document.write( "Question 1030814: Mary and Richard leave Edmonton to go to Jasper, a distance of 400 km. Mary drives at an average speed of 15 km/h faster than Richard. If Mary arrives in Jasper 40 minutes before Richard, find Richard's average speed, to the nearest km/h. \n" ); document.write( "
Algebra.Com's Answer #645638 by mananth(16946) You can put this solution on YOUR website! Mary and Richard leave Edmonton to go to Jasper, a distance of 400 km. Mary drives at an average speed of 15 km/h faster than Richard. If Mary arrives in Jasper 40 minutes before Richard, find Richard's average speed, to the nearest km/h.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Let Richard,s speed be x\r \n" ); document.write( "\n" ); document.write( "Mary's speed = x+15\r \n" ); document.write( "\n" ); document.write( "Time taken by Richard - time taken by Mary = 40 minutes = 2/3 hours\r \n" ); document.write( "\n" ); document.write( "400/x - 400/(x+15) = 2/3\r \n" ); document.write( "\n" ); document.write( "400(x+15)-400x = (2/3) *x(x+15)\r \n" ); document.write( "\n" ); document.write( "6000 = (2/3) *x(x+15)\r \n" ); document.write( "\n" ); document.write( "18000 = 2x^2 +30x \n" ); document.write( "/2 and rearrange\r \n" ); document.write( "\n" ); document.write( "x^2 +15x -9000 =0\r \n" ); document.write( "\n" ); document.write( "a= 1 , b= 15 c= -9000 \n" ); document.write( " \n" ); document.write( "b^2-4ac= 225 + 36000 \n" ); document.write( "b^2-4ac= 36225 \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "x1=( -15 + 190.33 )/ 2 \n" ); document.write( "x1= 87.66 \n" ); document.write( " \n" ); document.write( "x2=( -15 -190.33 ) / 2 \n" ); document.write( "x2= -102.66 \n" ); document.write( "Ignore negative value \n" ); document.write( " Richard's speed =87.66 mph \r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |