Question

pipe a and b fill a tank in 5 and 6 hour respectively . pipe c can empty it in 12 hour. if all three pipes are open together then the tank will be filled in

Answer 1

pipe a's 1 hour work 1/5
pioe b ine hour work 1/6
pipec one hour work is -1/12
work done by all pipes in 1 hour
$= \frac{1}{5} + \frac{1}{6} - \frac{1}{12} = \frac{12 + 10 - 5}{60} = \frac{17}{60}$
tank will be filed in 60/17 hours

Answer 2

$\huge{\mathfrak{\underline{Answer :}}} \\ \\ \underline{\boxed{\sf{Required \: time = 3\dfrac{9}{17}}}} \\ \\ \sf \huge{\mathfrak{\underline{Step-by-Step \: Explanation :}}} \\ \\ \textsf{Pipes A and B can fill the tank in 5 and 6 hours respectively.} \\ \\ \sf Therefore, \\ \textsf{Part filled by pipe A in 1 hour} = \sf \frac{1}{5} \\ \textsf{Part filled by pipe B in 1 hour} = \sf \frac{1}{6} \\ \\ \textsf{Pipe C can empty the tank in 12 hours.} \\ \sf Therefore, \\ \\ \textsf{Part emptied by pipe C in 1 hour} = \sf \frac{1}{12} \\ \\ \textsf{Net part filled by Pipes A, B, C together in 1 hour} \\ \qquad \sf \implies \frac{1}{5} + \frac{1}{6} - \frac{1}{12} \\ \qquad \sf \implies \frac{17}{60} \\ \\ \textsf{The tank can be filled in} \\ \qquad \sf \implies \frac{60}{17} \\ \qquad \sf \implies 3\frac{9}{17}$