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Question 644146: one pipe can fill a pool 1.25 times faster than a second pipe . When both pipes are opened , they fill the pool in five hours . How long would it take to fill the pool if only the slower pipe is used ? tanx
Answer by DrBeeee(684)   (Show Source):
You can put this solution on YOUR website! Let r1 = flow rate of slower pipe
Let r2 = flow rate of faster pipe
Let G = the capacity of the pool
Knowns:
1) faster rate = 1.25 slower rate.
2) It takes 5 hours to fill the pool with both pipes.
Unknowns:
a) The time, T. to fill pool at the slower flow rate.
Appropriate formula to use:
G = r*t, where G is capacity in gallons, r is the rate of flow in gal/hr and t is the time in hours.
Solution:
With both pipes flowing, we have
(1) G = (r1 + r2)*5
Substituting the knowns into (1) yields
(2) G = (r1 + 1.25*r1)*5
Simplifying (2), we get
(3) G = r1*(1 + 1.25)*5 or
(4) G = 2.25*5*r1 or
(5) G = 11.25*r1
Filling the pool with just at the slow rate, is given by
(6) G = r1*T
Setting G of (5) equal to the same G of (6) yields
(7) G = G or
(8) 11.25*r1 = r1*T
The r1 factor on LS cancels r1 on the RS, leaving only
(9) T = 11.25 hrs
We should always check our answer. Any idea of how can we check this value of time T? Let me ask you,"Is r2 = 1.25*r1?"
Let's find r1 and r2.
In (5) we have
(10) r1 = G/11.25
From (2) we have
(11) r1 + r2 = G/5
Now substitute (10) into (11) to get
(12) r2 = G/5 - G/11.25 or
(13) r2 = ((11.25 - 5)/(5*11.25))*G or
(14) r2 = (6.25/(5*11.25))*G or
(15) r2 = 125/1125*G or
(16) r2 = G/9
Now
Is (r2 = 1.25*r1)?
Is (G/9 = 1.25*(G/11.25))?
Is (G/9 = G/9)?
Is (9 = 9)? Yes
Answer: The pipe with the slower flow rate will take 11.25 hours to fill the pool.
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