Question

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?

• A. Three but not two
• B. Four but not three
• C. Three or four, depending on which drum had how much water initially
• D. Five but not four
• E. Five may be inadequate, depending on which drum had how much water initially

Answer 1

The Correct Option is D

To solve the problem, let's define the capacity of the smaller drums (the five drums with equal capacity) as \(x\). The sixth drum, which has double the capacity, would then have a capacity of \(2x\).

Now, let's determine the amount of water in each drum:

• Three drums are half full: \(\text{Volume} = 3 \times \frac{x}{2} = \frac{3x}{2}\)
• Two drums are two-thirds full: \(\text{Volume} = 2 \times \frac{2x}{3} = \frac{4x}{3}\)
• One drum is completely full: \(\text{Volume} = 2x\) (Assuming it is the drum with double capacity for maximum volume)

Adding all the water:

\[\frac{3x}{2} + \frac{4x}{3} + 2x\]

To simplify, find a common denominator (6):

\[= \frac{9x}{6} + \frac{8x}{6} + \frac{12x}{6} = \frac{29x}{6}\]

The total water to be stored is \(\frac{29x}{6}\).

Now let's determine how many smaller drums (each with capacity \(x\)) are needed to store this water:

\[\frac{29x}{6} \div x = \frac{29}{6} \approx 4.83\]

This results in just under 5 small drums being needed to store all the water. Therefore, 5 smaller drums are required because 4 won't be enough to store \(\frac{29}{6}x\) of water. Hence, the correct answer is:

Five but not four