Marla walked from York toward London at a constant rate. If she had gone 1/2 mph faster, then 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?
Solution
Jeepers! We don't know d, or r, or t. But we make our good old chart anyway. It looks really bad (see below).
Here's what we learn, row-by-row:
(1) d = rt,
(2) d = (r + 1/2) (4t/5) = 4rt/5 + 41/10,
(3) d = (r - 1/2) (t + 2.5) = rt + 5r/2 - t/2 - 5/4.
In (3), notice that we have d = rt + (other stuff). Now 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? That 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 \rightarrow 2t + 5 = 10 \cdot 2 2t = 15 t = 7.5
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.
