Solution
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.)