document.write( "Question 1150145: A minivan leaves Topeka at 8:00am. Half an hour later, a car leaves Topeka and follows the same route as the minivan. The car's speed is 5mph faster than the minivan's speed. If the car catches up to the minivan at 3:30 pm, how fast is each vehicle traveling? \n" ); document.write( "

Algebra.Com's Answer #771506 by ikleyn(52864) $\"\"$ $\"About$

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document.write( "Before the catching moment, the minivan was on the way 7.5 hours, from 8:00 am to 3:30 pm.\r\n" );
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document.write( "                            The car     was on the way half an hour less, i.e. 7.5-0.5 = 7 hours.\r\n" );
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document.write( "Let x be the average speed of the minivan, in miles per hpur.\r\n" );
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document.write( "Then the average speed of the car was (x+5) mph, according to the condition.\r\n" );
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document.write( "To the catching moment, they covered the same distance.  Hence,\r\n" );
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document.write( "    7.5*x = 7*(x+5).\r\n" );
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document.write( "From this equation,\r\n" );
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document.write( "    7.5x = 7x + 35\r\n" );
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document.write( "    7.5x - 7x = 35\r\n" );
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document.write( "    0.5x      = 35\r\n" );
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document.write( "       x      =  = 70.\r\n" );
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document.write( "ANSWER.  The minivan average speed was 70 mph.  That of car was 70+5 = 75 mph.\r\n" );
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