Answer / baljeet singh

minimum 2

Answer / ashok

min 1
max 3

Answer / revathi

two

Answer / ajay

Min 2
Max3
First div into 2grp of5-5 then weight it take that grp which weight is more
Then again divide more weighted grp into 3grp 2 2 1
Then weighed it if equal then took reaming two from three if reaming also equal then reaming only one definitely different
Which is required or we get it earlier any one of the step we get the req different ball

Answer / ashok timothy

1

Answer / sumit bakshi(b.c.roy engg. col

ans:1
At first we divided the 9 balls in two gr. let gr.1 & gr.2 .then simillarly divide gr.1 balls into gr.3 & gr.4.Again gr. 2 balls are divide into gr.5 & gr.6 .Then we get atleast one disimilar ball!

Answer / lava kumar

Max=4;
Min=1;

Answer / mopurizwarriors

anwer is zero!!

here is the explanation:
first divide the balls into almost two equal groups!
i.e.,5 and 5 balls in 2 groups.
then compare the weights of two groups:
again,divide the first and second groups into almost equal
groups!!(say them as 1,2,3,4 groups,where 1,2 are from first
group and 3,4 are from second group).
let group-1 contains 3 balls and group-2 contain 2 balls
find weights of group-1 and group-2.say x,y
so,average weight of balls in group-2 is x/3.
if 2*(x/3)=y,then that odd ball is not present in first
group(don't confuse group-1 with first group).
so,that odd ball may be present in either group-3 or group-4.
continue,the process until u get groups containing only one
ball.

IN THIS METHOD ,YOU HAVEN'T USED ANY WEIGHTS!!!

SO,THE ANSWER IS:zero

Box 1 cotain odd value X or Y the box 2 =X i.e 7 or 12 box 3= Y has value 19 if X is 7 or 31 if x is 12 get even value X or Y . Use any operator, steps allowed are 3.

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