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., how much time should p

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., how much time should p      Log On


   

Question 897728: 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.?
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
This solution will be in time quantities, and not points in time.

PIPE RATES FOR EMPTYING TANK UNITS of JOBS per HOUR.
Fast pipe, 1%2F4
Slow pipe, 1%2F11

Total time for the one job is to be 3 hours. The fast pipe starts first and the slow pipe is
later started and works with the large pipe until all 3 hours pass; and the 1 job is done.

Fast pipe works alone for t hours; and both pipes work together for 3-t hours.

highlight%28%281%2F4%29t%2B%281%2F4%2B1%2F11%29%283-t%29=1%29
Solve this equation for t, the amount of time that the fast pipe works alone before
the slow pipe is opened. (Lowest Common Denominator is 4%2A11, so multiply both sides of
the equation by this.)

11t%2B%2811%2B4%29%283-t%29=44
11t%2B15%283-t%29=44
11t%2B45-15t=44
-4t%2B45=44
-4t=-1
highlight%28t=1%2F4%29-------FIFTEEN MINUTES, a fourth of an hour.