Question 41096
VERY GOOD QUESTION.


If you are from India then you can refer to the Algebra book by K.C.Nag or an equally good book by K.P.Basu for these type of problems. You will learn lot from them. Now, let's come to the solution.


Let the volume of the vessel = x litres.

As {{{3/8}}}th part of it is full of water, so volume of water = {{{(3/8)*x}}} litres.
As {{{5/8}}}th part of it is full of syrup, so volume of syrup = {{{(5/8)*x}}} litres.


Let us suppose that the volume of mixture drawn = y litres = volume of water added.
In y litres of mixture drawn, there is {{{(3/8)*y}}} litres of water and {{{(5/8)*y}}} litres of syrup.
So after drawing y litres of mixture, the mixture in the vessel contains: 
{{{(3/8)*x - (3/8)*y}}} = {{{(3/8)*(x - y)}}} litres of water and
{{{(5/8)*x - (5/8)*y}}} = {{{(5/8)*(x - y)}}} litres of syrup.


Now, y litres of water is added.
So, the mixture in the vessel contains:
{{{3(x - y)/8 + y}}} = {{{(3x+5y)/8}}} litres of water and
{{{5(x - y)/8}}} litres of syrup.


[Check: Total volume of mixture finally in the vessel = {{{5(x - y)/8 + (3x+5y)/8}}} = x litres = volume of vessel. This happens because the volume of mixture taken out from the vessel has been substituted by an equal volume of water.]


Now, finally the volume of water in the vessel = volume of syrup in the vessel.
So {{{5(x - y)/8 = (3x+5y)/8}}}
or 5(x - y) = 3x + 5y
or 2x = 10y
or y = {{{(1/5)*x}}}


Thus, volume of mixture to be taken out and subsequently replaced by water is {{{1/5}}}th part of the volume of the vessel.
In other words {{{1/5}}}th of the mixture in the vessel is to be replaced by water.


I feel very happy to solve this type of problem with my own method. You won't find any approach as mine in any text book available in the market.