Can You Solve This
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10 Bags with 100 coins each Each coin is 10 gms in 9 bags Each coin is 9 gms in one bag (you dont know which ONE bag) You have a weiging machine which gives you the exact weight
You can use the weighing machine EXACTLY ONCE
How will you fnd which bag has coins of 9 gms
You have time till the evening Ask me if any doubt about the question
[DON"T SEE THE ANSWERS BELOW IF YOU SOLVED IT BY YOURSELF DO TELL BE HOW MUCH TIME IT TOOK] "Not everything that counts can be counted..." -Albert Einstein -- modified at 13:03 Monday 24th October, 2005
Wild guess here. Well, technically one weighing, using the scale once, would mean to have all 10 bags on the scale. So by putting one bag on the scale at a time,you would be able to see which bag does not keep the scale balance. Or if the incremental weight is not linear from one bag would also do the trick. i.e., 9 bags are 1000gms, 1 bag is 900gms. SO if I put 1 1000gms on a scale and then the next bag is the 900gms, it will make the scale off balance or the incremental weight will be 100gms less. Does this make any sense?
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The five dots are not a red herring... however my headache means I am not the best to predict the results this evening. Consider this: Arrange three pens or pencils on a table, or tooth picks if you prefer in a triangle shape:
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/ \This represents three of your points and one plane. Now hold a single pen/pencil or toothpic over the top of the triangle suspended in mid air. Imagine building up a connecting object by connecting all points to all points. The result is an object with six sides, but a very odd shape.
/\
/ \Each of those planes connects once at your object, with I believe 9 lines where the planes cross. However the planes are infinately long. So the question is, does any of those planes cross anywhere else other than at the shape in the center. That is where my headache cannot go beyond today. _________________________ Asu no koto o ieba, tenjo de nezumi ga warau. Talk about things of tomorrow and the mice in the ceiling laugh. (Japanese Proverb)
Jeffry J. Brickley wrote:
The five dots are not a red herring...
Yeah yeah. I misunderstood the original question. Like I said to Quartz, if this question had been on a test, I would have complained loudly to the prof that his question sucked and that he needed to clarify on the board. ;P And I think he probably meant that that planes are perfectly "straight", so they probably won't intersect anywhere but in the center.
Picture a huge catholic cathedral. In it there's many people, including a gregorian monk choir. You know, those who sing beautifully. Then they start singing, in latin, as they always do: "Ad hominem..." -Jörgen Sigvardsson
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sounds smarter than me, and follows along the number of shapes and planes I was counting as I add each additional combination of 3. _________________________ Asu no koto o ieba, tenjo de nezumi ga warau. Talk about things of tomorrow and the mice in the ceiling laugh. (Japanese Proverb)
Yea, but it is wrong!:( Rui A. Rebelo I don't smoke, don't gamble, don't sniff, don't drink and don't womanize. My only defect is that I lie just a little bit, sometimes. Tim Maia (brazilian pop singer)
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no more than three at a time can have a single flat surface pass through them :suss: ???? "Not everything that counts can be counted..." -Albert Einstein
18 [edit]Ooops it's 19 (forgot the vertical line)[/edit] The explanation is left as an exercise for the reader...:laugh: Ok: imagine 2 tetrahedrons (piramids) joined by one face (the whole thing looks like a diamond). Thats 9 lines. Now overlap it with 3 vertical planes sharing only one intersecting line which crosses the top and bottom vertices of the diamond above... Oh well, you'd need a CAD system... Rui A. Rebelo I don't smoke, don't gamble, don't sniff, don't drink and don't womanize. My only defect is that I lie just a little bit, sometimes. Tim Maia (brazilian pop singer) -- modified at 16:31 Monday 24th October, 2005
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For all the Mathematicians OUT THERE
Five dots are arranged in space so that no more than three at a time can have a single flat surface pass through them. If each group of three dots has a flat surface pass through it and extend an infinite distance in every direction, what is the maximum number of different lines at which these surfaces may intersect one another?
Don't worry, even from here I can clearly hear you say, "Huh?"
:)
"The most beautiful thing we can experience is the mysterious" -Albert Einstein -- modified at 13:25 Monday 24th October, 2005
Place 3 points in a plane in a equilateral triangluar arrangement. This is triangle ABC. Now place the other two points, D and E, exactly in the middle of triangle/plane ABC and move them outward in opposing directions from that plane. This describes a non-rectangular volume with six sides. Now, no more than three of the five points in this volume occupy the same plane. Count the lines formed by each of the triangles joined by points A, B, C, D, E.
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Wild guess here. Well, technically one weighing, using the scale once, would mean to have all 10 bags on the scale. So by putting one bag on the scale at a time,you would be able to see which bag does not keep the scale balance. Or if the incremental weight is not linear from one bag would also do the trick. i.e., 9 bags are 1000gms, 1 bag is 900gms. SO if I put 1 1000gms on a scale and then the next bag is the 900gms, it will make the scale off balance or the incremental weight will be 100gms less. Does this make any sense?
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Place 3 points in a plane in a equilateral triangluar arrangement. This is triangle ABC. Now place the other two points, D and E, exactly in the middle of triangle/plane ABC and move them outward in opposing directions from that plane. This describes a non-rectangular volume with six sides. Now, no more than three of the five points in this volume occupy the same plane. Count the lines formed by each of the triangles joined by points A, B, C, D, E.
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So what is the answer ?????? "Not everything that counts can be counted..." -Albert Einstein
The answer is 10 Lines AB, AD, AE, AC and BD, BE, BC and CD, CE and DE
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The answer is 10 Lines AB, AD, AE, AC and BD, BE, BC and CD, CE and DE
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Check Clue for the Clueless Below ! "Not everything that counts can be counted..." -Albert Einstein
Ok. It's 19, check my answer in Clue for the Clueless.:) Rui A. Rebelo I don't smoke, don't gamble, don't sniff, don't drink and don't womanize. My only defect is that I lie just a little bit, sometimes. Tim Maia (brazilian pop singer)
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Thats the same as the question "How many unique combinations of 5 items in groups of three", so the number of planes: 5!/(5-3)!/3! = (5*4*3*2*1)/(2*1)/3*2*1 = 5*4/2 = 20/2 = 10 -- modified at 17:19 Monday 24th October, 2005
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Thats the same as the question "How many unique combinations of 5 items in groups of three", so the number of planes: 5!/(5-3)!/3! = (5*4*3*2*1)/(2*1)/3*2*1 = 5*4/2 = 20/2 = 10 -- modified at 17:19 Monday 24th October, 2005
[Message Deleted]
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Ok. It's 19, check my answer in Clue for the Clueless.:) Rui A. Rebelo I don't smoke, don't gamble, don't sniff, don't drink and don't womanize. My only defect is that I lie just a little bit, sometimes. Tim Maia (brazilian pop singer)
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[Message Deleted]
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hahahaha good approach though "Not everything that counts can be counted..." -Albert Einstein
see my edited response
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18 [edit]Ooops it's 19 (forgot the vertical line)[/edit] The explanation is left as an exercise for the reader...:laugh: Ok: imagine 2 tetrahedrons (piramids) joined by one face (the whole thing looks like a diamond). Thats 9 lines. Now overlap it with 3 vertical planes sharing only one intersecting line which crosses the top and bottom vertices of the diamond above... Oh well, you'd need a CAD system... Rui A. Rebelo I don't smoke, don't gamble, don't sniff, don't drink and don't womanize. My only defect is that I lie just a little bit, sometimes. Tim Maia (brazilian pop singer) -- modified at 16:31 Monday 24th October, 2005
wrong!
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hahahaha good approach though "Not everything that counts can be counted..." -Albert Einstein
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hahahaha good approach though "Not everything that counts can be counted..." -Albert Einstein
ok the answer is 10 planes 5*4*3*2*1/3*2*1/2*1 = 5*4/2 = 20/2 = 10
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ok the answer is 10 planes 5*4*3*2*1/3*2*1/2*1 = 5*4/2 = 20/2 = 10
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Can you draw a picture of 10 faces and 10 intersection "Not everything that counts can be counted..." -Albert Einstein
In three dimensional space? Uh, no, I don't think so. By definition a face has to be visible and exposed to the "air" or vacuum as you may choose. Since some of the "faces" (or planes) would be embedded within the six sided figure I described earlier then it can't be drawn in three dimensional space. Unless there is a different three-dimensional figure which fits the problem.