How many mice are required to find the poisoned can?

If there are 30 cans out of them one is poisoned if a person tastes very little he will die within 14 hours so if there are mice to test and 24 hours to test, how many mice are required to find the poisoned can?

Questions by ganta.jyostna   answers by ganta.jyostna

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syed hannan yunus

  • Dec 30th, 2011

The number of mice is not related with the time and with any thing...
I think only ONE mice is required because if we take one mice and if the first can is poisoned then it will die or else we will test for 2 can and so on...and the death time of the mice is not mentioned..who knows it may die after taking poison.


  • Jan 29th, 2012


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  • Mar 29th, 2012

7 mice

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  • Jan 26th, 2013

Pick 5 mice
Assign binary codes to 30 cans. you need 5 bits to represent all 30. Pick 5 mice to represent each bit. Have a mice taste all the cans for which the corresponding bit is one. After 14 hours, some will die and some wont. Set 1 for the former and 0 for the latter. The corresponding bit representation is the poisoned can.


  • May 8th, 2013


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Nedyalko Marinov

  • May 13th, 2013

3 mice. Divide the cans in three groups with 10 cans each. Feed each mice from one can from the three groups. Wait one hour then feed them with the second can from each group. Continue 10 hours until all the three mice ate from each of the 10 cans in their group. Then wait and see which mice dies and when.


  • May 22nd, 2013

1 is requried

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  • Jun 23rd, 2013


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  • Jun 21st, 2014

1, because, if you let it eat from all cans, it will not die until it eats the poison

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  • Jul 2nd, 2014

3 mice: 3 mice will eat 3 cans together and simultaneously after each hr..if there is poison in any can it will die after 14 hrs


  • Jul 10th, 2014

equation is 2^x *(10)>=30
which gives x=2
hence 2 is correct answer

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  • Aug 22nd, 2014

for 30 cans @ 1 mice per can it needs 30*14 hours to get the result. so total minimum with 30 mice is 420 hours required. since we have 24 hours time left to find out . minimum mice required would be 420 hours/24 hours would be 17.5 mice or 18 mice to do the job in 24 hours

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  • Sep 15th, 2014


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  • Dec 13th, 2014

We can find it with 3 mice.
Initially, Give 3 mice one can each to taste. After 1 hour give them the next 3 cans. Continue it till the 10th hour and watch the time any mice die. Watching the time and subtracting 14 mins, and find which can was given to which mice gets the answer.
We can decide even with 2 mice or 1 mice if time taken for mice to die is exact 14. But since its said the time is less than 14 hours, assumes it will be between 13 and 14.

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  • Apr 23rd, 2015


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  • Jul 23rd, 2015


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Neha Jaiswal

  • Jul 27th, 2015

Lets assume that a mice also takes 14 hrs to die then only 1 mice is required. First label all the cans from 1st to 30 no. Take any time interval (it should be small, say 10 minute) starting from first can the same mice can taste next can in every 10 minutes sequentially by the label no. After 14 hrs we will start observing the mice and note its death time. 14 hrs back from the death time we can find at which time mice has tasted the poisoned and thus the labelled Can.

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  • Jul 28th, 2015

Take 29 mice. If all are alive after 14 hours means they remaining one is poisoned else the one which died consuming particular can is poisoned.

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Myron C Fernandes

  • Aug 1st, 2015

3 mice is required. Why ? Everything is based on Timing... Every 1 hours the mouse is fed the food from the can (10 cans are fed by each Mouse). we can calculate and know the poisonous can by( houroffeeding:::deathtime)1::14,2::15,3::16 goes on till 10::23 ( I Guess url understood my answer)

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  • Mar 18th, 2016

One(1) mouse.

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Zeeshan Khalid

  • Apr 26th, 2016

Assign 1 mice with 6 cans (remaining unused mice are 1)
One mice would die>>>those 6 cans can be separated
You are left with 5 mice
Now let each mice taste one can (one can would be in store (not to be used for now))
If mice die>> that can was poisonous
If no mice dies>>Stored can was poisonous

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Zeeshan Khalid

  • Apr 26th, 2016

1 mice.
Assign each can with a number (1-30)
Let the mice taste 3 cans a hour with 20 min gap and Note the exact time. Attach that time with can. Also calculate 14 hours from that time and attach with that can.
after 10 hours all cans would have been tasted.
when ever the mice dies>>match the time on a can
that matching can was the poisonous.

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  • May 13th, 2016

Why dont you test the poison in a Lab? Why should we need to kill a mouse? pity on it.

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Indranil Sikder

  • May 30th, 2016

Consider a matrix of 6*5. We mix up each elements of each row, there will be 6 mice required. Similarly we need 5 mice for each column. Finally we need 11 mice by adding. For a particular poisoned can we get a particular combination (R, C). By mixing cans we can find it.

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  • Jun 10th, 2016

Not mentioned about the mice death time

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  • Jun 18th, 2016

5 mice now we can covert 30 to binary ie. 11110 now we can see each binary no as a mice now specific pattern of death describe the can number. eg if 1 2 and 5 mice die then 11001 then its can no 25 then

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  • Jun 28th, 2016

We need only 1 mice, will make mice drink every can after every interval of 15 min., if the mice will die exactly after 14 hr when we started then its 1st can and so we can know which cans it is at the time of death.

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  • Jul 22nd, 2016

Only one will die as only one can poisoned.

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  • Sep 9th, 2016

5 mice as 2^5=32 which is nearest to and more than 30 cans

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  • Sep 15th, 2016

Assuming it takes 14 hours for a mice also to die(like a human being as mentioned in the question). we require 15 mice, making each mice to taste two cans. So one of 15 will die and in next 14hrs we can use other mice to test the two suspected can.

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  • Sep 18th, 2016

We need 29 mice. First 29 cans would be tested after 14 hrs. if no one died the 30th can would be poisoned.

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Alper Tunga

  • Oct 9th, 2016

Divide the cans into 3 x 10 groups.
3 mice is enough. I made them eat all the cans in every hour for 10 hours.
Starting from 14 hours one mice will die at a specific hour. E.g 2nd mice died at hour 20.
it means 7x3+1 = 22th can is poisoned :)

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  • Oct 19th, 2016

"die within 14 hours"
not exactly after 14...

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  • Sep 1st, 2017

Answer is 30...... it is because if we let each mice to get tested for each cans then each mice has 10 hrs left....with that 10 hrs it is not possible to test another can.

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  • Oct 4th, 2017

5 Rats. Answer lies in the magic of binary numbers. Give the cans numbers from 0-29 in binary, so the first bottle will be 0 0 0 0 0, second would be 0 0 0 0 1, eighth would be 0 0 1 1 1 , sixteenth would be 0 1 1 1 1, twenty ninth would be 1 1 1 0 0. Now put the 5 mice and give them number from 1-5 (M1, M2, M3, M4, M5) (Note that there are 5 digits for the bottles number in binary). Give the drink from cans according to the binary number of bottles now. The first bottle (0 0 0 0 0) would not be given to any mice, second bottle (0 0 0 0 1) would be given to the last mouse from left (M5), eleventh bottle (0 1 0 1 0) would be given to M2 and M4, sixteenth bottle (0 1 1 1 1) would be given to M2, M3, M4, and M5, twenty ninth bottle(1 1 1 0 0) would be given to M1, M2, M3.
Now after around 14 hours check who all died. If none of the mice died, then poisoned can was the first one, because no mice drank that. If M3, M4, M5 died, then the culprit was 0 0 1 1 1 i.e. eighth one, and so on

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Richard P

  • Oct 7th, 2017

3; each mouse tastes a different can every hour starting from 0 hours to 10 hours. At 14 hours if not dead, that can is good, 15 hours if not dead that can is good and so on, by 23 hours one mouse will be dead and that can identified

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  • Jan 3rd, 2018

I am assuming mice will die instantly after tasting the can, as per the question mice death time is not mentioned.
So we split the can in two halves 15 and 15. Initially we take 2 mice and put one mice in first half and another mice second half.
From here we get the idea in which half we have the poisoned can.
Now, we have 15 can to investigate and one mice is died. We take new mice and split the 15 can in two half, say 7 can and 8 can. From here we again get in which half we have a poisoned can. Till now we have experimented with 3 mice and two are dead and one mice is alive. Again we split say 4 can and 4 can (If it is 8) and take new mice. Then we left with 4 can and total 4 mice is used.
Again split 2 can and 2 can and new mice. Once mice will get dead. Till now we have used 5 mice. Now one remaining mice is enough to test the poisoned can as we have only 2 can to investigate. Similarly we do the same if the poisoned can would have found in 7 can half. so we can spilt like 3 & 4.
Then [1 can & 2 can (3 can)] and [2 can & 2 can (4can)].

Total no of mice required - 5

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  • Mar 26th, 2018

1. Only one mice is required to complete the task. The mice will eat or tastes the 1st can at time, lets assume 10.00 AM, after 15 minutes i.e, 10:15 AM move it to another can, do this for 30 cans. As the mice will die after 14 hours, note down the time. As we have performed this at regular intervals like for every 15 minutes, we can check the time to find the can which is poisoned with the help of mices death time.

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  • Aug 30th, 2018

Answer is one mice
Feed cans one by one with time gap of 20 min (20min*30= 10hrs)

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