A certain liquid passes through a drain at a rate of w/20 gallons ever

w/20 gallons every x seconds => w/20x gallons per second => (60w/20x) gallons per minute (we calculate this because the final answer is required to be in minutes).

So, we have

60w/20x gallons in 1 minute

y gallons in how many minutes?


Minutes = y*1 / (60w/20x) = xy/3w

Hence, (C)

The rate is (w/20)/x = w/(20x)

So y gallons will pass through the drain in y/(w/(20x)) = 20xy/w seconds, which is (20xy/w)/60 = 20xy/(60w) = xy/(3w) minutes

Answer: C

Hi All,

We're told that a certain liquid passes through a drain at a rate of W/20 gallons every X SECONDS. We're asked how many MINUTES it would take Y gallons to pass through the drain. This question can be solved in a number of different ways, including by TESTing VALUES.

IF....
W=20 and X=2, then 20/20 = 1 gallon passes through every 2 seconds
Y = 30 gallons --> which would take 2(30) = 60 seconds = 1 minute
so we're looking for an answer that equals 1 when W=20, X=3 and Y=30. There's only one answer that matches....

Final Answer:

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

Can be done using unitary method.

w/20 gallons every x seconds
therefore in 1 second W/20x gallons
in 60 seconds (w/20x)*60 = 1 minute
so
3w/x gallons pass in 1 minute
1 gallon in x/3w minutes
y gallons in xy/3w minutes

Rate of passing of liquid = w/20 gallons every x seconds = w/(20x) gallons per second
Time taken by y gallons to pass through the drain = y/[w/(20x)] seconds = 20xy/w seconds
in minutes = 20xy/(w*60) = xy/3w

Answer C

Excellent opportunity to use UNITS CONTROL , one of the most powerful tools of our method!

\(\frac{{\,\,\frac{w}{{20}}\,\,}}{x}\,\,\,\frac{{{\text{gallons}}}}{{\text{s}}}\,\,\, = \,\,\,\frac{w}{{20\,\,x}}\,\,\,\frac{{{\text{gallons}}}}{{\text{s}}}\)

\(?\,\,\,\min \,\,\, \leftrightarrow \,\,\,y\,\,\,{\text{gallons}}\,\)


\(?\,\,\, = \,\,\,y\,\,{\text{gallons}}\,\,\,\left( {\frac{{\,20\,\,x\,\,{\text{s}}\,}}{{w\,\,{\text{gallons}}}}\,\,\begin{array}{*{20}{c}}\\
\nearrow \\ \\
\nearrow \\
\end{array}} \right)\,\,\,\,\,\left( {\frac{{\,1\,\,\min \,}}{{60\,\,{\text{s}}}}\,\,\begin{array}{*{20}{c}}\\
\nearrow \\ \\
\nearrow \\
\end{array}} \right)\,\,\,\, = \,\,\,\,\frac{{20\,\,xy}}{{60\,\,w}}\,\,\, = \,\,\,\frac{{xy}}{{\,3w\,}}\,\,\,\,\,\,\,\,\left[ {\min } \right]\)

Obs.: arrows indicate licit converters.


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

Regards,
Fabio.

In every rate question, it's appropriate to first create the equilibrium of time -- all represent in the hour, minute, second and so forth. Then try to find the rate of this particular time:
w/60 gallon in x/60 minute or 60w/20x gallon per minute.
y:60w/20x or y:3w/x, which equals to xy/3w.

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