document.write( "Question 1098532: Abby drove from town A to B in 3 hours. Billy left town A at the same time, but drove 5km/h more slowly, and arrived at town B 20 minutes later than Abby. What is the distance between town A and B in km? \n" ); document.write( "

Algebra.Com's Answer #712913 by ikleyn(52785) $\"\"$ $\"About$

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document.write( "Let \"r\" be the Andy's rate, in kilometers per hour.\r\n" );
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document.write( "Then the Billy's rate was  (r-5) km/h.\r\n" );
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document.write( "Abby covered the distance  r*3  in three hours.\r\n" );
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document.write( "Billy covered the distance (r-5)*(3 + 1/3)   (notice that 1/3 = 1/3 of an hour = 20 minutes).\r\n" );
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document.write( "The distance is the same, which gives you an equation \r\n" );
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document.write( "3r = (10/3)*(r-5).\r\n" );
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document.write( "Simplify and solve for r. First step is to multiply both sides by 3 and open parentheses:\r\n" );
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document.write( "9r = 10r - 50  ====>   50 = 10r - 9r  ====>  r = 50.\r\n" );
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document.write( "Thus Andy's rate was 50 km/h.\r\n" );
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document.write( "Hence, the distance between the towns was 3*50 = 150 km.\r\n" );
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document.write( "Let D be the distance between the towns.\r\n" );
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document.write( "Andy's rate was   kilometers per hour.\r\n" );
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document.write( "Billy's rate was   kilometers per hour.    (Notice that  =  hours).\r\n" );
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document.write( "The difference of the rates was 5 km/h, according to the condition.\r\n" );
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document.write( "It gives you an equation\r\n" );
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document.write( " -  = 5,   or, which is the same\r\n" );
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document.write( " -  = 5.\r\n" );
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document.write( "Multiply both sides by 30. You will get\r\n" );
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document.write( "10D - 9D = 5*30  ====>  D = 150.\r\n" );
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document.write( "You got the same answer:  The distance is 150 kilometers.\r\n" );
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