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Lethal Solution


Best Answer CaptainEd, 17 December 2012 - 07:53 PM

Spoiler for OK, a=a solution=solution with=with no=no pouring=pouring at=at all'=all
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21 replies to this topic

#1 TimeSpaceLightForce

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Posted 17 December 2012 - 09:54 AM

On top of a digital weighing scale
are four identical glasses, filled with
water, weighs A grams each. Then,
somehow the poison X was sneaked
into one of them with no change in
color, odor  or volume.

  \       /    \       /     \       /    \      /
    \__/       \__/        \__/      \__/
==========================
               [4 A + x ]


Using the scale once.Can you find out
which of the glasses is poisoned?


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#2 David_Walker

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Posted 17 December 2012 - 11:28 AM

Spoiler for One possible way


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#3 TimeSpaceLightForce

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Posted 17 December 2012 - 04:00 PM

=========================

           _____||||_____

         /      (_____)       \

        [____________._]

           ''                   ''

 

Pls. use the weighing scale w/ LCD display  above,

 

not the beam balance below

 

============                = ============

          ||                                        ||

          "=========[  ]========="

                        ||||'''''''''''||||


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#4 jhawk

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Posted 17 December 2012 - 04:38 PM

I'm not sure I would accept this challange. But if I had to , place two glasses on scale. Either these two, or the other two weigh 2A+x. We now know that the poison is one of of the two heavier pairs. Now the hard part : drink one of these glasses. If you don't die, you know the other glass is poison. If you die, you go to your grave knowing you were also correct
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#5 curr3nt

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Posted 17 December 2012 - 05:29 PM

Spoiler for can you expand on the condition of use the scale once?

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#6 CaptainEd

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Posted 17 December 2012 - 05:40 PM


Spoiler for no change in volume...hmmm...

Interesting problem, TSLF! Thanks.
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#7 badhri

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Posted 17 December 2012 - 05:47 PM

Not sure!!!!

 

Take away all the glassess first (since i have to use it only once).

 

Let me have the glassess as g1,g2,g3,g4

 

pour one quater the amount of g2 to g1 say g1+g2/4

 

same half and three fourth of 3rd and 4th respectively

 

so the first glass will now have g1+g2/4+g3/2+g3*3/4 (Assuming we are able to measure g2/4 , g3/2 or g3*3/4)

 

now weigh that glass if the weight gonna be

CASE 1 :(A+X)+A/4+A/2+3A/4 then g1 is poisoned

CASE 2: A+(A+X)/4+A/2+3A/4 then g2 is poisoned

CASE 3 : A+A/4+(A+X)/2+3A/4 then g3 is poisoned

CASE 4: A+A/4+A/2+3(A+X)/4 then g4 is poisoned


Edited by badhri, 17 December 2012 - 05:53 PM.

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#8 TimeSpaceLightForce

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Posted 17 December 2012 - 06:11 PM

Not sure!!!!

 

Take away all the glassess first (since i have to use it only once).

 

Let me have the glassess as g1,g2,g3,g4

 

pour one quater the amount of g2 to g1 say g1+g2/4

 

same half and three fourth of 3rd and 4th respectively

 

so the first glass will now have g1+g2/4+g3/2+g3*3/4 (Assuming we are able to measure g2/4 , g3/2 or g3*3/4)

 

now weigh that glass if the weight gonna be

CASE 1 :(A+X)+A/4+A/2+3A/4 then g1 is poisoned

CASE 2: A+(A+X)/4+A/2+3A/4 then g2 is poisoned

CASE 3 : A+A/4+(A+X)/2+3A/4 then g3 is poisoned

CASE 4: A+A/4+A/2+3(A+X)/4 then g4 is poisoned

Spoiler for


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#9 TimeSpaceLightForce

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Posted 17 December 2012 - 06:17 PM

Spoiler for no change in volume...hmmm...

Interesting problem, TSLF! Thanks.

Right CatpatinEd in terms of Physics..but say we have a simple chemistry problem and 1:100  x to A ratio is observed..


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#10 CaptainEd

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Posted 17 December 2012 - 06:21 PM

I totally agree--Badhri's principle is much more practical in terms of masses.
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