Six drums are used to store water. Five drums are of equal capacity, while the sixth drum has double the capacity of each of these five drums. On one morning, three drums are found half full, two are found two-thirds full and one is found completely full. It is attempted to transfer all the water to the smaller drums. How many smaller drums are adequate to store the water?
Explanation:
Let the volume of 5 smaller drums be V ∴ Volume of the bigger drum = 2V
On one morning, three drums are found half full, two are found two-thirds full and one is found completely full.
Case 1: We will get the maximum water when the bigger drum is completely full ∴ Amount of water = 3 × V/2 + 2 × 2V/3 + 1 × 2V = 29V/6 = 4.83V
∴ Number of smaller drums required to store 29V/6 water = 5 Case 2: We will get the minimum water when the bigger drum is only half full ∴ Amount of water = 1 × 2V/2 + 2 × V/2 + 2 × 2V/3 + 1 × V = 13V/3 = 4.133V
∴ Number of smaller drums required to store 13V/3 water = 5
∴ In both cases we need at least 5 smaller drums to store the water.
Hence, option (c).
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