Lethal Solution

[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]

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]

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]

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 December 17, 2012 by badhri

[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

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]

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.

[TimeSpaceLightForce]

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

..too many moves and uncertainties

[famous1304]

If using the LCD screen, I would put all 4 on the scale, and remove one until one took off more weight than the others. only using the scale once. Or you could go the opposite route, and you would have to put them all on the scale one at a time as to not spill. So just watch the weights as you put a new cup on and you would know which one had the poison.

[CaptainEd]

Badhri's principle is to construct a situation with differing fractions of the four glasses, so that we can compute which glass it was that had added mass.

However, the OP says that the glasses are "full", which means there's no room to add from one into another.

So, here's an alternative:

(1) I gather that we are given:

* A, the weight of one glass full of water

* X, the weight of added poison.

(2) remove all glasses from the scale, without peeking at the scale.

(3) Using a volume measuring device, such as a pipet, measure the volume in G1, and empty it, call that volume V.

(4) measure V/6 from G2, transfer to G1

(5) measure V/3 from G3, transfer to G1

(6) measure V/2 from G4, transfer to G1

(7) At this point, G1 is full

(8) place G1 on the scale.

(9) If G1 weighs A, original G1 had the poison

--if G1 weighs A + X/6, original G2 had the poison

--if G1 weighs A + X/3, original G3 had the poison

--if G1 weighs A + X/2, original G4 had the poison

[TimeSpaceLightForce]

If using the LCD screen, I would put all 4 on the scale, and remove one until one took off more weight than the others. only using the scale once. Or you could go the opposite route, and you would have to put them all on the scale one at a time as to not spill. So just watch the weights as you put a new cup on and you would know which one had the poison.

Your solution is right unless making the machine do its job means using it

(in that case reading weight trice). It is resting if nothing is on it. If there is

a reading already using once means having a new reading..

Use the scale tray if you would like take them off or place them on the scale

Edited December 17, 2012 by TimeSpaceLightForce

[CaptainEd]

on the laboratory bench next to the digital scale, place a knife edge as a fulcrum, and balance a plank of length 2 meters on the knife edge, with one end touching the digital scale.

At this point, the plank is adding exactly no weight to the scale.

Now,

* place G1 on the plank directly over the scale,

* place G2 on the plank at a distance of 1/2 from the fulcrum

* place G3 on the plank at a distance of 1/3 from the fulcrum

* place G4 on the plank at a distance of 1/4 from the fulcrum

* read the scale

* Without the poison, the scale would read G1 + G2*1/2 + G3*1/3 + G4*1/4 will be (A+A/2+A/3+A/4) =

(12A + 6A + 4A + 3A)/12 = (25A/12)

However, the poison of weight X will either appear as an addition of X (if in G1), X/2 (if in G2), X/3 (if in G3) or X/4 (if in G4).

(Firefox doesn't seem to play well with this spoiler editor.)

Edited December 17, 2012 by CaptainEd

[TimeSpaceLightForce]

Badhri's principle is to construct a situation with differing fractions of the four glasses, so that we can compute which glass it was that had added mass.

However, the OP says that the glasses are "full", which means there's no room to add from one into another.

So, here's an alternative:

(1) I gather that we are given:

* A, the weight of one glass full of water

* X, the weight of added poison.

(2) remove all glasses from the scale, without peeking at the scale.

(3) Using a volume measuring device, such as a pipet, measure the volume in G1, and empty it, call that volume V.

(4) measure V/6 from G2, transfer to G1

(5) measure V/3 from G3, transfer to G1

(6) measure V/2 from G4, transfer to G1

(7) At this point, G1 is full

(8) place G1 on the scale.

(9) If G1 weighs A, original G1 had the poison

--if G1 weighs A + X/6, original G2 had the poison

--if G1 weighs A + X/3, original G3 had the poison

--if G1 weighs A + X/2, original G4 had the poison

if you have estimated half a glass and it weighs more than half the water weight, the additional weight can be water or poison's..thats why the volume must be precisely measured as done here..still that is troublesome.. the glass weight shall be considered too in that case

[TimeSpaceLightForce]

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

This is one solution if you have a goldfish in the bowl

Edited December 18, 2012 by TimeSpaceLightForce

[TimeSpaceLightForce]

on the laboratory bench next to the digital scale, place a knife edge as a fulcrum, and balance a plank of length 2 meters on the knife edge, with one end touching the digital scale.

At this point, the plank is adding exactly no weight to the scale.

Now,

* place G1 on the plank directly over the scale,

* place G2 on the plank at a distance of 1/2 from the fulcrum

* place G3 on the plank at a distance of 1/3 from the fulcrum

* place G4 on the plank at a distance of 1/4 from the fulcrum

* read the scale

* Without the poison, the scale would read G1 + G2*1/2 + G3*1/3 + G4*1/4 will be (A+A/2+A/3+A/4) =

(12A + 6A + 4A + 3A)/12 = (25A/12)

However, the poison of weight X will either appear as an addition of X (if in G1), X/2 (if in G2), X/3 (if in G3) or X/4 (if in G4).

(Firefox doesn't seem to play well with this spoiler editor.)

I can not say no to this one CaptainEd it is so resourceful and using mechanics is cool..using knife and bench and plank.

You might as well look at the corner cabinet for the poison scanner

[CaptainEd]

I can not say no to this one CaptainEd it is so resourceful and using mechanics is cool..using knife and bench and plank.

You might as well look at the corner cabinet for the poison scanner

Glad you like it, but now, in the absence of a poison scanner dog, what in the world was the solution you were envisioning?

[TimeSpaceLightForce]

actually the lever method is the best next to pipet solution ..

or

Take glasses 1 2 3 4 off the scale

Spill some of 2 then refill from 1

Weigh mixture 2 with 3. If reading:

a bit > A+A : glass1 has it

a bit < A+A+x :glass2

=A+A+x :glass3

=A+A :glass4

note: taking the 4 glasses at the same time with two hands needs caution

the purpose of spilling some of glass2 content is to make sufficient

difference on poison weight (e.i about 2% to 20% water volume).

The scale can read milligram anyway 2%gram=20mg 98%gram=980mg.

Warning: DO NOT try this at home !

[CaptainEd]

OK, I see, all these solutions (bodhri's solution, the pipet solution, the lever solution, and this solution) rely on the same principle at the highest level: construct one reading that adds together different fractions of the different glasses, so that the observed fraction of X identifies the glass. Makes sense to me...

[bonanova]

Kudos to TSLF and the solvers. Great puzzle.