document.write( "Question 837670: a man gets 20 minutes late to his office when he travels at a speed of 20 kmph and 25 mins. Early when he travels at a speed of 80 kmph. The distance he travels to reach his office is \n" ); document.write( "
Algebra.Com's Answer #504701 by josgarithmetic(39617)\"\" \"About 
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He begins at a time point on a time line. He can go slow and be \"t%2B1%2F3\" hours or he can go fast and be \"t-25%2F60=t-5%2F12\" hours. The variable, \"t\" is used as a reference to time on a number line.\r
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\n" ); document.write( "\n" ); document.write( "______________speed________time(hours)_____distance(km)
\n" ); document.write( "Slow,late_______20________t+1/3_____________d
\n" ); document.write( "Fast,early______80________t-5/12____________d\r
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\n" ); document.write( "\n" ); document.write( "The distance is the same value for both speeds used, early or late.\r
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\n" ); document.write( "\n" ); document.write( "\"Rate%2ATime=Distance\", basic concept.
\n" ); document.write( "\"highlight%2820%28t%2B1%2F3%29=80%28t-5%2F12%29%29\"
\n" ); document.write( "If everything makes sense to this extent, then you are ready to solve for \"t\". This is the value the man expects if to arrive ON-TIME.
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\n" ); document.write( "\"20t%2B20%2F3=80t-80%2A5%2F12\"
\n" ); document.write( "\"20%2F3%2B80%2A5%2F12=60t\"
\n" ); document.write( "\"80%2F12%2B400%2F12=60t\"
\n" ); document.write( "\"480%2F12=60t\"
\n" ); document.write( "\"40=60t\"
\n" ); document.write( "\"t=4%2F6\"
\n" ); document.write( "\"highlight%28t=2%2F3%29\" hour which is 40 minutes.
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\n" ); document.write( "Now, what is the distance?
\n" ); document.write( "Either direction's equation will work. Trying the slow form,
\n" ); document.write( "\"d=20%28t%2B1%2F3%29\"
\n" ); document.write( "\"d=20%282%2F3%2B1%2F3%29\"
\n" ); document.write( "\"d=20%2A1\"
\n" ); document.write( "\"highlight%28highlight%28d=20%29%29\" km
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