s = hours it takes the small pipe on its own.
L = hours it takes the large pipe on its own.
we know the small pipe can do it in 11 hours total, so s = 11, so, how much of the whole job has the small done in 1 hour only?
well, if it takes 11 hours to do the whole thing, in 1 hour it has done only 1/11 of the whole job.
we know both pipes working together can do it in 6 hours flat. So in 1 hour only, both of them have done only 1/6 of the whole thing.
in that 1 hour, how much has the large one done? well, it has only done 1/L of the job.
[tex] \bf \stackrel{\textit{how much has it been done by both in 1 hour}}{\stackrel{\stackrel{small}{rate}}{\cfrac{1}{s}}+\stackrel{\stackrel{large}{rate}}{\cfrac{1}{L}}~~=~~\stackrel{job}{\cfrac{1}{6}}}
\\\\\\
\cfrac{1}{11}+\cfrac{1}{L}=\cfrac{1}{6}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{66L}}{6L+66=11L}\implies 66=5L
\\\\\\
\cfrac{66}{5}=L\implies 13\frac{1}{5}=L\impliedby \textit{13 hours and 12 minutes} [/tex]
and you can round that up as needed.
