# Yorkshire to London

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 = 5r/2 – t/2 – 5/4 → 0 = 10r – 2t – 5 → 2t + 5 = 10r.

Now, (2). See the rt in 4rt/5? And rt is d.
So (2) becomes d = rt = 4rt/5 + 4t/10 → rt/5 = 4t/10 → 2rt/10 = 4t/10 → r = 2.

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