[mewminator]

This one is easy: you have 9 balls, one is heavier than the rest. You are provided with a scale balance.

How can you identify the heavier one by only using the balance twice.

N.B: the weight difference is not noticeable by hand, you cannot use any other tools.

Put 3 on each plate of the scale. The heavier side has the heavier ball, if both sides have equal weight then it is one of the excluded 3. Put one on each plate and the heavier one is the ball, if not then the heavy ball is the excluded one

[Guest]

you place 3 balls on both sides and if they balance out u take two from the remaining pile and weigh them.. if they balance then it's the remaining one if not then it's the heavier one... case 2 you place 3 on both sides if they don't balance u take the heavier side and take 2 from the pile and weigh.. like the first case: if they are the same then it's the remaining and if not then it's the heavier:)

.... just to point out this is an old riddle and i'm pretty sure it's been posted before so when u post smth try looking for it in the search to see if it was already posted that way the same riddles aren't posted twice or many times... no biggie just check before posting, and welcome to brainden:D!

Edited October 7, 2011 by guppy

[mewminator]

you place 3 balls on both sides and if they balance out u take two from the remaining pile and weigh them.. if they balance then it's the remaining one if not then it's the heavier one... case 2 you place 3 on both sides if they don't balance u take the heavier side and take 2 from the pile and weigh.. like the first case: if they are the same then it's the remaining and if not then it's the heavier:)

.... just to point out this is an old riddle and i'm pretty sure it's been posted before so when u post smth try looking for it in the search to see if it was already posted that way the same riddles aren't posted twice or many times... no biggie just check before posting, and welcome to brainden:D!

i only saw a similar one with 12 balls and using the balance twice

[mewminator]

i only saw a similar one with 12 balls and using the balance twice

sorry use the balance 3 times

[Guest]

Group the 9 balls into 3 smaller groups. (Group A, B & C)

Weighing No.1 : Weigh A & B, whichever is heavier the odd ball is in that group. If A & B are equal in weight definitely Group C has the heavier ball.

Weighing No.2 : For the remaining group (whatever it is) just do the same. Weigh 2 balls and whichever is heavier concludes to be the odd ball, if the first two balls are equal in weight, therfore the last remaining ball is the odd ball.

[Guest]

i only saw a similar one with 12 balls and using the balance twice

i'm sry i tried looking for it as well.. i saw many similar ones .. some with 10 and 8 but i couldn't find the nine, well i did but the objectives and rules were different.

i was so sure that i had seen it here before.. sry i really didn't mean anyth by it. guess no one here has posted it yet glad to see u really did check before posting

[Guest]

sorry use the balance 3 times

2 times only

lets mark them as A B C D E F G H I

first time- take A B C vs D E F if both are bal than heavier one is fro G H I

second time -take G Vs H if both are same than I , otherwise the havierone from GH

if ABC or DEF is not bal than u will find the heavier ball in ABC or DEF as the case may be, than go as per second take above for G H I