SOLUTION: 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., at what time should th
Algebra ->
Rate-of-work-word-problems
-> SOLUTION: 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., at what time should th
Log On
Question 499219: 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., at what time should the smaller pump be started so that the tank will be emptied at 4:00 P.M.? Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! 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., at what time should the smaller pump be started so that the tank will be emptied at 4:00 P.M.?
total time first pump is running:3 hours
Let x = time second pump is running
then
3(1/4) + x(1/11) = 1
multiply both sides by 44:
3(11) + x(4) = 44
33 + 4x = 44
4x = 11
x = 11/4
x = 2.75 hours
or
x = 2 and 3/4 hours
.
Answer: 1:15 pm
