## Two water pumps working together...

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### Two water pumps working together...

by yass20015 » Sat Aug 08, 2015 1:27 pm
Two water pumps, working together 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 its constant rate?

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by Brent@GMATPrepNow » Sat Aug 08, 2015 2:03 pm
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
Let's pick some nice numbers that adhere to the given information.

...the constant rate of one pump was 1.5 times the constant rate of the other
So, the fast pump has a pumping rate that's 1.5 faster then the slow pump.
So, let's say the SLOW pump pumps at 2 gallons per hour
This means the FASTER pump pumps at 3 gallons per hour

Note: we don't know the volume of the pool yet.

Two water pumps, working together at their respective constant rates, took exactly 4 hours to fill a certain swimming pool
Their COMBINED RATE = 2 + 3 = 5 gallons per hour
If it took 4 hours for both pumps to fill the pool, then the volume of the pool = (4)(5) = 20 GALLONS

How many hours would it have taken the faster pump to fill the pool if it had worked alone at its constant rate?
The pool holds 20 GALLONS and the FASTER pump pumps at 3 gallons per hour
Time = output/rate
= 20/3

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by GMATGuruNY » Sat Aug 08, 2015 2:07 pm
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 its constant rate?

A. 5
B. 16/3
C. 11/2
D. 6
E. 20/3
Let A = the slower pump and B = the faster pump.
Let A's rate = 2 liters per hour.
Since B's rate is 1.5 times A's rate, B's rate = (3/2)(2) = 3 liters per hour.
Combined rate for A+B working together = 2+3 = 5 liters per hour.
At a combined rate of 5 liters per hour, A and B take 4 hours to fill the pool, implying that the pool = rt = 5*4 = 20 liters.
At a rate of 3 liters per hour, the time for B to fill the 20-liter pool = w/r = 20/3 hours.

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by nikhilgmat31 » Wed Aug 12, 2015 3:23 am
1/x + 1/y = 1/4

lets say faster pump x = 1.5(1/y)

1.5/y + 1/y = 1/4
2.5/y = 4

1/y = 1/10

so rate of x x = 1.5/y = 1.5/10

hours taken by x alone = 10/1.5 = 20/3

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by [email protected] » Wed Aug 12, 2015 9:13 am
Hi yass20015,

This is an example of a Work Formula question. Any time you have two entities (people, machines, water pumps, etc.) working on a job together, you can use the following formula:

(AB)/(A+B) = Total time to do the job together

Here, we're told that the total time = 4 hours and that one machine's rate is 1.5 times the other machine's rate...

If B = 1.5A then we have...

(A)(1.5A)/(A + 1.5A) = 4

1.5(A^2)/2.5A = 4

(3/2)(A^2) = 10A

A^2 = 20A/3

A = 20/3 hours to fill the pool alone

With this info, we can plug back in and figure out the rate of the other pump (B = 10, so it would take 10 hours to fill the pool alone)

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by Matt@VeritasPrep » Sun Aug 16, 2015 10:01 pm
Another approach here:

A's rate = x
B's rate = (3/2)x

Since together they do the job in 4 hours, their joint rate is 1/4 of the job per hour.

x + (3/2)x = 1/4, or

(5/2)x = 1/4, or

x = 1/10

So A's rate = 1/10 and B's rate = (3/2)*(1/10) = 3/20.

Time is reciprocal to rate, as we saw before, so B's time = 20/3, and we're done!

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by Poisson » Sat Oct 01, 2016 4:34 am
[email protected] wrote:Hi yass20015,

This is an example of a Work Formula question. Any time you have two entities (people, machines, water pumps, etc.) working on a job together, you can use the following formula:

(AB)/(A+B) = Total time to do the job together

Here, we're told that the total time = 4 hours and that one machine's rate is 1.5 times the other machine's rate...

If B = 1.5A then we have...

(A)(1.5A)/(A + 1.5A) = 4

1.5(A^2)/2.5A = 4

(3/2)(A^2) = 10A

A^2 = 20A/3

A = 20/3 hours to fill the pool alone

With this info, we can plug back in and figure out the rate of the other pump (B = 10, so it would take 10 hours to fill the pool alone)

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I'm getting lost at the very end. Is A the faster pump? A = 20/3 or 6.66 hours and B = 10 hours. But we setup the equation so that B = 1.5A. Could someone please clarify? Many thanks

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by [email protected] » Sat Oct 01, 2016 10:18 am
Hi Poisson,

In these types of situations, you have to think (and translate the math) very carefully. I chose to make Pump A the 'faster' pump, which means that it takes LESS time to fill the swimming pool on its own (and that means that the number that represents "A" would be SMALLER than the number that represents the Pump B).

If you've read through the various explanations, you can see that it takes Pump A = 20/3 = 6.6666 hours to fill the pool and Pump B = 10 hours to fill the pool. Pump A IS the faster pump.

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