Rate problem

HPengineer wrote:Is it possible to apply weighted average formula to the following problem?

It took Ellen 6 hours to ride her bike a total distance of 120 miles. For the first part of the trip, her speed was constantly 25 miles per hour. For the second part of her trip, her speed was constantly 15 miles per hour. For how many miles did Ellen travel at 25 miles per hour?

a.) 60
b.) 62.5
C.) 66 2/3
D.) 75
E.) 90


I set the problem up like this... total distance/total time = avg speed for trip

120/6 = 20mph

So 25(x) + 15(120-x)/120 = 20 why wont this work??

The easiest approach is to recognize that the average for the whole trip (20mph) is halfway between the 2 rates (15mph and 25mph), so an equal amount of time (3 hours) must be spent going at each speed.

d = r*t = 25*3 = 75 miles

The correct answer is D.

If you want to set up a weighted average equation, it would look like this:

15x + 25y = 20(x+y)

x and y represent the time spent traveling at each speed. Since r*t = d:

15x = distance traveled at 15mph
25y = distance traveled at 25mph
20(x+y) = average speed for the whole trip * total time = total distance

Solving the equation above:
15x + 25y = 20x + 20y
-5x = -5y
x=y
Total time = x+y = 6, so x=3 and y=3.
Distance traveled at 25 mph = r*t = 25*3 = 75.