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by shibal » Sun Jul 12, 2009 1:31 pm
two water pumpss working together can fill a pool in 4 hrs. if the constant rate of one pump is 1.5 times the other, how many hours would take fot the faster pump to fill the pool alone?

oa [spoiler]20/3[/spoiler]
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by VP_Jim » Sun Jul 12, 2009 1:42 pm
Let's say that that pool has a capacity of 100 gallons. Now, we know that it takes both pumps working together 4 hours to fill the pool, so their combined rate is 25 gallons per hour.

We know that one pump's rate is 1.5x the other one's rate, so we can set up an equation:

Pump A's rate + Pump B's rate = Total rate
1.5x + x = 25
2.5 x = 25
x = 10

So, the slower pump's rate is 10 and the faster pump's rate is 15 (10 x 1.5).

Now, remember that we said the pool was 100 gallons. The faster pump will take 100/15 = 20/3 hours to fill it.

Plugging in values is a great way to do work and rate problems such as this, when the "job" - in this case, how big the pool is - is unknown.
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by shibal » Mon Jul 13, 2009 3:43 pm
is there a way of doing this w/out plugging in numbers?
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by ssmiles08 » Mon Jul 13, 2009 4:28 pm
shibal wrote:is there a way of doing this w/out plugging in numbers?
1/(3x/2) + 1/x = 1/4

(2+3)/3x = 5/(3x)



5/(3x) = 1/4

x = 20/3

the rate for the first one is 1/10 (1.5*20/3)
the rate for the second one is 1/(20/3) ( 1*20/3)

So the first machine pumps 1 tank in 10 hours
the second pumps 1 tank is 6.667 hours.

clearly the second tank has the faster rate, it pumps in 20/3 hours.
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