Question

A tank that holds 6000 gallons of water may be filled by two pipes. One pipe pours water into the tank at the rate of 80 gallons per minute. The other fills the tank at the rate of 120 gallons per minute.

(a) If the pipes run simultaneously, how long will it take to fill the tank?

(b) The small pipe is opened and water is run into the tank for a certain length of time. Then this pipe is shut off and the large one is used to finish filling the tank. How long was each pipe used if it took 60 minutes to fill the tank?

(c) The large pipe alone is opened for a certain time and then both pipes together are used to finish filling the tank. If it takes 34 minutes to fill the tank in this manner, how long was the small pipe used?

Answer 1

Below.

Explanation:

Working together the rate is 80 + 120 = 200 gallons per minute.

(a) Time to fill tank = 6000 / 200 = 30 minutes.

(b) Let length of time the small pipe was used be x minutes , the the large pipe was used for is 60-x minutes.

Volume delivered by small pipe = y gallons, and by large pipe = 6000-y gallons.

Rate for small pipe = 80 = y/x.

Rate for large pipe = 120 = (6000-y) / 60-x.

y/x = 80

(6000-y) / 60-x = 120

From first equation y = 80x so:

6000 - 80 x = 7200 - 120x

40x = 1200

x = 30

So time for small pipe = 30 and time for large on = 30 minutes also.

C. In a similar way to part B:

y/x = 120

(6000-y) / 34 - x = 200

(6000-120x) / 34 - x = 200

6000 - 120x = 6800 - 200x

80x = 800 so the large pipe on its own took 10 minutes and the small pipe was used for 34 - 10 = 24 minutes.