document.write( "Question 937753: A cyclist has 5 and 1/4 hours to complete today´s training. How long can he ride away from his house if he plans to average 24 mph riding with the wind on the way out and 18 mph against the wind on the way back ? How far away from his house will he go ? \n" ); document.write( "

Algebra.Com's Answer #571190 by MathTherapy(10552) $\"\"$ $\"About$

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\n" ); document.write( "A cyclist has 5 and 1/4 hours to complete today´s training. How long can he ride away from his house if he plans to average 24 mph riding with the wind on the way out and 18 mph against the wind on the way back ? How far away from his house will he go ?
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Note that the distance traveled away from the house will be the same as the distance traveled back to the house\r
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document.write( "Let distance traveled away from the house, be D
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document.write( "Then time taken to travel D miles from house = 
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document.write( "Time taken to travel D miles, back to house = 
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document.write( "Thus, we now have:  
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document.write( "3D + 4D = 21(18) ------ Multiplying by LCD, 72
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document.write( "7D = 378
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document.write( "D, or distance traveled away from the house = , or  miles \n" );
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