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Working alone at its constant rate, pump x pumped out 1/3 of the water
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alandizzle
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Working alone at its constant rate, pump x pumped out 1/3 of the water [#permalink] New post  22 Jan 2017, 13:57
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Working alone at its constant rate, pump x pumped out 1/3 of the water in a pool in 4 hours. Then pump Y started working and the two pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 6 hours. How many hours would it have taken pump Y, working alone at its constant rate, to pump out all the water that was pumped out of the pool?

A: 18
B: 24
C: 36
D: 48
E: 52
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chetan2u
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Re: Working alone at its constant rate, pump x pumped out 1/3 of the water [#permalink] New post  22 Jan 2017, 15:13
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alandizzle wrote:
Working alone at its constant rate, pump x pumped out 1/3 of the water in a pool in 4 hours. Then pump Y started working and the two pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 6 hours. How many hours would it have taken pump Y, working alone at its constant rate, to pump out all the water that was pumped out of the pool?

I'm having a tough time with this question, can someone help me break it down? I guessed correctly but not sure how to get there within 2 minutes.

A: 18
B: 24
C: 36
D: 48
E: 52


Hi

Three steps that will get you to answer..
1) time of pump x..
1/3 in 4 hrs, so complete in 4*3=12 hrs

2) combined time of x and y..
(1-1/3) or 2/3 in 6 hrs, so complete in 6*3/2=9 hrs..

3) finally our answer or pump y..
1/(combined time) - 1/(time of pump x)= 1/9-1/12=1/36..
So 36hrs

Hope it helps
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Bunuel
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Re: Working alone at its constant rate, pump x pumped out 1/3 of the water [#permalink] New post  23 Jan 2017, 00:16
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alandizzle wrote:
Working alone at its constant rate, pump x pumped out 1/3 of the water in a pool in 4 hours. Then pump Y started working and the two pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 6 hours. How many hours would it have taken pump Y, working alone at its constant rate, to pump out all the water that was pumped out of the pool?

I'm having a tough time with this question, can someone help me break it down? I guessed correctly but not sure how to get there within 2 minutes.

A: 18
B: 24
C: 36
D: 48
E: 52


Very similar questions to practice:
working-at-constant-rate-pump-x-pumped-out-half-of-the-water-in-a-flo-223089.html
working-alone-at-its-constant-rate-pump-x-pumped-out-of-the-water-216023.html
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alandizzle
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Re: Working alone at its constant rate, pump x pumped out 1/3 of the water [#permalink] New post  23 Jan 2017, 11:12
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Bunuel wrote:
alandizzle wrote:
Working alone at its constant rate, pump x pumped out 1/3 of the water in a pool in 4 hours. Then pump Y started working and the two pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 6 hours. How many hours would it have taken pump Y, working alone at its constant rate, to pump out all the water that was pumped out of the pool?

I'm having a tough time with this question, can someone help me break it down? I guessed correctly but not sure how to get there within 2 minutes.

A: 18
B: 24
C: 36
D: 48
E: 52


Very similar questions to practice:
working-at-constant-rate-pump-x-pumped-out-half-of-the-water-in-a-flo-223089.html
working-alone-at-its-constant-rate-pump-x-pumped-out-of-the-water-216023.html


Thank you Bunuel! You are such an amazing help!
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gracie
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Re: Working alone at its constant rate, pump x pumped out 1/3 of the water [#permalink] New post  23 Jan 2017, 11:40
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Working alone at its constant rate, pump x pumped out 1/3 of the water in a pool in 4 hours. Then pump Y started working and the two pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 6 hours. How many hours would it have taken pump Y, working alone at its constant rate, to pump out all the water that was pumped out of the pool?

A: 18
B: 24
C: 36
D: 48
E: 52

rate of x=1/12
rate of x and y=1/9
rate of y=1/9-1/12=1/36
36 hours
C
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Re: Working alone at its constant rate, pump x pumped out 1/3 of the water [#permalink] New post  25 Jan 2017, 09:16
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alandizzle wrote:
Working alone at its constant rate, pump x pumped out 1/3 of the water in a pool in 4 hours. Then pump Y started working and the two pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 6 hours. How many hours would it have taken pump Y, working alone at its constant rate, to pump out all the water that was pumped out of the pool?

A: 18
B: 24
C: 36
D: 48
E: 52


We are given that pump X pumped out 1/3 of the water in a pool in 4 hours. Since rate = work/time, the rate of pump X is (1/3)/4 = 1/12. Also, since 1/3 of the water was pumped out of the pool, 2/3 was left to be removed.

Since pump X and Y pumped out 2/3 of the water in 6 hours, the combined rate of pumps X and Y is (2/3)/6 = 2/18 = 1/9.

Finally, since rate of pump X + rate of pump Y = the combined rate of pumps X and Y, the rate of pump Y is:

1/9 - 1/12 = 4/36 - 3/36 = 1/36

Thus, pump Y can pump all the water out of the pool in 36 hours.

Answer: C
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Re: Working alone at its constant rate, pump x pumped out 1/3 of the water [#permalink] New post  14 Apr 2017, 13:03
alandizzle wrote:
Working alone at its constant rate, pump x pumped out 1/3 of the water in a pool in 4 hours. Then pump Y started working and the two pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 6 hours. How many hours would it have taken pump Y, working alone at its constant rate, to pump out all the water that was pumped out of the pool?

I'm having a tough time with this question, can someone help me break it down? I guessed correctly but not sure how to get there within 2 minutes.

A: 18
B: 24
C: 36
D: 48
E: 52


1/X (amount of work done by X in 1h)= (1/3)/4=1/12
Now, there are 2/3 of work left to be done.
(1/X)+(1/Y)=(2/3)/6
(1/12)+(1/Y)=1/9
(1/Y)=(12/108)-(9/108)=3/108=1/36
Time = reciprocal of rate (work done in one hour) = 36h. C
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Re: Working alone at its constant rate, pump x pumped out 1/3 of the water [#permalink] New post  04 Sep 2018, 05:21
We can also use the percentage method to solve this question

we know that pump A did 1/3 of work in 4 hours, which means that 1/12 or 8.33% of work was done by pump A in one hour.

Now 2/3 or 66.6% of work is left. This work was done by pump A and B in 6 Hours. which means that 11.1% work was done in 1 hour. Of the 11.1% work 8.33% was done by A and 11.1 - 8.33 = 2.67% of work was done by Pump B.

If we round 2.67 to 3 It will take pump B 33.3 hours to do the work. The closest is 36 which is the answer.

Thanks
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kitipriyanka
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Re: Working alone at its constant rate, pump x pumped out 1/3 of the water [#permalink] New post  20 Apr 2019, 10:42
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Say pool contains 12l of of water (multiple of 3 and 4 for ease of calculation)

Then X pumped out 1/3= 4l in 4 hours, speed of X = 1l/hr

Then Y joined X and both took 6 hours to take out restof water which is 12-4(pumped out by X) = 8l

X working at is constant speed of 1l/hr took out 6l and remaining 2l was pumped out by Y in 6hrs.
So speed of Y= 2/6l/he

Y working alone to take out 12l of water at above speed would take 12/(2/6)= 36 hrs.

Please share correct answer.

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Re: Working alone at its constant rate, pump x pumped out 1/3 of the water [#permalink] New post  20 Apr 2019, 11:34
Simple formula based question that must be complete in 1-1.5 min. Make sure to write your work on the paper, and follow the steps.

Given,
Pump X pumped out 1/3 of the water in a pool in 4 hours. Therefore, the pump will take 12 hours to completely empty the pool.
Pumps X and Y pumped out 2/3 of the water in a pool in 6 hours. Therefore, both pumps will take 9 hours to completely empty the pool.

From the above two statements, we can calculate that Pump Y alone will take 36 hours to completely empty the pool.

Therefore, the correct answer is answer choice C. 36

Important Formula
If Pump/Pipe X can fill a tank in A hours and Pumps/Pipes X and Y can together fill it in B hours, then Pump/Pipe Y alone can fill the tank in AB/(A-B)
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Re: Working alone at its constant rate, pump x pumped out 1/3 of the water [#permalink] New post  19 Jul 2020, 10:09
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1/3 water by first pump in 4hrs
let total pump value be x :
work=1/3x
time =4 hrs
therefore, rate of first machine = 1/3*(total pump vol=x)*4. {As rate*Time =work}
rate of first and second machine combined = 2x/3*6
since rate of first machine + rate of second machine = rate of combined machines
x/12+ rate of second machine= x/9
rate of second machine= x/36
hence 36 hours
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Re: Working alone at its constant rate, pump x pumped out 1/3 of the water [#permalink]
19 Jul 2020, 10:09
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