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Two vessels contain equal quantities of 40% alcohol. Sachin changed the concentration of the first vessel to 50% by adding extra quantity of pure alcohol. Vivek changed the concentration of the second vessel to 50% replacing a certain quantity of the solution with pure alcohol. By what percentage is the quantity of alcohol added by Sachin more/less than that replaced by Vivek ?

Two vessels contain equal quantities of 40% alcohol. Sachin changed the concentration of the first vessel to 50% by adding extra quantity of pure alcohol. Vivek changed the concentration of the second vessel to 50% replacing a certain quantity of the solution with pure alcohol. By what percentage is the quantity of alcohol added by Sachin more/less than that replaced by Vivek ? Correct Answer $$20$$% more

Let each vessel contain 100 litres of 40% alcohol
Suppose Sachine added x litres of pure alcohol.
Then,
$$\eqalign{ & \Leftrightarrow \frac{{40 + x}}{{100 + x}} = \frac{{50}}{{100}} \cr & \Leftrightarrow \frac{{40 + x}}{{100 + x}} = \frac{1}{2} \cr & \Leftrightarrow 80 + 2x = 100 + x \cr & \Leftrightarrow x = 20 \cr} $$
Suppose Vivek replaced y litres.
Then, alcohol in y litres = 40% of y = $$\frac{2y}{5}$$
$$\eqalign{ & \therefore \frac{{40 - \frac{{2y}}{5} + y}}{{100}} = \frac{{50}}{{100}} \cr & \Rightarrow \frac{{40 - \frac{{2y}}{5} + y}}{{100}} = \frac{1}{2} \cr & \Rightarrow 80 + \frac{{6y}}{{25}} = 100 \cr & \Rightarrow y = \frac{{20 \times 5}}{6} \cr & \Rightarrow y = \frac{{50}}{3} \cr} $$
Required percentage :
$$\eqalign{ & = \left\% \cr & = \left( {\frac{{10}}{3} \times \frac{3}{{50}} \times 100} \right)\% \cr & = 20\% \cr} $$

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