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)\"\" \"About 
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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?
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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" );
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document.write( "With algebra:\r\n" );
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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" );
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document.write( "           rate × time = distance\r\n" );
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document.write( "Train A     25  × t+.5 = 25(t+.5) \r\n" );
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document.write( "Train B     35  ×  t   =    35t\r\n" );
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document.write( "Set the distances equal:\r\n" );
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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( "           \"12.5%2F10\" = t\r\n" );
document.write( "           1.25 = t\r\n" );
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document.write( "1 hour and a quarter or 1 hour and 15 minutes.\r\n" );
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document.write( "Edwin
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