Answer / santosh

2-try

Answer / vysali

well first we place 3 in each side of the weighing machine.
the ball having diff weight will move up or down.
now take 2 one on each side..
if these two are same the third ball is heavier or lighter.

Answer / madhu

min 2,
weight 2balls on eachside,
if weights are same then consider the remaining 2. Between
these 2 we can weight and know the diff wieght ball.

if not same then consider the low wieght balls and weight
them again.

Answer / sandip

consider Ball wt low
1 1 1 1 1 ?
Comparing 2 sets:=> 1-try
we get low wt set:=>
1 1 ?
Compare 1st 2 ball:=>If equal than go next other wise we
will get lower wt ball:=> 2-try
if(1==1)
{
goto next comparision:
}
else{
compare last 2 ball , we will get answer.
}


Answer:=>Atleast:=>2-try
At Most:=>3-try

Answer / lion251

Three tries will always be enough.
Start by weighing balls 1,2 together (left) against balls 3,4 (right).
If the balance is equal, then we know that balls 1,2,3,4 are normal. Then weigh ball 5 against a 'normal' ball (one of 1,2,3,4). If the balance is equal, then 6 is the different ball, else 5 is different.
If the balance tipped was unequal on the first measurement, then we know that either one of the two balls on the lower side is too heavy, or one of the two balls on the higher side is too light.
Then, weigh the two balls that were on the lower side against each other. If one side goes down, then that is the different ball, and it is too heavy. If the scales balance, then weigh the balls that were on the higher side in the first measurement against each other. The side that goes up holds a ball that is too light.
Thus, three measurements will always be enough.
It is interesting to know that, in the same three measurements, you can even find out a different ball from a group of 12(!). That is quite a bit more difficult, but even then, three measurements will be enough in all cases (provided you choose your weighings extremely carefully)

Answer / m

Lets start by keeping 2 balls on each side of the weighing balance.

If the balance is equal then these 4 balls are the ones which have equal weight and the ball which is of a different weight is amongst the two balls which are left....So take one of the 4 weighed balls on one side of the balance again and keep a new ball(which has still not been kept on the balance) on another side.This would result in a maximum of 3 tries to find the ball with a different weight.

If the balance is not equal, then any one of these two sides can have a ball with different weight as we are not given whether the ball is heavier or lighter than the other balls!
So, remove the balls from one side of the balance and now weigh the other two balls which were earlier on the same side, against each other.If the balance is not equal then compare each ball by weighing with one the 2 balls which were not at all kept on the balance.So here also we will require atleast 3 tries in the best case!

All in the best case we require 3 tries and in the worst case we will require 5 tries.

Answer / jyotimamuni.s

6 trys

Answer / suda

first keep 3 balls on each side and weigh.
the ball having diff weight will move up r down..consider
that side and weigh again with 2 on one side and 1 on other
side...
It will take 2 or 3 times to know the weight...

Answer / amruta

wel one can find dat d ball has a diff weight in 6!(six
factorial) ways,

Answer / karthik

NO!! It will surely take a minimum of four or more tries!!
not 2 or 3....

(According to question)


Let us take 2 balls for instance of which one is different
weight.. different in the sense it may be heavier or
lighter, as it was also specified in the question..

now if we put both of them in weighing balance, any of the
side may go to either up or down.. we never know which is
the wrong one. so we need to compare with the left out once
also..

i can solve it but need to type a lot... sorry.. but in a
simple way i can say it takes atleast 4 not 2 or 3..

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