) since we have assumed 1 gram....so it must weigh 55 gram(1+2+3+....10).
  now check for the defective one as....
  5)(note: i have assumed defective box means it contains ball of 2 grams)
  if it weighs 56 one more than total weigh than the defect is from first box...if it weighs 60 it is from 5th box(because we have taken 5 balls from 5th box....
  6) like this we can find the defective box.
  
  
• 10 years agoHelpfull: Yes(28) No(1)
• do balls have weights?I mean what are the weights of normal and defective balls.they must have given this option.
• 10 years agoHelpfull: Yes(1) No(1)
• minimum two chances first we keep 3 boxes in left and 3 boxes in right if in that defective then we will check like this out of three 1 in left and 1 in right if no defect left out 1 is defective and in the first attempt left out 4 are there we can check with 2 in left and 2 in right if defect another check so total 2 chances.
• 10 years agoHelpfull: Yes(1) No(2)
• If weights are known then we can solve it using one weighing option
• 10 years agoHelpfull: Yes(1) No(0)
• ya condition is that if nay box have defective balls then the weight of each ball should be double or half ; the weight of ball is 2 kg then you have to find out the box number which have defective balls in only one time
• 10 years agoHelpfull: Yes(0) No(0)
• find ur solution here http://www.mathsisfun.com/puzzles/weighing-10-bags-solution.html
• 10 years agoHelpfull: Yes(0) No(0)
• good question i think worth asking
• 8 years agoHelpfull: Yes(0) No(0)