Interview Questions And Answers 
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You have eight balls all of the same size. 7 of them weigh the same, and one of them weighs slightly more. How can you fine the ball that is heavier by using a balance and only two weighings?
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Put 4 balls on either of pan, you will get to know which pan is heavier. Now take the four balls from heavier pan and split them in group of two and place in each pan of balance to get the heavier group. And you can physically feel and can distinguish the heavier of the 2 balls.
choose 6 balls and weigh 3 against 3
- if they weigh the same, you have another weighing for the remaining 2 balls and you can find the heavier one
- if they don’t weigh the same, from the group of 3 which was heavier, choose any 2 balls and weigh them:
- if they weigh the same, the remaining ball is the heavier one; otherwise you just found the heavier one by weighing the 2 chosen balls
1. Out of the 7 balls, take 6 balls and weigh 3 against 3.
2. Possibility 1: If they weigh the same, then the one ball that was left out is the heavier one. (Problem solved in 1 weighing)
3.Possibility 2: If one of the group of 3 balls is heavier, then take 2 balls out of that and weigh 1 against another.
4. Possibility1:
If one of the balls is heavier then the problem ends here.
5. Possibility2: If both the balls weigh the same, then the ball that was left out is the heavier one.
First go: Place four balls on either side, remove one from either side. If the scale balances you know you have the heavier ball in your hand, if not keep removing one ball from either side at teh same time.
Second go (Optional): Double check the two balls you removed to make the scales balance.
Put three balls in each pan and weigh.
1- If they are equal then the heavier ball is between the two balls that are left un-weighed. Put each of the two unweigthtd balls on each pan and weigh, the heavier ball is the culprit.
2- If they are un-equal. Take the two balls from the heavier pan and weigh them against each other. the heavier is the culprit. In case they are equal then the one left unweighted from the first heavier pan is the culprit.
1./ pan 1: 3 balls, pan 2: 3 balls
if 1 = 2 then put every else into 2 pans to get know which heavier.
if 1 > 2 (or 2>1) then choose 2 random balls in 1 group to put into pans.
if they’re equal => the other is the wanted ball.
if they’re not equal => the wanted ball is the heavier ball
Put two balls in each pan for the first weighing. If the scale shows both sides are equal then none of the balls in the pans is the heavier one. Then for the second weighing replace one ball in each pan with one of the balls that was left out, if the scale turns up even again then the last unused ball is the heavier one. If during the second weighing the scale doesn’t balance then the newly added ball on the lower side is one you are looking for.
But, if during the first weighing the scale shows one side lower than the other then replace one ball on each side with one of the unused balls and make note of which removed ball came out of each side. Then if the scale does balance you will know that the removed ball on that side was heavier. If it does not balance the ball that was not replaced was heavier.
Ok, you can do it in plenty of modes using only the balls. It’s really simple, but I do suppose you have to use the weights once they are given. How would you do it then? There’s nothing said about the weights.
simple take 6 balls 3 each and weigh
if amongst them 1 side is heavier than
take 1 ball each in the pan (from the side which was heavier )heavier ball will be known but if the balls have same weight than the ball left is heavier
this was first case that if the balls in the pan has a heavier ball amongst them but if they doesnot have the heavier ball them check out for heavier ball amongst the balls left behind in the begining
simple!!!!!!!!!!!!!!!!
finding out the heavier ball is easy. but what if we dont know whether one of the 8 balls is heavier or lighter??
n we can weigh them only 3 times..
ducu25 is right. But the question must be 9 balls instead of 8.
The best way is :
Take 4 balls first and make two groups of 2 balls each.
Weigh them against each other(putting them on two sides of the weight measure instrument). If they are equal means the 4 balls are of equal weight and ignore them. Now take the two balls from the remaining three and weigh them against each other. if they equals mean the remaining one ball is the heaviest.
Let say in initial weight of four balls one side is heavier than other mean the heavy ball is in that side (among the two balls that makes that side). Now weigh that ball putting them on two sides of a weight machine and get the heaviest balls.
Was easy Right ?
first divide the 8 balls into 3,3,2 .
Weigh first two sets of 3 balls.
if they are of same weight, the required ball will be in the remain two .In this case we can weigh and find it by weighing remaining 2 balls.So in this case we got in two weighings.
If the two sets of 3 balls are not of same weight, we can consider one of them whose weight is more.So the required ball is in this set of 3 balls. we can leave the remaining 5 balls.
Now divide this balls into 2, 1.Weigh first set of 2 balls.If they are same remaining one ball is the result otherwise the ball which weighs more is the result.
Are any of you guys blind?
You can do TWO WEIGHINGS -_-
Most of the solutions above have 3 or more weighings…
The one that has at most two weighing assumes that the difference of weight of the two balls is big enough so you can measure with your hands. But if that was true, you could do with no weighing… Just pick them up and see how is heavier -_-
The difference of weight would be 1 gram. So now you can’t measure physically.
This problem is not solvable with only 2 weighings..
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