Chemerical71 wrote:

2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days . In how many days can 2 man and 1 boy do the work?

A. 15

B. 18

C. 12.5

D. 10

E. 16

Let´s repeat EXACTLY the same approach we used here:

https://gmatclub.com/forum/if-12-men-an ... 11-20.html\(\begin{array}{*{20}{c}}

{\left( {\text{I}} \right)} \\

{\left( {{\text{II}}} \right)}

\end{array}\,\,\begin{array}{*{20}{c}}

{\left[ {2\,\,{\text{men}}\,\,\, \cup \,\,\,{\text{3}}\,\,{\text{boys}}} \right]\,\,\, - \,\,{\text{1}}\,\,{\text{work}}\,\,\, - \,\,\,10\,\,\,{\text{days}}} \\

{\left[ {3\,\,{\text{men}}\,\,\, \cup \,\,\,{\text{2}}\,\,{\text{boys}}} \right]\,\,\, - \,\,{\text{1}}\,\,{\text{work}}\,\,\, - \,\,\,8\,\,\,{\text{days}}}

\end{array}\)

\({\text{?}}\,\,\,{\text{:}}\,\,\,\left[ {2\,\,{\text{men}}\,\,\, \cup \,\,\,1\,\,{\text{boy}}} \right]\,\,\, - \,\,{\text{1}}\,\,{\text{work}}\,\,\, - \,\,\,?\,\,{\text{days}}\)

Let "task" be the fraction of this 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 boy can do in 1 day (where k may be between 0 and 1, or equal to 1, or greater), hence:

\(1\,{\text{boy}}\,\,\, - \,\,\,1\,\,{\text{day}}\,\,\, - \,\,\,k\,\,\,{\text{task}}\)

Now the long-lasting benefit of this "structure": everything else becomes easy and "automatic":

\(\left( {\text{I}} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\left[ {2\,\,{\text{men}}\,\,\, \cup \,\,\,{\text{3}}\,\,{\text{boys}}} \right]\,\, - \,\,{\text{10}}\,\,{\text{days}}\,\,\, - \,\,\,2 \cdot 10 \cdot 1 + 3 \cdot 10 \cdot k\,\,\,{\text{tasks}}\,\, = \,\,\,1\,\,{\text{work}}\,\)

\(\left( {{\text{II}}} \right)\,\,\, \Rightarrow \,\,\,\,\left[ {3\,\,{\text{men}}\,\,\, \cup \,\,\,{\text{2}}\,\,{\text{boys}}} \right]\,\, - \,\,{\text{8}}\,\,{\text{days}}\,\,\, - \,\,3 \cdot 8 \cdot 1 + 2 \cdot 8 \cdot k\,\,\,{\text{tasks}}\,\, = \,\,\,1\,\,{\text{work}}\)

Therefore: \(20 + 30 \cdot k = 24 + 16 \cdot k\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,k = \frac{2}{7}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,1\,{\text{work}}\,\,{\text{ = }}\,\,\,{\text{28}}\frac{4}{7}\,\,\,{\text{tasks}}\)

\(?\,\,\,\,:\,\,\,\,\left[ {2\,\,{\text{men}}\,\,\, \cup \,\,\,1\,\,{\text{boy}}} \right]\,\,\,\, - \,\,\,{\text{1}}\,\,{\text{day}}\,\,\, - \,\,\,2 \cdot 1 + 1 \cdot \frac{2}{7} = 2\frac{2}{7}\,\,{\text{tasks}}\)

And we finish in "high style", using UNITS CONTROL, one of the most powerful tools of our method:

\(?\,\,\, = \,\,\,28\frac{4}{7}\,\,\,{\text{tasks}}\,\,\,\left( {\frac{{1\,\,{\text{day}}}}{{2\frac{2}{7}\,\,{\text{tasks}}}}\begin{array}{*{20}{c}}

\nearrow \\

\nearrow

\end{array}} \right)\,\,\,\, = \,\,\,\,\frac{{7 \cdot 28 + 4}}{{2 \cdot 7 + 2}} = \frac{{200}}{{16}} = \frac{{80 + 16 + 4}}{8} = 12\frac{1}{2}\,\,\,\,\,\left[ {{\text{days}}} \right]\)

Obs.: arrows indicate

licit converter.

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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