There are 100 can out of of them one is poisoned .if a person tastes very little of this he will die within 14 hours so they decided to test with a mice .given that a mouse dies in 24 hours and you have 24 hours in all to find out the poisoned can , how many mice are required to find the poisoned can?

• it will be 7 mice
  we have a formula for this which is 2^m > b, where m=mice,b=bottle
  so according to this 2^7=128 immediaately greater than 100.
  reason is that to find the poisoned bottle we'll definitely need combinations of mice coz we dont hv the EXACT time of death do we??? like if 10 ppl go to a party and 1 has limca,2 hv pepsi, 3 have coke nd 4 hv dew, the one who had limca also had coke and one person from pepsi group also had dew.now the group of 4 plus the one frm pepsi group who had dew fall sick, then we'll easily know that dew was the reason of sickness and its identifier will be the GROUP, or rather the COMBINATION of ppl who had it.... now see frm the above xample, that only ONE particular group was the identifier for ONE particular drink... draw the analogy from here for this case too that we'll need to have a unique grup of mice fr each bottle which will b the identifier for it... so the total no. of grups of mice will be nothing but the total no. of bottles..
  we'll get the grups simply by combination of mice, and the total no. of bottles as said earlier will b equal to the sum of those combinations.
  now we know, sum of combinations nC0+nC1+...+nCn=2^n
  we said total no of combinations, ie sum of combinations=no. of bottles(B)
  so nC0+nC1+...+nCn=2^n=B
  and we cant take nC0 as it wud mean combi of n MICE taking 0 at a time whch is impossible. so 2^n will always have to be greater than B and that too the immediate next greater. now fr this prob, its 100 bottles so immediate higher power of 2 is 128=2^7 so the no of mice=27...
  plss dont consider combination of bottles as v r determining the poisoned BOTTLE and NOT the poisoned mouse.... :P
  hope this helps... the xplanation is fr undrstanding..jus remember the formula..
• 13 years agoHelpfull: Yes(23) No(3)
• another easy method :
  consider the same prob for 8 bottle
  let take 7 bottles and keep one bottle separately
  thn make the following consideration
  bottle no. rat1 rat 2 rat3
  1 0 0 1
  2 0 1 0
  3 0 1 1
  4 1 0 0
  5 1 0 1
  6 1 1 0
  7 1 1 1
  if the value of rat is 1 for thn tat is given to tat rat ...
  if rat 3 alone die thn bottle one is poison
  if rat 2 & 3 die thn bottle 3 is poison
  lik wise go wit all options ...
  so for this 100 bottle prob nearest 2 power is considered
  so 2^6=128
  so wit six rats u can do the combination and easily find the bottle with using minimum number of rats :-) :-)
• 13 years agoHelpfull: Yes(15) No(7)
• 5 mice is required
  
  
• 13 years agoHelpfull: Yes(6) No(13)
• one mice
  we have 24 hrs to test...one mice dies in 24 hr
  so we can find the poisioned can using one mice only..becoz in 24 we have totest all the cans so we can use the same mice 24 hr(life tym)
• 13 years agoHelpfull: Yes(4) No(16)
• its 7 mice
• 13 years agoHelpfull: Yes(4) No(2)
• one mice.
  
  we assign a no to each can like 1,2,3,..,99,100.
  beggining from time =0 we gave sample from each can in ascending order to
  the mice in EVERY 5 MINUTES..
  and write the time of testing of each bottle on it.
  and note the time of the death of the mice and SUBTRACT 14 from it..
  match the time found with the TIMES NOTED ON CANS.
  the can which has the same time which is found after subtraction is POISONOUS.
  By this way we can carry out the test in 22 hour and 15 minutes.
• 13 years agoHelpfull: Yes(2) No(8)
• no chance for one mouse.
  if the same mouse tastes all cans, it will die within 24 hours.
  then how to find out the poisoned can?
  5 mice also not.
  so, i think 99 mice.with 99 mice,we can test 99 cans.
  if no mouse dies,the last can is the poisoned??
  
• 13 years agoHelpfull: Yes(1) No(10)
• ans is 10...
  divide 100 into 20 each.....then divide each 20 into 10 each....for every 10 can...put 1 mice to test the poisoned can....for first set of 10 can...make one mice to drink of all 10 can at interval of 1 hr....similar for other set of 10 cans.....if mice dies at 14th hr,1st can is poisoned,if at 15th hr,then 2nd one is poisoned...and likewise...and ans =10 mice...
• 12 years agoHelpfull: Yes(1) No(4)