A bucket is filled with water such that the weight of bucket alone is 25 its weight when it is fille...........

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A bucket is filled with water such that the weight of bucket alone is 25% its weight when it is filled with water. Now some of the water is removed from the bucket and now the weight of bucket along with remaining water is 50% of the original total weight. What part of the water was removed from the bucket?

2/5

1/4

2/3

1/2

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Explanation:

Let original weight of bucket when it is filled with water = x
Then weight of bucket = (25/100) * x = x/4
Original weight of water = x – (x/4) = 3x/4
Now when some water removed, new weight of bucket with remaining water = (50/100) * x = x/2
So new weight of water = new weight of bucket with remaining water – weight of bucket = [(x/2) – (x/4)] = x/4
So part of water removed = [(3x/4) – (x/4)]/(3x/4) = 2/3