\(5R=\frac{1}{4} \implies 3R=\frac{20}{3}\) (10 sec)
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Rates Pool/Hour; Both of their rate is 1/4 (Since they fill 1 pool in 4 hours)

Fast - X;
Slow - X/1.5 = 10X/15 = 2x/3

X + 2X/3 = 1/4 => X= 3/20 Pool per Hour

Since we marked faster one with the X if we want it's time we just need it's reciprocal 20/3 H.
(Work=timexRate => Time => Work/Time => Time = 1/(3/20) = 20/3)

I read quite a few solutions and they all have the following equation set up: 1/x+1/1.5x=1/4. But I have 1/x+1/1.5x=4. I don't understand why "x" is time. "x" to me is rate because the question states "constant rate of one pump was 1.5 times the constant rate of the other".

Can Bunuel or someone else please help? Thanks!

x is an unknown and you can denote by x either time or rate, it's your choice. Doesn't really matter.

If you say x is the rate, then you get: x + 2x/3 = 1/4. My solution here uses x as the rate: https://gmatclub.com/forum/two-water-pu ... l#p1245761

If you say x is the time, then you get: 1/x + 2/(3x) = 1/4.

In any case notice that we sum the rates.
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Maybe thinking in terms of numbers may help Irising


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Thank you for your reply! Please bare with me as I am still trying to understand the problem. When you say "we sum rates", do you also imply that we cannot sum time? In my equation, I set x as rate. Then I thought 1/x must be time (work/rate=time). And based on that, I sum time to solve for x: 1/x+1/(1.5x)=4, time of the slow pump+time of the fast pump=total time. There must be something wrong with my logic here but I can't seem to figure it out.

Thank you for your suggestion. Unfortunately, I am very bad with picking the "perfect" numbers that just work. It'll take me forever to come up with your solutions using numbers like 200, 20 and 30

Yes, we can sum rates but not times.

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.

17. Work/Rate Problems

• Theory
  Work Rate Problems - All in One Topic!
  Work Word Problems Made Easy
  How to Solve Relative Rate of Work Questions on the GMAT
  
  • Questions
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    PS Questions

On other subjects:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread
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Wow, thank you very much for your help Bunuel

Given: Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool.
Asked: If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate?

Let us assume that faster pump will take x hours to fill the pool working alone at its constant rate .
Slower pump will take 1.5x hours to fill the pool working alone at its constant rate.

1/x + 1/1.5x = 1/4
3/3x + 2/3x = 1/4
5/3x = 1/4
x = (5/3) * 4 = 20/3

IMO E
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slow pump works at a rate of 1x, fast pump works at a rate of 1.5x
slow pump does 1/3 of the job in 4 hours, fast pump does 2/3 of the job in 4 hours
Thus, for fast pump to accomplish the remaining 1/3 of the job, it would take 6 hours.
Where does this go wrong?

jasminelin this is where you went astray:


Based on your inference, the fast pump is twice as fast as the slow pump (2/3 is twice as much as 1/3).
But, we were told that the fast pump is only 1.5 times as fast as the slow pump. So, in any given amount of time, the fast pump should do three-halves (3/2) as much work as the slow pump does. Not twice as much.
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