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# Weighing VI.

### #11

Posted 15 March 2008 - 06:32 AM

### #12

Posted 21 March 2008 - 04:52 PM

### #13

Posted 23 March 2008 - 01:52 AM

and i also thought the other variable was wind resistance

### #14

Posted 26 March 2008 - 07:14 PM

Select 6 balls and Weigh 3 on one side and 3 on the other. Select the pile that is heaviest. of they are the same weight then the Heavy ball is in the pile you didn't weigh.

Now you have 3 balls one of witch is the heavy ball Select 2 balls adn compare there weight. if nether of them are the heavy ball the one not weighed was.

Yupp!! thas the right answer.. groups of 3 and proceed.. bout its interesting to note that the question mentioned one ball to be bit heavier. Had it been lighter or heavier.. then finding it out in 2 chances is not possible, unless u get lucky in the first try.

\m/

### #15

Posted 29 March 2008 - 05:48 AM

Yupp!! thas the right answer.. groups of 3 and proceed.. bout its interesting to note that the question mentioned one ball to be bit heavier. Had it been lighter or heavier.. then finding it out in 2 chances is not possible, unless u get lucky in the first try.

\m/

Ex-squeeze me? Baking powder? What are you trying to say there?

If you don't get the baking powder reference click here.

### #16

Posted 22 April 2008 - 05:22 AM

put the 2 groups on the scale, if they are of equal weight, then it has to be the last ball. If they are different weight, split the heavier group into 2 groups of 2, then with the heavier group; instead of using the scale; grab the heavier group and its only a matter of weighing them yourselves (use your hands and see which one is heavier) to find the ball; or guess... you have a 50 50 chance.

It said nothing against using yourself as a weight device.

The divide by 3 method is the best because it gives you the answer, but 2 groups of 4 was my first guess so i felt like making it my final and only guess.

### #17

Posted 09 October 2008 - 04:52 AM

Whichever trio is heavier (the left out set if the two sides of the scale were equal), take 2 of those balls (one on each side of the scale). The heaviest will be either one of those balls, or the left out ball if the two on the scale are equal...

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