The weight of an empty bucket is 25\% of the weight of the bucket when filled with some liquid. Some of the liquid has been removed. Then, the bucket, along with the remaining liquid, weighed three-fifths of the original weight. What percentage of the liquid has been removed?
(a)40\%
(b)$62 \frac{1}{2} \%$
(c)$56 \frac{2}{3} \%$
(d)$53 \frac{1}{3} \%$
Answer
Answer (as printed): D
Explanation
Let the weight of the bucket when it is full, be 1 kg. Then, weight of empty bucket = 25\% of 1 kg = $\frac{1}{4} \mathrm{kg}$.Weight of liquid in the bucket $=\left(1-\frac{1}{4}\right) \mathrm{kg}=\frac{3}{4} \mathrm{kg}$. On removing the liquid, Weight of (bucket + liquid) $=\frac{3}{5} \mathrm{kg}$. Weight of liquid in the bucket $=\left(\frac{3}{5}-\frac{1}{4}\right) \mathrm{kg}=\frac{7}{20} \mathrm{kg}$. Weigh of liquid removed $=\left(\frac{3}{4}-\frac{7}{20}\right) \mathrm{kg}=\frac{8}{20} \mathrm{kg}=\frac{2}{5} \mathrm{kg}$. Hence, required percentage $=\left(\frac{2}{5} \times \frac{4}{3} \times 100\right) \%$ $=\frac{160}{3} \%=53 \frac{1}{3} \%$.