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

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A certain liquid passes through a drain at a rate of w/20 gallons ever  [#permalink]

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New post 16 Nov 2015, 03:19
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A
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C
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66% (01:45) correct 34% (01:53) wrong based on 297 sessions

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A certain liquid passes through a drain at a rate of w/20 gallons ever  [#permalink]

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New post Updated on: 16 Nov 2015, 04:52
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Time needed for w/20 gallons of liquid to pass through a drain = x seconds
Time needed for w gallons of liquid to pass through a drain = 20x seconds

Time needed for y gallons of liquid to pass through a drain = (20x/w)*y = 20xy/w seconds
= (20xy/w )/60 = xy/(3w) mins

Answer C
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Re: A certain liquid passes through a drain at a rate of w/20 gallons ever  [#permalink]

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New post 16 Nov 2015, 03:59
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
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Re: A certain liquid passes through a drain at a rate of w/20 gallons ever  [#permalink]

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New post 16 Nov 2015, 04:00
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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)

Bunuel wrote:
A certain liquid passes through a drain at a rate of w/20 gallons every x seconds. At that rate, how many minutes will it take y gallons to pass through the drain?

A. y/(1200xy)
B. 20xy/w
C. xy/(3w(
D. w/(3xy)
E. 3y/(wx)

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Re: A certain liquid passes through a drain at a rate of w/20 gallons ever  [#permalink]

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New post 09 Apr 2018, 21:54
Bunuel wrote:
A certain liquid passes through a drain at a rate of w/20 gallons every x seconds. At that rate, how many minutes will it take y gallons to pass through the drain?

A. y/(1200xy)
B. 20xy/w
C. xy/(3w(
D. w/(3xy)
E. 3y/(wx)


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
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Re: A certain liquid passes through a drain at a rate of w/20 gallons ever  [#permalink]

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New post 09 Apr 2018, 23:38

Solution



Given:
    • The liquid passes through the drain at the rate of \(\frac{w}{20}\) gallons in every x seconds

To find:
    • In how many minutes, y gallons of liquid can pass through the drain.

Approach and Working:

In every x seconds, \(\frac{w}{20}\) gallons of liquid passes.
    • In x seconds => \(\frac{w}{20}\) gallons pass through the drain.
    • Therefore, in 1 seconds= \(\frac{w}{(20x}\) gallons pass through the drain.
    • Therefore, in 1 minute=\(\frac{(w*60)}{(20*x)}\) = \(\frac{3w}{x}\) gallons pass through the drain.

Since \(\frac{3w}{x}\) gallons pass through the drain in 1 minute. Then, 1 gallon will pass through the drain in:
    • 1-gallon pass through the drain = \(\frac{x}{3w}\) minutes

Hence, y gallon will pass thorough the drain in \(\frac{xy}{3w}\)minutes.

Hence, the correct answer is option C.

Answer: C
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Re: A certain liquid passes through a drain at a rate of w/20 gallons ever  [#permalink]

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New post 14 Sep 2018, 07:21
Bunuel wrote:
A certain liquid passes through a drain at a rate of w/20 gallons every x seconds. At that rate, how many minutes will it take y gallons to pass through the drain?

A. y/(1200xy)
B. 20xy/w
C. xy/(3w)
D. w/(3xy)
E. 3y/(wx)


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.
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A certain liquid passes through a drain at a rate of w/20 gallons ever  [#permalink]

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New post 17 Sep 2018, 14:26
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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Re: A certain liquid passes through a drain at a rate of w/20 gallons ever  [#permalink]

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New post 18 Sep 2018, 18:45
Bunuel wrote:
A certain liquid passes through a drain at a rate of w/20 gallons every x seconds. At that rate, how many minutes will it take y gallons to pass through the drain?

A. y/(1200xy)
B. 20xy/w
C. xy/(3w(
D. w/(3xy)
E. 3y/(wx)


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
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Re: A certain liquid passes through a drain at a rate of w/20 gallons ever  [#permalink]

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New post 27 Sep 2018, 20:28
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:

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Re: A certain liquid passes through a drain at a rate of w/20 gallons ever   [#permalink] 27 Sep 2018, 20:28
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