document.write( "Question 618505: A tourist can bicycle 33 miles in the same time as he can walk 12 miles. If he can ride 7 mph faster than he can walk, how much time should he allow to walk a 30-mile trail? \n" ); document.write( "
Algebra.Com's Answer #388960 by swincher4391(1107)\"\" \"About 
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This is a proportion.\r
\n" ); document.write( "\n" ); document.write( "Let b be how fast a tourist can ride on is bicycle.\r
\n" ); document.write( "\n" ); document.write( "Then b-7 is how fast he can walk.\r
\n" ); document.write( "\n" ); document.write( "Recall that d = rt.\r
\n" ); document.write( "\n" ); document.write( "Then 33 = b*t and 12 = (b-7)*t\r
\n" ); document.write( "\n" ); document.write( "We need to solve for b.\r
\n" ); document.write( "\n" ); document.write( "Since 12 = bt - 7t but recall that 33=bt so we have that:\r
\n" ); document.write( "\n" ); document.write( "12 = 33-7t\r
\n" ); document.write( "\n" ); document.write( "-21 = -7t\r
\n" ); document.write( "\n" ); document.write( "t = 3 hours.\r
\n" ); document.write( "\n" ); document.write( "33miles = b* 3hours\r
\n" ); document.write( "\n" ); document.write( "b = 11 miles per hour.\r
\n" ); document.write( "\n" ); document.write( "That means b-7 = 4 miles per hour\r
\n" ); document.write( "\n" ); document.write( "Then following d =rt\r
\n" ); document.write( "\n" ); document.write( "30 = 4*t\r
\n" ); document.write( "\n" ); document.write( "30/4 = t\r
\n" ); document.write( "\n" ); document.write( "t = 7.5 hours\r
\n" ); document.write( "\n" ); document.write( "So he should allow himself 7.5 hours.\r
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