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?


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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.)

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