## College Algebra (10th Edition)

$9.45$ a.m.
The main pump empties the tank in $4$ hours but has only $3$ hours now so it will empty $0.75$ of the tank, so $0.25$ should be emptied by the auxiliary pump, which does this in $\frac{0.25}{\frac{1}{9}}=\frac{9}{4}$ hours, so to finish at noon, it has to start $\frac{9}{4}$ hours earlier, which is at $9.45$ a.m.