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)\"\" \"About 
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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?
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\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
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\n" ); document.write( "Write a time equation: Time = \"distance%2Fspeed\"
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\n" ); document.write( "\"d%2F60\" = \"%28%28d-30%29%29%2F40\"
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\n" ); document.write( "Cross mutiply:
\n" ); document.write( "60(d-30) = 40d
\n" ); document.write( ":
\n" ); document.write( "60d - 1800 - 40d
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\n" ); document.write( "60d - 40a = 1800
\n" ); document.write( ":
\n" ); document.write( "20d = 1800
\n" ); document.write( "d = \"1800%2F20\"
\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 = \"90%2F60\"
\n" ); document.write( "time = 1.5 hrs
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\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
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