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# Puzzle asked in an interview...plz answer??

1.  You have a weighing machine, and you have 27 balls in which 26 balls are of same weight and 1 ball is of different weight. How many times/steps needed to identify the 1 ball which is different from other ball?

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### Replies to This Discussion

1. 2^(n-1), n is he total number of ball. here n27.

Sorry mr. Suseentheran

we have to find out by weight not by formula

> implying "step" is a formation of logical grouped interactions.

assuming a "step" is one use of the weighing machine you would need 3-26 steps (or if you know that the defect ball is lighter/heavier only 2-25).

My friend it is only 4 steps process

as i stated earlier it is.

But the method mentioned by Vineet would take the 3-26 or 2-25 steps.

yes sorry i did'nt see your first post.have u seen my method(first post)? i think our method is same?
I think so too. I just wanted to summarize the steps needed by this method

take 13-13(If weight is balanced the rest is the one which we are searching) else take the 13 which weighs more,

now make it as 6-6(If weight is balanced the rest is the one which we are searching) else take the 6 which weighs more,

now make it as 3-3. Obviously there will be diff in one side. so take that THREE

now make it as 1-1, If weight is balanced the rest is the one which we are searching.

> implying you know that the defect ball is heavier, not lighter ;)

I dont think so...whether heavier or lighter it will work.

consider the weight of 6 balls as 2,2,2,2,2,3 or 1...now it will work with what I said.

I  should not have said it as Heavier specially :)

yeah but you won't know which group of 13 balls you have to proceed with when one group is lighter, as you don't know if the defect ball is in the lighter group (->the defect means light) or other way around.

given these 2 groups of a different weight and the knowledge of one defect doesn't let you decide which group conains the defect ball.

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