electron1212

Branch Unspecified

15 Oct 2008

**How Much Alcohol :P**

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.

Crack this and also mention the steps you have taken.

durga ch

Communications

15 Oct 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

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

electron1212

Branch Unspecified

15 Oct 2008

sorry thats not the correct answer.

durga ch

Communications

15 Oct 2008

oh okie!!😔

will give it a second go!!

will give it a second go!!

electron1212

Branch Unspecified

15 Oct 2008

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

mech_guy

Branch Unspecified

08 Jul 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

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

pradeep_agrawal

Branch Unspecified

13 Jul 2009

Solution looks good to me, good work mech_guy.

-Pradeep

-Pradeep

Anil Jain

Civil

13 Jul 2009

Looks fine..mech_guyLets 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

-CB

jhbalaji

Branch Unspecified

26 Jul 2009

need clue mate...

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