document.write( "Question 37145: HELP!!! I have worked this problem and can't get it.\r
\n" ); document.write( "\n" ); document.write( "Problem:\r
\n" ); document.write( "\n" ); document.write( "Jim can run 5 miles per hour on level ground on a still day. One windy day, he runs 10 miles with the wind, and in the same amount of time runs 4 miles against the wind. What is the rate of the wind?
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Algebra.Com's Answer #22828 by mbarugel(146)\"\" \"About 
You can put this solution on YOUR website!
Hello!
\n" ); document.write( "We must use the following relationship in order to answer this:\r
\n" ); document.write( "\n" ); document.write( "If an object is moving at a speed of V miles per hour, then the time (in hours) it takes to travel a distance D (in miles) is D/V.\r
\n" ); document.write( "\n" ); document.write( "Now, we know that Jim can run at 5 mph without wind. Let's call X to the speed of the wind. So Jim's speed is 5 + X mph when he goes with the wind, and 5 - X mph when he goes against it.\r
\n" ); document.write( "\n" ); document.write( "Using the aforementioned relationship, and the fact that he takes the same time to make 10 miles with the wind and 4 miles against the wind, we get the following equation:\r
\n" ); document.write( "\n" ); document.write( "\"10%2F%285%2BX%29+=+4%2F%285-X%29\"\r
\n" ); document.write( "\n" ); document.write( "So we just isolate X:\r
\n" ); document.write( "\n" ); document.write( "\"50+-+10X+=+20+%2B+4X\"
\n" ); document.write( "\"30+=+14X\"
\n" ); document.write( "\"X+=+30%2F14+=+2.1428\"\r
\n" ); document.write( "\n" ); document.write( "So the speed of the wind is 2.1428... mph. \r
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\n" ); document.write( "\n" ); document.write( "I hope this helps!
\n" ); document.write( "Get more answers at Online Math Answers.com!
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