document.write( "Question 257967: A bicyclist and a hiker leave the same place at the same time and travel in the same direction. The bicyclist travels three times as fast as the hiker. At the end of 3 hr they are 24 miles apart. How fast does the hiker travel? \n" ); document.write( "

Algebra.Com's Answer #189834 by Edwin McCravy(20059) $\"\"$ $\"About$

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document.write( "The other tutor's solution is wrong for if the hiker had gone\r\n" );
document.write( " mph, then in 3 hours he would have gone 8 miles and\r\n" );
document.write( "the biker would have gone 24 miles and they would be 16 miles\r\n" );
document.write( "apart, not 24 miles apart.  \r\n" );
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document.write( "Here's the correct solution:\r\n" );
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document.write( "Let r = the rate of the hiker\r\n" );
document.write( "Then 3r = the rate of the biker\r\n" );
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document.write( "Then after 3 hours, using d=rt, the hiker went r*(3) or 3r miles\r\n" );
document.write( "And after the same 3 hours, the biker went (3r)*(3) or 9r miles.\r\n" );
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document.write( "Then 9r is 24 more than 3r\r\n" );
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document.write( "9r = 3r + 24\r\n" );
document.write( "6r = 24\r\n" );
document.write( " r = 4 mi/hr\r\n" );
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document.write( "To analyze the problem further as a check, the biker went \r\n" );
document.write( "3 times as fast or 12 mi/hr and after 3 hours the biker \r\n" );
document.write( "had gone 36 miles and the hiker had gone 4*3 or 12 miles, and \r\n" );
document.write( "they were therefore 36-12 or 24 miles apart after 3 hours. \r\n" );
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document.write( "Edwin

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