document.write( "Question 663753: Train A leaves a station traveling at 25 mph. One half hour later Train B leaves the same station traveling the same direction at 35 mph. How does it take for Train B to catch up to Train A? \n" ); document.write( "
Algebra.Com's Answer #412936 by Edwin McCravy(20054)![]() ![]() You can put this solution on YOUR website! Train A leaves a station traveling at 25 mph. One half hour later Train B leaves the same station traveling the same direction at 35 mph. How does it take for Train B to catch up to Train A? \n" ); document.write( " \r\n" ); document.write( "Without algebra (in your head):\r\n" ); document.write( "When B starts, A is 12.5 miles down the track (half of 25). B's catch-up \r\n" ); document.write( "rate is 35-25 or 10 mph. In 1 hour B has caught up 10 of the 12.5 miles.\r\n" ); document.write( "The remaining 2.5 miles is one quarter of 10 miles, so he'll need another\r\n" ); document.write( "quarter of an hour or 15 minutes. Answer: 1 hour 15 minutes.\r\n" ); document.write( "\r\n" ); document.write( "With algebra:\r\n" ); document.write( "\r\n" ); document.write( "Let t = train A's time. Train A's time is .5 hour more \r\n" ); document.write( "than train B's time, or t+.5 \r\n" ); document.write( "\r\n" ); document.write( " rate × time = distance\r\n" ); document.write( "---------------------------------\r\n" ); document.write( "Train A 25 × t+.5 = 25(t+.5) \r\n" ); document.write( "---------------------------------\r\n" ); document.write( "Train B 35 × t = 35t\r\n" ); document.write( "\r\n" ); document.write( "Set the distances equal:\r\n" ); document.write( "\r\n" ); document.write( " 25(t+.5) = 35t\r\n" ); document.write( " 25t+12.5 = 35t\r\n" ); document.write( " 12.5 = 10t\r\n" ); document.write( "\n" ); document.write( " |