There are two inlets and one outlet to a cistern. One of the

This is right where I got stuck. I knew that it should have taken only 1/2 an hour to fill the remaining 1/4 of the cistern. Instead it took double the time.

Can you explain your last equation? I'm confused about exactly why 1/4 and 1/2 have this relationship. Thank you.

Edit:

Is this a good explanation? If it takes double the time, that means that really only 1/4 per hour is filled. The other 1/4 is "stolen" by the outlet. So that must be the rate that the outlet pumps. Therefore, it would take 4 hours.

Similarly, if instead we had inlets pumping at a rate of 1/2 per hour and they were slowed down to 1/3 per hour, the outlet would be stealing 1/6, so the full time it would take the outlet is 6 hours? Is that correct?

Recall that we can sum/subtract the rates.

Inflow rate of the two inlets is 1/2 cistern/hour;
Outflow rate of the outlet is x cistern/hour;

Net inflow is 1/2 - x cistern/hour, hence 1/2 - x = 1/4.

Hope it helps.

P.S. Yes, your approach is correct.

Thank you so much! I have seen others use this approach on other Rate/Work problems, but I'm not sure I find all the formulas intuitive. I get this one now. Is there a comprehensive explanation of why we can use these shortcuts? I find that I have to reason through each individual problem, which is certainly inefficient.

THEORY
There are several important things you should know to solve work problems:

1. Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance in rate problems.

\(time*speed=distance\) <--> \(time*rate=job \ done\). For example when we are told that a man can do a certain job in 3 hours we can write: \(3*rate=1\) --> \(rate=\frac{1}{3}\) job/hour. Or when we are told that 2 printers need 5 hours to complete a certain job then \(5*(2*rate)=1\) --> so rate of 1 printer is \(rate=\frac{1}{10}\) job/hour. Another example: if we are told that 2 printers need 3 hours to print 12 pages then \(3*(2*rate)=12\) --> so rate of 1 printer is \(rate=2\) pages per hour;

So, time to complete one job = reciprocal of rate. For example if 6 hours (time) are needed to complete one job --> 1/6 of the job will be done in 1 hour (rate).

2. We can sum the rates.

If we are told that A can complete one job in 2 hours and B can complete the same job in 3 hours, then A's rate is \(rate_a=\frac{job}{time}=\frac{1}{2}\) job/hour and B's rate is \(rate_b=\frac{job}{time}=\frac{1}{3}\) job/hour. Combined rate of A and B working simultaneously would be \(rate_{a+b}=rate_a+rate_b=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\) job/hour, which means that they will complete \(\frac{5}{6}\) job in one hour working together.

3. For multiple entities: \(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+...+\frac{1}{t_n}=\frac{1}{T}\), where \(T\) is time needed for these entities to complete a given job working simultaneously.

For example if:
Time needed for A to complete the job is A hours;
Time needed for B to complete the job is B hours;
Time needed for C to complete the job is C hours;
...
Time needed for N to complete the job is N hours;

Then: \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}+...+\frac{1}{N}=\frac{1}{T}\), where T is the time needed for A, B, C, ..., and N to complete the job working simultaneously.

For two and three entities (workers, pumps, ...):

General formula for calculating the time needed for two workers A and B working simultaneously to complete one job:

Given that \(t_1\) and \(t_2\) are the respective individual times needed for \(A\) and \(B\) workers (pumps, ...) to complete the job, then time needed for \(A\) and \(B\) working simultaneously to complete the job equals to \(T_{(A&B)}=\frac{t_1*t_2}{t_1+t_2}\) hours, which is reciprocal of the sum of their respective rates (\(\frac{1}{t_1}+\frac{1}{t_2}=\frac{1}{T}\)).

General formula for calculating the time needed for three A, B and C workers working simultaneously to complete one job:

\(T_{(A&B&C)}=\frac{t_1*t_2*t_3}{t_1*t_2+t_1*t_3+t_2*t_3}\) hours.

Some work problems with solutions:
time-n-work-problem-82718.html?hilit=reciprocal%20rate
facing-problem-with-this-question-91187.html?highlight=rate+reciprocal
what-am-i-doing-wrong-to-bunuel-91124.html?highlight=rate+reciprocal
gmat-prep-ps-93365.html?hilit=reciprocal%20rate
questions-from-gmat-prep-practice-exam-please-help-93632.html?hilit=reciprocal%20rate
a-good-one-98479.html?hilit=rate
solution-required-100221.html?hilit=work%20rate%20done
work-problem-98599.html?hilit=work%20rate%20done
hours-to-type-pages-102407.html?hilit=answer%20choices%20or%20solve%20quadratic%20equation.%20R



Hope this helps.

In the case of tank eventually filling, the formula is 1/ f1 + 1/f2 - 1/d = 1 /F
where f1 and f2 are the time taken by pipes 1 and 2 resp to fill, d is the time taken to drain and F is the total time taken to Fill.

Here f1 and f2 happen for 2.5 hrs and d happens for 1 hr. So the formula becomes,

1/f1 + 1/f2 - 1/(2.5 d) = 1/F
1/3 + 1/6 - 1/ 2.5d = 1/2.5
1/ 2.5d = 1/10
d= 4hrs

working together : I1 and I2 take 2 hrs to complete

Let the rate of 3rd Pipe be I3

2.5*(1/2) - (I3 * 1) = 1

I3 = 1/4

Ans 4 hrs

I approached it this way.
at 10.30am the inlets had completed 3/4 of the work of filling the cistern. 1/4 cistern is empty.
they were supposed to finish it by 11.00am.

However at 10.30am the outlet is turned on.

the result is that the cistern will now be full by 11.30am.
So the inlets end up working 30 mins more to nullify the effect of outlet working for 1 hour from 10.30am to 11.30am.

so the work done by outlet in 1 hr = work done by both inlets in 30 mins. = emptying / filling 1/4 of the cistern.
work done by outlet in 1 hr = 1/4 of the cistern.
hence outlet takes 4 hours to complete the work

3/2 hour*1/2 combined rate=3/4 of cistern filled by 2 inlets together by 10:30
with no outlet running, 2 inlets would need another half hour to fill last 1/4
with outlet running from 10:30, 2 inlets need another hour to fill last 1/4
thus, outlet empties 1/4 of cistern in one hour,
and entire cistern in 4 hours
E

Hi All,

Here, we're told that one inlet takes 3 hours to fill a cistern while another takes twice as long (meaning it takes 6 hours to fill a cistern).

Using the Work Formula, we can figure out how long it takes the two inlets, working together, to fill the cistern:

Work = AB/(A+B) = (3)(6)/(3+6) = 18/9 = 2 hours

This means that the two inlets will completely fill 1/2 the cistern per hour.

Starting at 9am, the cistern would be filled at 11am. However, the outlet removes water at such a rate that at 10:30am, it takes a full hour to fill the cistern (as opposed to the 1/2 hour that it would take if there was no outlet at all).

So, at 10:30am, the cistern is 3/4 full, but then the outlet is turned on and it starts draining water….an hour later, the cistern is full. Since the cistern will be 1/2 full after an hour, the outlet must be removing 1/4 of the tank during that time (3/4 + 1/2 - 1/4 = 1 = full).

Thus, the outlet removes 1/4 of the water in 1 hour.

The question asks how long it will take the outlet to empty the cistern when it's full. Since the outlet removes 1/4 of the water in 1 hour, it will remove the entire volume of water in 4 hours.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

I figured this one out eventually, but it look 8 minutes Gotta be able to hit these in a more timely manner

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.