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
_________________

Because we are computing ratio of the volumes of the two types of bottles, whether we find it for 24 hours or 1 second, the answer will the same.

Let the number of bottles of soda Q bottled every second be 'n'
So, the number of bottles of soda V bottled every second will be '2n'

Each second 500 litres of soda Q are bottled.
If in each second 'n' bottles of soda Q are bottled, volume of each bottle of soda Q = \(\frac{500}{n}\) litres. ---- (1)

Each second 300 litres of soda V are bottled.
Each second '2n' bottles of soda V are bottled.
So, volume of each bottle of soda V = \(\frac{300}{2n} = \frac{150}{n}\) litres. ---- (2)

Answer to question ratio of (1) and (2)
= \(\frac{500}{n}\)/\(\frac{150}{n}\)
= \(\frac{500}{n} * \frac{n}{150}\)
= \(\frac{500}{150} = \frac{10}{3}\)

Choice 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 \(Q:V\) -->\(\frac{1}{3}/\frac{1}{10}\) \(-->\) \(\frac{1}{3}*\frac{10}{1}\) \(-->\) \(\frac{10}{3}\)
_________________

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
_________________

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________