document.write( "Question 669964: A lady drove to a city 30 miles away to shop and returned home in the evening. She spent15 minutes longer driving the return trip than in the going, and she drover three-fourths as fast when returning as she did when going to the city. How long did it take her to drive to the city? \n" ); document.write( "
Algebra.Com's Answer #416712 by ankor@dixie-net.com(22740) You can put this solution on YOUR website! A lady drove to a city 30 miles away to shop and returned home in the evening. \n" ); document.write( " She spent15 minutes longer driving the return trip than in the going, and she \n" ); document.write( " drover three-fourths as fast when returning as she did when going to the city. \n" ); document.write( " How long did it take her to drive to the city? \n" ); document.write( ": \n" ); document.write( "Find the speed first \n" ); document.write( "let s = speed to the city \n" ); document.write( "then \n" ); document.write( ".75s = return speed \n" ); document.write( ": \n" ); document.write( "Change 15 min to .25 hr \n" ); document.write( ": \n" ); document.write( "Write a time equation; time = dist/speed \n" ); document.write( " \n" ); document.write( "multiply by 3s, to clear the denominators, results \n" ); document.write( "3(30) + 3s(.25) = 4(30) \n" ); document.write( "90 + .75s = 120 \n" ); document.write( ".75s = 120 - 90 \n" ); document.write( ".75s = 30 \n" ); document.write( "s = 30/.75 \n" ); document.write( "s = 40 mph to the city \n" ); document.write( ": \n" ); document.write( "Find the time \n" ); document.write( "30/40 = .75 hrs or 45 min to drive to the city\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |