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

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

Mohammadmo · Accepted Answer

(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

Posted from my mobile device

MathRevolution · Answer

=>

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

eakabuah · Answer

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.

Posted from my mobile device

Kinshook · Answer

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

Gmat2Cracker · Answer

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 ?

sharathnair14 · Answer

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

Talha95 · Answer

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.

Posted from my mobile device

Kinshook · Answer

Given: 
1. A swimming pool is full of water. 
2. Pump A takes 6 hours to empty the pool completely, and pump B takes 7 hours to completely empty the pool. 
3. 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?

Total time to drain the pool = 6PM - 1PM = 5 hours
Let the time when Phil starts pump B be x PM

Pump A worked for 5 hours 
Pump A drained = 5/6 of pool

Remaining 1/6 of pool is drained by Pump B
Pump B worked for = 7/6 hours = 1 hours 10 mins

Pump B was started at = 6PM - 1hour 10 mins = 4:50 PM

IMO D

Tiburcio1987 · Answer

Thanks for this, can you please explain where do you get 6-x hours from?

GMATGuruNY · Answer

Let the pool = the product of the times for A and B = 6*7 = 42 gallons.

Since A takes 6 hours to empty the 42-gallon pool, A's rate  = \frac{work}{time} = \frac{42}{6 }= 7  gallons per hour.
Since B takes 7 hours to empty the 42-gallon pool, B's rate  = \frac{work}{time} = \frac{42}{7} = 6  gallons per hour.

Since A's rate = 7 gallons per hour, the amount emptied by A in the 5 hours from 1pm to 6pm = rate*time = 7*5 = 35 gallons.
Remaining work = (total pool) - (A's work) = 42-35 = 7 gallons.
Since B's rate = 6 gallons per hour, the time for B to empty the remaining 7 gallons  = \frac{work}{rate} = \frac{7}{6}  hours = 70 minutes.
Since B works for the last 70 minutes before the completion of the job at 6pm, the time B is started = 6pm - 70 minutes = 4:50pm.

D

AnuragNair · Answer

Let us work in terms of efficiency of pumps to avoid usage of fractions.

Let the capacity of tank be 42 units(LCM of 6,7)
Pump A efficiency = 42/6 = 7 units/hour
Pump B efficiency = 42/7 = 6 units/hour

Now we know Pump A works for entire 5 hours(1pm to 6pm)
Work done by Pump A = 7*5 = 35 units

Remaining work to be done by Pump B = 42-35 = 7 units

Time required by Pump B = 7/6 hours = 70 minutes

Hence, 4:50, Pump B should be switched on.
Option D.

Snehaaaaa · Answer

The total time the pumps have to empty the swimming pool is 6 PM - 1 PM = 5 hours, and since Pump A works from start to end, it works for 5 hours.
Now, Pump B works for x hours and it starts the work at (6 PM- x), hence (6-x) is used.

bumpbot · Answer

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

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