- Have any questions?
- [email protected]
- Home
- A bucket is filled with water such that the weight of bucket alone is 25% its weight when it is filled with water.
Question 1:
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? A. 2/5
B. 1/4
C. 2/3
Feedback 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)
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)
D. 1/2
These questions are from this test. Would you like to take a practice test?