Question 1186798
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If speed is increased by 20 percent, a person arrives 15 minutes {{{highlight(cross(early))}}} <U>earlier</U>. 
What is the {{{highlight(cross(initial))}}} <U>regular</U> time taken by the person.
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<pre>
Let  "d"  be the distance, let "t" be the regular (= the scheduled) travel time (in minutes), and let "u"  be the regular speed. 

Then the increased speed is  1.2u, and reduced traveled time is  t-15 minutes.


We have two equations

    {{{d/u}}} = t              (1)

and

    {{{d/(1.2u)}}} = t - 15    (2)


From equation (2), we have

    {{{d/u}}} = 1.2*(t-15).


In this equation, replace its left side  {{{d/u}}}  by  t, based on equation (1).  
You will get then simple linear equation for the single unknown t


    t = 1.2*(t-15).


Simplify and find t

    t = 1.2t - 18

    18 = 1.2t - t

    18 = 0.2t

     t = {{{18/0.2}}} = {{{180/2}}} = 90 minutes.


<U>ANSWER</U>.  The regular (=the scheduled) travel time is 90 minutes.
</pre>

Solved.