To solve the problem, we need to calculate the total time taken by both pipes A and B to fill the cistern with the condition that pipe B is turned off after 5 minutes.
First, let's determine their individual rates of filling the cistern:
When both pipes are opened together for the first 5 minutes, their combined rate is:
\(\frac{1}{15} + \frac{1}{30} = \frac{2}{30} + \frac{1}{30} = \frac{3}{30} = \frac{1}{10}\)
So, in 5 minutes, they fill:
\(\frac{1}{10} \times 5 = \frac{5}{10} = \frac{1}{2}\) of the cistern.
After 5 minutes, pipe B is turned off, and only pipe A continues to fill the cistern. At this point, half of the cistern is already filled, so half remains. Pipe A will take additional time to fill the remaining cistern:
Pipe A fills \(\frac{1}{15}\) of the cistern per minute. To fill the remaining \(\frac{1}{2}\), it requires:
\(\frac{1}{2} \div \frac{1}{15} = \frac{1}{2} \times 15 = 7.5\) minutes.
Therefore, the total time= 5 + 7.5= 12.5 minutes