document.write( "Question 308296: It takes Meihua two hours more to complete a 50-km journey than it takes Ailin to complete a 40-km journey.If the average speed of Meihua for the journey is 5 km/h less than ailin, calculate the average speed for each girl. \n" ); document.write( "
Algebra.Com's Answer #220419 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
It takes Meihua two hours more to complete a 50-km journey than it takes Ailin to complete a 40-km journey.If the average speed of Meihua for the journey is 5 km/h less than ailin, calculate the average speed for each girl.\r
\n" ); document.write( "\n" ); document.write( "Let ailin's speed be x\r
\n" ); document.write( "\n" ); document.write( "mehua's speed will be x-5 \r
\n" ); document.write( "\n" ); document.write( "Time taken by ailin to cover 40 km= 40/x
\n" ); document.write( ".
\n" ); document.write( "Time taken by mehua to cover 50 km = 50/x-5
\n" ); document.write( "Mehua takes 2 hours more
\n" ); document.write( ".
\n" ); document.write( "50/x-5 - 2 = 40/x\r
\n" ); document.write( "\n" ); document.write( "50/x-5 -40/x = 2\r
\n" ); document.write( "\n" ); document.write( "50x-40(x-5) / x(x-5)=2\r
\n" ); document.write( "\n" ); document.write( "50x-40x+200= 2x(x-5)\r
\n" ); document.write( "\n" ); document.write( "50x-40x+200= 2x^2 -10x\r
\n" ); document.write( "\n" ); document.write( "2x^2-20x-200=0\r
\n" ); document.write( "\n" ); document.write( "x^2-10x-100=0\r
\n" ); document.write( "\n" ); document.write( "x1,x2 the roots of the equation\r
\n" ); document.write( "\n" ); document.write( "will be x1 = 10+sqrt(100+400) / 2
\n" ); document.write( "x1= 16.18 mph the speed of ailin\r
\n" ); document.write( "\n" ); document.write( "x2 =-6.18 which is absurd.\r
\n" ); document.write( "\n" ); document.write( "Mehua's apeed will be 11.18 mph\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
\n" );