## Basic College Mathematics (10th Edition)

Each car needs $\frac{2}{3}$ gallon of lubricant. To determine how man cars can be serviced from a 50 gallon drum of lubricant you must divide 50 by $\frac{2}{3}$. $50 \div \frac{2}{3}$ = $50 \times \frac{3}{2}$ = $\frac{150}{2}$ = $\frac{150 \div 2}{2 \div 2}$ = $\frac{75}{1}$ = 75
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