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jasminelin this is where you went astray:
jasminelin
slow pump does 1/3 of the job in 4 hours, fast pump does 2/3 of the job in 4 hours

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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Hi experts, could anyone help to clarify why is faster pump's rate is x and not 1.5x here?
Also not sure why answer is the same regardless of faster pump rate is x or 1.5x? Thanks for your time in advanced.
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Two water pumps, working simultaneously at their respective constant rates, took exactly four 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

[spoiler=]Work rate problems are my weakest area. For whatever reason I get mixed up on these problem. Even knowing the answer to this question I can't figure out how it comes out to that answer. Any help would be great, thanks
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imagine pump 1 has rate of 4, while the second pump has 1.5 the rate of pump 1, so 6. pump 1 +pump 2 = 4+6 =10 rate per hour. In 4 hours it will complete 40 total.

as we need time of the the greatest pump which has the rate of 6, hence 40/6=20/3.

Answer 20/3
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If speed of faster one is 1.5 times the slower one, then time taken will be inverse, i.e. x and 1.5x.
Combined time formula = (x*1.5x)/(x+1.5x) = 1.5x/2.5 = 3x/5
This value is given as 4
x = 20/3

kmasonbx
Two water pumps, working simultaneously at their respective constant rates, took exactly four 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

Work rate problems are my weakest area. For whatever reason I get mixed up on these problem. Even knowing the answer to this question I can't figure out how it comes out to that answer. Any help would be great, thanks!
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Let the rate of slower pump=x then the rate of faster pump=1.5x

Their combined rate= x + 1.5x = 2.5x

Now, they take 4 hrs to complete the work, therefore, total work = R*T = 2.5x * 4 = 10x

Time taken by faster pump to complete the same work = 10x/1.5x = 20/3
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Rate of slower pump = a
Rate of faster pump = \(\frac{3}{2}\)a

Together, they filled the swimming pool in 4h.

Work Done to fill the pool = (a + \(\frac{3}{2}\)a) * 4 = \(\frac{5}{2}\)a * 4 = 10a.

How long will the faster pump take to do the same work alone?

\(\frac{10a}{(3/2)a}\) = 20/3 hours (Choice E).


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Let A be the constant rate of one pump and B the constant rate of the other pump.

We are told that the rate of one of the pumps is 1.5 faster than the other one.

Let's say that the rate of B is is 1.5 faster than the rate of A.

We can re write it: A + 1.5A = 1/4

Solve for A:

2.5A = 1/4
A = (1/4)/(5/2)

A = 1/4 * 2/5
A = 2/20
A = 1/10

Let's substitute A by its rate 1/10

1/10 + B = 1/4

B = 1/4 - 1/10
B = 10/40 - 4/40
B = 6/40
B = 3/20

Remember B is the faster pump, and we want to know how long it would take to fill the pool alone

Let's use the formula Time = Work/Rate

Time = 1/ (3/20)
Time = 1/1 * 20/3
Time = 20/3

kmasonbx
Two water pumps, working simultaneously at their respective constant rates, took exactly four 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

Work rate problems are my weakest area. For whatever reason I get mixed up on these problem. Even knowing the answer to this question I can't figure out how it comes out to that answer. Any help would be great, thanks!
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