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Bunuel
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I solved it this way:

First identify that it was emptied for 10 hours. From 13:00 to 23:00 hrs.

Second multiply the emptying ratio (1/4)*10 hrs. Resulting in 10/4 or 5/2. Corresponding to the emptying of 2 and a half pools.

Third, I subtract the envelope of the emptying of one pool from step two. 5/2-2/2 = 3/2.

Fourth, the 3/2 corresponds to the filling ratio by the number of hours "x" that it was on. 3/2 = (1/5)* x ---> 7,5 = X

Fifth, I subtract from 23:00 hrs the 7.5 hours that the filling was on. Result 15:30.
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Solution 1: imagine pipe and drain kept open for 4 hours. Pipe will fill the tank but drain will empty only 4/5th of the tank. Thus if they are both kept open 4/5th of the tank is filled in 4 hours or full tank in 20 hours. We are told the tank was filled in 10 hours. How could this be ? If 10 hours only half the tank can be filled by them together. So more than half the tank must have been filled.
If pipe was open for 2 hours half the tank will be filled and remaining half will taken 10 hours. A total of 12 hours . If pipe was open for 3 hours 3/4th of the tank will be filled and remaining 1/4th will take 5 hours. A total of 8 hours. So for a total of 10 hours it must have taken pipe exactly between 2 and 3 hours or 2.5 hours.

I don’t like solution 1 . It’s needlessly complicated and in the thick of the exam I don’t think there will be time to come up with all this logic.

Solution 2: is easier. We are told pipe was kept open for 10 hours. So pipe would have filled the tank 2.5 times over. So drain did the work for emptying 1.5 tank. How long would that take ? 5*1.5= 7.5 hours. Pipe alone was open for 10-7.5= 2.5 hours
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well, please explain -
10*1/4 - x*1/5 = 1
why this equation is equals to 1,
thanks.

Posted from my mobile device
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Smriti5645
well, please explain -
10*1/4 - x*1/5 = 1
why this equation is equals to 1,
thanks.

Posted from my mobile device
­1 there represents the job done. Specifically, 1 pool is filled while water equal to 10*1/4 = 2.5 of the pool's capacity is poured in, and the amount of water equal to x*1/5 of the pool's capacity is drained out.
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Thanks, Bunuel ! It’s much clear now.
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Smriti5645
well, please explain -
10*1/4 - x*1/5 = 1
why this equation is equals to 1,
thanks.

Posted from my mobile device
­1 there represents the job done. Specifically, 1 pool is filled while water equal to 10*1/4 = 2.5 of the pool's capacity is poured in, and the amount of water equal to x*1/5 of the pool's capacity is drained out.
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Bunuel sorry for the stupid question, but after calculating that the drain valve was open for 7.5 hrs, can we not think that maybe it was opened at 2pm and then closed at 9:30pm? (Same logic for Options A, B, C as it will not exceed the 11pm time). Can you please let me know the error in my thought process, thanks.
Bunuel
Official Solution:

A filling pipe can fill an empty swimming pool in 4 hours, while a drain valve can empty the full pool in 5 hours. To fill the empty pool, the filling pipe was opened at 1:00 pm, and after some time, a drain valve was also opened. If the pool was filled at 11:00 pm, when was the drain valve opened?

A. at 2:00 pm
B. at 2:30 pm
C. at 3:00 pm
D. at 3:30 pm
E. at 4:00 pm


The rate of the filling pipe is given as \(\frac{1}{4}\) pool/hour and the rate of drain valve is given as \(\frac{1}{5}\) pool/hour. The filling pipe was opened for 10 hours, from 1:00 pm to 11:00 pm. Let x denote the time for which the drain valve was opened, then we'd have \(10*\frac{1}{4} - x*\frac{1}{5} = 1\), which gives \(x = 7.5\) hours. Hence, the drain valve was opened for 7.5 hours, and therefore was opened at 11:00 pm - 7.5 hours = 3:30 pm.


Answer: D
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ragnarok13
Bunuel sorry for the stupid question, but after calculating that the drain valve was open for 7.5 hrs, can we not think that maybe it was opened at 2pm and then closed at 9:30pm? (Same logic for Options A, B, C as it will not exceed the 11pm time). Can you please let me know the error in my thought process, thanks.
Bunuel
Official Solution:

A filling pipe can fill an empty swimming pool in 4 hours, while a drain valve can empty the full pool in 5 hours. To fill the empty pool, the filling pipe was opened at 1:00 pm, and after some time, a drain valve was also opened. If the pool was filled at 11:00 pm, when was the drain valve opened?

A. at 2:00 pm
B. at 2:30 pm
C. at 3:00 pm
D. at 3:30 pm
E. at 4:00 pm


The rate of the filling pipe is given as \(\frac{1}{4}\) pool/hour and the rate of drain valve is given as \(\frac{1}{5}\) pool/hour. The filling pipe was opened for 10 hours, from 1:00 pm to 11:00 pm. Let x denote the time for which the drain valve was opened, then we'd have \(10*\frac{1}{4} - x*\frac{1}{5} = 1\), which gives \(x = 7.5\) hours. Hence, the drain valve was opened for 7.5 hours, and therefore was opened at 11:00 pm - 7.5 hours = 3:30 pm.


Answer: D

There’s no indication that the drain valve was ever closed after it was opened. Introducing a closing time adds an extra unknown not supported by the information given, making the problem unsolvable under that assumption. Therefore, the only valid interpretation is that once opened, the drain valve stayed open until the pool was filled.
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Bunuel, sorry for bothering you a little more on this, but just for improving my understanding, what extra unknown would be added if we were to assume that the valve closed after a certain time, that would make the problem unsolvable.
Rate of filling = (1/4)th pool/hr and draining = (1/5)th pool/hr. Filling pipe is open for 10 hours and let drain pipe be open for x hours. In that case also this logic works : 10∗1/4−x∗1/5=1, because we know that the pool was filled at 11pm.
Can you please explain it a little further, thanks and sorry again!
Bunuel
ragnarok13
Bunuel sorry for the stupid question, but after calculating that the drain valve was open for 7.5 hrs, can we not think that maybe it was opened at 2pm and then closed at 9:30pm? (Same logic for Options A, B, C as it will not exceed the 11pm time). Can you please let me know the error in my thought process, thanks.
Bunuel
Official Solution:

A filling pipe can fill an empty swimming pool in 4 hours, while a drain valve can empty the full pool in 5 hours. To fill the empty pool, the filling pipe was opened at 1:00 pm, and after some time, a drain valve was also opened. If the pool was filled at 11:00 pm, when was the drain valve opened?

A. at 2:00 pm
B. at 2:30 pm
C. at 3:00 pm
D. at 3:30 pm
E. at 4:00 pm


The rate of the filling pipe is given as \(\frac{1}{4}\) pool/hour and the rate of drain valve is given as \(\frac{1}{5}\) pool/hour. The filling pipe was opened for 10 hours, from 1:00 pm to 11:00 pm. Let x denote the time for which the drain valve was opened, then we'd have \(10*\frac{1}{4} - x*\frac{1}{5} = 1\), which gives \(x = 7.5\) hours. Hence, the drain valve was opened for 7.5 hours, and therefore was opened at 11:00 pm - 7.5 hours = 3:30 pm.


Answer: D

There’s no indication that the drain valve was ever closed after it was opened. Introducing a closing time adds an extra unknown not supported by the information given, making the problem unsolvable under that assumption. Therefore, the only valid interpretation is that once opened, the drain valve stayed open until the pool was filled.
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ragnarok13
Bunuel, sorry for bothering you a little more on this, but just for improving my understanding, what extra unknown would be added if we were to assume that the valve closed after a certain time, that would make the problem unsolvable.
Rate of filling = (1/4)th pool/hr and draining = (1/5)th pool/hr. Filling pipe is open for 10 hours and let drain pipe be open for x hours. In that case also this logic works : 10∗1/4−x∗1/5=1, because we know that the pool was filled at 11pm.
Can you please explain it a little further, thanks and sorry again!

For example, we could have been told that the drain valve was stopped once, from 4:00 pm to 5:00 pm. That would mean the drain valve’s first opening occurred at 2:30 pm. However, I think you are overthinking and overcomplicating the problem. We are not told that the valve was ever stopped, and if you assume otherwise, you cannot solve the problem. Thus, the only valid interpretation is that once opened, the drain valve stayed open until the pool was filled.
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Bunuel Thank you so much :)
Bunuel
ragnarok13
Bunuel, sorry for bothering you a little more on this, but just for improving my understanding, what extra unknown would be added if we were to assume that the valve closed after a certain time, that would make the problem unsolvable.
Rate of filling = (1/4)th pool/hr and draining = (1/5)th pool/hr. Filling pipe is open for 10 hours and let drain pipe be open for x hours. In that case also this logic works : 10∗1/4−x∗1/5=1, because we know that the pool was filled at 11pm.
Can you please explain it a little further, thanks and sorry again!

For example, we could have been told that the drain valve was stopped once, from 4:00 pm to 5:00 pm. That would mean the drain valve’s first opening occurred at 2:30 pm. However, I think you are overthinking and overcomplicating the problem. We are not told that the valve was ever stopped, and if you assume otherwise, you cannot solve the problem. Thus, the only valid interpretation is that once opened, the drain valve stayed open until the pool was filled.
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I solved it this way....

Capacity of pool be = 20...
Filling rate is 5 litres/hour..Empty rate is 4 litres/hour
When together..1 litre/hour

Now...5(x) {Time when drain was not open} + 1(10-x) {Time when drain and filling both were open} = 20
5x +10 -x = 20
x = 2.5 hours Means..Drain started 2.5 hours post 1 pm...Answer D 3:30
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I like the solution - it’s helpful.
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I like the solution - it’s helpful.
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I had a doubt, I approached it as following
If the pool can be filled in 5 hrs when filling valve is open then the rate is 1/4 to do the work. Similarly for drain valve the rate is -1/5 and when both valves are functioning the rate at which the pool gets filled would be 1/20.

so if filling valve is open for say x hrs and then the drain valve is opened so for the rest y hrs both the valves are functioning.This would mean 2 things
1.x+y=10
2.x/4 of the pool will be filled in x hrs by filling valve and then 3x/4 of the pool will be filled while both the valves are working(Y hrs).=>Which helps me to arrive to the soln eqn. Y=15x and subtituting this in the 1st eqn which gave me x=5/8. So I thought 5/8 hrs after 1:00pm would be the time when drain valve should have be opened. 1+(5/8 *60) some 1:37 odd hrs.Which is no where close to the answer choices.

Bunuel can you please help me out where am I going wrong here?Anyone any inshights would help me,thanks!
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SachinNayak
I had a doubt, I approached it as following
If the pool can be filled in 5 hrs when filling valve is open then the rate is 1/4 to do the work. Similarly for drain valve the rate is -1/5 and when both valves are functioning the rate at which the pool gets filled would be 1/20.

so if filling valve is open for say x hrs and then the drain valve is opened so for the rest y hrs both the valves are functioning.This would mean 2 things
1.x+y=10
2.x/4 of the pool will be filled in x hrs by filling valve and then 3x/4 of the pool will be filled while both the valves are working(Y hrs).=>Which helps me to arrive to the soln eqn. Y=15x and subtituting this in the 1st eqn which gave me x=5/8. So I thought 5/8 hrs after 1:00pm would be the time when drain valve should have be opened. 1+(5/8 *60) some 1:37 odd hrs.Which is no where close to the answer choices.

Bunuel can you please help me out where am I going wrong here?Anyone any inshights would help me,thanks!

It should be x/4 + y/20 = 1 and x + y = 10, which gives x = 2.5 and y = 7.5.
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