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 ?

Answer: Option D

Let the weight of the bucket when it is full, be 1 kg.
Then, weight of empty bucket
= 25% of 1 kg
= $$\frac{1}{4}$$ kg
Weight of liquid in the bucket
= 1 - $$\frac{1}{4}$$ kg
= $$\frac{3}{4}$$ kg
On removing the liquid, weight of (bucket + liquid) = $$\frac{3}{5}$$ kg
Weight of liquid in the bucket
= $$\frac{3}{5}$$ - $$\frac{1}{4}$$ kg
= $$\frac{7}{20}$$ kg
Weight of liquid removed
= $$\frac{3}{4}$$ - $$\frac{7}{20}$$
= $$\frac{8}{20}$$
= $$\frac{2}{5}$$ kg
Hence, required percentage
= $$\frac{2}{5}$$ × $$\frac{4}{3}$$ × 100%
= $$\frac{160}{3}$$%
= $$53\frac{1}{3}$$%