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 in 24 hours, how many mice’s are required to find the poisoned can?

• there is 3 mice....
  A,B,C
  first we drink one cane to every rat...
  after one hour again drink one cane to every rat...
  and so on till 10 hour... In 10 hour every rat drink 10 cans so all rat drink 30 cans...
  if A OR B OR C died from poison in 14th hour so they died from 1st cane OR 2nd cane OR 3rd cane respectively
  if died in 15th hour they died from 4th,5th,6th cans respectively....so on
  In 24th hour they died from 28th,29th,30th cans respectively...
  so we can determine which cane is poisoned..
• 10 years agoHelpfull: Yes(31) No(9)
• 5 mice
  
  Let A, B, C, D, E be the mice
  Give
  Can 1 to E (00001)
  Can 2 to D (00010)
  Can 3 to E,D (00011)
  .............
  Can 29 to A,B,C,E (11101)
  Can 30 to A,B,C,D (11110)
  
  Poisoned CAN can be found after 14 hrs by looking which mice are dead. ie CAN number corresponding to binary output.
  
  example
  If A,C,D are dead ===>CAN no 22 is poisoned (10110 is binary of 22)
  and if B is dead ===>CAN no 8 is poisoned (01000 is binary of 8)
  
• 10 years agoHelpfull: Yes(25) No(11)
• @ Sheetal Kothari
  Since there are 30 cans to find which can is poisoned minimum number of mice required is n where 2^n > 30
  
  Since our aim is to find minimum number of mice required, take the minimun value for n which satisfies the condition.
  
  2^5 = 32 > 30
  So here n = 5
  
  ie if there are 100 cans
  2^n > 100
  2^7 = 128
  ===> 7 mice are required
• 10 years agoHelpfull: Yes(12) No(3)
• testing capacity of 1 mice = (maximum time given - dying time)
  here max time given is 24 hrs,dying time = 14 hrs, so 24-14 = 10. one mice can test 10 cans. we have 30 cans .... so 30/10 = 3 . 3 mice is the correct answer.
• 9 years agoHelpfull: Yes(8) No(0)
• At least 3 mice.
  There are 30 cans, divide it into 3 groups, say group 1-10, group 11-20, group 21-30. mouse 1 will be drinking from group 1-10, mouse 2 from group 11-20 and mouse 3 from group 21-30 after every 1 hour. so each mouse will complete drinking their group cans within 10 hours.
  Now 1st check which mouse dies, and what is the time. when it dies, surely 14 hrs back from then, it drank the poisonous can. Say, mouse 3 dies at 24th hour. its group was group 21-30. Again, it dies at 24th hour, so, 24-14=10th hour was the time when it drank the poisonous can. so, for sure, 30th can is poisonous.
• 9 years agoHelpfull: Yes(5) No(0)
• please send easy solution
  
• 10 years agoHelpfull: Yes(3) No(0)
• please anybody post the correct answer with explanation..
• 9 years agoHelpfull: Yes(3) No(0)
• only 1 mouse is required. Say experiment began at 10:00 A.m then feed can 1 to the mouse in 1st min, can 2 in 2nd min, can 3 in 3rd min, similarly can 30 in 30th min . If the mouse dies at 12:00 am can 1 is poisoned , if he dies at 12:01 can 2 is poisoned and so on..
• 9 years agoHelpfull: Yes(3) No(0)
• correct ans is 5 mice
• 9 years agoHelpfull: Yes(2) No(0)
• it will b 1 mouse at least........
• 9 years agoHelpfull: Yes(1) No(4)
• one mice is enough
• 9 years agoHelpfull: Yes(1) No(0)
• correct ans is 5mice
• 9 years agoHelpfull: Yes(0) No(0)
• 1 mice s ans......
• 9 years agoHelpfull: Yes(0) No(1)
• 5 mice required
  
• 9 years agoHelpfull: Yes(0) No(0)