document.write( "Question 125959: A red sports car leaves Madison traveling to Green Bay at 60 mi/hr. At the same time, a green minivan leaves Columbus traveling 40 mi/hour. Columbus is 30 miles closers to Green Bay than Madison. When will the sports car catch the minivan and at what distance from Madison? \n" ); document.write( "
Algebra.Com's Answer #92326 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! car leaves Madison traveling to Green Bay at 60 mi/hr. At the same time, a green minivan leaves Columbus traveling 40 mi/hour. Columbus is 30 miles closers to Green Bay than Madison. When will the sports car catch the minivan and at what distance from Madison? \n" ); document.write( ": \n" ); document.write( "Find the distance from Madison first \n" ); document.write( ": \n" ); document.write( "Let d = distance from Madison \n" ); document.write( "then \n" ); document.write( "(d-30) = distance from Columbus \n" ); document.write( ": \n" ); document.write( "Since they left at the same time we know their travel times will be equal \n" ); document.write( ": \n" ); document.write( "Write a time equation: Time = \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( "Cross mutiply: \n" ); document.write( "60(d-30) = 40d \n" ); document.write( ": \n" ); document.write( "60d - 1800 - 40d \n" ); document.write( ": \n" ); document.write( "60d - 40a = 1800 \n" ); document.write( ": \n" ); document.write( "20d = 1800 \n" ); document.write( "d = \n" ); document.write( "d = 90 mi is the distance from Madison when the car catches the van \n" ); document.write( ": \n" ); document.write( "Find the time: \n" ); document.write( "Time = \n" ); document.write( "time = 1.5 hrs \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check solution using the van's speed which travels 90 - 30 = 60 mi \n" ); document.write( "time = 60/40 \n" ); document.write( "time = 1.5 hrs also \n" ); document.write( " |