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?
Answers were Sorted based on User's Feedback
Answer / mahender reddy lakkadi
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...
| Is This Answer Correct ? | 234 Yes | 83 No |
Answer / abhishek jain
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.
| Is This Answer Correct ? | 33 Yes | 7 No |
Answer / ann
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.(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)
| Is This Answer Correct ? | 22 Yes | 5 No |
Answer / rock
Look the answer for this is 5..... it says within 14 hours
so it can be 13 or 12 or anything that the person die of
poison. So if 30 can are there then we will think it as in
binary no. which will give us some combination of "1" and
"0" i.e 11110(total 5 in numbers). let 1 be the mice and 0
be none. so we will be using 5 mice to carry out the work...
where for detecting in this case we have used only four. If
i would have been 33 cans then we would have used 6 mice.
and if 15 then only 4. and so on..... so concept is : think
in term of binary, how could you express the given no in
binary and in total how many 1's and 0's are required. that
will give u the required answer.
| Is This Answer Correct ? | 9 Yes | 3 No |
Answer / raghav
Answer 6 is wrong.
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.
| Is This Answer Correct ? | 11 Yes | 6 No |
Answer / a.v.vysali
the question gave only the time for humans...(14 hrs) not
mice.....asuming that mice will die much faster... we need
only one...
| Is This Answer Correct ? | 41 Yes | 37 No |
Answer / kaifi arshu
Ans is 5
let as take can as 1 2 3 4 5 6 ... 30 and mice as A,B,C,D & E
make mice A to drink 1-5 cans
B to 6-10
C to 11-15
D to 16-20
E to 21-25
aft 12hrs check which one is dead A, B, C, D, or E.
If no1 is dead upto 25, then check 26 to 30..
A-26, B-27, C-28, D-29, E-30
-----------------------------------------------------------
If any1 is dead, then check in this way..
for eg.:
if C dies, then take 11-15 cans
A to 11
B to 12
D to 13
E to 14
leave the 15th can..
if A dies then 11 is poisoned, if B dies then 12 is poisoned, if D dies then 13 is poisoned, if E dies then 14 is poisoned, if no1 is dead then 15 is poisoned one..
------------------------------------------------------------
| Is This Answer Correct ? | 6 Yes | 3 No |
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