Question 1011876
Two arrangements of travel are described.  The individual time quantities are not given but overall time quantities are given.  Descriptions of rates are not given.  Let d be the trip's distance.


<pre>
Arrangement A
                 rate    time    distance
TRAIN             R              60
CAR               r              d-60
Total                    4


Arrangement B
                rate     time     distance
TRAIN            R                 100
CAR              r                 d-100
Total                    4&1/6
</pre>

Fill the missing time slots according to y=mx, x=y/m.


<pre>
Arrangement A
                 rate    time        distance
TRAIN             R      60/R        60
CAR               r      (d-60)/r    d-60
Total                    4


Arrangement B
                rate     time        distance
TRAIN            R       100/R        100
CAR              r       (d-100)/r    d-100
Total                    4&1/6
</pre>
Note that 10 minutes is {{{1/6}}} of an hour.


You have two time-sum equations to make.  The system of equations is {{{system(60/R+(d-60)/r=4,100/R+(d-100)/r=4&1/6)}}}.


Do you see ANYTHING else?
<i>MAYBE</i> the time difference of the two arrangements?
{{{60/R+(d-60)/r+1/6=100/R+(d-100)/r}}}


A few steps of work on this time difference equation gives simplified {{{40/R-40/r=1/6}}}



Return now to the system of two equations with the three variables, r, R, d, and solve each of them for d; and then equate the expressions!  I will not include showing those steps here, but you may get an equation consistent with  {{{r/6-100r/R+60r/R=0}}}
or
{{{r/6-40r/R=0}}}


You have from this another system of two equations in just r and R:
{{{system(40/R-40/r=1/6,r/6-40r/R=0)}}}.



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I did not try to solve this further.