If Bofa can bike from his house to Oberlin College, miles in hours, at what rate must he make the return trip in such a way that the average speed for the whole journey becomes miles per hour?
Solution
We've got a lot of variables already, but we fearlessly go ahead and let represent the rate for the return trip. Then,
We solve this equation for .
The answer is (d).
Hmm: That answer is a fraction, and fractions may not have denominator zero. What if , meaning that our answer doesn't work? We try some numbers. Suppose the distance miles and the time hours, so the rate going is miles per hour. Substituting those numbers in, so , the desired average speed, cannot be 10 mi/hr.
Ah ha! Substituting again, going miles at an average speed of mi/hr will take hours, and going already took those hours, so there is no time to get back. That is, the return trip must take zero time, which is of course impossible. It appears that we can't ask that the average speed be double the rate going.
Now we notice that if , then is negative! Another impossibility. We can't ask that the average speed be greater than double the going rate either.