# Poison in a Glass

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To determine the glass with poison:

1) take them all out of the weighing scale

2) label them 1 , 2 , 3 , 4 respectively

3) pour small part of 1 to mix with 2

4) weigh 2 with 3 and read:

if the reading is a bit > A+ A : glass 1

if the reading is a bit < A + A + x : glass 2

if the reading is = A + A + x : glass 3

if the reading is = A + A : glass 4

\_/ \_/ \_/ \_/

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

.......( 4A )

On top of a digital weighing scale

are four identical glasses filled

with water , weighs A+A+A+A

\_/ \_/ \_/ \_/

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

....( 4A + x )

Poison x was mixed with one of them.

No odor,color or volume change.

If the weighing scale is to be used just once.

How can you identify the glass with poison?

Warning : Do not try this at home!

Edited by TimeSpaceLightForce

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To determine the glass with poison:

1) take them all out of the weighing scale

2) label them 1 , 2 , 3 , 4 respectively

3) pour small part of 1 to mix with 2

4) weigh 2 with 3 and read:

if the reading is a bit > A+ A : glass 1

if the reading is a bit < A + A + x : glass 2

if the reading is = A + A + x : glass 3

if the reading is = A + A : glass 4

\_/ \_/ \_/ \_/

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

.......( 4A )

On top of a digital weighing scale

are four identical glasses filled

with water , weighs A+A+A+A

\_/ \_/ \_/ \_/

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

....( 4A + x )

Poison x was mixed with one of them.

No odor,color or volume change.

If the weighing scale is to be used just once.

How can you identify the glass with poison?

Warning : Do not try this at home!

The scale can only be used once, but since all the cups are already on the scale, you can keep removing cups until you get a mass that is a multiple of A, and when you do you will know the cup you just removed has to be the poison because before it wasn't a perfect multiple of A, or in the event that it's the last cup, the last mass will not be a multiple of A. It all counts as one use because your not putting anything on the scale or resetting the scale.

Also, dang, I was so planning on trying this at my house until you said to not try.

Edited by EquisDeXD

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Weigh two of them.

If they are more than 2A drink one to determine which of the two is poison.

If it is exactly 2A then they are OK...drink one of the others to determine which of the other 2 is poison.

Assumes the poison will produce a noticeable result (don't try it at home)

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Pour out all but 1/4 of the first glass, 1/2 the second, 3/4 the third, and leave the 4th glass alone. Weigh; let the weight be B. Then (B - 2.5A)*4/x is the number of the glass.

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Also, dang, I was so planning on trying this at my house until you said to not try.

It should be performed in a lab where scale tray,gloves and wiping cloth

are available for precaution.

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.

All solutions out numbers all the problems

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A mad scientist would let different rats drink the different glasses of water and then would wait to see which dies . Or you could be boring and do what Equis said.

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pour the contents of the first cup into the second cup and the contents onf the third cup into the fourth cup. Remove the two empty cups and the remaining cup that weighs the most will contain the poison

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Good work making a glass with a poison.

Better yet pour out empty all the glasses and declare any glass as containing the poison. No argument.

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spoler

1/2 chance of not dieing

if the peson presenting it knows that the safe one is left then pick right theyll say fine then distract one bye dropping somthing

then swap cuz they say yes you can tell you picked wrong so no matter what swap them while they cant see and then you drink itll still be on the right but the left one is now right and since left was right and you have now picked the correct drink you win so pick the one they want you to swap them while theyre distracted then drink or just dont drink any of them and bring your own drink of non-poison

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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

a bit < A+A+x :glass2 has it

=A+A+x :glass3 has it

=A+A :glass4 has it

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.

Edited by TimeSpaceLightForce

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The scale can only be used once, but since all the cups are already on the scale, you can keep removing cups until you get a mass that is a multiple of A, and when you do you will know the cup you just removed has to be the poison because before it wasn't a perfect multiple of A, or in the event that it's the last cup, the last mass will not be a multiple of A. It all counts as one use because your not putting anything on the scale or resetting the scale.

Also, dang, I was so planning on trying this at my house until you said to not try.

What? No way. Your method violates the rules. By removing a second glass....If you need to since the first one was not one fourth the total weight showed on the scale, you use the scale a second time. How can you think this is allowed?

Right, OP?

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What? No way. Your method violates the rules. By removing a second glass....If you need to since the first one was not one fourth the total weight showed on the scale, you use the scale a second time. How can you think this is allowed?

Right, OP?

Doesn't make sense to me.

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What? No way. Your method violates the rules. By removing a second glass....If you need to since the first one was not one fourth the total weight showed on the scale, you use the scale a second time. How can you think this is allowed?

Right, OP?

Taking all glasses off the scale does not count as using it once.