Discussion
Two pipes A and B can fill a tank in 6 hrs and 4 Hrs respectively. If they are opened on alternate hrs and Pipe A is opened first, in how many hours shall the tank be full ..?
Answer: C
Run the hours in pairs, since A and B alternate with A first. Each pair of hours fills
\[ \frac{1}{6} + \frac{1}{4} = \frac{2 + 3}{12} = \frac{5}{12} \]
After two full pairs, that is 4 hours:
\[ 2 \times \frac{5}{12} = \frac{5}{6} \]
leaving \( \tfrac{1}{6} \). The fifth hour belongs to A, which fills exactly \( \tfrac{1}{6} \) in an hour, so the tank is full at the end of it:
\[ \text{total} = 5 \text{ hours} \]
The remainder happening to match A's hourly output is what makes the answer a whole number here.
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