Question

John is trying to drain a swimming pool. He has two pumps, but he can only use one at a time. He knows that the time a pump takes to drain the pool varies inversely with the power in watts of the pump. His old pump is a 80 watt pump, and it can drain the pool in 5 hours. How long would the job take if he uses his new 100 watt pump?

Answer 1

`\bf \begin{array}{llllll} \textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\ \textit{something}&=&\cfrac{{{\textit{some value}}}}{}&\cfrac{}{\textit{something else}}\\ \quad \\ y&=&\cfrac{{{\textit{k}}}}{}&\cfrac{}{x} \\ &&y=\cfrac{{{ k}}}{x} \end{array}\\\\ -----------------------------\\\\`


`\bf \textit{the time a pump drains the pool varies inversely with the wattage } \\\\\\ thus\qquad t=\cfrac{k}{w}\qquad \begin{cases} t=time\\ k=\textit{constant of variation}\\ w=\textit{wattage power} \end{cases} \\\\\\ \textit{now, we know that with the old pump } \begin{cases} w=80\\ t=5 \end{cases}\implies 5=\cfrac{k}{80} \\\\\\ 5\cdot 80=k\implies 400=k\qquad thus\implies \boxed{y=\cfrac{400}{w}}`

now, how long will it take with the new pump that does 100watts?

well, set w = 100, to get the "t" value

Answer 2

it would take 6
`(1)/(4)` hours or 6.25 hours if John uses the 100 watt pump.