A tank can be filled by two pipes A and B in 18 minutes and 24 minutes respectively. Another tap C can empty the full tank in 36 minutes. If the tap C is opened 6 minutes after the pipes A and B are opened, the tank will become full in a total of :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 12 minutes
In one minute A fills \(\frac{1}{18}\) of the tank, B fills \(\frac{1}{24}\) and C empties \(\frac{1}{36}\).
For the first 6 minutes only A and B work:
\( \displaystyle 6\left(\frac{1}{18} + \frac{1}{24}\right) = 6 \times \frac{4 + 3}{72} = \frac{42}{72} = \frac{7}{12} \)
So \(\frac{5}{12}\) of the tank is still empty when C is opened. With all three open, the net filling rate is
\( \displaystyle \frac{1}{18} + \frac{1}{24} - \frac{1}{36} = \frac{4 + 3 - 2}{72} = \frac{5}{72} \)
Time to fill the remaining part:
\( \displaystyle \frac{5}{12} \div \frac{5}{72} = \frac{5}{12} \times \frac{72}{5} = 6 \text{ minutes} \)
Total time \(= 6 + 6 = 12\) minutes.
CBSE Applied Mathematics (Class XII):
Pipes and Cisterns
