Question 102118
YOU ARE ON THE RIGHT TRACK!!!! SOLVE YOUR EQUATIONS FOR t1 AND t2; ADD THEM TO GET TOTAL TIME AND OUR CALCULATIONS BEGIN TO AGREE.  

distance(d) equals rate(r) times time(t) or d=rt;r=d/t and t=d/r

Let S=average speed on this trip
S=(Total distance travelled)/(Total time required)

(Note: Here's another problem that sort of underscores what average speed is all about:  "You are going on a 60-mile trip.  The first 30 miles, you travel at the rate of 30 mph.  How fast must you travel the remaining 30 miles in order to average 60 mph???"  ANSWER:  Impossible!!! Why???  In order to average 60 mph, you must travel the entire 60 miles in an hour.  Unfortunately, you have already used up that hour in the first 30 miles.)

Now back to your problem:
Let d=total distance travelled

Let d1=distance travelled for the first half of the trip
And let d2=distance travelled for the second half of the trip
*****Now we know that d1=d2. So total distance (d)=d1+d2=2d1 (or 2d2)
Let t=total time required
Let t1=time to travel the first half and this equals d1/30
Let t2=time to travel the second half and this equals d1/60
*****So Total time (t)=t1+t2=(d1/30+d1/60)=(2d1/60+d1/60)=(3d1/60)=d1/20

So S=2d1/(d1/20)---Now we multiply numerator and denominator by (20/d1) to get rid of the complex fraction and we get:

S=(2d1(20/d1))/(d1/20*20/d1)=(40d1/d1)/1 or 40 mph  (the d1's cancel)

S=40 mph-------------------------------ans


Hope this helps---ptaylor