there are 8 identical balls out of which 1 is defective u hav been given a weighing machine How many

first we can take 3 in one side and 3 in another side ,(remaining 2 is grounded).... if any one side can show defect take that 3 balls and again calculate 1 ball in one side and 1 ball in another side... if both r same 3rd ball is defective(weight).., if 3 in one side , 3 in another r same , take remaining 2 grounded balls & calculate 1 side in one ball and another side in another ball......
10 years agoHelpfull: Yes(2) No(0)
i want an optimised ans. 3 is not optimised. n pls state ur method also
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weigh three balls against another three balls. if both weigh the same , then just weighing the remain two (one against one) will show the lighter ball. if the sets of three do not weigh equal, then weigh any two balls in the lighter set, one against the other . the balance will show if the lighter one is on the balance,if not the remaining one is the lighter one.

8= (3 + 3 ) + 2

(the numbers in the brackets are balls on either side of the balance)

if both are equal, then

2= (1 + 1) done.

else, from the lighter set of 3

3= (1 + 1) + 1 done.
10 years agoHelpfull: Yes(0) No(0)

there are 8 identical balls. out of which 1 is defective. u hav been given a weighing machine. How many times you will weight (loops) to find the defective ball. ans shud b optimised