document.write( "Question 918081: the time taken by a person to cover 150km was 2.5 hours more than the time taken in the return of the journey. if he returned at a speed of 10km per hour, more than the speed of going , what was the speed per hour in each direction in km \n" ); document.write( "
Algebra.Com's Answer #556902 by mananth(16946) You can put this solution on YOUR website! Forward speed x \n" ); document.write( "return speed x + 10 \n" ); document.write( " \n" ); document.write( " Distance = same 150 miles 1 \n" ); document.write( " original time – time with increased speed = 2.5 hour \n" ); document.write( " t=d/t \n" ); document.write( " \n" ); document.write( " 150 / x - 150 / ( x + 10 ) = 2.50 \n" ); document.write( " LCD= x ( x + 10 ) \n" ); document.write( " multiply by LCD \n" ); document.write( " 150 ( x + 10 ) - 150 x = 2.5 x ( x + 10 ) \n" ); document.write( " 150 x + 1500 - 150 x = 2.5 X^2 + 25 x \n" ); document.write( " 1500 = 2.5 X^2 + 25 x \n" ); document.write( " 2.5 X^2 + 25 x - -1500 = 0 \n" ); document.write( " Find roots of the quadratic equation \n" ); document.write( " a= 2.5 b= 25 c= -1500 \n" ); document.write( " \n" ); document.write( " x1= ( -25 + sqrt( 625 + 15000 )) / 5 \n" ); document.write( " x1=( -25 + 125 )/ 5 \n" ); document.write( " x1= 20 \n" ); document.write( " \n" ); document.write( " x2= ( -7 - sqrt( 49 - 20 ) / 2 \n" ); document.write( " x2=( -25 - 125 )/ 5 \n" ); document.write( " x2= -30 \n" ); document.write( "Ignore negative\r \n" ); document.write( "\n" ); document.write( "forward speed = 20 kmph \n" ); document.write( "return speed = 30 kmph \n" ); document.write( " |