1/3 ed of the contents of a container evaporated on the 1st day. 3/4 th of the remaining contents of the container evaporated the second day. what part of the contents of the container are left at the end of the second day.

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Raghavendran

  • Jul 7th, 2005
 

Answer is 1/6. 

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chandu

  • Oct 16th, 2005
 

its not 1/6. the answer is 50%

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Varun

  • Oct 29th, 2005
 

chandu Wrote:

its not 1/6. the answer is 50%

i think 1/6th is right . 50% of total is 3/4th of 2/3rd which evaporated leaving behind abt 16.6% i.e. 1/6th


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kavita

  • Nov 12th, 2005
 

come to think of it logically 
i'll represent original content
with this line below
------------------(no evap)
------------      (rem 2/3rd after one day)
---               (rem 1/4th of previous 2/3rd
                    at end of 2nd day i.e. 1/6 is left)

i'd say the result doesn't resemble 50% ?
the 1/2 u r calculating is actually the amt evaporated
on 2nd day =2/3 - [(3/4)*(2/3)] = 1/2 of 'original' content.

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Rama

  • Apr 1st, 2006
 

1/6 is right.....

First day it is 1/3 tat gets evaporated...then after 1st day we have 1-1/3 remaining ie.,2/3 and on the 2nd day 3/4 of the remaining (2/3) gets evaporated ie.(2/3*3/4))

lets calculate the content evaporated first.....

                  1/3 + (2/3*3/4) = 1/3 + 1/2 = 5/6.......

Therefore the remainig content is 1 - 5/6 = 1/6....

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