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Working alone at its constant rate, pump x pumped out 1/3 of the water

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Working alone at its constant rate, pump x pumped out 1/3 of the water [#permalink]

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New post 22 Jan 2017, 14:57
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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?

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
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Re: Working alone at its constant rate, pump x pumped out 1/3 of the water [#permalink]

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New post 22 Jan 2017, 16:13
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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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Re: Working alone at its constant rate, pump x pumped out 1/3 of the water [#permalink]

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New post 22 Jan 2017, 16:43
chetan2u 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


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



Wow thank you so much!
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Re: Working alone at its constant rate, pump x pumped out 1/3 of the water [#permalink]

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New post 23 Jan 2017, 01:16
2
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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Re: Working alone at its constant rate, pump x pumped out 1/3 of the water [#permalink]

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New post 23 Jan 2017, 12:12
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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Re: Working alone at its constant rate, pump x pumped out 1/3 of the water [#permalink]

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New post 23 Jan 2017, 12:40
1
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]

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New post 25 Jan 2017, 10:16
1
3
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]

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New post 14 Apr 2017, 14:03
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]

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Re: Working alone at its constant rate, pump x pumped out 1/3 of the water   [#permalink] 30 May 2018, 18:32
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