document.write( "Question 1145617: carlo left their house at 9:00 am riding his bicycle at a rate of 10kph. three hours later, his brother Jaypee also left their house in a car at an average speed of 80kph heading in the direction of carlo. After how many hours will jaypee overtake carlo? \n" ); document.write( "
Algebra.Com's Answer #766855 by ankor@dixie-net.com(22740)\"\" \"About 
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carlo left their house at 9:00 am riding his bicycle at a rate of 10kph.
\n" ); document.write( " three hours later, his brother Jaypee also left their house in a car at an average speed of 80kph heading in the direction of carlo.
\n" ); document.write( " After how many hours will jaypee overtake carlo?
\n" ); document.write( ":
\n" ); document.write( "let t = JP's travel time
\n" ); document.write( "C left 3 hrs earlier, therefore
\n" ); document.write( "(t+3) = C's travel time
\n" ); document.write( ":
\n" ); document.write( "Write a distance equation, dist = speed * time
\n" ); document.write( "When JP catches C, they will have traveled the same dist
\n" ); document.write( "80t = 10(t+3)
\n" ); document.write( "80t = 10t + 30
\n" ); document.write( "80t - 10t = 30
\n" ); document.write( "70t = 30
\n" ); document.write( "t = 30/70
\n" ); document.write( "t = 3/7 hrs or \"3%2F7\"*60 = 27.7 minutes, for JP to catch C
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "Check, (3/7 = .429)
\n" ); document.write( "80 * .429 = 34.3 mi
\n" ); document.write( "10 * 3.429 = 34.3 mi \n" ); document.write( "
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