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, how many mices are
required to find the poisoned can?

1 mice & 14hrs 28mins are enough to get the result...
At 1st min the mice tastes 1st can & dies at 14hr
At 2nd min the mice tastes 2nd can & dies at 14hr.1Min
At 3rd min the mice tastes 3rd can & dies at 14hr.2Min
.
.
.
At 29th min the mice tastes 29th can & dies at 14hr.28Min
If the mice is till alive after 14hr.28Min then the poison will be in the 30th Can....

So no death and result achieved in the last case...

Because u forgot to see that we only have 24 hours.
even if you use that GIVE EACH MOUSE 5 CANS procedure. You
will need 4 more hours to find out in which CAN the poison is.

5 mice are needed.
let them be A,B,C,D,E.
let the drinking pattern be as follows:

A to E drink alone frm bottle 1-5.
all the possible combination of two i.e 5C2 number of bottles are drunk by A&B, B&C, A&E...and all possible 5C2 combinations.
all the possible 5C3 bottle nos. are drank by all posible three numbered groups i.e ABC, ABD and so on.
all possible 4 mice combinations drink from 5C4=5 bottles.
so total no. of bottles drank from = 5+ 5C2+ 5C3+ 5C4= 5+10+10+5=30

depending on the group of mice died, we can find the bottle no. which is poisned.

we cannot use one mouse split time technique because mouse dies within 14 hrs NOT after exactly 14 hrs. so ambiguity may arise as at an instant of time, more than on bottle can be held responsible for death.

only one yaar.
mice will taste 1st can in first minute.
mice will taste 2nd can in second min.
mice will taste 3rd can in third min.
.
.
.
.
mice will taste 30th can in 30th min.
if mice die in 14 hours then fist can is poisoned.
if mice die in 14 hours 1 min then second one is poisoned.
if mice die in 14 hours 2 min then third one is poisoned.
.
.
.
if mice die in 14 hours 29 min then last one is poisoned.
finally we need 1 mice and 14 hours 30 min for testing thats it.

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.(take 1 if the mouse is dead and 0 if it is alive and dont forget the order)

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)

but guys if a single mouse drinks multiple times it is bound die earlier .the question says"if a person takes little it dies in 14 hours" .so 1 cant be the answer

two trains are separated by 200km.one leaves at 6:00am from
Delhi and reaches Merrut at 10.00 am.another train leaves from
Merrut at 8.00 am and reaches Delhi at 11.30am.at what time
two trains meet each other?

In a survey, 20% of a town has a car, and 10% have a bike.
1/3rd of the car owners also own a bike. What proportion of
the town does not own either a car or a bike?

There are 100 doors all are closed. The below action is
taken
1) all are opened
2)all multiple of 2's are togelled ie if
open close or viceversa
.....100 times.
What is the status of the door after 100th.

(Words not repeated exactlybut question is this)
Digit sliding means if there is a number 1627, after digit
sliding it will be 7162. Say a lowest positive integer with
unit digit as z, after digit sliding a new number is
acquired which four times the original number before digit
sliding. How many digits do the original number have?