Question 1203653
.
Jay traveled from his house to his office at an average speed of 25mph. 
by traveling 5mph faster, he took 30 minutes less to return home. 
How far is Jay's office from his house
~~~~~~~~~~~~~~~~~~~~~~


<pre>
Let d be the distance from the house to the office, in miles.

Traveling at 25 mph, Jay spends {{{d/25}}}  hours.

Traveling back at the speed 25 mph + 5 mph = 30 mph, he spends {{{d/30}}}  hours.


The difference is 30 minutes, or  {{{1/2}}} of an hour;

so we write this "time equation"

    {{{d/25}}} - {{{d/30}}} = {{{1/2}}}  of an hour.


To solve it, multiply all the terms by 150.  you will get then

    6d - 5d = {{{(1/2)*150}}}

       d    =   75.


<U>ANSWER</U>.  The distance from the house to the office is 75 miles.


<U>CHECK</U>.   Time to the office  {{{75/25}}} = 3 hours.

         Time returning back is  {{{75/30}}} = 2.5 hours.

         The difference is  3 - 2.5 = 0.5 of an hour, correct.
</pre>

Solved.