algebra
posted by Saige .
Pump A can fill a swimming pool in 6 hours and pump B can fill the same pool in 4 hours. The two pumps began working together to fill an empty pool, but after 40 minutes pump B broke down, and pump A had to complete the job alone. The pool was scheduled to open 4 hours after the pumping started. Was it full by then, or did the opening have to be delayed?

rewording to indicate the rate of completion,
a = 1/6 pool/hr
b = 1/4 pool/hr
they work together for 2/3 hour, so they filled
(1/6 + 1/4)*2/3 = 5/18 pool
So, how long does it take A to fill the other 13/18 pool?
13/18 / 1/6 = 13/3 hour
So, the whole pool took 2/3 + 13/3 = 15/3 = 5 hours to fill, finishing an hour late. 
T = t1*t2/(t1+t2) = 6*4/(6+4) = 2.4h = 144 Min. to fill the tank.
40min./144min. * 1 = 5/18 of a tank in 40 min.
15/18 = 18/18  5/18 = 13/18 to be filled by pump A.
13/18 * 6h = 4 1/3h = Time required for pump A to finish the job. But this is greater than the 4hour limit to do the job. So the opening had to be delayed.
4 1/3 + 40/60 = 13/3 + 2/3 = 5h = Tot.
to do the job.
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