ABC26GN2476 · Percentage

Subject: General Aptitude · Chapter: Percentage · Exam: · Marks: · Difficulty:

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?
(a)$11 \frac{1}{9} \%$ less
(b)$11 \frac{1}{9} \%$ more
(c)$16 \frac{2}{3} \%$ less
(d)20\% more
Answer
Answer (as printed): D
Explanation
Let each vessel contain 100 litres of 40\% alcohol. Suppose Sachin added $x$ litres of pure alcohol. Then, $\frac{40+x}{100+x}=\frac{50}{100}=\frac{1}{2} \Rightarrow 80+2 x=100+x \Rightarrow x=20$. Suppose Vivek replaced $y$ litres. Then, alcohol in $y$ litres $=40 \%$ of $y=\frac{2 y}{5}$ litres. $$\begin{aligned} & \therefore \quad \frac{40-\frac{2 y}{5}+y}{100}=\frac{50}{100}=\frac{1}{2} \Rightarrow 80+\frac{6 y}{5}=100 \\ & \Rightarrow \quad y=\frac{20 \times 5}{6}=\frac{50}{3} . \end{aligned}$$ Required percentage $=\left[\frac{\left(20-\frac{50}{3}\right)}{\left(\frac{50}{3}\right)} \times 100\right] \%$ $$=\left(\frac{10}{3} \times \frac{3}{50} \times 100\right) \%=20 \% .$$

Explanation as extracted from the printed page; notation may be imperfect.

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