Two jars contain equal quantities of 40 % alc | vedclass

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(B) Let the initial volume of solution in each jar be $100 \ ml$. Thus,the initial amount of alcohol is $40 \ ml$ and water is $60 \ ml$.For the first

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$20$

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(B) Let the initial volume of solution in each jar be $100 \ ml$. Thus,the initial amount of alcohol is $40 \ ml$ and water is $60 \ ml$.For the first jar (Swati): Let $y \ ml$ of pure alcohol be added. The new concentration is $\frac{40+y}{100+y} = \frac{50}{100} = \frac{1}{2}$.Solving this: $80 + 2y = 100 + y \Rightarrow y = 20 \ ml$.For the second jar (Sonali): Let $x \ ml$ of the solution be replaced by $x \ ml$ of pure alcohol. The amount of alcohol removed is $0.4x$. The new amount of alcohol is $40 - 0.4x + x = 40 + 0.6x$. The total volume remains $100 \ ml$. The new concentration is $\frac{40+0.6x}{100} = \frac{50}{100} = 0.5$.Solving this: $40 + 0.6x = 50 \Rightarrow 0.6x = 10 \Rightarrow x = \frac{10}{0.6} = \frac{100}{6} = \frac{50}{3} \ ml$.The percentage by which Swati's added quantity $(20 \ ml)$ is more than Sonali's replaced quantity $(\frac{50}{3} \ ml)$ is:$\frac{20 - 50/3}{50/3} 	imes 100 = \frac{60/3 - 50/3}{50/3} 	imes 100 = \frac{10/3}{50/3} 	imes 100 = \frac{10}{50} 	imes 100 = 20 \%$.

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