document.write( "Question 733600: A cyclist and a jogger start in a town and want to end 12 miles away. The cyclists rate is 2xs the jogger. He arrives 1.2 hours before the jogger. What is the rate of the cyclist. \n" ); document.write( "
Algebra.Com's Answer #448542 by ankor@dixie-net.com(22740) You can put this solution on YOUR website! A cyclist and a jogger start in a town and want to end 12 miles away. \n" ); document.write( " The cyclists rate is 2xs the jogger. He arrives 1.2 hours before the jogger. What is the rate of the cyclist.\r \n" ); document.write( "\n" ); document.write( ": \n" ); document.write( "let r = rate of the jogger \n" ); document.write( "then \n" ); document.write( "2r = rate of the cyclist \n" ); document.write( ": \n" ); document.write( "Write a time equation: time = dist/rate \n" ); document.write( ": \n" ); document.write( "jogger time - cyclist time = 1.2 hr \n" ); document.write( " \n" ); document.write( "reduce the fraction \n" ); document.write( " \n" ); document.write( "multiply by r \n" ); document.write( "12 - 6 = 1.2r \n" ); document.write( "6 = 1.2r \n" ); document.write( "r = 6/1.2 \n" ); document.write( "r = 5 mph rate of the jogger \n" ); document.write( "then \n" ); document.write( "2(5) = 10 mph rate of the cyclist \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "check this; find the actual time of each \n" ); document.write( "12/5 = 2.4 hrs \n" ); document.write( "12/10= 1.2 hrs \n" ); document.write( "--------------- \n" ); document.write( "differ: 1.2 hrs \n" ); document.write( " |