CAMANISHPARMAR wrote:

Working alone at a constant rate, machine P produces \(a\) widgets in 3 hours. Working alone at a constant rate, machine Q produces \(b\) widgets in 4 hours. If machines P and Q work together for \(c\) hours, then in terms of \(a, b,\) and \(c\) how many widgets will machines P and Q produce?

A) \(\frac{3ac + 4bc}{12}\)

B) \(\frac{4ac + 3bc}{12}\)

C) \(\frac{4ac + 3bc}{6}\)

D) \(4ac + 3bc\)

E) \(\frac{ac + 2bc}{4}\)

\(P \cup Q\,\,\, - \,\,\,c\,\,{\text{h}}\,\,\, - \,\,\,\boxed{? = f\left( {a,b,c} \right)}\,\,\,{\text{widgets}}\)

\(P\,\,\, - \,\,\,3\,\,{\text{h}}\,\,\, - \,\,\,a\,\,{\text{widgets}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,P\,\,\, - \,\,\,12\,\,{\text{h}}\,\,\, - \,\,\,4a\,\,{\text{widgets}}\)

\(Q\,\,\, - \,\,\,4\,\,{\text{h}}\,\,\, - \,\,\,b\,\,{\text{widgets}}\,\,\,\,\, \Rightarrow \,\,\,\,\,Q\,\,\, - \,\,\,12\,\,{\text{h}}\,\,\, - \,\,\,3b\,\,{\text{widgets}}\)

Now let´s apply the UNITS CONTROL, one of the most powerful tools of our method!

\({\text{?}}\,\,\,\,{\text{ = }}\,\,\,\,{\text{c}}\,\,{\text{h}}\,\,\,\left( {\frac{{4a + 3b\,\,\,{\text{widgets}}}}{{12\,\,\,{\text{h}}}}\,\,\begin{array}{*{20}{c}}

\nearrow \\

\nearrow

\end{array}\,\,} \right)\,\,\,\,\, = \,\,\,\,\,\,\frac{{\left( {4a + 3b} \right)c}}{{12}}\,\,\,\,\left[ {{\text{widgets}}} \right]\)

Obs.: arrows indicate

licit converter.

The right answer is therefore (B).

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

Regards,

Fabio.

_________________

Fabio Skilnik :: http://www.GMATH.net (Math for the GMAT)

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