document.write( "Question 881687: a man reaches his office from his home 40 minutes too late.if he walks at 3 km an hour and 30 minutes too soon.if he walks 4 km an hour.how far is his office frm his home? \n" ); document.write( "
Algebra.Com's Answer #532407 by ankor@dixie-net.com(22740)\"\" \"About 
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a man reaches his office from his home 40 minutes too late if he walks at 3 km an hour.
\n" ); document.write( " And 30 minutes too soon, if he walks 4 km an hour.
\n" ); document.write( "how far is his office from his home?
\n" ); document.write( ":
\n" ); document.write( "Let d = distance from his home to his office
\n" ); document.write( "the difference between being 40 min late and 30 min to soon, is 70 min
\n" ); document.write( "70 min \"7%2F6\" hrs
\n" ); document.write( ":
\n" ); document.write( "Write time equation; time = dist/speed
\n" ); document.write( "Late time - early time = 70 min
\n" ); document.write( "\"d%2F3\" - \"d%2F4\" = \"7%2F6\"
\n" ); document.write( "multiply equation by 24
\n" ); document.write( "24*\"d%2F3\" - 24*\"d%2F4\" = 24*\"7%2F6\"
\n" ); document.write( "Cancel the denominators and you have
\n" ); document.write( "8d - 6d = 4(7)
\n" ); document.write( "2d = 28
\n" ); document.write( "d = 28/2
\n" ); document.write( "d = 14 km from home to work
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\n" ); document.write( "See if that checks out, find the time at each speed
\n" ); document.write( "14/3 = 4.67 hrs
\n" ); document.write( "14/4 = 3.5 hrs
\n" ); document.write( "---------------
\n" ); document.write( "time dif: 1.17 hrs which is about 70 min
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