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You are given 4 identical glasses, completely filled with transparent, odorless liquids. Three of the liquids are pure water, and the fourth is poison, which is slightly heavier. If the water glasses weigh 250 grams each, and the poisoned glass weighs 260 grams, how can you figure out which one is which, using a measuring scale just once?

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    $\begingroup$ Do you know the value of A? $\endgroup$
    – Gintas K
    Oct 6, 2016 at 18:57
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    $\begingroup$ I could also take small sips from all four glasses over a period of years and build up an immunity. $\endgroup$
    – Chris Cudmore
    Oct 6, 2016 at 19:21
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    $\begingroup$ "None of the glasses is poisoned. (It's the contents.)" $\endgroup$
    – Arkku
    Oct 6, 2016 at 19:28
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    $\begingroup$ The currently accepted answer doesn't use lateral-thinking much, so I think that tag should be removed from the question. $\endgroup$
    – Buffer Over Read
    Oct 6, 2016 at 22:23
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    $\begingroup$ @ArturKirkoryan: You don't have to ask my permission. I think you should go ahead and edit it - your phrasing is better IMO. TSLF, are you okay with that? $\endgroup$
    – Deusovi ♦
    Oct 7, 2016 at 4:01

14 Answers 14

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Let's call our glasses $1,2,3$ and $4$. We are trying to find out which glass is poisoned, so we don't need the poisoned water!

Initially we have 4 glass of waters as shown below;

enter image description here

  • Pour $1$ away and empty $1$ totally.

enter image description here

  • Put a little water from $2$ to $1$ around less than half of the glass (very little amount of $2$) and put $2$ away,

enter image description here

  • Likely, fill the rest of $1$ with $3$ (there will be much more $3$ in $1$ than $2$'s) and put $3$ away.

enter image description here

  • Weight $1$ (filled with $2$ and $3$) and $4$ only.

As a result;

If the weight is exactly $500$:

$1$ is poisoned

If the weight is exactly $510$:

: $4$ is poisoned

If the weight is between $500$ and $505$:

$2$ is poisoned

If the weight is between $505$ and $510$:

$3$ is poisoned

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  • $\begingroup$ Your criteria for glass 2 and glass 3 being poisoned are the same $\endgroup$
    – dcfyj
    Oct 6, 2016 at 19:07
  • $\begingroup$ why? I put only a little from 2. $\endgroup$
    – Oray
    Oct 6, 2016 at 19:08
  • $\begingroup$ "If the weight is 2A+(<0.5)x: 2 is poisoned" and If the weight is 2A+(<0.5)x: 3 is poisoned" both the weights are the same. $\endgroup$
    – dcfyj
    Oct 6, 2016 at 19:09
  • $\begingroup$ @dcfyj I always do little typos and you always find them! $\endgroup$
    – Oray
    Oct 6, 2016 at 19:10
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    $\begingroup$ This is all fine and dandy if we know what X is. $\endgroup$
    – John
    Oct 6, 2016 at 20:19
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Yes.

I take the glasses off the scale, pick up the scale and threaten to beat my 4 assistants with it unless they each drink a glass of water. The one that dies drank the poison one.

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    $\begingroup$ Welcome to lateral thinking! hahaha $\endgroup$
    – dcfyj
    Oct 6, 2016 at 19:01
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    $\begingroup$ Some will complain you only need 3 assistants, but I say 4 is more elegant. $\endgroup$
    – deep thought
    Oct 6, 2016 at 19:03
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    $\begingroup$ Actually, you only need 1 assistant. He'll only die if/when he drinks the poisoned glass. $\endgroup$
    – dcfyj
    Oct 6, 2016 at 19:04
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    $\begingroup$ OP made no reference as to how fast Poison X works. $\endgroup$
    – Chris Cudmore
    Oct 6, 2016 at 19:05
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    $\begingroup$ I would put small fishes with your idea. Anyways your answer is funny +1 $\endgroup$
    – Hemant Rupani
    Oct 6, 2016 at 19:28
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It is very simple:

  1. Put all the water from the 4 glasses in a bucket.

  2. Mix all the water evenly.

  3. Put the water back in the glasses, evenly distributing it.

  4. Put the empty bucket in the scale, in order to pass the requirement of using the scale.

Now we know the solution:

  1. All the 4 glasses are poisoned.

With that, you can proceed to the plot twist:

  1. Grab 4 of your enemies to a death game where each one should drink a glass, telling them that one of them are poisoned. Watch all of them die.

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Going with Deep Thought's idea:

If we put all the glasses on the scale at once and consider the scale to no longer be used once the are all removed.

So

Put all the glasses on the scale and remove them one at a time until the measured weight is no longer a multiple of A grams.

Once you've done that you'll know which glass was poisoned.

Alternatively:

Pour all the glasses into a large beaker and you'll know where the poison is.

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If the solution is homogeneous then we can identify the correct glass. Throw 1/5 liquid from the 1st glass, 1/4 from the 2nd, 1/3 from the 3rd and 1/2 from the 4th. Now ideally we should have a total weight of 2.71A but since the poison solution is homogeneous we'll have a slight change in the weight. If {2.71-observed weight} has a ratio of 4:5 with X then the 1st glass is poisonous, if the ratio is 3:4 then the 2nd one and so on.

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I slowly drink from each of the glasses in turn. The one that poisons me is the poison one. I don't need the scale.

EDIT: @Chris Cudmore beat me to this general idea by 1 minute!

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    $\begingroup$ @ChrisCudmore's answer does use the scale, and doesn't result in his death. $\endgroup$
    – deep thought
    Oct 6, 2016 at 19:29
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Digital scales work by testing electrical resistance. Use it once to weigh all the water just to satisfy that requirement. Then rip the components out to test the electrical resistance of the 4 fluids.

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If we don't know what X is, we can take four equal size (small?) water balloons and drop them into each solution.

Which ever one ultimately settles just higher than bottom has the poison, provided there's not an amount so small as to be mostly unobservable. In that case, you may be able to see which one sinks the slowest.

Finally,

Take all the glasses off the scale and measure the one that you think has the poison to verify your observation. :)

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Yes there is a way to find the poison by using the scale just once!

Odds are you will fail though.

I just realized how important it is to ask for "a fail proof way" to do something.

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Drink the glasses, wait to feel like I'm dying. Repeat. Zero uses of the scale.

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  • $\begingroup$ I like your way of thinking. +1 for you. Nothing was said about staying alive. They did say however that you should use the scale one time. $\endgroup$
    – stack reader
    Oct 7, 2016 at 1:21
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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, if a bit < A+A+x :glass2, if=A+A+x :glass3, =A+A :glass4 *note: taking the 4 glasses at the same time with two hands needs caution

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    $\begingroup$ If the reading is greater than A+A, then it could be 1. But it could be 2. Or it could be 3. All you've determined is that it's not 4. $\endgroup$
    – John
    Oct 6, 2016 at 20:25
  • $\begingroup$ 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. Throwing the contents of glass1 ..is losing the evidence. $\endgroup$
    – TSLF
    Oct 8, 2016 at 12:34
  • $\begingroup$ I understand now (based on the accepted answer) that the amount of poison was intended to be measurable. That was based on the original question where the amount of poison was X. With the revision of the question where the weight difference is made clear (10 grams), my comment no longer applies. $\endgroup$
    – John
    Oct 10, 2016 at 13:34
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Its the old bags of gold and scale riddle. With a new twist. But same solution applies. 1 cubic cm of water weighs 1 gram. Assuming i can zero the scale to elininate the weight of the vessel holding liquid and get a true weight of the liquid. Take one cm3 out if glass 1. Dump the rest. Put glass on scale and zero scale. (Not a weighing but reading scale by zering vessil) Return the cm3 back to glass 1. Take 2 cm3 out of glass 2 and add to glass 1. Take 3 cm3 out if glass 3. And add to 1. Take 4 cm3 out of glass 4 and add to glass 1. You now have 10 cm3 in glass 1. Take your only weighing. You should have 10 grams and something. Your something is the poison. The something will be 4,3,2 or 1 unit of x over 10 grams. Which ever it is, the corresponding glass holds the poison. Dont drink it.

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Well it is marked "lateral-thinking" so if we stretch the meaning of

"use the scale once"

a little bit, then it's not hard.

Specifically, once we've used the scale, we take the glasses off the scale, of course. But we take them off in a certain order.

I'm sure everyone can figure it out now, but if you are impatient:

Put two glasses on each side. Now, one side is heavier. First, take a glass off the lighter side. Then take a glass off the heavier side. If it is now balanced, the glass you took off the heavier side is poisoned. If it is still unbalanced, the glass still on the heavier side is poisoned.

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    $\begingroup$ It's a digital scale, not a balance. $\endgroup$
    – Chris Cudmore
    Oct 6, 2016 at 18:56
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    $\begingroup$ it is a digital scale showing as number, not sides on it. $\endgroup$
    – Oray
    Oct 6, 2016 at 18:56
  • $\begingroup$ Ahh, missed that point. $\endgroup$
    – deep thought
    Oct 6, 2016 at 18:57
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    $\begingroup$ using scale means reading the digital display..once $\endgroup$
    – TSLF
    Oct 6, 2016 at 19:04
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    $\begingroup$ Or, you could leverage the use of the word "Can" in the question. Yes, you can by just putting 1 glass on the scale, but maybe not 100% of the time. $\endgroup$
    – David Starkey
    Oct 6, 2016 at 20:46
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My one:

Boils each 4 glasses at 100ºC .A posion in a glass doesnot vapourize

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    $\begingroup$ "poison does not vapourize" wat $\endgroup$
    – deep thought
    Oct 6, 2016 at 19:32
  • $\begingroup$ @deepthought Answers are correct. Generalities are fun. $\endgroup$
    – George Cummins
    Oct 6, 2016 at 21:54

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