this is the last time i'm gonna post this riddle..i hope i get a correct answer this time?

Question

you have 12 balls same size and weight except one that is different, may be lighter may be heavier you don't know..
you have a balance scale where you can compare the weight of the balls, you're allowed to use it only 3 times to figure out the different ball.. can you? no tricks i assure you...

remember that you dont know whether the odd ball is heavier or lighter, you just know that it's different...maybe lighter maybe heavier.

excellent , dscougar!!!!
i was gonna say so

a.amitkumar: super job! this is the correct answer, you got the idea right but it can be simpler:
if a&b come different and a is heavier, take 125 vs 346
see which group is heavier and weigh the two possible heavy balls together, if different it's the heavier if same it's the lighter from the other group"5"or "6"
if 125vs346 come same then weigh 7vs8 for the light one
if at the beginning a&b came same weigh any two of c if same weigh one of them vs a new one if different it's the new on if same it's the last one left..
if the first two of c come different weigh one of them vs any old ball if different that's it if same it's the other one...

Answer 1

Label the balls {1,2,3,4,5,6,7,8,9,10,11,12}. Weigh A={1,2,3,4} against B={5,6,7,8}. Use A< B to dentoe A is lighter than.

If A< B (a similar situation if A> B) let C={1,2,3,5} and D={4,9,10,11}.
If C< D, the faulty (light) ball is in {1,2,3}. Weigh 1 vs. 2 to resolve.
If C=D, the faulty (heavy) ball is in {6,7,8}. Weigh 6 vs. 7 to resolve.
If C> D, then either 5 is heavy or 4 is light. Weigh 5 vs. 12 to resolve.

If A=B then you have the remaining 4 balls {9,10,11,12} and 2 weighings.
Let C={9,10,11} and D={1,2,3} and weigh C vs. D.
If C< D, weigh 9 vs. 10 to determine which ball is light.
If C> D, weigh 9 vs. 10 to determine which ball is heavy.
If C=D, weigh 12 vs. 1 to determine whether it is heavy or light.

You have fun and your riddles are welcome... but should give preference for posting a solution so soon :-)

Answer 2

But what if the first set of 4 and 4 turns out to be unequal? Then you have to remove one and test the last one...let's say you leave the heavier one on the balance and find that it's also heavier than the last set; then you know it has the defect, but now you're on step 2 of Richmond Girl's idea but step 3 in reality! It would only work in 3 if you happen to pick the two sets of 4 without the defect, right?

Answer 3

Part of this test is whether you're confident enough too take a risk on the answer.

No you can't do it in three weighings. You could if you knew whether the errant ball was lighter or heavier. Then you would sort four balls into three groups. Weigh two groups. The first weighing would eliminate two groups or eight balls. The remaining four balls would be weighed eliminating two balls leaving two. You weigh the remaining two and you find your errant ball by identifying it as heavy or light.

Answer 4

Six on one side, six on other. O wait. Got to go read Sherlock Holmes to family for night. I'll ask him. (Maybe I'll figure it out meantime.

Oh, wait. It's coming to me. Weigh four against 4. If they are same the odd is in remaining 4. If not, swap twoozies.

Answer 5

But wait- rich girl, if the first two groups of 4 are uneven, and you don't know whether the oddball is lighter or heavier, how do you know which of the two groups contains the oddball?

Answer 6

Take 4,4,4 balls .
step1: check for 4 one side & 4 other side in the pan .
{ case : if both are equal , defect in the last 4 .
step2: check last 4 , in the pan like 2 & 2 both the sides .
then u wil get the 2 balls with the defect one.
step 3: then again check for the last 2 like 1 & 1 .
u wil get the last one .
}
if both the 4 & 4 are not equal , then take the defect one , and repeat it from Step 2.

Answer 7

number d balls from 1 to 12
now divide d balls in 4 equal groups:
A(1,2,3)
B(4,5,6)
C(7,8,9)
D(10,11,12)

weigh A & B
if A>B or A if A=B then it means that either C or D group has d odd ball
in dis way u can find any 2 groups one of which has d odd ball.
now, suppose d odd ball lies in either group A or B
den weigh any one of dem (say A) with C or D (say C)
if A=C den B contains d odd ball
if AC den A contains d odd ball weighing less or more resp. than d other balls
since A contains ball no. 1,2,3
so compare d weights of 1 & 2
if 1 =2 den 3 is d odd one
if 1<2 or 1>2 and d odd ball weighs less or more as judged earlier den 1 or 2 resp. is d odd ball in case of less
or 2 or 1 resp. is d odd ball in case of more.
similarly group B can also b checked in case of B containing d odd ball.

Answer 8

Really? This is the last time? The very last time? The very VERY last time.

Yes?

Well, that IS a relief, dearie.

Answer 9

pick up each ball. The one you think is lightest measure it with one you think is the heaviest. Then vis-versa....idk lol

Answer 10

richmond girl got it. my tennis instructor taught that one to me, except we did it with tennis balls, and we used the racket as a scale. lol.