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A swimming pool is full of water. Pump A takes 6 hours to empty the po

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A swimming pool is full of water. Pump A takes 6 hours to empty the po  [#permalink]

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New post 12 Aug 2019, 01:08
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[GMAT math practice question]

A swimming pool is full of water. Pump A takes 6 hours to empty the pool completely, and pump B takes 7 hours to completely empty the pool. Phil starts to use pump A to drain the pool at 1 PM. Some time later, he completes the job using both pumps A and B together. If he wants to empty the pool completely by 6 PM, at what time should Phil start pump B?

A. 3:30 PM
B. 4 PM
C. 4:20 PM
D. 4:50 PM
E. 5 PM

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Re: A swimming pool is full of water. Pump A takes 6 hours to empty the po  [#permalink]

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New post 12 Aug 2019, 02:53
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1
(5/6 + X/7)=1
(35+6X)/42 =1
6X=7==》X=7/6 or 1&1/6 which means 1 hour and 10 minutes.

So pump B has to start the work at 4:50 such that after 1 hour and 10 minutes it completes the work at 6 pm.
6:00- 1:10= 4:50
Option D

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A swimming pool is full of water. Pump A takes 6 hours to empty the po  [#permalink]

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New post 13 Aug 2019, 16:26
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Very nicely done by Mohammadmo. I really like your solution due to its simplicity. I will surely stick to your approach in future on the actual GMAT exams.

Here is my very lengthy version of the solution.

Let the time Pump A alone will work be x and the time both Pumps will work be y.
There is a time constraint of emptying the tank by 6pm after commencing at 1pm.
This means that our first equation will be
x+y=5 ..........(1)
If Rab is the rate of operating both pumps, and Ra is the rate of operation of Pump A, then we can write the second equation as follows:
xRa+yRab=1 ..........(2)
But Ra=1/6, and Rab = 1/6 + 1/7 = 13/42
So putting Ra=1/6 and Rab=13/42 into (2) yields x/6 + 13y/42 = 1
7x + 13y = 42 .........(3)
(1)*7 yields 7x + 7y = 35 .....(4)
(3)-(4) yields 6y=7
y=7/6
Meaning both pumps has to be operated for a period of 7/6hrs which is equivalent to 1hr 10 minutes.
Pump B has to be switched on at 6:00-1:10 = 5:60-1:10 = 4:50pm.

So answer is D.

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Re: A swimming pool is full of water. Pump A takes 6 hours to empty the po  [#permalink]

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New post 15 Aug 2019, 18:47
=>

Suppose Phil starts pump \(B\) at time \(x.\)

Pump \(A\) works for \(5\) hours, and pump \(B\) works for \(6 – x\) hours.

Pump \(A\) empties \(\frac{1}{6}\) of the pool in \(1\) hour and pump \(B\) empties \(\frac{1}{7}\) of the pool in \(1\) hour.

So,
\((\frac{1}{6})*5 + (\frac{1}{7})(6-x) = 1\), and, after multiplying both sides by \(42\) we obtain \(35 + 36 – 6x = 42.\)

Thus, \(x = 4 + \frac{5}{6}\) and pump \(B\) begins work at 4:50 pm.

Therefore, the answer is D.
Answer: D
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A swimming pool is full of water. Pump A takes 6 hours to empty the po  [#permalink]

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New post 15 Aug 2019, 22:51
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MathRevolution wrote:
[GMAT math practice question]

A swimming pool is full of water. Pump A takes 6 hours to empty the pool completely, and pump B takes 7 hours to completely empty the pool. Phil starts to use pump A to drain the pool at 1 PM. Some time later, he completes the job using both pumps A and B together. If he wants to empty the pool completely by 6 PM, at what time should Phil start pump B?

A. 3:30 PM
B. 4 PM
C. 4:20 PM
D. 4:50 PM
E. 5 PM


Given: A swimming pool is full of water. Pump A takes 6 hours to empty the pool completely, and pump B takes 7 hours to completely empty the pool. Phil starts to use pump A to drain the pool at 1 PM. Some time later, he completes the job using both pumps A and B together.

Asked: If he wants to empty the pool completely by 6 PM, at what time should Phil start pump B?

Let Phil starts pump B x hours after 1 PM.

Work completed by that time = x/6
Work balance = 1 - x/6 = (6-x)/6

Rate = 1/6 + 1/7 = 13/42
Time needed to complete balance work =\(\frac{(6-x)}{6} / \frac{13}{42} = \frac{6-x * 42}{6 * 13} = \frac{7 (6-x)}{13}\)

Total time needed = x + \frac{7 (6-x)}{13} = 6 PM - 1PM = 5 hours
\(\frac{13x + 42 - 7x}{13} = 5\)
6x + 42 = 65
x = 23/6 = 3 5/6 hours = 3 hours 50 mins

Let Phil starts pump B 3 hours 50 mins after 1 PM. at 4: 50 PM

IMO D
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Re: A swimming pool is full of water. Pump A takes 6 hours to empty the po  [#permalink]

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New post 19 Aug 2019, 12:37
MathRevolution wrote:
=>

Suppose Phil starts pump \(B\) at time \(x.\)

Pump \(A\) works for \(5\) hours, and pump \(B\) works for \(6 – x\) hours.

Pump \(A\) empties \(\frac{1}{6}\) of the pool in \(1\) hour and pump \(B\) empties \(\frac{1}{7}\) of the pool in \(1\) hour.

So,
\((\frac{1}{6})*5 + (\frac{1}{7})(6-x) = 1\), and, after multiplying both sides by \(42\) we obtain \(35 + 36 – 6x = 42.\)

Thus, \(x = 4 + \frac{5}{6}\) and pump \(B\) begins work at 4:50 pm.

Therefore, the answer is D.
Answer: D



please explain how 6-x is used here?
I understood for pump A work done in 5 hours. But for B, I am not able to understand. Shouldn't the time for B= 5-x ?
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A swimming pool is full of water. Pump A takes 6 hours to empty the po  [#permalink]

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New post 25 Aug 2019, 10:56
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This can be solved through a fairly simple way
Since the time for A to complete the work is 6 hrs & the time for B to complete the work is 7 hours,
Assume the pool capacity to be 4200 units.
Rate of A&B will then be 700 units/hour and 600 units/hour respectively.

We know in total A will work for 5 hours and will empty 3500 units of the pool.
So B will only need to empty the remaining which is 700 units of the pool.
therefore, \(700=600 * x\) where x is the time taken for B to empty 700 units.

Simplifying, you'll get x ≈ 1 hour and 10 mins

Subtract \(6 hours -1 hour and 10 mins\)

You'll get \(4:50\)

4:50 PM
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Re: A swimming pool is full of water. Pump A takes 6 hours to empty the po  [#permalink]

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New post 13 Sep 2019, 17:40
Remaining work after doing A is 1/6.
B can complete the whole work in 7 hours.
So 1/6 part is done by 1/6*7=1.10
That means B started 1 hour 10 minutes before the work is completed.

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Re: A swimming pool is full of water. Pump A takes 6 hours to empty the po   [#permalink] 13 Sep 2019, 17:40
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