2440.41 – York to London

Jeepers! We don't know $d$, or $r$, or $t$. But we make our good old chart anyway. It looks really bad.

Here's what we learn, row-by-row:

1. $d = rt$

2. $d = \left(r + \dfrac{1}{2}\right) \left(\dfrac{4t}{5}\right) = \dfrac{4rt}{5} + \dfrac{41}{10}$

3. $d = \left(r - \dfrac{1}{2}\right) (t + 2.5) = rt + \dfrac{5r}{2} - \dfrac{t}{2} - \dfrac{5}{4}$

In (3), notice that we have $d = rt \ +$ (other stuff). Now we know $d$ equals $rt$. Ha! So (3) becomes

$$\begin{align*}0 &= \dfrac{5r}{2} - \dfrac{t}{2} - \dfrac{5}{4} \\[1em]0 &= 10r - 2t - 5 \\[1em]2t + 5 &= 10r\end{align*}$$

Now, (2). See the $rt$ in $\dfrac{4rt}{5}$? That $rt$ is $d$. So (2) becomes

$$\begin{align*}d = rt &= \dfrac{4rt}{5} + \dfrac{4t}{10} \\[1em]\dfrac{rt}{5} &= \dfrac{4t}{10} \\[1em]\dfrac{2rt}{10} &= \dfrac{4t}{10} \\[1em]r &= 2\end{align*}$$

And now, going back to (3),

$$\begin{align*}2t + 5 &= 10r \\2t + 5 &= 10 \cdot 2 \\2t &= 15 \\t &= 7.5\end{align*}$$

So Marla walked 7.5 hours at a (slow) rate of 2 miles per hour and she walked 15 miles. (Normal walking speed is $\approx$ 3 miles per hour.)

It's highly unlikely that she didn't stop along the way. The problem probably stops the clock running during any breaks.