Working continuously 24 hours a day, a factory bottles Soda Q at a rat

Let the volume of a bottle of Soda Q be q and the volume of a bottle of Soda V be v. Note that if twice as many bottles of Soda V are filled each day, the same must be true for the number of bottles filled each second; so we will compare those quantities instead.

In one second, 500/q bottles of Soda Q and 300/v bottles of Soda V are filled. Thus, we can create the equation:

2 x 500/q = (300/v)

Multiplying both sides of the equation by q/300, we have:

1000/300 = q/v

q/v = 10/3

Answer: E

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

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

(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.

Hi All,

We're told that 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 and that TWICE as many bottles of Soda V as of Soda Q are filled at the factory each day. We're asked for the ratio of the volume of a bottle of Soda Q to a bottle of Soda V. This question is all about ratios, so you can approach the math in a variety of different ways. You might also find it useful to TEST VALUES.

The fact that this scenario takes place in a 24-hour day is actually irrelevant to the math involved. Since the number of bottles created is an unknown, we can TEST VALUES to define the overall ratios involved.

IF... 1 bottle of Soda Q is created each second, then 2 bottles of Soda V are created each second.
Thus, 1 bottle of Soda Q is 500 liters and the 2 bottle of Soda V TOTAL 300 liters (meaning that each bottle is 300/2 = 150 liters).

The ratio of a bottle of Soda Q to a bottle of Soda V = 500:150 = 500/150 = 10/3

Final Answer:

GMAT assassins aren't born, they're made,
Rich

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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