Working continuously 24 hours a day, a factory bottles Soda

BTGmoderatorLU wrote:Source: Manhattan Prep

Working continuously 24 hours a day, a factory bottles Soda Q at a rate of 500 liters per second and Soda V at a rate of 300 liters per second. If twice as many bottles of Soda V as of Soda Q are filled at the factory each day, what is the ratio of the volume of a bottle of Soda Q to a bottle of Soda V?

A. 3/10
B. 5/6
C. 6/5
D. 8/3
E. 10/3

\[Q:\,\,\,\,\frac{{500\,\,{\text{liters}}}}{{1\,\,{\text{second}}}}\,\,\,\,\,\,\,;\,\,\,\,\,\,\frac{{k\,\,{\text{bottles}}}}{{\left( {{\text{any}}\,\,{\text{time}}\,\,{\text{fixed,}}\,\,{\text{say}}} \right)\,\,\,\,1\,\,\,{\text{second}}}}\]
\[V:\,\,\,\,\frac{{300\,\,{\text{liters}}}}{{1\,\,{\text{second}}}}\,\,\,\,\,\,\,;\,\,\,\,\,\,\frac{{2k\,\,{\text{bottles}}}}{{\left( {{\text{same}}\,\,{\text{time}}\,\,{\text{fixed}}} \right)\,\,\,\,1\,\,\,{\text{second}}}}\]
\[? = \frac{{{\text{volume}}\,\,{\text{bottle}}\,\,Q}}{{{\text{volume}}\,\,{\text{bottle}}\,\,V}}\]

(We could explore a particular case, say k=1, it doesn´t matter. We left "k" so that you will have a "better feeling" of the whole structure!)

Let´s use UNITS CONTROL, one of the most powerful tools of our method!
\[Q:\,\,\,\,\,\,\,\frac{{500\,\,{\text{liters}}}}{{1\,\,{\text{second}}}}\,\,\,\left( {\frac{{1\,\,\,{\text{second}}}}{{\,k\,\,{\text{bottles}}\,}}\,\,\begin{array}{*{20}{c}}
\nearrow \\
\nearrow
\end{array}} \right)\,\,\, = \,\,\,\frac{{500\,\,\,{\text{liters}}}}{{k\,\,\,{\text{bottles}}}} = \frac{{\frac{{500}}{k}\,\,\,{\text{liters}}}}{{1\,\,\,{\text{bottle}}}}\]
\[V:\,\,\,\,\,\,\,\frac{{300\,\,{\text{liters}}}}{{1\,\,{\text{second}}}}\,\,\,\left( {\frac{{1\,\,\,{\text{second}}}}{{\,2k\,\,{\text{bottles}}\,}}\,\,\begin{array}{*{20}{c}}
\nearrow \\
\nearrow
\end{array}} \right)\,\,\, = \,\,\,\frac{{300\,\,\,{\text{liters}}}}{{2k\,\,\,{\text{bottles}}}} = \frac{{\frac{{150}}{k}\,\,\,{\text{liters}}}}{{1\,\,\,{\text{bottle}}}}\]
Obs.: arrows indicate licit converters.

\[? = \frac{{\,\,\,\frac{{500}}{k}\,\,\,}}{{\frac{{150}}{k}}} = \frac{{50}}{{15}} = \frac{{10}}{3}\]


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

Regards,
Fabio.