Discussion
A large tanker can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tanker from empty if B is used for half the time, and A and B fill it together for the other half?
Answer: D
Let the total time be \( t \). For the first half of it B works alone at \( \tfrac{1}{40} \); for the second half both work, at \( \tfrac{1}{60} + \tfrac{1}{40} \):
\[ \frac{t}{2}\cdot\frac{1}{40} + \frac{t}{2}\left(\frac{1}{60} + \frac{1}{40}\right) = 1 \]
\[ \frac{t}{80} + \frac{t}{2}\cdot\frac{5}{120} = 1 \quad\Longrightarrow\quad \frac{t}{80} + \frac{t}{48} = 1 \]
Over a denominator of 240:
\[ \frac{3t + 5t}{240} = 1 \quad\Longrightarrow\quad \frac{8t}{240} = 1 \quad\Longrightarrow\quad t = 30 \text{ minutes} \]
The phrase half the time means half of the unknown answer, which is why the equation has to be set up in \( t \) rather than solved forwards.
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