GATE Mechanical Engineering (ME) 2019 Shift 2 Solved Paper

1 hour work of $\mathit{P}=1/6$
1 hour work of $\mathit{Q}=1/9$
1 hour work of $\mathit{R}=-1/12$ (negative sign indicates that its emptying the tank)
Tank filled when $\mathit{P}$ and $\mathit{R}$ works initially for 4 hours $=4\left(1/6-1/12\right)=1/3$
Tank filled by when $\mathit{Q}$ and $\mathit{R}$ works for another 6 hours $=6\left(1/9-1/12\right)=1/6$
Total tank filled in these 10 hours $=1/3+1/6=0.5$
So, the remaining half tank will be filled by $\mathit{Q}$ alone in $4.5$ hours (Since $\mathit{Q}$ can fill the complete tank in 9 hours)
Therefore, total time required to fill the tank $=4+6+4.5=14.5$ hours