Two inlets P and Q are working on alternate hour for filling an empty

\(job\,\, = {\text{fill}}\,\,{\text{the}}\,\,{\text{tank}}\)

\(?\,\,\,:\,\,\,{\text{time}}\,\,\,{\text{for}}\,\,{\text{job}}\,\,\,\left( {{\text{1 - h}}\,\,{\text{alternating}}} \right)\)

(1) Insufficient:

Let´s imagine one job is defined by 60 identical tasks (60 = LCM(10,12)). From this statement, we know that:

P does 6 tasks/h
Q does 5 tasks/h

In two hours 11tasks are done, therefore in 5*2 = 10 hours we have 5*11 = 55 tasks done without taking into account who started.

After that, there are still 5 tasks to be completed.

> If P starts (meaning P started at first hour), we need less than 1 additional hour to finish the job.
> If Q starts (meaning Q started at first hour), we need exactly 1h to finish the job.

Obs.: check this slightly different problem when you finish reading this one...
https://gmatclub.com/forum/two-inlets-p ... l#p2145015


(2) Insufficient:

If each inlet fills the tank in 2h, then ? equals 2h.
If each inlet fills the tank in 3h, then ? equals 3h.


(1+2) Sufficient: as explained in (1), after 10h there is still 5 tasks left, and it´s time for P to work (taking into account statement (2)):

\(? = 10{\text{h}} + 5\,{\text{tasks}} \cdot \left( {\frac{{1\,\,{\text{h}}}}{{6\,\,{\text{tasks}}}}\,\,\begin{array}{*{20}{c}}\\
\nearrow \\ \\
\nearrow \\
\end{array}} \right)\,\,\, = \,\,\,10\frac{5}{6}h\)

Obs.: arrows indicate licit converter.


The correct answer is therefore (C).


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.