document.write( "Question 489182: a freight train leave a for b 175 miles away and travels at the rate or 31.5 mph. after 1.5 hours a train leaves B for A traveling at 21.5 mph. How many miles from B will they meet? \n" ); document.write( "
Algebra.Com's Answer #333592 by Edwin McCravy(20060)\"\" \"About 
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document.write( "A freight train leave A for B 175 miles away and travels at the rate or 31.5\r\n" );
document.write( "mph. after 1.5 hours a train leaves B for A traveling at 21.5 mph. How many\r\n" );
document.write( "miles from B will they meet?\r\n" );
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document.write( "Distance = rate×time.  Therefore:\r\n" );
document.write( "When the train leaves B, the train that left A is already (31.5)(1.5) or\r\n" );
document.write( "47.25 miles down the track toward B, so they are only 175-47.25 or 127.75 miles\r\n" );
document.write( "apart then.  \r\n" );
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document.write( "Their approach rate is the sum of their rates or 31.5+21.5 or 53 mph.\r\n" );
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document.write( "Since \"time=distance%2Frate\", they will meet \"127.75%2F53\" hours after the train that left B started. \r\n" );
document.write( "So, since \"distance+=+rate%2Atime\", it will have traveled 21.5×\"127.75%2F53\" or\r\n" );
document.write( "51.82311321 miles from B.\r\n" );
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document.write( "Checking:  The train that left A will have traveled an additional \r\n" );
document.write( "31.5×\"127.75%2F53\" miles or 75.92688679 miles and when we add those\r\n" );
document.write( "we get 127.75 miles.\r\n" );
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document.write( "Answer: 51.8 miles approximately\r\n" );
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document.write( "Here is another way to solve it.  Make this chart, filling in the\r\n" );
document.write( "rates:\r\n" );
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document.write( "                  distance    rate    time\r\n" );
document.write( "train leaving A               31.5         \r\n" );
document.write( "train leaving B               21.5     \r\n" );
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document.write( "Let the time traveled by the train that left B be t.\r\n" );
document.write( "Then the train that left A will have traveled t+1.5\r\n" );
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document.write( "                  distance    rate    time\r\n" );
document.write( "train leaving A               31.5    t+1.5     \r\n" );
document.write( "train leaving B               21.5     t\r\n" );
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document.write( "Fill in the distances using distance = rate×time\r\n" );
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document.write( "                  distance    rate    time\r\n" );
document.write( "train leaving A  31.5(t+1.5)  31.5    t+1.5     \r\n" );
document.write( "train leaving B    21.5t      21.5     t\r\n" );
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document.write( "The sum of the distances must equal 175, so\r\n" );
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document.write( "        31.5(t+1.5) + 21.5t = 175\r\n" );
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document.write( "      31.5t + 47.25 + 21.5t = 175\r\n" );
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document.write( "                53t + 47.25 = 175\r\n" );
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document.write( "                        53t = 127.75\r\n" );
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document.write( "                          t = \"127.75%2F53\"\r\n" );
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document.write( "                          t = 2.410377358 hours\r\n" );
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document.write( "The answer is the distance traveled by the train leaving B.\r\n" );
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document.write( " 21.5t = 21.5(2.410377358) =  51.8231132 miles.\r\n" );
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document.write( "Edwin
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