Pipe A can fill an empty cistern in 6 hours and pipe B can fill the same empty cistern in 8 hours. Both the pipes are opened simultaneously but after two hours, pipe A is closed. How many hours will B take to fill the remaining part of the cistern?

$3 \frac{1}{3}$

A (+) = 6 hrs, B (+) = 8 hrs,

⇒ A + B opened for 2 hours, then A was closed = (4 + 3) x 2 = 7 x 2 = 14 units,

⇒ Work remaining = 24 - 14 = 10 units,

⇒ Time required to complete 10 units alone = \(\frac{10}{3}\) = \( { 3}_{3 }^{1 } \) hours.