Three pipes A, B and C can fill a cistern in 6 hours. After working together for 2 hours, pipe C stopped working and (A+B) filled the cistern in 8 hours. Find the time in which cistern can be filled by pipe C alone?

12 hours

Let total work = 6

Efficiency of (A + B+ C) = 1

Work done in first two hours = 2

Remaining work = 6 - 2 = 4

Remaining work done by A + B in 8 hours, therefore,

Efficiency of A + B = \(\frac{4}{8}\) = \(\frac{1}{2}\)

Efficiency of C = 1 - \(\frac{1}{2}\) = \(\frac{1}{2}\)

So,

Pipe C takes = \(\frac{6}{\frac{1}{2}}\) = 12hrs to fill the cistern