Question 837670
He begins at a time point on a time line. He can go slow and be {{{t+1/3}}} hours or he can go fast and be {{{t-25/60=t-5/12}}} hours.  The variable, {{{t}}} is used as a reference to time on a number line.



______________speed________time(hours)_____distance(km)
Slow,late_______20________t+1/3_____________d
Fast,early______80________t-5/12____________d



The distance is the same value for both speeds used, early or late.


{{{Rate*Time=Distance}}}, basic concept.
{{{highlight(20(t+1/3)=80(t-5/12))}}}
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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{{{20t+20/3=80t-80*5/12}}}
{{{20/3+80*5/12=60t}}}
{{{80/12+400/12=60t}}}
{{{480/12=60t}}}
{{{40=60t}}}
{{{t=4/6}}}
{{{highlight(t=2/3)}}} hour which is 40 minutes.
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Now, what is the distance?
Either direction's equation will work.  Trying the slow form,
{{{d=20(t+1/3)}}}
{{{d=20(2/3+1/3)}}}
{{{d=20*1}}}
{{{highlight(highlight(d=20))}}} km