Solution

Given:

• 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
• Twice as many bottles of Soda V as of Soda Q are filled at the factory each day

To find:

• The ratio of the volume of a bottle of Soda Q to a bottle of Soda V

Approach and Working:
We assume that,

• Number of bottles of Soda Q = n
• Number of bottles of Soda V = 2n

Therefore, we can write:

• Volume of Soda per bottle of Q = V1 = \(\frac{500 * 3600 * 24}{n}\)
• Volume of Soda per bottle of V = V2 = \(\frac{300 * 3600 * 24}{2n}\)

Hence, the ratio V1: V2 = \(\frac{500 * 3600 * 24}{n}: \frac{300 * 3600 * 24}{2n}\) = 500: 150 = 10: 3

Hence, the correct answer is option E.

Answer: E
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\(Q:\,\,\,\,\frac{{500\,\,{\text{liters}}}}{{1\,\,{\text{second}}}}\,\,\,\,\,\,\,;\,\,\,\,\,\,\frac{{k\,\,{\text{bottles}}}}{{\left( {{\text{any}}\,\,{\text{time}}\,\,{\text{fixed,}}\,\,{\text{say}}} \right)\,\,\,\,1\,\,\,{\text{second}}}}\)

\(V:\,\,\,\,\frac{{300\,\,{\text{liters}}}}{{1\,\,{\text{second}}}}\,\,\,\,\,\,\,;\,\,\,\,\,\,\frac{{2k\,\,{\text{bottles}}}}{{\left( {{\text{same}}\,\,{\text{time}}\,\,{\text{fixed}}} \right)\,\,\,\,1\,\,\,{\text{second}}}}\)

(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!)

\(? = \frac{{{\text{volume}}\,\,{\text{bottle}}\,\,Q}}{{{\text{volume}}\,\,{\text{bottle}}\,\,V}}\)

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.
_________________

Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net

Different way of looking at this problem:

The ratio of liters per day will alwys be \(\frac{Q}{V} = \frac{5}{3}\).

Lets assume there are only \(5\) liters of \(Q\) and \(3\) liters of \(V\). To make things easier lets use the LCM as the number of bottles of \(Q\).

So \(15\) bottles of \(Q\), therefore \(30\) bottles of \(V\).

Liters per bottle for \(Q\): \(\frac{5}{15}\) \(-->\) \(\frac{1}{3}\)

Liters per bottle for \(V\): \(\frac{3}{30}\) \(-->\) \(\frac{1}{10}\)

Ratio bottles volume \(Q:V\) -->\(\frac{1}{3}/\frac{1}{10}\) \(-->\) \(\frac{1}{3}*\frac{10}{1}\) \(-->\) \(\frac{10}{3}\)
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