YART: Yet Another Riddle Thread

[GeneValgene]

friend asked me this...and it's killing me:

I have 10 machines, which produce ball-bearings of 10 grams each.

But one of the machines is defective and produces ball-bearings of 9 grams.

If the machines are numbered from 1 to 10, then how will you identify the defective machine, provided

1. You can take any number of ball-bearings from any machine

2. You are given a weight balance and you can use it only once (remember only once)

 

[JDrake]

Guesstimate w/ the remaining five?

 

[wvtalbot]

Take 1 from #1, 2 from #2, 3 from #3 and so on.....weigh them all, it is simple to figure out the answer.

 

[isasir]

Originally posted by: wvtalbot
Take 1 from #1, 2 from #2, 3 from #3 and so on.....weigh them all, it is simple to figure out the answer.

Yup. If you're off by 1 gram it's the first machine, 2 grams (assuming you put 2 weights from #2) machine 2, etc.

 

[tvbi]

Take one ball from each machine and label them then put 5 on each side.. the lighter size will have the defected ball. Start take one ball off on each size at the same time.. when the scale balance (it wasn't ballanced before because of the defected ball).. when this happen, the ball that you just took off on the lighter side is the defective one.

 

[Schfifty Five]

Ohh, I heard this one before but forgot the answer >_<

edit: Sweet, someone posted the answer, now i remmeber lol

 

[dullard]

Originally posted by: wvtalbot
Take 1 from #1, 2 from #2, 3 from #3 and so on.....weigh them all, it is simple to figure out the answer.

 

[GeneValgene]

Originally posted by: isasir

Originally posted by: wvtalbot
Take 1 from #1, 2 from #2, 3 from #3 and so on.....weigh them all, it is simple to figure out the answer.

Yup. If you're off by 1 gram it's the first machine, 2 grams (assuming you put 2 weights from #2) machine 2, etc.

dayum

you are genius...i feel really stupid now

 

[QED]

Originally posted by: dullard

Originally posted by: wvtalbot
Take 1 from #1, 2 from #2, 3 from #3 and so on.....weigh them all, it is simple to figure out the answer.

I thought you were supposed to use a weight balance, not a scale...

If you really only can make one use of the weight scale (i.e. just one comparision between two quantities of ball bearings) than this becomes a more difficult problem.

 

[dullard]

Originally posted by: MathMan
I thought you were supposed to use a weight balance, not a scale...

If you really only can make one use of the weight scale (i.e. just one comparision between two quantities of ball bearings) than this becomes a more difficult problem.

Put all ball-bearings on one side. Put 550 grams on the other side. Problem solved.

 

[tvbi]

so ..was my solution wrong?

 

[ucdbiendog]

Originally posted by: MathMan

Originally posted by: dullard

Originally posted by: wvtalbot
Take 1 from #1, 2 from #2, 3 from #3 and so on.....weigh them all, it is simple to figure out the answer.

I thought you were supposed to use a weight balance, not a scale...

If you really only can make one use of the weight scale (i.e. just one comparision between two quantities of ball bearings) than this becomes a more difficult problem.

i was thinking the same thing

 

[QED]

Originally posted by: dullard

Originally posted by: MathMan
I thought you were supposed to use a weight balance, not a scale...

If you really only can make one use of the weight scale (i.e. just one comparision between two quantities of ball bearings) than this becomes a more difficult problem.

Put all ball-bearings on one side. Put 550 grams on the other side. Problem solved.

The point of a balance is that you only know whether the items on the left weigh more than, less than, or equal to the items on the right-- but you don't know by how much they differ, though-- and that is key to the proposed solution.

 

[dullard]

Originally posted by: MathMan
The point of a balance is that you only know whether the items on the left weigh more than, less than, or equal to the items on the right-- but you don't know by how much they differ, though-- and that is key to the proposed solution.

Doh. I was thinking of a similarly worded problem and didn't read. I take it all back.

 

[Mo0o]

This kind of problem pops up quite frequently in riddle lists

 

[wvtalbot]

Originally posted by: MathMan

Originally posted by: dullard

Originally posted by: MathMan
I thought you were supposed to use a weight balance, not a scale...

If you really only can make one use of the weight scale (i.e. just one comparision between two quantities of ball bearings) than this becomes a more difficult problem.

Put all ball-bearings on one side. Put 550 grams on the other side. Problem solved.

The point of a balance is that you only know whether the items on the left weigh more than, less than, or equal to the items on the right-- but you don't know by how much they differ, though-- and that is key to the proposed solution.

I am pretty sure he meant a scale, by using your definition of a balance there is no solution to the problem.