SOLUTION: A water tank can be emptied by using one pump for 6 hours. A second, smaller pump can empty the tank in 10 hours. If the larger pump is started at 1:00 P.M., how much time should p

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Question 832354: A water tank can be emptied by using one pump for 6 hours. A second, smaller pump can empty the tank in 10 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 5:00 P.M.?

Answer by josgarithmetic(39615) About Me  (Show Source):
You can put this solution on YOUR website!
One job is, "empty the tank".
Regular pump, 1%2F6 tank per hour
Small pump, 1%2F10 tank per hour
Combined, 1%2F6%2B1%2F10 tank per hour

The combined rate is 5%2F30%2B3%2F30=8%2F30=4%2F15 tank per hour

Your fundamental formula fact: Rate%2Atime=job.

Ignoring time points but just dealing with time lengths,
Total time to empty tank is 4 hours.
Let t = time that pumps operate together.
Regular tank alone for 4-t hours.
The pumps work together for t hours.
'
ACCOUNT FOR THE ONE JOB: highlight%28%281%2F6%29%284-t%29%2B%284%2F15%29%28t%29=1%29
Solve for t.
Remember, t is a time quantity, NOT a time point.

I am stopping here, so you do the arithmetic steps yourself. Start by multiplying left and right members of the equation by 30, which will clear the denominators, since lowest common denominator is 30.