document.write( "Question 1132807: A salesman drives from Ajax to Barrington, a distance of 147 mi, at a steady speed. He then increases his speed by 15 mi/h to drive the 190 mi from Barrington to Collins. If the second leg of his trip took 5 min more time than the first leg, how fast was he driving between Ajax and Barrington? \n" ); document.write( "
Algebra.Com's Answer #749945 by josgarithmetic(39618)![]() ![]() ![]() You can put this solution on YOUR website! r, speed of first leg of trip\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "- \n" ); document.write( " \r\n" ); document.write( " SPEED TIME DISTANCE\r\n" ); document.write( "Ajax to Barrington r 147/r 147\r\n" ); document.write( "Barrington to Collins r+15 190/(r+15) 190\r\n" ); document.write( "\r\n" ); document.write( "Difference 5/60=1/12\r\n" ); document.write( "\r \n" ); document.write( "\n" ); document.write( ". \n" ); document.write( ". \n" ); document.write( ". \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "- \n" ); document.write( " |