Two inlets P and Q are working on alternate hour for filling an empty tank and only one inlet works at any point of time. How much time will it take to fill the tank completely?

1) Inlet P can fill the tank in 10 hours while working alone while Inlet Q takes 12 hours to fill the tank alone
2) Inlet P works for first hour to fill the tank

Source: www.GMATinsight.com
_________________

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

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

(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.
_________________