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...
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.
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.
Alice and Bob play the following coins-on-a-stack game. 20
coins are stacked one above the other. One of them is a
special (gold) coin and the rest are ordinary coins. The
goal is to bring the gold coin to the top by repeatedly
moving the topmost coin to another position in the stack.
Alice starts and the players take turns. A turn consists of
moving the coin on the top to a position i below the top
coin (0 = i = 20). We will call this an i-move (thus a 0-
move implies doing nothing). The proviso is that an i-move
cannot be repeated; for example once a player makes a 2-
move, on subsequent turns neither player can make a 2-move.
If the gold coin happens to be on top when it's a player's
turn then the player wins the game. Initially, the gold
coinis the third coin from the top
a certain no of bullets were shared by 3 ppl equally.each of
them fired 4 of the bullets and the sum of the remaining
bullets are equal to the initial share each had got.what was
the initial number of bullets?