### Puzzle 3: Weigh ’em Up!!

This one is simple…

You have eight balls and a balancing pan (the one on which you have two pans and can compare weights), one of the eight balls is either heavier or lighter than the rest. The remaining 7 have the same weight. How many least no of weight comparisons will it take to find out the defective ball and also weather it is heavier or lighter.

Also tell how you achieved at the number… mail your detailed reply’s to **gurus [dot] comments [at] gmail [dot] com **and put up the least number you required here as a comment ğŸ™‚

Solution (by ross):

put 2 balls on each side**(1st weigh)**, if they balance put them to one side and mark as OK if they don’t balance then mark the other 4 as OK.

take 3 of the 4 NOT OK balls, put 2 on one side and 1 on the other with 1 of the balanced balls **(2nd weigh)**. if they balance the only ball that hasn’t yet been on the scales is DEFECTIVE, weight it against one of the OK balls**(3rd weigh)** to see if it’s heavier or lighter…

if the balls don’t balance on the second weigh then take the known OK ball and the one that you put with it (let’s call it ball A) off then weigh **(3rd weigh)** the other 2 balls (call them B & C). If B & C balance then A is the defective ball. We now know that B & C are OK balls so if A & OK were heavier on the 2nd weigh then A is HEAVY DEFECTIVE, if A & OK were lighter on the 2nd weigh then A is LIGHT DEFECTIVE…

If B & C don’t balance on the **(3rd weigh)** then we know one of them is DEFECTIVE so we know A is OK. If B & C were heavier than A & OK **(on the 2nd weigh)** then we know the defective ball must be HEAVY so which ever is heavier out of B & C is the HEAVY DEFECTIVE ball. If B & C were lighter than A & OK **(on the 2nd weigh)** then we know the defective ball must be LIGHT so which ever is lighter out of B & C is the LIGHT DEFECTIVE ball.

Only 4 are required.

I can do it in 3

ok, 87.5% i can do it in 3. 12.5% of the time I’d need an extra go.

[…] Puzzle 3: Weigh ‘em Up!! [image] [image] […]

I can do it in 2 ğŸ˜‰

I can also do it in 2 [:p]

I can also do it in 2 ğŸ˜€

If I did not make it clear enough it is

notknown if the last ball is heavier or lighter then the remaining 7.None of you got the right answer yet… (You have to mail the detailed solution at gurus.comments[at]gmail[dot]com )

i think the trick in these kind of questions is not to think symmetrically (cudn’t find a good word to put it)

ok, i sent the mail… check it ğŸ™‚

wrong answer dark…

You are assuming that the 8th ball is heavy whereas it can also be lighter you have to find both the defective ball and also if it is heavier or lighter

Change of plan, I can do it in 3

well 3 is the right answer…

just tell how

mail sent.

man i hate puzzles.. why use unnecessarily…:(

3 is right… Assume that one ball is weighs less.. Place 4 balls on either side.. Take the 4 which weighs on the lesser side.. seperate into twos… weigh the same.. Again select the two which weighs less.. weigh the two.. You have got the ball with the minimum weight. The reverse process if the ball weighs more than the others.

Oops… I am basically wrong… Even at the first instance we do not know whether the odd ball weighs more or less than the others.

hope this solves the puzzle…

say the balls are ABCDEFGH

1st weigh –> 2nd weigh –> 3rd weigh

(ABC == DEF) –> (G == A) –> (H (H > A)……H is heavier

–> (G (G > A)………………………..G is heavier

(ABC > DEF) –> (ADG == BEH) –> (C == A)……F is lighter

–> (C != A)……C is heavier

–> (ADG > BEH) –> (A == C)……E is lighter

–> (A != C)……A is heavier

–> (ADG (D == C)……B is heavier

–> (D != C)……D is lighter

(ABC (ADG == BEH) –> (C == A)……F is heavier

–> (C != A)……C is lighter

–> (ADG > BEH) –> (D == C)……B is lighter

–> (D != C)……D is heavier

–> (ADG (A == C)……E is heavier

–> (A != C)……A is lighter

sorry for the formating above…

say the balls are ABCDEFGH

1st weigh –> 2nd weigh –> 3rd weigh

————————————————————————–

(ABC == DEF) –> (G == A) –> (H (G == A) –> (H > A)……H is heavier

————————————————————————–

(ABC == DEF) –> (G (G > A)…………………..G is heavier

————————————————————————–

(ABC > DEF) –> (ADG == BEH) –> (C == A)……F is lighter

(ABC > DEF) –> (ADG == BEH) –> (C != A)……C is heavier

————————————————————————–

(ABC > DEF) –> (ADG > BEH) –> (A == C)……E is lighter

(ABC > DEF) –> (ADG > BEH) –> (A != C)……A is heavier

————————————————————————–

(ABC > DEF) –> (ADG (D == C)……B is heavier

(ABC > DEF) –> (ADG (D != C)……D is lighter

————————————————————————–

(ABC (ADG == BEH) –> (C == A)……F is heavier

(ABC (ADG == BEH) –> (C != A)……C is lighter

————————————————————————–

(ABC (ADG > BEH) –> (D == C)……B is lighter

(ABC (ADG > BEH) –> (D != C)……D is heavier

————————————————————————–

(ABC (ADG (A == C)……E is heavier

(ABC (ADG (A != C)……A is lighter

Hey i cud do it 3.. u gotta put 3 – 3 balls at a time and if they balance its gonna be one one of the remaining 2.. and then one of those and one of the first 6..

and if they don balance.. its one one of the 3 in any one.. and if that also does not balance replace one with the third one

3rd posibility wud be 3 -3 in first trial, they dont balance and then 1-1 and they balance and then 3rd one with any of the remaining ones..

am i right fiery shadow??

3 is correct but your last part explanation is not complete you also have to tell weather it is heavier than the others or lighter…