Marla walked from Yorkshire toward London at a constant rate. If she had gone 1/2 mph faster she would have walked the distance in 4/5 of the time. If she had gone 1/2 mph slower she would have taken 2.5 hours longer. How many miles did she walk?

Jeepers, we don’t know *d*, or *r*, or *t*. But we make our good old chart anyway.

actual | d |
r |
t |
(1) d = rt |

if faster | d |
r + 1/2 |
4t/5 |
(2) d = (r + 1/2)(4t5) = 4rt/5 + 4t/10 |

if slower | d |
r – 1/2 |
t + 2.5 |
(3) d = (r – 1/2)(t + 5/2) = rt + 5r/2 – t/2 – 5/4. |

This looks really bad.

But! In (3), notice that we have *d* = *rt* + …, and we know *d* equals *rt*. Ha!

So (3) becomes 0 = 5*r*/2 – *t*/2 – 5/4 → 0 = 10*r* – 2*t* – 5 → 2*t* + 5 = 10*r*.

Now, (2). See the *rt* in 4*rt*/5? And *rt* is *d*.

So (2) becomes *d* = *rt* = 4*rt*/5 + 4*t*/10 → *rt*/5 = 4*t*/10 → 2*rt*/10 = 4*t*/10 → *r* = 2.

And now, going back to (3), 2*t* + 5 = 10*r* → 2*t* + 5 = 10 · 2 → 2*t* = 15 → *t* = 7.5.

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