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

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A pool can be filled in 4 hours and drained in 5 hours. The [#permalink]

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01 Apr 2009, 11:50
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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 some time 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?

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

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

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[Reveal] Spoiler: OA

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01 Apr 2009, 16:16
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Economist wrote:
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 some time 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?
*at 2:00 pm
* at 2:30 pm
* at 3:00 pm
* at 3:30 pm
* at 4:00 pm

Don't have the OA I am interested in the approach.
I did:
10/4 - x/5 = 1/10 ( x = number of hours drain pipe is opened ). But I cant find the 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.
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01 Apr 2009, 20:21
Thanks Ian.
Here you added the work in the last equation. Can we solve the same problem by adding rates instead?

IanStewart wrote:
Economist wrote:
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 some time 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?
*at 2:00 pm
* at 2:30 pm
* at 3:00 pm
* at 3:30 pm
* at 4:00 pm

Don't have the OA I am interested in the approach.
I did:
10/4 - x/5 = 1/10 ( x = number of hours drain pipe is opened ). But I cant find the 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.

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01 Apr 2009, 21:27
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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

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03 Apr 2009, 07:47
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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.
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03 Apr 2009, 08:25
IanStewart wrote:
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.

Great explanation !!! Thanks Ian

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16 Aug 2009, 05:13
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 some time 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?
*at 2:00 pm
* at 2:30 pm
* at 3:00 pm
* at 3:30 pm
* at 4:00 pm

Here is how I solved the problem:

let A = the number of hours the filling valve is open by itself
let B = the number of hours the filling and draining valve are open together

Filling rate: 1 pool / 4 hrs
Draining rate: 1 pool / 5 hrs

therefore,

(1 pool / 4 hrs)*(A+B hrs) - (1 pool / 5 hrs)*(B hrs) = 1 pool

and we know that the pool was filled in 10 hours:

A + B = 10

so now we have two equations and two unknowns; solve for A:

(A+B)/4 - B/5 = 1
5*(A+B) - 4*5 = 20
5*A + B = 20

substitute B for 10-A:

5*A + 10 - A = 20
A = 2.5

So the pool was filled 2 and half hours after 1 PM or 3:30 PM

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26 Oct 2009, 10:41
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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

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30 Jun 2010, 12:15
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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

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06 Jul 2010, 05:38
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

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07 Jul 2010, 11:58
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bibha wrote:
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

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

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08 Jul 2010, 05:02
oh WOW!!! great . thanks for the explanation

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17 Dec 2011, 12:44
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Filled rate= 1/4, Drained rate = 1/5
Total hours took = 10 [Given]
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
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22 Dec 2011, 02:18
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?
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22 Dec 2011, 06:56
siddharthmuzumdar wrote:
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?

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.
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23 Dec 2011, 23:27
IanStewart wrote:
siddharthmuzumdar wrote:
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?

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.

Thank you for the advice Sir. Will keep a tab on the easier ones and try to finish them off (correctly) in as less as time as possible so that I have sufficient time for the tougher problems.

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26 Dec 2011, 05:49
after x hrs B was opened
so work done by A = 10 hrs
and work done by B = 10-X hrs
so as per question 10/4 - (10-x)/5 =1 gives x=2.5

so time of opening B = 1+2.5 =3.5 or u may say 3.30 PM

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22 Apr 2014, 22:23
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Let total units =4*5=20
In 1 hour A fills = 20/4=5 units
In 1 hour B empties =20/5= 4 units
So, when both are opened they fill =5-4=1 unit
Let in X hours B will be opened
5x+(10-x)=20, so, x=2.5

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A pool can be filled in 4 hours and drained in 5 hours. The [#permalink]

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12 Dec 2014, 02:32
Economist wrote:
Thanks Ian.
Here you added the work in the last equation. Can we solve the same problem by adding rates instead?

IanStewart wrote:
Economist wrote:
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 some time 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?
*at 2:00 pm
* at 2:30 pm
* at 3:00 pm
* at 3:30 pm
* at 4:00 pm

Don't have the OA I am interested in the approach.
I did:
10/4 - x/5 = 1/10 ( x = number of hours drain pipe is opened ). But I cant find the 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.

I preferred a slightly different approach, perhaps the way I perceived it - of how long the drain had to be left open. So, my equation became (given 1 to be the unit of a single pool):
(10/4) - (x/5) = 1
=>50-4x = 20
=> x = 30/4 = 7.5 hours

(11 pm - 7.5 hours) or (1 pm + 2.5 hours) = 3.30 pm.

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Re: A pool can be filled in 4 hours and drained in 5 hours. The [#permalink]

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12 Dec 2014, 04:53
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Economist wrote:
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 some time 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?

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

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

m20 q32

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 some time 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?

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

When only the valve is open, the pool is filling at a rate of $$\frac{1}{4}$$ an hour. When both the valve and the drain are open, the pool is filling at a rate of $$\frac{1}{4} - \frac{1}{5} = \frac{1}{20}$$ an hour. Let $$x$$ denote the time when only the valve was open. Then both the valve and the drain were open for $$11 - 1 - x = 10 - x$$hours. Now we can compose the equation $$\frac{1}{4}x + \frac{1}{20}(10 - x) = 1$$ which reduces to $$\frac{x}{5} = 0.5$$ from where $$x = 2.5$$. Thus, the drain was opened at $$1:00 + 2:30 = 3:30$$pm.

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Re: A pool can be filled in 4 hours and drained in 5 hours. The   [#permalink] 12 Dec 2014, 04:53

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