You have 13 balls which all look identical. All the balls
are the same weight except for one. Using only a balance
scale, can find the odd one out with only 3 weighings?
Is it possible to always tell if the odd one out is heavier
or lighter than the other balls?
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Answer / guest
It is always possible to find odd ball in 3 weighings and in
most of the cases it is possible to tell whether the odd
ball is heavier or lighter. Only in one case, it is not
possible to tell the odd ball is whether heavier or lighter.
1. Take 8 balls and weigh 4 against 4.
o If both are not equal, goto step 2
o If both are equal, goto step 3
2. One of these 8 balls is the odd one. Name the balls on
heavier side of the scale as H1, H2, H3 and H4. Similarly,
name the balls on the lighter side of the scale as L1, L2,
L3 and L4. Either one of H's is heavier or one of L's is
lighter. Weigh (H1, H2, L1) against (H3, H4, X) where X is
one ball from the remaining 5 balls in intial weighing.
o If both are equal, one of L2, L3, L4 is lighter. Weigh L2
against L3.
? If both are equal, L4 is the odd ball and is lighter.
? If L2 is light, L2 is the odd ball and is lighter.
? If L3 is light, L3 is the odd ball and is lighter.
o If (H1, H2, L1) is heavier side on the scale, either H1 or
H2 is heavier. Weight H1 against H2
? If both are equal, there is some error.
? If H1 is heavy, H1 is the odd ball and is heavier.
? If H2 is heavy, H2 is the odd ball and is heavier.
o If (H3, H4, X) is heavier side on the scale, either H3 or
H4 is heavier or L1 is lighter. Weight H3 against H4
? If both are equal, L1 is the odd ball and is lighter.
? If H3 is heavy, H3 is the odd ball and is heavier.
? If H4 is heavy, H4 is the odd ball and is heavier.
3. One of the remaining 5 balls is the odd one. Name the
balls as C1, C2, C3, C4, C5. Weight (C1, C2, C3) against
(X1, X2, X3) where X1, X2, X3 are any three balls from the
first weighing of 8 balls.
o If both are equal, one of remaining 2 balls is the odd
i.e. either C4 or C5. Weigh C4 with X1
? If both are equal, C5 is the odd ball. But you can not
tell whether it is heavier or lighter.
? If C4 is heavy, C4 is the odd ball and is heavier.
? If C4 is light, C4 is the odd ball and is lighter.
o If (C1, C2, C3) is heavier side, one of C1, C2, C3 is the
odd ball and is heavier. Weigh C1 and C2.
? If both are equal, C3 is the odd ball and is heavier.
? If C1 is heavy, C1 is the odd ball and is heavier.
? If C2 is heavy, C2 is the odd ball and is heavier.
o If (C1, C2, C3) is lighter side, one of C1, C2, C3 is the
odd ball and is lighter. Weigh C1 and C2.
? If both are equal, C3 is the odd ball and is heavier.
? If C1 is light, C1 is the odd ball and is lighter.
? If C2 is light, C2 is the odd ball and is lighter.
| Is This Answer Correct ? | 49 Yes | 10 No |
Answer / rahul khanduri
Little difficult, I found the soln then I forgot, and I am too lazy to right it here, so basically you have to use little brain, you have to think that if one side is heavy this can't be because of this and that, I think op got my ans.
| Is This Answer Correct ? | 0 Yes | 2 No |
Answer / atul saxena
Yes it is always possible to tell if the odd ball is heavier or lighter than the other balls with only three weighings.
| Is This Answer Correct ? | 1 Yes | 5 No |
Devide these 13 balls into 3 categories i.e 6,6,1. Then measure both 6 and 6 . You will find the result. If both are equal then off one is the left out ball. If not equal then make another separate to the less weight 6 ball into 3 and 3. Same way measure these 3 and 3 . Find out the less weight side. Again separate this three into 1,1,1. Then measure any 2. You will find the needed result.
| Is This Answer Correct ? | 2 Yes | 6 No |
Answer / kamran khan
13 balls are divide into four group (3,3,3,3,1)
1)First measure one pare of (3,3) . If it is equal then measure an other part
if it is also equal then last one is defective .
you suppose to first measure is unequal then unequal portion take and pick up on ball i.e(measure 1,1) and one is taken out if both are same then taken out ball is defective otherwise between two any one is defective
| Is This Answer Correct ? | 1 Yes | 8 No |
Answer / vinay tyagi
first of all it is necessary that we must know that
the odd ball is heavy or light than others.
Ans. suppose odd ball is heavy.
divide balls among three group abruptly
I 6 balls
II 6 balls
III 1 ball
now measure group I against II
1. if equal then odd ball is III grop
2 if not equal then odd ball is in heavier side
3. now from the heavier side 6 ball , just divide them into
two groups of 3 ball each
4. measure these groups and find the heavier one
5 now we have 3 ball , just choose two balls and measure
against each other, we can find the heavier one if found equal then third ball is the odd one
| Is This Answer Correct ? | 5 Yes | 13 No |
Answer / mohit kaushik
divide 13 in 4(I) , 4(II) , 4(III) , 1
1.) weigh I against II if equal
2.) weigh I against III if equal , the ball left is defective.
after performing step 1. and 2. in either of the step the
scale is not balanced.
Suppose it is not balanced in case 1 .( You can consider
case 2. if u like )
took any two balls from case 1 & weigh against each other.
if equal defective ball must be in group of other two.
take any ball from other group if scale is equal the ball
left is defective or else the chosen ball from the group of
two is defective.
By : mohit
Provide feedback
| Is This Answer Correct ? | 6 Yes | 16 No |
Answer / nitin verma
The question is incomplete
at least one of the thirteen balls should be given as known correct weighing ball separately.
Then only it can be solved correctly.
And it is actually much more difficult than it look to be.
And there is only one way to start I.e keeping 5 balls on both sides with one known ball on any side and rest 3 aside.
Your question is incomplete or wrong
| Is This Answer Correct ? | 2 Yes | 12 No |
Answer / nitin phunsukh wangru verma
How simple
you guys are such a JACK ASS
read the question carefully
the right answer is
No it not possible to catch the faulty ball all the time with exact fault I.e light or heavy
but it is a good question and can be asked in a better way
as I said earlier.
| Is This Answer Correct ? | 2 Yes | 17 No |
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