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

Question

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

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

ScottTargetTestPrep · Accepted Answer

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

chetan2u · Answer

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

Bunuel · Answer

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

alandizzle · Answer

Thank you Bunuel! You are such an amazing help!

gracie · Answer

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

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

Cez005 · Answer

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

helas92 · Answer

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

kitipriyanka · Answer

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.

Posted from my mobile device

DisciplinedPrep · Answer

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)

rishabh1331 · Answer

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

mcelroytutoring · Answer

Attached is a visual that should help.

https://www.drivecms.com/uploads/mcelroytutoring.com/Screen%20Shot%202023-09-07%20at%2010.16.45%20AM.png

Paras96 · Answer

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

Solution:- rate of work done by x = (1/3)/4 = 1/12 => 12 hours for full tank
 rate of work done by x and y both = 6/(2/3) => 9 hours for full tank
 work done in one hour by x and y =>
 (1/x)+(1/y)=1/9
 (1/12) + (1/y) = 1/9
 1/y=1/36
so 36 hours to empty the full tank by y

hence C

Praful_0810 · Answer

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

Given that pump x pumped out 1/3 of water in a pool in 4 hours. 
That means it can pumped out all water in 12 hours

Assume total water is 12 units
So Rate of Pump x = 1 unit per hr.

Pump x work for 4 hours and pumped out 1*4 = 4 units. So balance is 8 units
Then Pump x and y pumped our 8 units in 6 hours. So rate of Pump x + y = 8/6 = 4/3

So Rate(x + y) - Rate (x) = Rate (y)
Rate of y = 4/3-1 = 1/3 units/ hr.

So pump y will take = 12 units / (1/3) = 36 hrs.

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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My Rewards