document.write( "Question 1160587: \"It took Emily 25 min to ride her bicycle to the repair shop and 1h 15 min to walk back home. If Emily can ride her bicycle 8 km/h faster than she can walk, how far is the repair shop from her house?\"\r
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Algebra.Com's Answer #783903 by ikleyn(52776) $\"\"$ $\"About$

You can put this solution on YOUR website!
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document.write( "Let \"w\" be her rate walking, in km/h.\r\n" );
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document.write( "Then the biking rate is (w+8) km/h.\r\n" );
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document.write( "From the condition you have this equation\r\n" );
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document.write( "     = \r\n" );
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document.write( "saying that the distance is the same in both directions.\r\n" );
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document.write( "From the equation (canceling 60 in the denominators)\r\n" );
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document.write( "    25*(w+8) = 75w\r\n" );
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document.write( "    w + 8 = 3w\r\n" );
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document.write( "    8 = 3w - w = 2w\r\n" );
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document.write( "    w = 8/2 = 4 km/h.\r\n" );
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document.write( "So, the walking rate is 4 km/h;  then the (walking) distance is   = 5 kilometres.    ANSWER\r\n" );
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document.write( "Let \"d\" be the distance under the question.\r\n" );
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document.write( "Then from condition you have this equation, connecting rates\r\n" );
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document.write( "     -  =   kilometres per minute.\r\n" );
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document.write( "Simplify by canceling factor 5 in denominators\r\n" );
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document.write( "     -  =  = .\r\n" );
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document.write( "Multiply both sides by 15\r\n" );
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document.write( "    3d - d = 10\r\n" );
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document.write( "    2d     = 10\r\n" );
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document.write( "     d     = 10/2 = 5 kilometres,\r\n" );
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document.write( "and you get the same answer for the distance.\r\n" );
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\n" ); document.write( "\n" ); document.write( "Solved two times by different methods for your better understanding.\r
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