21 replies to this topic

[TimeSpaceLightForce]

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?

[David_Walker]

Spoiler for One possible way

Say that the poisons are labeled i-iv
put i on one side of the scale and iv on the other. If there is a difference, the heaviest has had the poison added. Then place ii with i and iii with iv. If a difference only appears now, the lowest side on the balance/scales contains the poisoned sample, which would be, in the case where the first two on were evenly balanced, one of the last two on the heaviest side.
Hope it makes sense, it does somewhat to me (and I guess it would only count as using it once).

[TimeSpaceLightForce]

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

 

not the beam balance below

 

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[jhawk]

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

[curr3nt]

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

Technically we have already used the scale once to get the value of 4A+x.

 

If we are allowed to I'd take three of the glasses off the scale one at a time.

 

g1 + g2 + g3 + g4 = w4

Remove g4 : g1 + g2 + g3 = w3

Remove g3 : g1 + g2 = w2

Remove g2 : g1 = w1

 

If 4 * w1 > w4 then g1 is poisoned

ElseIf 2 * w1 != w2 then g2 is poisoned

ElseIf 3 * w1 != w3 then g3 is poisoned

Else g4 is poisoned

 

If we can only retrieve two weight values where one is the total weight... not sure yet.

 

[CaptainEd]

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

I guess we are assuming that some mass has been added. So, if there's no change in volume, the vessel with the poison must have a higher density than the others. So, without even using the scale, we could perhaps measure density in each vessel, and declare the one with the highest as the poisoned one. Since we don't know what mass of poison has been added, I don't know whether this is a realistic approach or not. For example, if only one molecule of the poison has been added, we won't be able to measure the change in density.

Interesting problem, TSLF! Thanks.

[badhri]

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.

[TimeSpaceLightForce]

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

That is a good solution :(Assuming we are able to measure g2/4 , g3/2 or g3*3/4) 

Let weight be B. Then (B - 2.5A)*4/x is the number of the glass ..but less practical solution

[TimeSpaceLightForce]

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

I guess we are assuming that some mass has been added. So, if there's no change in volume, the vessel with the poison must have a higher density than the others. So, without even using the scale, we could perhaps measure density in each vessel, and declare the one with the highest as the poisoned one. Since we don't know what mass of poison has been added, I don't know whether this is a realistic approach or not. For example, if only one molecule of the poison has been added, we won't be able to measure the change in density.

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..

[CaptainEd]

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