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Working alone at its constant rate, pump x pumped out 1/3 of the water
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22 Jan 2017, 13:57
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84% (02:04) correct 16% (02:23) wrong based on 273 sessions
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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
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22 Jan 2017, 15:13
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.. (11/3) 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)= 1/91/12=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
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22 Jan 2017, 15:43
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.. (11/3) 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)= 1/91/12=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
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23 Jan 2017, 00:16



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Re: Working alone at its constant rate, pump x pumped out 1/3 of the water
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23 Jan 2017, 11:12
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: workingatconstantratepumpxpumpedouthalfofthewaterinaflo223089.htmlworkingaloneatitsconstantratepumpxpumpedoutofthewater216023.htmlThank 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
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23 Jan 2017, 11:40
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=1/91/12=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
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25 Jan 2017, 09:16
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
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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
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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.
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