Question 102404
In these kind of problems, bikes, trains, boats, etc. it 
really helps if the first question you ask is "what is the
same for both?" In this case it's the distance from school to home,
and also, they both start at 3 PM.
Notice everything else is different.
{{{d = r*t}}} holds true for both
For Mary,
{{{t[M] = t[J] - 15}}} in minutes
{{{r[M] = 12}}} in mi/hr
For Jocelyn, 
{{{r[J] = 9}}} in mi/hr
Now write equations
{{{d[M] = d[J]}}} because they're going the same distance, so call
them both {{{d}}}
{{{d = r[M]*t[M]}}}
{{{d = r[M]*(t[J] - .25)}}} (Note I converted 15 minutes to .25 hrs)
{{{d = 12*(t[J] - .25)}}}
Now for Jocelyn,
{{{d = r[J]*t[J]}}}
{{{d = 9*t[J]}}}
Now set the equations equal to eachother
{{{12*(t[J] - .25) = 9*t[J]}}}
{{{12*t[J] - 3 = 9*t[J]}}}
{{{3*t[J] = 3}}}
{{{t[J] = 1}}} in hours
Mary got home a quarter hour before Jocelyn, so
{{{t[M] = .75}}} in hours
Or, Mary's time was 45 minutes
You should chck these answers by putting them back in the
original equations
{{{d = 12*(t[J] - .25)}}}
{{{d = 12*(1 - .25)}}}
{{{d = 12*(3/4)}}}
{{{d = 9}}} in miles
and
{{{d = 9*t[J]}}}
{{{d = 9*1}}}
{{{d = 9}}} in miles
So, you get the same answer for school - home distance