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24 February, 12:59

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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  1. 24 February, 15:13
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    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.
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