document.write( "Question 1048117: A tour bus averaged 50 miles per hour between two cities on the first leg of a trip and 45 miles per hour on the return trip. The return trip took 1/2 hour longer. Find the distance between the two cities. \n" ); document.write( "
Algebra.Com's Answer #663736 by ikleyn(52858)\"\" \"About 
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\n" ); document.write( "A tour bus averaged 50 miles per hour between two cities on the first leg of a trip and 45 miles per hour on the return trip.
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\n" ); document.write( "\n" ); document.write( "Solution 1\r
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document.write( "Let \"t\" be the time spent on the first leg at the average speed of 50 miles per hour.\r\n" );
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document.write( "Then (t+0.5) is the time spent on the returning trip at the average speed of 45 miles per hour.\r\n" );
document.write( "   (Here o.5 is 1/2 of an hour)\r\n" );
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document.write( "Since the distance is the same in both directions, you have an equation\r\n" );
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document.write( "50*t = 45*(t+0.5).\r\n" );
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document.write( "Simplify and solve it for \"t\".\r\n" );
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document.write( "50t = 45t + 22.5,\r\n" );
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document.write( "5t = 22.5  --->  t = \"22.5%2F5\" = 4.5.\r\n" );
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document.write( "Thus we found the time spent at the speed 50 mph. It is 4.5 hours.\r\n" );
document.write( "Then the distance between the cities is 50*4.5 = 225 miles.\r\n" );
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document.write( "Check. The returning trip takes \"225%2F45\" = 5 hours, which is 1/2 hour  longer than 4.5 hours.  The solution is checked and is correct.\r\n" );
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document.write( "Answer.   The distance between the cities is 50*4.5 = 225 miles.\r\n" );
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\n" ); document.write( "\n" ); document.write( "Solution 2. \r
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document.write( "Let D be the distance between the cities. Then the condition directly says that\r\n" );
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document.write( "\"D%2F45+-+D%2F50\" = 0.5.\r\n" );
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document.write( "The first term in the left side is the time spent on the returning trip.\r\n" );
document.write( "The second term in the left is the time spent on the first leg trip.\r\n" );
document.write( "The difference of the terms on the left is that 0.5 hour.\r\n" );
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document.write( "Multiply both sides of the equation by 450. You will get\r\n" );
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document.write( "10D - 9D = 225, or\r\n" );
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document.write( "D = 225.\r\n" );
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document.write( "You got the same answer for the distance.\r\n" );
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document.write( "Congratulations !\r\n" );
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\n" ); document.write( "\n" ); document.write( "There is a bunch of lessons on Travel and Distance in this site:\r
\n" ); document.write( "\n" ); document.write( "    - Travel and Distance problems \r
\n" ); document.write( "\n" ); document.write( "    - Wind and Current problems \r
\n" ); document.write( "\n" ); document.write( "    - More problems on upstream and downstream round trips \r
\n" ); document.write( "\n" ); document.write( "    - Wind and Current problems solvable by quadratic equations \r
\n" ); document.write( "\n" ); document.write( "    - Using fractions to solve Travel problems \r
\n" ); document.write( "\n" ); document.write( "    - Unpowered raft floating downstream along a river\r
\n" ); document.write( "\n" ); document.write( "    - Had a car move faster it would arrive quicker\r
\n" ); document.write( "\n" ); document.write( "    - How far do you live from school? \r
\n" ); document.write( "\n" ); document.write( "    - Selected Travel and Distance problems from the archive \r
\n" ); document.write( "\n" ); document.write( "    - Selected problems from the archive on the boat floating Upstream and Downstream \r
\n" ); document.write( "\n" ); document.write( "    - Selected problems from the archive on a plane flying with and against the wind \r
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