2440.62 – Bike Ride to Oberlin

If Bofa can bike from his house to Oberlin College, mm miles in tt hours, at what rate must he make the return trip in such a way that the average speed for the wholewhole journey becomes kk miles per hour?

  1. 2k−mt\frac{2k-m}{t}
  2. 2m−ktkm\frac{2m-kt}{km}
  3. 2m2m+kt\frac{2m}{2m+kt}
  4. 2m2m−kt\frac{2m}{2m-kt}
  5. 2mt−kmkt\frac{2mt-km}{kt}
Solution

We've got a lot of variables already, but we fearlessly go ahead and let xx represent the rate for the return trip. Then,

k=2mt+mx.k = \frac{2m}{t+\frac{m}{x}}.

We solve this equation for xx.

k(t+mx)=2mt+mx=2mkmx=2mk−t=2m−ktkxm=k2m−ktx=km2m−kt\begin{aligned} k(t+\frac{m}{x})&=& 2m \\ t+\frac{m}{x} &=& \frac{2m}{k} \\ \frac{m}{x} &=& \frac{2m}{k}-t = \frac{2m-kt}{k} \\ \frac{x}{m} &=& \frac{k}{2m-kt}\\ x &=& \frac{km}{2m-kt} \end{aligned}

The answer is (d).

Hmm: That answer is a fraction, and fractions may not have denominator zero. What if 2m=kt2m = kt, meaning that our answer doesn't work? We try some numbers. Suppose the distance m=10m = 10 miles and the time t=2t = 2 hours, so the rate going is 10/2=510/2 = 5 miles per hour. Substituting those numbers in, 2(10)=k⋅22(10) = k \cdot 2 so kk, the desired average speed, cannot be 10 mi/hr.

Ah ha! Substituting again, going 2m=202m = 20 miles at an average speed of 1010 mi/hr will take 22 hours, and going already took those 22 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 kt>2mkt > 2 m, then xx is negative! Another impossibility. We can't ask that the average speed be greater than double the going rate either.