TimeSpaceLightForce Posted October 20, 2012 Posted October 20, 2012 (edited) 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? (please use spoiler for your solution) Warning : Do not try this at home! Edited October 20, 2012 by TimeSpaceLightForce
EquisDeXD Posted October 20, 2012 Posted October 20, 2012 (edited) 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? (please use spoiler for your solution) 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 October 21, 2012 by EquisDeXD 1
J.C.MacSwell Posted October 21, 2012 Posted October 21, 2012 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) 1
uncool Posted October 21, 2012 Posted October 21, 2012 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. 1
TimeSpaceLightForce Posted October 22, 2012 Author Posted October 22, 2012 Also, dang, I was so planning on trying this at my house until you said to not try. I'm glad you are safe. 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
Axioms Posted October 22, 2012 Posted October 22, 2012 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. 1
Jesussm0kedweed Posted November 15, 2012 Posted November 15, 2012 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 1
TimeSpaceLightForce Posted November 15, 2012 Author Posted November 15, 2012 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.
Albert Einstine Posted February 25, 2017 Posted February 25, 2017 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 1
TimeSpaceLightForce Posted March 5, 2017 Author Posted March 5, 2017 (edited) 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 March 5, 2017 by TimeSpaceLightForce
Velocity_Boy Posted March 20, 2017 Posted March 20, 2017 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?
Raider5678 Posted March 20, 2017 Posted March 20, 2017 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.
TimeSpaceLightForce Posted March 30, 2017 Author Posted March 30, 2017 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.
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