smcgrath12 wrote:
If 12 men and 16 women can do a piece of work in 5 days and 13 men and 24 women can do it in 4 days, how long will 7 men and 10 women take to do it?
(A) 4.2 days
(B) 6.8 days
(C) 8.3 days
(D) 9.8 days
(E) 10.2 days
\(7\,\,{\text{men}}\,\, \cup \,\,\,{\text{10}}\,\,{\text{women}}\,\,\, - \,\,\,1\,\,{\text{work}}\,\,\,\, - \,\,\,?\,\,{\text{days}}\)
Is there a systematic way of dealing with this kind of problem, to be able to do it in a few minutes "naturally"?
Certainly! Let´s do it:
Let "task" be the fraction of this (piece of) work that one man can do in 1 day, hence:
\(1\,\,{\text{man}}\,\,\, - \,\,\,1\,\,{\text{day}}\,\,\,\, - \,\,\,1\,\,{\text{task}}\)
Let k (k>0) be the fraction of the "task" defined above that one woman can do in 1 day (where k may be between 0 and 1, or equal to 1, or greater), hence:
\(1\,\,{\text{woman}}\,\,\, - \,\,\,1\,\,{\text{day}}\,\,\, - k\,\,{\text{tasks}}\,\)
Now the long-lasting benefit of this "structure": everything else becomes easy and "automatic":
\(\left. \begin{gathered}\\
{\text{12}}\,\,{\text{men}} - \,\,\,5\,\,{\text{days}}\,\,\, - \,\,\,\,12 \cdot \,5 \cdot 1\,\,\,{\text{tasks}}\, \hfill \\\\
{\text{16}}\,\,{\text{women}} - \,\,\,5\,\,{\text{days}}\,\,\, - \,\,\,\,16 \cdot \,5 \cdot k\,\,\,{\text{tasks}}\,\,\, \hfill \\ \\
\end{gathered} \right\}\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{question}}\,\,{\text{stem}}} \,\,\,\,5\left( {12 + 16k} \right)\,\,{\text{tasks}}\,\,\,\, = \,\,\,\,1\,\,{\text{work}}\,\,\,\,\left( * \right)\)
\(\left. \begin{gathered}\\
{\text{13}}\,\,{\text{men}} - \,\,\,4\,\,{\text{days}}\,\,\, - \,\,\,\,13 \cdot \,4 \cdot 1\,\,\,{\text{tasks}}\, \hfill \\\\
{\text{24}}\,\,{\text{women}} - \,\,\,4\,\,{\text{days}}\,\,\, - \,\,\,\,24 \cdot \,4 \cdot k\,\,\,{\text{tasks}}\,\,\, \hfill \\ \\
\end{gathered} \right\}\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{question}}\,\,{\text{stem}}} \,\,\,\,4\left( {13 + 24k} \right)\,\,{\text{tasks}}\,\,\,\, = \,\,\,\,1\,\,{\text{work}}\,\,\,\,\left( {**} \right)\)
\(\left( * \right) = \left( {**} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,5\left( {12 + 16k} \right) = 4\left( {13 + 24k} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,k = \frac{1}{2}\)
\(?\,\,\,\,:\,\,\,\,\,\left. \begin{gathered}\\
\boxed{{\text{7}}\,\,{\text{men}}} - \,\,\,1\,\,{\text{day}}\,\,\, - \,\,\,\,7\,\,\,{\text{tasks}}\, \hfill \\\\
\boxed{{\text{10}}\,\,{\text{women}}} - \,\,\,1\,\,{\text{day}}\,\,\, - \,\,\,\,10 \cdot k = 5\,\,\,{\text{tasks}}\,\,\, \hfill \\ \\
\end{gathered} \right\}\,\,\,\,\,\mathop \Rightarrow \limits^{\boxed{{\text{FOCUSED - GROUP}}}} \,\,\,\,\frac{{12\,\,{\text{tasks}}}}{{1\,\,\,{\text{day}}}}\,\,\,\,\left( {***} \right)\)
\(\left( * \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,5\left( {12 + 16 \cdot \frac{1}{2}} \right) = 100\,\,{\text{tasks}}\,\,{\text{ = }}\,\,{\text{1}}\,\,{\text{work}}\,\)
And we finish in "high style", using UNITS CONTROL, one of the most powerful tools of our method!
\(\left( {***} \right)\,\,\,\,?\,\,\, = \,\,\,100\,\,{\text{tasks}}\,\,\,\,\left( {\frac{{1\,\,\,\,{\text{day}}}}{{12\,\,{\text{tasks}}}}\begin{array}{*{20}{c}}\\
\nearrow \\ \\
\nearrow \\
\end{array}} \right)\,\,\,\, = \,\,\,\,\frac{{100}}{{12}} = \frac{{25}}{3} = \frac{{24 + 1}}{3} = 8\frac{1}{3}\,\,\,\,\left[ {{\text{days}}} \right]\)
Obs.: arrows indicate licit converter.
This solution follows the notations and rationale taught in the GMATH method.
Regards,
fskilnik.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)