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  • How Much Alcohol :P

    electron1212

    Member

    Updated: Oct 24, 2024
    Views: 1.3K
    A 20 litre vessel is filled with alcohol. Some of the alcohol is poured out into another vessel of an equal capacity, which is then completely filled by adding water. The mixture thus obtained is then poured into the first vessel to capacity. Then (20/3) litres is poured from the first vessel into the second. Both vessels now contain an equal amount of alcohol. How much alcohol was originally poured from the first vessel into the second ?

    Crack this and also mention the steps you have taken.
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  • durga ch

    Member • Oct 15, 2008

    initial volumn 1st vesel=20lts-filled
    initial volumn 2nd vessel=20lts-empty

    volumn transferred=xlts

    volumn in first vessel=20-xlts
    and volumn in seoncd vessel=xlts

    amount of volumn of water added =20-x in second vessel

    now what ever amount of seoncd vessels liquid is transferred, the volumn solution in first vessel will be = 20

    but since content to be transfered is x... now the total volumn in seocnd vessel is 20-x where as volumn in vessel 1 is 20

    transfering 20/3 lts to second vessel makes the volumnequal.

    20-20/3=20-x

    x=20/3lts


    I have not considered the concentrations though
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  • electron1212

    Member • Oct 15, 2008

    sorry thats not the correct answer.
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  • durga ch

    Member • Oct 15, 2008

    oh okie!!😔
    will give it a second go!!
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  • electron1212

    Member • Oct 15, 2008

    44+ views but still no correct answer/explanation and i thought this one was easy.:sshhh:
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  • mech_guy

    Member • Jul 8, 2009

    Lets say, Alcohol transferred from first vessel to second is "V"

    So, alcohol in first vessel = 20 - V
    alcohol in 2nd vessel = V

    Now, we add (20-V) lt of water in second vessel and mix it well.
    To completely fill first vessel V lt. solution should be taken from vessel two.
    Alcohol in this V lt of solution will be = V/(20/V) = V*V/20

    Amount of alcohol in first vessel now = (20 - V + V*V/20) = (20 - X) in 20 lt solution
    Amount of alcohol in second vessel now = (V - V*V/20) = X

    where, V - (V*V/20) = X

    Now (20/3) lt of solution is taken from first vessel.
    Alcohol in this (20/3) lt of solution will be = (20 - X)/3

    So, now alcohol left in first vessel = (20 - x) - [(20 - x)/3} = 2*(20 - X)/3
    and alcohol in second vessel = X + (20 - X)/3 = (20 + 2*x)/3

    equating both will give X = 5
    i.e. V - (V*V/20) = 5
    which gives V = 10 lts

    So originally 10 lts of alcohol was transferred from first vessel to second.

    Regards
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  • pradeep_agrawal

    Member • Jul 13, 2009

    Solution looks good to me, good work mech_guy.

    -Pradeep
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  • Anil Jain

    Member • Jul 13, 2009

    mech_guy
    Lets say, Alcohol transferred from first vessel to second is "V"

    So, alcohol in first vessel = 20 - V
    alcohol in 2nd vessel = V

    Now, we add (20-V) lt of water in second vessel and mix it well.
    To completely fill first vessel V lt. solution should be taken from vessel two.
    Alcohol in this V lt of solution will be = V/(20/V) = V*V/20

    Amount of alcohol in first vessel now = (20 - V + V*V/20) = (20 - X) in 20 lt solution
    Amount of alcohol in second vessel now = (V - V*V/20) = X

    where, V - (V*V/20) = X

    Now (20/3) lt of solution is taken from first vessel.
    Alcohol in this (20/3) lt of solution will be = (20 - X)/3

    So, now alcohol left in first vessel = (20 - x) - [(20 - x)/3} = 2*(20 - X)/3
    and alcohol in second vessel = X + (20 - X)/3 = (20 + 2*x)/3

    equating both will give X = 5
    i.e. V - (V*V/20) = 5
    which gives V = 10 lts

    So originally 10 lts of alcohol was transferred from first vessel to second.

    Regards
    Looks fine..

    -CB
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  • jhbalaji

    Member • Jul 26, 2009

    need clue mate...
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