A cyclist bikes x distance at 10 mph and returns over the same path at 8 mph. What is the cyclist's average rate for the round trip in miles per hour. First, I'll tell you what it is NOT. It it NOT the average of the two rates, or it is NOT 9 mph, but a little bit less. Since she travels a distance of x miles at 10 mph then, using TIME = DISTANCE/RATE, it took her x/10 hours to travel the x miles going. Since she then travels a distance of x miles at 8 mph then, using TIME = DISTANCE/RATE, it took her x/8 hours to travel the x miles returning. Her total distance was 2x miles and her total time was x/10 + x/8. To simplify that we must get an LCD of 40 x/10 + x/8 = 4x/40 + 5x/40 = 9x/40 AVERAGE RATE = (TOTAL DISTANCE TRAVELED)÷(TOTAL TIME TRAVELED) so AVERAGE RATE = (2x)÷(9x/40) = (2x)·[40/(9x)] = (80x)/(9x) = 80/9 = 8 8/9 mph, which is a little less than the average of her two speeds. Edwin