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The steamboat River Queen travels at the rate of 30km/h in still water. If it can travel 45 km up stream in the same amount of time that it takes to go 63 km down stream what is the rate of the current.



Answer :

[tex]x-the\ rate\ of\ the\ current\\t-time\\\\ \left\{\begin{array}{ccc}(30-x)t=45\\(30+x)t=63\end{array}\right\\ \left\{\begin{array}{ccc}t=\frac{45}{30-x}\\t=\frac{63}{30+x}\end{array}\right\\\\\frac{45}{30-x}=\frac{63}{30+x}\\\\cross\ multiply\\\\63(30-x)=45(30+x)\\1890-63x=1350+45x\\-63x-45x=1350-1890\\-108x=-540\ \ \ \ /:(-108)\\x=5\\\\Answer:The\ rate\ of\ the\ current\ is\ 5\ km/h.[/tex]

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