document.write( "Question 766096: A jogger starts a course at a steady rate of 8kph, five minute later, a second jogger start the same force at 10kph. How long will it take the second jogger to catch the first? \n" ); document.write( "
Algebra.Com's Answer #466654 by mananth(16946)\"\" \"About 
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let them meet after distance x km\r
\n" ); document.write( "\n" ); document.write( "let time taken by first jogger be t
\n" ); document.write( "distance = x
\n" ); document.write( "speed = 8kmph\r
\n" ); document.write( "\n" ); document.write( "x= 8*t \r
\n" ); document.write( "\n" ); document.write( "time taken by jogger II = t-5 hours
\n" ); document.write( "distance = x km
\n" ); document.write( "speed = 10 kmph\r
\n" ); document.write( "\n" ); document.write( "x= 10(t-5/60)\r
\n" ); document.write( "\n" ); document.write( "8t= 10(t-1/12)multiply equation by 12\r
\n" ); document.write( "\n" ); document.write( "96t=120t-10\r
\n" ); document.write( "\n" ); document.write( "24t= 10
\n" ); document.write( "t=10/24
\n" ); document.write( "t=5/12\r
\n" ); document.write( "\n" ); document.write( "t= 5/12 hours
\n" ); document.write( "5/12 * 60 = 25 minutes\r
\n" ); document.write( "\n" ); document.write( "the Jogger II will catch up after 25-5 = 20 minutes\r
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