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# Two pumps are connected to a certain empty container at the

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Joined: 21 Oct 2013
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Two pumps are connected to a certain empty container at the  [#permalink]

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27 Jul 2014, 09:15
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Difficulty:

55% (hard)

Question Stats:

71% (02:42) correct 29% (02:45) wrong based on 296 sessions

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Two pumps are connected to a certain empty container at the same time. Pump X fills the container water at a constant rate, while pump Y drains water out of the container at a constant rate. The two pumps finish filling the container in four times the duration it would take pump X alone to fill the container. If pump Y alone can empty a whole container in 48 minutes, then how many minutes does it take pump X alone to fill the container?

A. 24
B. 36
C. 48
D. 50
E. 64
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Posts: 58381
Re: Two pumps are connected to a certain empty container at the  [#permalink]

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27 Jul 2014, 16:09
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goodyear2013 wrote:
Two pumps are connected to a certain empty container at the same time. Pump X fills the container water at a constant rate, while pump Y drains water out of the container at a constant rate. The two pumps finish filling the container in four times the duration it would take pump X alone to fill the container. If pump Y alone can empty a whole container in 48 minutes, then how many minutes does it take pump X alone to fill the container?

A. 24
B. 36
C. 48
D. 50
E. 64

Say the rate of pump X is x pool/minute and the rate of pump Y is y pool/minute. The combined rate is therefore {filling rate} - {emptying rate} = x - y pool minute.

The two pumps finish filling the container in four times the duration it would take pump X alone to fill the container: (x - y)*4t = tx --> x = 4/3*y.

Since, pump Y alone can empty a whole container in 48 minutes, then the rate of Y is 1/48 pool/minute (rate is a reciprocal of time).

Hence, x = 4/3*1/48 = 1/36 --> time is a reciprocal of rate, hence it takes 36 minutes pump X alone to fill the container.

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Re: Two pumps are connected to a certain empty container at the  [#permalink]

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27 Jul 2014, 16:16
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Bunuel wrote:
goodyear2013 wrote:
Two pumps are connected to a certain empty container at the same time. Pump X fills the container water at a constant rate, while pump Y drains water out of the container at a constant rate. The two pumps finish filling the container in four times the duration it would take pump X alone to fill the container. If pump Y alone can empty a whole container in 48 minutes, then how many minutes does it take pump X alone to fill the container?

A. 24
B. 36
C. 48
D. 50
E. 64

Say the rate of pump X is x pool/minute and the rate of pump Y is y pool/minute. The combined rate is therefore {filling rate} - {emptying rate} = x - y pool minute.

The two pumps finish filling the container in four times the duration it would take pump X alone to fill the container: (x - y)*4t = tx --> x = 4/3*y.

Since, pump Y alone can empty a whole container in 48 minutes, then the rate of Y is 1/48 pool/minute (rate is a reciprocal of time).

Hence, x = 4/3*1/48 = 1/36 --> time is a reciprocal of rate, hence it takes 36 minutes pump X alone to fill the container.

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Hope this helps.
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Re: Two pumps are connected to a certain empty container at the  [#permalink]

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27 Jul 2014, 17:07
Bunuel: I set my equation up as : (x-y)t = 4tx because the two rates are four times as longer. Ex:the two rates together take 40 mins, x alone takes 10, the relationship is (x-y)t=4tx correct?
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Re: Two pumps are connected to a certain empty container at the  [#permalink]

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27 Jul 2014, 17:13
bankerboy30 wrote:
Bunuel: I set my equation up as : (x-y)t = 4tx because the two rates are four times as longer. Ex:the two rates together take 40 mins, x alone takes 10, the relationship is (x-y)t=4tx correct?

No.

The two pumps finish filling the container in four times the duration it would take pump X alone to fill the container, so the the two pipes take longer. If X takes t then two piped take four times of that, so 4t: (x - y)*4t = tx.
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Re: Two pumps are connected to a certain empty container at the  [#permalink]

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09 Sep 2014, 12:44
Using ratio approach:

Since Rate * Time = Work

We know that Pipe x (in pipe) can complete the task x minutes. But after opening Pipe y (out pipe) its time to complete the same task (filling the tank) is four times higher.

So, from the equation above, rate must be four times lower (due the effect of 2nd Pipe). Alternatively, we can say that rate of out pipe is 3/4 times the In-pipe (so the net rate is 1/4 of original. since 1-3/4 = 1/4)

So the ratio between rates of In and Out pipe rates is: In : Out --> 1: 3/4 --> 4:3

For times, the ratio will be reverse In : Out --> 3:4

From 48 mins for out pipe, we know that multiplier is 12. So rate for out pipe is 36
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Re: Two pumps are connected to a certain empty container at the  [#permalink]

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17 Oct 2015, 16:00
rate of x+y -> 1/4x, where X is the time pump X needs to fill out the pool
rate of pump x is 1/x
rate of pump y is 1/48
now we have:
1/x+1/48 = 1/4x
1/48=1/4x - 1/x
1/48 = -3/4x
cross multiply:
-3*48 = 4x
simplify by 4
-3*12 = x
-36 = x

answ. is clearly 36
but why do I get a negative time?
where did I do the mistake?
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Re: Two pumps are connected to a certain empty container at the  [#permalink]

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18 Oct 2015, 10:30
mvictor wrote:
rate of x+y -> 1/4x, where X is the time pump X needs to fill out the pool
rate of pump x is 1/x
rate of pump y is 1/48
now we have:
1/x+1/48 = 1/4x
1/48=1/4x - 1/x
1/48 = -3/4x
cross multiply:
-3*48 = 4x
simplify by 4
-3*12 = x
-36 = x

answ. is clearly 36
but why do I get a negative time?
where did I do the mistake?

It should be 1/x - 1/48 = 1/(4x). We are told that pump Y drains water out of the container, so you should subtract the rate of Y to get the net gain.
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Two pumps are connected to a certain empty container at the  [#permalink]

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Updated on: 10 Aug 2018, 11:06
goodyear2013 wrote:
Two pumps are connected to a certain empty container at the same time. Pump X fills the container water at a constant rate, while pump Y drains water out of the container at a constant rate. The two pumps finish filling the container in four times the duration it would take pump X alone to fill the container. If pump Y alone can empty a whole container in 48 minutes, then how many minutes does it take pump X alone to fill the container?

A. 24
B. 36
C. 48
D. 50
E. 64

let t=minutes for pump X to fill container alone
1/t-1/48=1/4t
t=36
B

Originally posted by gracie on 24 Dec 2016, 13:13.
Last edited by gracie on 10 Aug 2018, 11:06, edited 1 time in total.
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Re: Two pumps are connected to a certain empty container at the  [#permalink]

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21 Oct 2018, 12:14
goodyear2013 wrote:
Two pumps are connected to a certain empty container at the same time. Pump X fills the container water at a constant rate, while pump Y drains water out of the container at a constant rate. The two pumps finish filling the container in four times the duration it would take pump X alone to fill the container. If pump Y alone can empty a whole container in 48 minutes, then how many minutes does it take pump X alone to fill the container?

A. 24
B. 36
C. 48
D. 50
E. 64

$$?\,\,\,:\,\,\,\# \,\,\,{\text{minutes}}\,\,X\,\,{\text{fills}}\,\,{\text{container}}$$

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

$$X\,\,:\,\,\,\,\frac{{x\,\,{\text{gallons}}}}{{1\,\,{\text{minute}}}}\,\,\,\,\,\left( {{\text{filling}}} \right)$$

$$Y\,\,:\,\,\,\,\frac{{y\,\,{\text{gallons}}}}{{1\,\,{\text{minute}}}}\,\,\,\,\,\left( {{\text{draining}}} \right)\,\,$$

The filling TIME ratio 4:1 (for any given volume) of water is inversely proportional to the filling VOLUME ratio (for any given time), hence:

$${\text{Relative}}\,\,{\text{volume}}\,\,\left( {{\text{filling}}} \right)\,\,{\text{rate}}\,\,\,:\,\,\,\,\frac{1}{4} = \frac{{\frac{{\left( {x - y} \right)\,\,{\text{gallons}}}}{{1\,\,{\text{minute}}}}}}{{\frac{{x\,\,{\text{gallons}}}}{{1\,\,{\text{minute}}}}}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\, \ldots \,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\frac{x}{y} = \frac{4}{3}\,\,\,$$

$$y = 3\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\{ \begin{gathered} \,\,x = 4\, \hfill \\ 48\,\,{\text{minutes}}\,\,\,\left( {\frac{{3\,\,{\text{gallons}}}}{{1\,\,{\text{minute}}}}} \right)\,\,\, = \,\,\,48 \cdot 3\,\,{\text{gallons}}\,\,\, = \,\,\,{\text{Volume}}\,\,{\text{container}} \hfill \\ \end{gathered} \right.$$

$${\text{?}}\,\,\,{\text{ = }}\,\,\,48 \cdot 3\,\,{\text{gallons}}\,\,\left( {\frac{{1\,\,{\text{minute}}}}{{4\,\,{\text{gallons}}}}} \right)\,\,\,\, = \,\,\,\,36\,\,{\text{minutes}}$$

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

Regards,
Fabio.
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Re: Two pumps are connected to a certain empty container at the  [#permalink]

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22 Oct 2018, 02:44
goodyear2013 wrote:
Two pumps are connected to a certain empty container at the same time. Pump X fills the container water at a constant rate, while pump Y drains water out of the container at a constant rate. The two pumps finish filling the container in four times the duration it would take pump X alone to fill the container. If pump Y alone can empty a whole container in 48 minutes, then how many minutes does it take pump X alone to fill the container?

A. 24
B. 36
C. 48
D. 50
E. 64

Note that when the two pumps are working together, the tank gets filled so rate of filling pump is more than the rate of draining pump. Since pump Y takes 48 mins alone to drain the container, pump X will take less than 48 mins to fill the container alone (since its rate of work is higher). So only options (A) and (B) are possible.

Assuming option (A), if pump X takes 24 mins alone to fill the container and pump Y we know takes 48 mins, the time taken is in the ratio 1:2 which means rate of work is 2:1. Then the rate of pump X would halve when pump Y would be open too. So time taken when both are open would double. But we are given that time taken becomes 4 times when both are open. Hence this option is not correct.

Let's confirm (though not required during the actual test) - If pump X takes 36 mins working alone, and pump Y takes 48 mins working alone, their rate of work is in the ratio 4:3. When both pumps are open, the total rate of work is only 1 now (4 - 3) i.e. 1/4th of rate of work of X alone. Hence time taken will be 4 times. This matches.
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Re: Two pumps are connected to a certain empty container at the  [#permalink]

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22 Oct 2018, 23:03
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Attachment:

Screen Shot 2018-10-23 at 11.29.13 AM.png [ 14 KiB | Viewed 891 times ]

$$X - Y = (X - Y)$$

$$\frac{1}{x} - \frac{1}{48} = \frac{1}{4x}$$

$$\frac{48 - x}{48x} = \frac{1}{4x}$$

$$48 - x = 12$$

$$-x = 12 - 48$$

$$x = 36$$
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Re: Two pumps are connected to a certain empty container at the   [#permalink] 22 Oct 2018, 23:03
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