Question 1207746
<pre>
Two cars enter the Florida Turnpike at Commercial Boulevard at 8:00 AM, each heading for Wildwood.One car’s average speed is 10 miles per hour more than the other’s.The faster car arrives at Wildwood at 11:00 AM, (1/2) hour before the other car.What was the average speed of each car? How far did each travel?

Let distance each traveled (from Commercial Blvd to Wildwood), be D
Let average speed of faster car, in mph, be S
Then average speed of slower car, in mph, is S - 10

Time faster car took to travel the distance: {{{D/S}}}
Also, faster car took 3 hours (8:00 a.m. - 11:00 a.m.) to reach Wildwood 
Then, faster car’s TIME equation is: {{{matrix(1,3, D/S, "=", 3)}}} 
                                      D = 3S ------ eq (i)

Time slower car took to travel the distance, {{{D/(S - 10)}}}
And, slower car took 3½ hrs (½ hr more than faster car, or from 8:00 a.m. - 11:30 a.m.) to reach Wildwood 
Slower car’s TIME equation is then: {{{matrix(2,3, D/(S - 10), "=", 3&1/2, D/(S  -  10), "=", 7/2)}}}
                                         2D = 7(S - 10) --- Cross-multiplying
                                         2D = 7S - 70 ----- eq (ii)
                                      2(3S) = 7S - 70 ----- Substituting 3S for D in eq (ii) 
                                         6S = 7S - 70
                                    6S - 7S = - 70
                                        - S = - 70
  <font color = red><font size  = 4><b>Average speed of faster car</font></font></b>, or {{{highlight_green(matrix(1,5, S, "=", (- 70)/(- 1), "=", highlight(matrix(1,2, 70, mph))))}}}
 <font color = red><font size  = 4><b>Average speed of slower car</font></font></b>: S - 10 = 70 - 10 = <font color = red><font size  = 4><b>60 mph</font></font></b>

<font color = red><font size  = 4><b>Distance traveled (Commercial Blvd to Wildwood)</font></font></b>: 3S = 3(70) = <font color = red><font size  = 4><b>210 miles</font></font></b></pre>