document.write( "Question 628313: A boy leaves a town at noon on a bicycle traveling west at rate of 15 mph at 4pm. A girl leaves town in a car in the same direction at the rate of 45 mph.How many hours will it take to watch a boy and What time? \n" ); document.write( "
Algebra.Com's Answer #395578 by ankor@dixie-net.com(22740)\"\" \"About 
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A boy leaves a town at noon on a bicycle traveling west at rate of 15 mph.
\n" ); document.write( "At 4pm a girl leaves town in a car in the same direction at the rate of
\n" ); document.write( " 45 mph.
\n" ); document.write( "How many hours will it take to catch the boy and What time?
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\n" ); document.write( "Let t = travel time of the girl in the car
\n" ); document.write( "then
\n" ); document.write( "(t+4) = travel time of the boy on the bike (left 4 hrs earlier)
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\n" ); document.write( "When the girl catches the boy, they will have traveled the same distance
\n" ); document.write( "Write a distance equation; dist = speed * time
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\n" ); document.write( "45t = 15(t+4)
\n" ); document.write( "45t = 15t + 60
\n" ); document.write( "45t - 15t = 60
\n" ); document.write( "30t = 60
\n" ); document.write( "t = 60/30
\n" ); document.write( "t = 2 hrs travel time of the girl in the car to catch the boy on the bike
\n" ); document.write( "She left at 4 PM, therefore time of catch = 6 PM
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\n" ); document.write( "Confirm our solution by finding the distance of each, should be equal
\n" ); document.write( "45(2) = 90 mi
\n" ); document.write( "15(2+4) = 90 mi \n" ); document.write( "
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