Question 1014461
More than one way to analyze and handle the description.  The slow cyclist at rate r went distance
d.  The fast cyclist in the same amount of time, instead of returning to A covering the full
 round trip, went d miles less than the full found trip; so that he met the slow cyclist on the
way back.
<pre>
                      rate          time          distance

slow cyclist          r             2&2/5          d

fast cyclist          (3/2)r        2&2/5          2*120-d

total(one way)                                      120
</pre>

24 minutes is {{{2/5}}} hour;
"one and five tenths" is the same as {{{3/2}}}, about the factor relating their two speeds or rates.


When the tabulated data makes sense, two equations will be ready to formulate.


{{{system(r*(2&2/5)=d,(3/2)r(2&2/5)=2*120-d)}}}
Simplify this system and solve for r and d.


{{{system((12/5)r=d,(3/2)(12/5)r=240-d)}}}


Substitute for d into the second equation.
{{{(36/10)r=240-(12/5)r}}}
{{{(36/10+12/5)r=240}}}
{{{(36/10+24/10)r=240}}}
{{{(60/10)r=240}}}
{{{6r=240}}}
{{{r=240/6}}}
{{{highlight(r=40)}}}------the rate or speed of the slower mototcyclist.  You can find the
distance which each motorcyclist had traveled from this.