Interview Questions And Answers 
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You are given 9 stainless steel balls (same size, but only one weighs slightly more than the other 8). You have a scale that allows you to balance things on either side. You are only allowed to use this scale twice. How do you identify which ball weighs more?
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1. You keep aside 3 balls and balance 3 and 3 balls on each pane.
2. If the two panes are equal, you take up the kept aside 3 balls and keep 1 and 1 ball on each pane. if one side is heavier than that is the heavy ball and if they are equal the third one is the heavy ball.
3. If the weight of the 3 and 3 pane is same then
4. Take the 3 balls from the heavier side and follow step 2.
You are done.
ha tats gr8 solution….if every step goes into your favor this solution holds good…
but think (first step) if both sides are unequal….
you gonna loose one chance….right
If both sides are unequal( in the first step),
2nd Step : take the 3 balls in the heavier side.Keep one ball aside. weigh the remaining two balls.
If they are equal then the 3rd ball is the heaviest.
If they are unequal then the side which goes down has the heavier ball.
note:
This iswhat said in the answer. But there is a typo error which makes it confusing.
9999 — if the the sides are unequal on the first weigh — take the heavier one and weigh 2 balls of them — u r done ..
I can see how it can be done in three weighs but not two. If you work in groups of three, what happens if your first two groups of three are equal? So then you add the third group of three which will tell you which group the heavier ball is in but it won’t tell you which one it is but your two uses of the scale are gone!! And you might need two more to work it out!
If you work with groups of 4 you end up with two that might be it in two weighs (half split rule!). Maybe you just make an assessment yourself by weighing the two balles in your hands??
I think that answer was said in the first reply itself. First weigh, keep 3 on one side and 3 on other. here are the cases,
STEP1-> (a) If both sides are balancing equal, then you know that these 6 does not have the one which we are searching for. So discard them. Now take the 3 remaining balls and pass it to STEP2.
(b) If both sides don’t balance equal, then take the three balls which weigh more than the other and pass it to STEP2.
STEP2-> (you have 3 balls to consider in this step)
In weigh two, place one ball on each side. If they weigh same, then the 3rd ball outside the weigh is the one you are searching. Or else, which ever ball sinks in the weigh is the one you are searching.
See you got the one you are searching for within 2 weighs. The problem gets little more complicated when you do not know whether that one ball weighs more or less. think of this case….
I think this problem requires minimum 3 chances. First step: keep 3 on oneside and 3 on other side.
a) if they are of same weight then consider the left 3.
b) if they are of unequal weight then weigh the one of the previous 3 and the left 3 and choose the right 3
Second step: now choose the 2 balls from the 3 balls and weigh them
a) if they are of same weight then the third one is the required one
b) if they are of unequal weight then the hevier one is the required one.
thus we need atleat 3 chances to find out the answer.
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