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Categories >> Aptitude Questions >> General Aptitude


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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?
 Question Submitted By :: General-Aptitude
I also faced this Question!!     Answer Posted By  
# 1
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 ?    207 Yes 59 No
Mahender Reddy Lakkadi
# 2
Only one 
Is This Answer Correct ?    75 Yes 24 No
# 3
I am sure only one Mice ...... 
Is This Answer Correct ?    34 Yes 20 No
# 4
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 28 No
# 5
only 1 mice s needed 2 predict 
Is This Answer Correct ?    29 Yes 20 No
# 6
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.
Is This Answer Correct ?    15 Yes 7 No
Sudhakar Singh
# 7
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 ?    10 Yes 3 No
# 8
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 ?    3 Yes 1 No
Abhishek Jain
# 9
5 mice

Let A, B, C, D, E be the mice
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)

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 ?    1 Yes 0 No
# 10
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 
Is This Answer Correct ?    2 Yes 2 No

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