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Firstly 10 boxes of ball kept in a queue and mark it as box1,box2,box3.........box10
then from 1st box we have to take 1 ball,from 2nd box 2 ball..........from 10th box we kept 10 balls. and weight the total ball....
so if total nos ball is :n(n+1)/2 (n= total nos of box that is 10)=55
if all the ball is 10 gm.then total weight will be=550gm.
suppose total weight is 546gm.
so the difference is(550-546=4gm).
then the d defective ball is in box no:4.
similarly we can try for all possible box difference gives us the box number which contains the defective balls
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