A bucket is filled with water such that the weight of bucket alone is ...
C) 2/3
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)
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A bucket is filled with water such that the weight of bucket alone is ...
C) 2/3
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)
A bucket is filled with water such that the weight of bucket alone is ...
Given:
- Weight of bucket alone = 25% of weight of bucket when filled with water
- Weight of bucket + remaining water = 50% of original total weight
To find:
- Part of water removed from the bucket
Solution:
Let's assume the weight of the bucket when empty = x
Weight of bucket when filled with water = x + y (where y is the weight of water)
Given, weight of bucket alone = 25% of weight of bucket when filled with water
=> x = 0.25(x + y)
=> 0.75x = 0.25y
=> y = 3x
Now, weight of bucket + remaining water = 50% of original total weight
=> x + (3x - z) = 0.5(x + 3x)
=> x + 3x - z = 2x + 3x
=> z = 4x
Therefore, the weight of water removed from the bucket = 3x - z = 3x - 4x = -x
Since weight cannot be negative, it means that no water was removed from the bucket.
Hence, the correct answer is option (c) 2/3.