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: Option D
Solution(By Examveda Team)
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%
= $$\frac{160}{3}$$%
= $$53\frac{1}{3}$$%
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Comments ( 2 )
Related Questions on Percentage
A. $$\frac{1}{4}$$
B. $$\frac{1}{3}$$
C. $$\frac{1}{2}$$
D. $$\frac{2}{3}$$
Let total weight is 100
Then, Bucket =25 , liquid =75
After removing liquid , remaining weight of bucket with liquid = 100×⅗= 60
bucket's weight will be same so, liquid = 35, bucket = 25
Required percentage = 40/75×100= 53⅓%
Any shortcut ?