Work Rate Problem

jk2010 wrote:Two water pumps working simultaneously at respective constant rates took exactly 4 hours to fill a certain swimming pool. 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 alone?

[spoiler]20/3

[/spoiler]

Can you please confirm the source as well as provide the answer choices from the original question?

Thanks!
Whit

jk2010 wrote:Unfortunately, I don't have the other answer choices. The source, however, was a GMAC practice test.

Looks like you are correct, this should be the complete question:

Two water pumps, working simultaneously at their respective constant rates, took exactly 4 hours to fill a certain swimming pool. 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?

A. 5
B. 16/3
C. 11/2
D. 6
E. 20/3

METHOD #1
One way to think about Work/Rate problems is to remember that when 2 (or more) things are working together, we can add their rates (where the rate is just job/time). This would mean we could have the following equation:



So we can start to plug in what we know. If the rate for the fast pump is 1.5 times the rate of the slow pump, then



So we can substitute in:

1.5*(1/t) + 1/t = 1/4

Solving, we get:

1.5/t + 1/t = 1/4
2.5/t = 1/4
t = 2.5*4 = 10 = time it takes for the slow pump to fill the pool.

Now we can go back to the substitution equation we used:



1/Time(for fast) = 1.5 *(1/10)

1/Time(for fast) = 1.5/10

Time(for fast) = 10/1.5 = 100/15 = 20/3.


***METHOD 2****
An alternate way to think about the 1.5 is to consider that IF the faster pump has a rate 1.5 times that of the slow pump, then for the same job, the faster pump can finish the job in the slower pump's time divided by 1.5 --> we just have to remember that if something is working FASTER, then it should finish the job in a SHORTER amount of time (hence the division by 1.5). The work from above should help you understand how this works out mathematically.

Now, let's call s the time for the slow pump and f the time for the fast pump.

f = s/1.5, or 1.5f = s. Now we can plug this in to the original formula:



1/f + 1/s = 1/4

1/f + 1/(1.5f) = 1/4

1.5/1.5f + 1/1.5f = 1/4

2.5/1.5f = 1/4
10 = 1.5f

f = 10/1.5 = 100/15 = 20/3

Hope this helps!!


Whit