XAT 2020 Question 72
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XAT
Let's five small drums have a capacity of 1 unit capacity each and one bigger drum has that of 2 units capacity. We need to consider two
cases here. One with minimum volume and the other with maximum volume.
Case 1: Minimum value is possible if the bigger drum is half filled. So, total volume of water = 1 + 2 * (1/2) + 2 * (2/3) + 1 = 26/6 ~ 4.3
Case 2: Maximum value is possible if the bigger drum is completely full. So, total volume of water = 2 + 3 * (1/2) + 2 * (2/3) = 29/6 = 4.833
In any case volume of water is more than 4 units and less than 5 units. Hence, exactly 5 smaller drums are adequate to store the water.