Q). If there are 254 barrels out of which 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 barrel ?

1) 3
2) 2
3) 6
4) 8

• 2^8=256>254.....ans=8
• 10 years agoHelpfull: Yes(14) No(3)
• ans is 1...
  
  lets start with 1 mice...
  
  mice taste 1st bottle at 1st min.....he will die if it is poisoned after 14 hrs.
  again same mice taste at 2nd min....he will...............after 14hrs 1 min.
  again same mice taste at 3rd min....he will...............after 14hrs 2 min.
  again same mice taste at 4th min....he will...............after 14hrs 3 min.
  again same mice taste at 5th min....he will...............after 14hrs 4 min..
  .
  .
  .
  ..
  .
  .
  .
  .
  .
  again same mice taste at 253th min....he will...............after 14hrs 252 min.
  and if still mice didnt die then last 254th barrel is poised...
  
  acc. to time we can judge which barrel is poisoned...
  so minimum mice requirement is 1....answer is 1
• 10 years agoHelpfull: Yes(13) No(7)
• suppose there are 4 barrels
  suppose we have 4 bottle
  we will allow first mouse to drink from first 2 bottle and 2nd mouse from last 2 bottle.
  if first mouse died then first bottel is poisoned and 2nd died then 3th bottle is poisoned. if both died then poison is in 3 bottle. and if no one dies poisen is in 4th bottle.
  
  so for 4bottle 2mouse are required 2^n is the number of bottle and n is the number of mouse 4bottle ,2mouse
  
• 10 years agoHelpfull: Yes(4) No(4)
• this on is correct
  suppose we have 4 bottle
  we will allow first mouse to drink from first 2 bottle and 2nd mouse from 2nd and 3rd bottle
  if first mouse died then first bottel is poisoned and 2nd died then 3th bottle is poisoned. if both died then poison is in 2nd bottle. and if no one dies poisen is in 4th bottle.
  
  so for 4bottle 2mouse are required 2^n is the number of bottle and n is the number of mouse 4bottle ,2mouse
  
• 10 years agoHelpfull: Yes(3) No(1)
• suppose there are 4 barrels
  suppose we have 4 bottle
  we will allow first mouse to drink from first 2 bottle and 2nd mouse from 2nd and 3rd bottle.
  if first mouse died then first bottel is poisoned and 2nd died then 3th bottle is poisoned. if both died then poison is in 3 bottle. and if no one dies poisen is in 4th bottle.
  
  so for 4bottle 2mouse are required 2^n is the number of bottle and n is the number of mouse 4bottle ,2mouse
• 10 years agoHelpfull: Yes(1) No(0)
• vaibhav agarval solution is absolutely correct...:)
• 10 years agoHelpfull: Yes(1) No(5)
• can u plz explain the logic
• 10 years agoHelpfull: Yes(0) No(1)
• pls explain..
• 10 years agoHelpfull: Yes(0) No(0)
• this on is correct
  suppose we have 4 bottle
  we will allow first mouse to drink from first 2 bottle and 2nd mouse from last 2 bottle.
  if first mouse died then first bottel is poisoned and 2nd died then 3th bottle is poisoned. if both died then poison is in 2nd bottle. and if no one dies poisen is in 4th bottle.
  
  so for 4bottle 2mouse are required 2^n is the number of bottle and n is the number of mouse 4bottle ,2mouse
• 10 years agoHelpfull: Yes(0) No(1)
• pls explain it in detail
• 10 years agoHelpfull: Yes(0) No(0)
• data is not sufficient
• 9 years agoHelpfull: Yes(0) No(0)
• Its 8.
  
  An old puzzle
  
  Call barrel 1 as 00000001
  Barrel 2 as 00000010
  Barrel 3 as 00000011
  .
  .
  Barrel 254 as 11111110
  
  Now, call mouse 8 as 2^7, mouse 7 as 2^6, mouse 6 as 2^5, ...... and mouse 1 as 2^0
  
  Mouse 1 should drink from all those cans having digit at place 2^0 as 1, mouse 2 should drink from all those cans having digit at place 2^1 as 1, similarly for mouse 3, 4, 5, 6, 7 and 8
  
  Now, suppose mice 1, 2, 3 die, then barrel 00000111 is poisoned, i.e, barrel 7 is poisoned
• 9 years agoHelpfull: Yes(0) No(0)
• Please explain logic in detail
• 9 years agoHelpfull: Yes(0) No(0)
• 2^8=256>254
  so 8
  256-254=2
  8-2=6
  so ans =6
• 9 years agoHelpfull: Yes(0) No(0)
• 2^8=256
  
  so ans is 8
• 9 years agoHelpfull: Yes(0) No(0)