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 mices are required to find the poisoned can?
a) 3
b) 2
c) 6
d) 1

• only 1 mice is enough
  already solved by dipin..find it
• 14 years agoHelpfull: Yes(42) No(24)
• 6
• 14 years agoHelpfull: Yes(8) No(32)
• 1
• 14 years agoHelpfull: Yes(7) No(19)
• m1 m2
  1st hour: c1,c11 c1,c21
  2nd hour:c2,c12 c2,c22
  ..... ......
  10th hour:c10,c20 c10,c30
  If c1 is poisoned can, then m1 and m2 will die
  If c11 is poisoned can, then m1 will die
  If c21 is poisoned can, then m2 will die
  So,only two mice are enough....
  
• 14 years agoHelpfull: Yes(4) No(37)
• the answer is 2 :
  the two mice will taste the can after every 5 minutes...
  Ex : mice A at 12:00 mice B at : 12:05 ,mice A again at 12:10 and mice B again at 12:10 and so on...........it will take totally 24 hours
  So both the mice will die at some 1 can...
  and their time of death will let us know the common can....because there will be the diffrence of 5 minutes in their death...
  thats the rite 1 ...........
• 13 years agoHelpfull: Yes(2) No(29)
• i think total mice needed will be three(3) as if we divide 30 cans into 6 groups then for firdt three groups three mice and for
  4th can- mice 1 &2
  5th can- mice 2 &3
  6th can- mice 3 &1
  a is the answer
• 12 years agoHelpfull: Yes(2) No(10)
• Exact Ans will be 5.
  
  Reason why it cant be 1: It is told in the question that the mice will die WITHIN 14 hours not exactly on the 14th hour.
  Now lets consider any time (say 12 noon), at which the 1st can is given
  1st can @ 12noon
  2nd can @ 12:15 (with a 15mins gap)
  3rd can @ 12.30 and so on...
  
  Now if the mice die at some hour say 5 pm which is
  5 hours from 12 noon and
• 10 years agoHelpfull: Yes(2) No(1)
• If mouse would die exactly after 14 hrs then definitely 1 will be the answer but here the given condition says that mouse will die within 14 hrs.
  so this question will be solved by set theory.
  answer is following
  total beer bottles = 6
  drinking strategy
  bottle1 : mouse1
  bottle2 : mouse2
  bottle3 : mouse3
  bottle4 : mouse1 and mouse2
  bottle5 : mouse2 and mouse3
  bottle6 : mouse3 and mouse1
  on the basis of death of mice set we can decide the poisoned bottle
  
  in this question upto 2 bottle answer will be 1 because 2^1 = 2
  similarly upto 4 bottles answer will be 2 because 2^2 = 4
  for upto 8,16,32,64,128 bottles answer will be 3,4,5,6,7
  
  Read more at http://www.m4maths.com/643-If-there-are-30-cans-out-of-them-one-is-poisoned-if-a-person-tastes-very-little-he-will-die.html#Hr0jrFOCPbEPAgJp.99
• 9 years agoHelpfull: Yes(1) No(0)
• ans will be 3 100%. here the solution
  ,
  let at 1 p.m 3 mice take beer,2 pm 3 mice take beer which is same who drink beer.
  at 10pm 3 mice taste all 30 cans .if mice got poisson at 1 pm beer we would know same process will be repeat and end at 1 am which is exactly 24 hours.
• 11 years agoHelpfull: Yes(0) No(4)
• If mouse would die exactly at 14 hours then definitely 1 will be the answer but here the given condition says that mouse will die within 14 hours.
  so this question can be solved by set theory.
  
  There is a Formula to use for these types of problems :
  Consider there are b cans to be tested. Then the minimum value of m such that 2^m > = b gives the minimum number of mice required.
  In our case b = 30.
  The minimum value of m for 2^m > 30 is 3. (2^1=2, 2^2 = 4, 2^3 = 8. Only when m = 5, 2m exceeds 30.)
  
  
  So answer is definitely 5
• 7 years agoHelpfull: Yes(0) No(0)