document.write( "Question 289706: George jogged downhill at 6 mi/h and then jogged back up at 4mi/h. If the total jogging time was 1.25 h, how far did he jog in all? \n" ); document.write( "

Algebra.Com's Answer #209785 by ankor@dixie-net.com(22740) $\"\"$ $\"About$

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George jogged downhill at 6 mi/h and then jogged back up at 4mi/h.
\n" ); document.write( " If the total jogging time was 1.25 h, how far did he jog in all?
\n" ); document.write( ":
\n" ); document.write( "Let d = dist up the hill
\n" ); document.write( "Same distance down the hill
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\n" ); document.write( "Write a time equation: Time = dist/speed
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\n" ); document.write( "down time + up time = 1.25 hr
\n" ); document.write( " $\"d%2F6\"$ + $\"d%2F4\"$ = 1.25
\n" ); document.write( "Multiply by 12 to get rid of the denominators, results
\n" ); document.write( "2d + 3d = 12(1.25)
\n" ); document.write( "5d = 15
\n" ); document.write( "d = 3 mi up and 3 mi down
\n" ); document.write( ":
\n" ); document.write( "Total dist = 6 mi
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\n" ); document.write( "Check solution, find the time
\n" ); document.write( "3/6 + 3/4 =
\n" ); document.write( ".5 + .75 = 1.25 \n" ); document.write( "

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