SOLUTION: One pipe can fill a cistern in 12 hours and a smaller pipe can cill the cistern in 15 hours. If both pipes are used, how many hours are needed?

Question 638573: One pipe can fill a cistern in 12 hours and a smaller pipe can cill the cistern in 15 hours. If both pipes are used, how many hours are needed?
Answer by DrBeeee(684) (Show Source): You can put this solution on YOUR website!
In general, the number of gallons delivered is the rate of delivery in gallons per hour, times the number of hours.
Let G = total number of gallons of water that the cistern holds.
The rate that the large pipe can deliver G gallons is
(1) rl = G/12
The rate that the smaller pipe delivers water is
(2) rs = G/15
When we fill the cistern with both pipes, the rate is the sum of the two,
(3) R = (rl + rs|
Using the general flow rate fomula gives, for both pipes filling the cistern
(4) G = (rl + rs)*T, where T is the new time to fill the cistern when both pipes are being used.
Solve (4) for T, yields
(5) T = G/(rl + rs)
Now substitute (1) and (2) into (5) yields
(6) T = G/[G/12 + G/15]
Now factor G from the denominator of (6)
(7) T = G/[G(1/12 + 1/15)]
Cancel G from (7) yields the final equation for T
(8) T = 1/(1/12 + 1/15)
Evaluate (8) gives us the value of T
T = 20/3 hrs
Answer: It will take 6 hours and 40 minutes to fill the cistern when we use both pipes. Note: we did not need to know the capacity, G, of the cistern
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