If you have an engine that supplies 6 hp but is only 90% efficient, what's the power input?

\(\text { Efficiency }=\frac{\text { Power }_{\text {out }}}{\text { Power }_{\text {in }}} \times 100\)
\(E=\frac{P_{o}}{P_{i}} \times 100\)
So
\(P_{i}=\frac{P_{o}}{\frac{E}{100}}\)
\(P_{i}=\frac{6}{\frac{90}{100}}=\frac{6}{0.9}=6.67 \mathrm{hp}\)