College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.7 - Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications - 1.7 Asses Your Understanding - Page 143: 49


$9.45$ a.m.

Work Step by Step

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.
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