there r 9 ball of same size,weight.one of them is defective in weight. we can use weighing m/c only 3 times?

Question

we can use the weighing machine only 3 times and olt one ball is defective

Answer 1

weight first 3 and 3 if it is equal
second weight remaining 1 by 1 if its also = the the remaining is the defective

if1st step is not equal then the reaming 3 is good so cross check with one of the 2 good one if you find both are = the the remaining is defective

Answer 2

This is not possible in just three weighing, but if we know whether the defective ball is heavier or lighter than the normal ball, we can find it out in just two weighing; otherwise, we will require four weighing.

Suppose we have the information that the defective ball is lighter then we can go as follows:

First make 3 groups of 3 balls each. Now take any two groups and weigh them. Now there are two options:
a) The groups are equal.
b) The groups are unequal.

a) IF THEY ARE EQUAL: This means that the defective ball is in the third group. Now take 2 balls from the third group. Now again two options:
i) The balls are equal.
ii) The balls are not equal.

i) THE BALLS ARE EQUAL: This means that the ball which we did not weigh is the defective ball.
ii) THE BALLS ARE NOT EQUAL: This means the lighter one is defective.

b) IF THEY ARE NOT EQUAL: This means that the ball is in the lighter group. Now do the same operation on this group that we did in the previous step.

Answer 3

It can be done this way:
first weight 3 and 3, if it is equal
second weight remaining 1 by 1 if its also equa then the remaining is the defective
OR
if the 1st step is not equal then the remaining 3 are good so cross check with one of the 2 good one if you find both are equa then the remaining is defective

Answer 4

This can actually be done for 12 balls with only 3 weighings. To find the ball which weighs differently, and whether or not it is heavier or lighter. But the procedure is far too long to put in an answer here.

Answer 5

If we know the defective one has more weight or less weight then rest then we can do that in 2 steps, otherwise we need 4 steps

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Let A1,A2,A3, B1,B2,B3, C1,C2,C3 be the 9 balls

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If we dont know the odd one is heavier or lighter.

Comparision #1 [{A1,A2,A3} , {B1,B2,B3}]
if equal go to # 2
if unequal go to # 4

#2 [{C1},{C2}]
if equal C3 is the odd one
if unequal go to #3

#3[{C1},{C3}]
if equal C2 is the odd one
if unequal C1 is the odd one

#4 [{A1,A2,A3} , {C1,C2,C3}]
if equal go to #2 and compare for the balls B
if unequal go to #2 and compare for the balls A

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If we know the defective one is heavier then (lighter one also the same case)

Comparision #1 [{A1,A2,A3} , {B1,B2,B3}]
if equal go to # 2
if unequal go to # 2 with the heavier group (if lighter one is the odd one then go with the lighter group)

#2 [{C1},{C2}]
if equal C3 is the odd one
if unequal then the heavier one is the odd one.(in case of lighter one the lighter one is the odd one)

Answer 6

Duh. Weigh 4 at a time (per side of balance), then 2 then 1.

Answer 7

okay so what is the question?