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# M20-32

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Math Expert
Joined: 02 Sep 2009
Posts: 44639

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16 Sep 2014, 01:09
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Difficulty:

65% (hard)

Question Stats:

63% (01:43) correct 37% (01:37) wrong based on 150 sessions

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A pool can be filled in 4 hours and drained in 5 hours. The valve that fills the pool was opened at 1:00 pm and some time later the drain that empties the pool was also opened. If the pool was filled by 11:00 pm and not earlier, when was the drain opened?

A. at 2:00 pm
B. at 2:30 pm
C. at 3:00 pm
D. at 3:30 pm
E. at 4:00 pm
[Reveal] Spoiler: OA

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Joined: 02 Sep 2009
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16 Sep 2014, 01:09
Expert's post
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Official Solution:

A pool can be filled in 4 hours and drained in 5 hours. The valve that fills the pool was opened at 1:00 pm and some time later the drain that empties the pool was also opened. If the pool was filled by 11:00 pm and not earlier, when was the drain opened?

A. at 2:00 pm
B. at 2:30 pm
C. at 3:00 pm
D. at 3:30 pm
E. at 4:00 pm

When only the valve is open, the pool is filling at a rate of $$\frac{1}{4}$$ an hour. When both the valve and the drain are open, the pool is filling at a rate of $$\frac{1}{4} - \frac{1}{5} = \frac{1}{20}$$ an hour. Let $$x$$ denote the time when only the valve was open. Then both the valve and the drain were open for $$11 - 1 - x = 10 - x$$hours. Now we can compose the equation $$\frac{1}{4}x + \frac{1}{20}(10 - x) = 1$$ which reduces to $$\frac{x}{5} = 0.5$$ from where $$x = 2.5$$. Thus, the drain was opened at $$1:00 + 2:30 = 3:30$$pm.

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Joined: 22 Jul 2013
Posts: 19
Location: United States
Concentration: Technology, Entrepreneurship
Schools: IIM A '15
GMAT 1: 650 Q46 V34
GMAT 2: 720 Q49 V38
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01 Dec 2014, 19:06
2
KUDOS
The Valve that Fills is working at a rate of 1/4 per hour
The Drain is working at a rate of 1/5 per hour

Now The valve is open for 10 hrs ( between 1 Pm to 11 Pm)
Let the drain be open for x hrs.

so work done is (10)/4 (Valve's work)
Drains work is (x)/5

==> (10/4 ) - (x/5) = 1 ( Assuming work is 1)

gives us x= 7.5 hrs ., Drain was open for 7.5 hrs which is 2.5 hrs from 1 Am = 3:30 Am

Hope that helps.

Regards,
Arun

Bunuel wrote:
Official Solution:

A pool can be filled in 4 hours and drained in 5 hours. The valve that fills the pool was opened at 1:00 pm and some time later the drain that empties the pool was also opened. If the pool was filled by 11:00 pm and not earlier, when was the drain opened?

A. at 2:00 pm
B. at 2:30 pm
C. at 3:00 pm
D. at 3:30 pm
E. at 4:00 pm

When only the valve is open, the pool is filling at a rate of $$\frac{1}{4}$$ an hour. When both the valve and the drain are open, the pool is filling at a rate of $$\frac{1}{4} - \frac{1}{5} = \frac{1}{20}$$ an hour. Let $$x$$ denote the time when only the valve was open. Then both the valve and the drain were open for $$11 - 1 - x = 10 - x$$hours. Now we can compose the equation $$\frac{1}{4}x + \frac{1}{20}(10 - x) = 1$$ which reduces to $$\frac{x}{5} = 0.5$$ from where $$x = 2.5$$. Thus, the drain was opened at $$1:00 + 2:30 = 3:30$$pm.

Intern
Joined: 20 Jun 2014
Posts: 9
Schools: CBS '18

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19 Feb 2015, 16:34
1
KUDOS
Bunuel wrote:
Official Solution:

A pool can be filled in 4 hours and drained in 5 hours. The valve that fills the pool was opened at 1:00 pm and some time later the drain that empties the pool was also opened. If the pool was filled by 11:00 pm and not earlier, when was the drain opened?

A. at 2:00 pm
B. at 2:30 pm
C. at 3:00 pm
D. at 3:30 pm
E. at 4:00 pm

When only the valve is open, the pool is filling at a rate of $$\frac{1}{4}$$ an hour. When both the valve and the drain are open, the pool is filling at a rate of $$\frac{1}{4} - \frac{1}{5} = \frac{1}{20}$$ an hour. Let $$x$$ denote the time when only the valve was open. Then both the valve and the drain were open for $$11 - 1 - x = 10 - x$$hours. Now we can compose the equation $$\frac{1}{4}x + \frac{1}{20}(10 - x) = 1$$ which reduces to $$\frac{x}{5} = 0.5$$ from where $$x = 2.5$$. Thus, the drain was opened at $$1:00 + 2:30 = 3:30$$pm.

Bunuel,

can you please explain why we have $$\frac{1}{4}x + \frac{1}{20}(10 - x) = 1$$ equalling to 1 instead of 1/10?

Thanks
Current Student
Joined: 12 Aug 2015
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Concentration: General Management, Operations
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WE: Management Consulting (Consulting)

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26 Jan 2016, 22:22
Combined approach: where back-solving meets algebra half-way.

Combined rate is 1/20 as correctly noted above. Assume that hose 1 had worked for 2 hours thus completing half of the job. Hence hose 2 joined in at 3 pm. By testing this value we will be able to cut off 2 options that are either too high or too low.

So rt=d formula helps us here: 1/20 * t = 1/2 ---> t = 10. 10+2 is way too much because the work was done in 10 hours as per prompt. Hence hose 1 worked for longer than 2 hours and completed more than half a job to fit the 10 hour requirement. We can safele eliminate A and B.

I don't like ugly numbers so I will test E - hose 1 worked for 3 hours hence having completed 3/4 of the job. 1/20 * t = 1/4 ---> t = 5. 5+2 is way too low.. So E is also wrong and D is the answer.
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KUDO me plenty

Intern
Joined: 10 May 2016
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13 Jul 2016, 20:15
1
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Work done by 1st for 10 hours = 10/4 = 2.5W
Total work done = w
Work for drain pipe = 1.5 W
Time for drain pipe = 1.5 * 5 = 7.5 hours
Intern
Joined: 07 Mar 2011
Posts: 31

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25 Jul 2016, 07:57
Just a thought without equations.

Given that
1. in 4 hrs the pool can be filled.
2. in 5 hrs the pool can be drained.

in 10 hrs the valve can fill 2.5 times the pool [ 10/4 = 2.5 ].
So it means in 10 hrs 1.5 times of the pool water can be drained out if together both fill and drain valves are kept open and the pool is full. Finally, to drain out 1.5 times of the pool it requires 7.5 hrs [= 5hrs + 2.5 hrs ]. So drain valve would have kept opened at 11pm- 7.5 hrs = 3.5 hrs => 3:30Pm
Intern
Joined: 20 Jan 2013
Posts: 9
GMAT 1: 660 Q49 V30

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04 Sep 2016, 03:27
Let total work = 100%

Work completed each hour for filling A =25% (Total work completed in 4 hours, therfore work completed in 1 hour = 25%)

Work by Drain (B) = -20% (negative because it is acting against natural force of filling tank, finding 20% same as above)

Let time taken for filling = x , therfore time taken by A and B together = (10- x)

Now work = time * rate

100% = 25*x + (25-20=5) *(10-x)

Solving, x=2.5 hours

Therfore drain was opened 2.5 hours after A was opened. Therefore drain was opened at 1:00 p.m.+2.5 hrs =3:30 p.m.
Senior Manager
Joined: 08 Jun 2015
Posts: 478
Location: India
GMAT 1: 640 Q48 V29

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26 Oct 2017, 08:18
+1 for option D.
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Intern
Joined: 20 Oct 2014
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14 Nov 2017, 05:13
Bunuel,

can you please explain why we are equating it to 1 instead of 1/10?

Thanks
Re: M20-32   [#permalink] 14 Nov 2017, 05:13
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# M20-32

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