Question 1161456
<pre>
10 minutes is 10/60th or 1/6th of an hour.

The way the problem is stated it should be worked this way:

                  distance   rate     time
going to office      d        50        t    
returning home       d        60      t-1/6

{{{system(d=50t, d=60(t-1/6))}}}

This will give the correct answer for d also.

The way the practice book did it, requires interpreting it as if were
worded this way:

Corrine returns d miles from her office at an average speed of 60 miles per
hour.  When she left from home, she traveled by the same route and averaged
50 miles per hour.  If her trip to the office took her 10 minutes longer
than her trip home her office, what is the value of d?

The practice book reinterpreted the problem.  The difference is that the
practice book considered t to be the time returning so the 10 minutes 
1/6 of an hour had to be added to make the time going longer.  But the
way the problem is stated, we should let t be the time going, so we would
subtract the 1/6th of an hour to make the time retuning be shorter by 1/6th
of an hour.

Both ways are correct, but the best way is to solve it as the problem is
stated literally.  Since the problem states that the time is shorter
returning, then you should pick t as the time going, for that's the time
that we are to make shorter by subtracting.  IOW, why change the problem
from one where you make the time shorter to a problem to one where you make
the time longer, when the one that was given was the one where you make the
time shorter?

The moral of the story is:  Don't assume the practice book always gives the
most obvious, and best, way to work a problem.

Edwin</pre>