A water tank can be emptied by using one pump for 4 hours. A second, smaller pump can empty the tank in 11 hours. If the larger pump is started at 1:00 P. M., how much time should pass until the smaller pump is started so that the tank will be emptied at 4:00 P. M.?

Question

Faith Weeks

Mathematics

8 March, 04:47

A water tank can be emptied by using one pump for 4 hours. A second, smaller pump can empty the tank in 11 hours. If the larger pump is started at 1:00 P. M., how much time should pass until the smaller pump is started so that the tank will be emptied at 4:00 P. M.?

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Answers (1)

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Bautista · Answer 1

The second pump should be started at 13:15 hs.

Step-by-step explanation:

Pump number can empty the tank in 4 hours.

If it started at 1:00 PM then, at 5:00 PM would empty the tank.

Pump number 2, can empty the tank in 11 hours.

Pump 1, drains 1/4 of the tank each hour.

Pump 2, drains 1/11 of the tank each hour.

I need to complete the drainage in 3 hours.

The equation is. 3 hours using the pump one at a rate of 1/4 per hour plus unknown time using the second pump at a rate of 1/11 per hour is equal to the whole tank. (whole tank is 1)

3*1/4 + X * 1/11 = 1

We are going to clear the equation and leave X.

So it is X = 11/4. That is the time needed to have the large pump active.

That is 2 hours and 45 min.

So the second pump should be started at 4 pm - 2.45hs = 13:15 hs.