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.
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Fabio Skilnik :: GMATH method creator (Math for the GMAT)