A pool can be filled in 4 hours and drained in 5 hours. The valve that

Question

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

I did:
10/4 - x/5 = 1/10 ( x = number of hours drain pipe is opened ). But I cant find the answer.

M20-32

cheryl3007 · Accepted Answer

My solution is similar to h2polo's:

Filling rate : ¼
Draining rate: 1/5 
From 1.00 pm to 11 pm → 10 hours
Let’s say X is the hours that Draining valve works
→ 10/4 – 1/5 * X = 1 (pool is filled)
→ X = 7,5
→ The filling valve works 10-7,5 = 2,5 hours before the draining valve is started
1.00 pm + 2.5 = 3.30 pm → D

Cheers,
Cheryl

IanStewart · Answer

I sent this via PM to Economist, but thought it might be useful to others:

When you have workers working in sequence, you'd want to add the work done by each. It's when you have workers working simultaneously that you'd want to add their rates.

I can illustrate with three different questions, all based on the same initial setup:

Say, worker A can do 1 job in 10 hours, and worker B can do 1 job in 15 hours.

a) If they work together for 30 hours, how many jobs will they complete?

Here, you'd add their rates to determine how many jobs they can finish together. You should find they can complete one job every six hours working together, so in 30 hours they'd finish five jobs.

b) If worker A works alone for 15 hours, and then stops, and then worker B works alone for 15 hours, how many jobs will they complete in the 30 hours?

Here, you don't want to add their rates, because they never work simultaneously. Instead you'd want to add together the work each does during the 15 hours each works - in 15 hours, A will do 1.5 jobs, and in 15 hours, B will do 1 job, so combined they'll complete 2.5 jobs.

c) If worker A works alone for 15 hours, and then B and A work together for 15 hours, how many jobs will they complete in the 30 hours?

This question most resembles the one from the post above. A will complete 1.5 jobs in the first 15 hours. For the remaining 15 hours, A+B work together. We need to work out their combined rate, and we find they will complete 1 job every 6 hours, and will therefore complete 2.5 jobs in 15 hours. In total, over the 30 hours, 4 jobs will therefore be completed.

_______________________

There are other approaches to the problems above, of course, but I've confined the discussion to work/rate principles to illustrate the differences between each question setup.

Bunuel · Answer

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

IanStewart · Answer

Imagine both the drain and the valve are on at the same time. You can use a formula here; I just convert each to the same amount of time (20 hours):

valve will fill 5 pools in 20 hours
drain will empty 4 pools in 20 hours
valve+drain will fill 1 pool in 20 hours

So we really have two 'workers' here; the valve, which fills one pool in 4 hours, is on for the first t hours, and the valve+drain combo, which fills one pool every 20 hours, which was on for the remaining 10 - t hours. While the valve was on, t/4 of the pool was filled, and while the valve+drain combo were on, (10-t)/20 of the pool was filled. Adding these, we must get 1 pool:

t/4 + (10-t)/20 = 1
5t + 10 - t = 20
4t = 10
t = 2.5

So the drain was turned on 2.5 hours after 1pm, or at 3:30pm.

pleonasm · Answer

A different approach :

So in one hour V1 will fill 1/4th of the pool
V2 will empty 1/5th of the pool.

If the valves were opened at the same time: 1/20th of the pool will be full in 1 hour ( Difference of 1/4 and 1/5 )

So if both the valves were opened at 1 PM it would take 20 hours to fill the pool. That's not the case. Work from the answer choices.

If V2 was opened at 3 PM. In 2 hours V1 would have filled : 1/2 of the tank. To fill the remaining half it would take 10 hours. 3 + 10 = 13. So the time should be after 3.

If V2 was opened at 3.30 : in 2.5 hours, V1 would have filled 1/4*2.5 : 5/8th of the tank. From 3.30 to 11 : V1 and V2 have 7.5 hours to fill the remaining 3/8th of the tank.

Operating together : 1/20 can be filled in 1 hour
3/8 can be filled in 3/8 *20 = 7.5 hours. Bingo

-pradeep

tingle15 · Answer

We know that Rate x Time = Work done

Here,
Ri = Rate of the inlet valve
Ro = Rate of the outlet valve
Ti = Time for which the inlet valve was open
To = Time for which the outlet valve was open
x = Time after which the outlet valve is open

The time taken for pool to fill to capacity is 10 hours.

RiTi - RoTo = 1
(1/4)(10) - (1/5)(10-x) = 1

Solving the above equation gives x = 2.5 hours

Hence the outlet valve was opened 2.5 hours after 1pm i.e., 3:30pm

bibha · Answer

Ok.....why is it that i can't understand any of the explanations??????? Guys, i am really weak with these kind of work problems. Would anyone care to provide a solution to this problem once again??? please :-(

GMATBLACKBELT720 · Answer

You can fill 1 pool in 4 hours. However we have a drain that drains at a rate of 1 pool in 5 hours.

Lets set up a basic rate equation.

we know the rate of the pump is 1pool/4hours. We also know this pump operated from 1PM to 11PM, thus 10 hours. Thus the pump filled 1pool/4hours*10hours or 10/4pools.

But the drain drained the pool at 1pool/5hours * t hours. t is our unknown value and is the number of hours the drain was on.

The total work was one pool, or represented as 10/4+1/5t = 1. Now we just solve for t.

t=15/2 = 7.5 hours or the number of hours the drain was on. Thus we have to make one final step of 10-7.5hrs = 2.5hrs, which is the number of hours the drain was turned on after the pump.

1PM +2.5hrs = 3:30PM

D

MHIKER · Answer

Filled rate= 1/4, Drained rate = 1/5
Total hours took = 10  
Let drained valve work for x hours
so, (1/4)(10) - (1/5)(10-x) = 1
x = 2.5 hours

Thus, the outlet valve was opened 2.5 hours after 1pm i.e., 3:30pm
Ans. D

siddharthmuzumdar · Answer

D is the answer.
Nice problem. Thanks a ton. I got the solution but it took me about 2.5 minutes to get to the correct answer. Is that reasonable for such problems or is it a little over the limit?

IanStewart · Answer

This is not an official question, so you shouldn't be so concerned about solving it in under 2 minutes. I think it is a bit of a lengthy problem anyway. On the real test, you need to average 2 minutes per question - if you can solve some questions in 1.5 minutes, then it's okay to spend 2.5 minutes sometimes. Normally I wouldn't recommend ever spending more than 3 minutes on a question (and if you do, you better be sure you'll get the right answer) unless you know you have extra time.

mahrah · Answer

Bunuel

If the value was open for x hours, then why is it (11-1-x) hours when both the valve and the drain were open and not just (11−x)hours?

Thanks!

Bunuel · Answer

The valve was opened from 1:00 pm to 11:00 pm, so for 11 - 1 = 10 hours. If x is the amount of time when  only  the valve was open, then 10 - x is the amount of time when both the valve and the drain were open.

EgmatQuantExpert · Answer

Solution

Given:  
 • The pool can be filled in 4 hours
• The pool can be drained in 5 hours. 
• Filling valve was opened at 1:00 pm
• The Pool was filled at 11:00 pm

Working out:

We need to find out the time when the drain valve was opened.

Per our conceptual knowledge, we know that the total volume of the pool can be taken as the LCM of 4 and 5. 
 Thus, let’s assume that the total volume of the pool = 20 Liters.

Since the pool can be filled in 4 hours, in 1 hour, the amount of water filled in it = 5 litres. 
Similarly, the amount of water drained in 1 hour = 4 litres.

Let’s also assume that the time after which the drain valve was opened = x hours. 
So, from 1 pm to (1+x) pm, the pool was being filled at the rate of 5 liters per hour. 
And from (1+x) pm to 11 pm, the pool was filled at a rate of (5-4) = 1 liter per hour.

Thus, 
 • 5*x+ 1(10-x) = 20 
• Or, 5x + 10 -x = 20 
• Or, 4x = 10
• Or, x =2.5 hours.

Thus, the drain valve was opened at 1+ 2:30 hours = 3:30 pm.

Answer: Option D

GmatPoint · Answer

A pool can be filled in 4 hours and drained in 5 hours. The valve that fills the pool was opened at 1:00 pm and sometime later the drain that empties the pool was also opened. If the pool was filled by 11:00 pm and not earlier, when was the drain opened?.
﻿Considering the total work required to fill the pipe = W.
﻿The rate of work done only when the inlet is working =  \frac{W}{4} 
﻿The rate of work done when both the inlet and the outlet are working =  \frac{W}{4}-\frac{W}{5}=\ \frac{W}{20}

The total time taken to get the work done is equal to 10 hours.

For the 10 hours complete period the inlet pipe was working. The total work done by the inlet =  \frac{10W}{4}

For a certain period of time assuming to be x, the outlet was draining : =  -x\left(\frac{W}{5}\right)

The work combination amounts to total work done:  \frac{10W}{4}-\ \frac{xW}{5}=\ W  
  \frac{3W}{2}=\ \frac{xW}{5} 
﻿x = 7.5 hours.
﻿Hence the outlet pipe started 2 and half hours later the inlet pipe started working.
﻿Hence it was opened at 3: 30 pm.
﻿﻿﻿

adewale223 · Answer

Rate for filling

1/4

Rate for draining

1/5

Effective rate

1/4 - 1/5

= 1/20

Pool was filled for some hours before draining pipe opened.

Total time 10 hours.

Let a be time filling pipe operated and b when draining pipe opened.

a + b = 10.

So b = 10-a

Let y be fraction of pool filled before drain was opened.

1/4 = y/a

y = a/4

And let z be remained work done after drain was opened.

1/20 = z / 10-a

20z = 10-a

z = 10-a / 20

Therefore total work done

a / 4 + 10-a/ 20 = 1

5a + 10 - a = 20

4a = 10

a = 2.5

Which means drain was opened after 2 hours 30 minutes.

Answer is 3:30pm.

Posted from my mobile device

Bunuel · Answer

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

e3thekid · Answer

We are given that the pool was filled at 11 pm which is 10 hours after the fill valve was opened at 1 pm. Total time = 10 hours. Knowing the rates of both valves, we can set up a matrix.

Work type : r x t = w
 Fill 1/4 x 10 = 5/2
 Drain 1/5 x 10-t = 2 - t/5

5/2 - (2 -t/5) = 1
10 (5/2 -2 + t/5 = 1) 10

25 - 20 + 2t = 10
5 + 2t = 10
2t = 5
t = 2.5

1 pm + 2.5 hours = 3:30 pm

Option D

Posted from my mobile device

GMAT695 · Answer

Total Work = 20 units.

A- 5 units/hr
B- -4 units/hr 
T- 1units/hr

5 X 10 - 4x = 20
x= 30/4 = 7.5.

The drain valved was opened at  3:30 PM ( 11pm - 7.5 hr )

A pool can be filled in 4 hours and drained in 5 hours. The valve that

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