Question 955979
<pre>
          SPEED           TIME          DISTANCE
Monday     45             t-(1/60)        d
Regularly  R              t               d
Tuesday    40             t+(1/60)        d
</pre>
Same distance, d, in the three tabulated situations.

{{{d=highlight_green(45(t-1/60)=40(t+1/60))}}}


{{{45t-3/4=40t+2/32}}}

{{{5t=2/3+3/4}}}

{{{60t=8+9}}}

{{{60t=17}}}

{{{t=17/60}}}--------this means normal expected trip time would be 17 minutes.


Looking at Tuesday's trip,  time was {{{17/60+1/60=18/60}}}, or 18 minutes.
{{{d=40*(18/60)}}}
{{{d=18(2/3)}}}
{{{highlight(d=12)}}}, drive distance from work, 12 miles.


The question, what speed needed to arrive 5 minutes early?

{{{12/(17/60-5/60)}}}

{{{12/(12/60)}}}

{{{12/(1/5)}}}

{{{12*5=highlight(60)}}}