document.write( "Question 1194658: Janet leaves home at 10:15a jogging 8mph.
\n" ); document.write( "Sue leaves the same home at 10:35am biking 18mph.
\n" ); document.write( "What time do they meet and at what mile?\r
\n" ); document.write( "\n" ); document.write( "solved for
\n" ); document.write( "their ratio of speed S4:J9 and distance of J9:S4
\n" ); document.write( "and at 10:25 Janet would have gone 2 2/3 miles.
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Algebra.Com's Answer #826897 by josgarithmetic(39617)\"\" \"About 
You can put this solution on YOUR website!
Sue leaves 20 minutes or one-third hour after Janet. In that time, the distance between the two is \"8%28miles%2Fhour%29%2A%281%2F3%29%2Ahour=highlight_green%28%288%2F3%29%2Amiles%29\".\r
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\n" ); document.write( "\n" ); document.write( "If Janet catches up in x hours, then \"%2818-8%29x=8%2F3\";
\n" ); document.write( "\"10x=8%2F3\"
\n" ); document.write( "\"x=8%2F30\"\"hours\";\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "For number of minutes, \"%288%2F30%29%2Ahour%2A60%28minutes%2Fhour%29=highlight_green%2816%2Aminutes%29\"
\n" ); document.write( "and you can compute the time on the clock.\r
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\n" ); document.write( "\n" ); document.write( "What time met?
\n" ); document.write( "\"35%2B16\"
\n" ); document.write( "\"51\"
\n" ); document.write( "10:51AM\r
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\n" ); document.write( "\n" ); document.write( "What distance when met?
\n" ); document.write( "18 miles per hour for 16 minutes
\n" ); document.write( "\"18%28miles%2Fhour%29%2A%2816%2F60%29hour\"
\n" ); document.write( "\"%2818%2A16%29%2F60\"\"miles\"
\n" ); document.write( "\"%283%2A2%2A3%2A4%2A4%29%2F%282%2A3%2A2%2A5%29\"
\n" ); document.write( "\"%282%2A3%2A4%29%2F5\"
\n" ); document.write( "\"24%2F5\"\r
\n" ); document.write( "\n" ); document.write( "\"highlight%284%264%2F5%29\"\"highlight%28miles%29\"
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