document.write( "Question 148362: After riding at a steady speed for 40 miles, a bicylist had a flat tire and walked 5 miles to a repair shop. The cycling rate was 4 times faster than the walking rate. If the time spent cycling and walking was 5 hours, at what rate was the cyclist riding? \n" ); document.write( "

Algebra.Com's Answer #108749 by ptaylor(2198) $\"\"$ $\"About$

You can put this solution on YOUR website!
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
\n" ); document.write( "Let r=walking rate
\n" ); document.write( "Then 4r=cycling rate\r
\n" ); document.write( "\n" ); document.write( "Time spent walking=5/r
\n" ); document.write( "Time spent cycling=40/4r=10/r\r
\n" ); document.write( "\n" ); document.write( "Now we are told that the sum of the above times is 5 hours, so:\r
\n" ); document.write( "\n" ); document.write( "5/r + 10/r=5 multiply each term by r
\n" ); document.write( "5+10=5r
\n" ); document.write( "15=5r divide each side by 5
\n" ); document.write( "r=3 mph------------------walking rate
\n" ); document.write( "4r=4*3=12 mph----------------cycling rate\r
\n" ); document.write( "\n" ); document.write( "ck
\n" ); document.write( "40/12 + 5/3=5
\n" ); document.write( "40/12+20/12=5
\n" ); document.write( "60/12=5
\n" ); document.write( "5=5\r
\n" ); document.write( "\n" ); document.write( "Hope this helps------ptaylor \n" ); document.write( "

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