PROBLEM: A Tangled Tail

A problem from Lewis Carroll --

- A man walked for 5 hours, first along a level road, then up a hill,
and then he turned around and walked back to the starting point along the
same path. He walks 4 mph on the level, 3 mph uphill, and 6 mph downhill.
Find the distance he walked.

Since the man walked up a hill and then came back down the same hill,
we may assume that he followed the same path both times, thus he traveled
the same distance both times. We can then calculate the man's average rate
while climbing the hill and descending from the hill. In order to use the
distance formula **d=rt**, we need to know the total time it took for
the man to walk up and down the hill. The time it took the man to walk up
the hill would be t_{u}=d/3 (where d is the distance up the hill)
and the time it took the man to walk down the hill would be t_{d}=d/6.
The total distance the man travels over this period of time is 2d. Now,
we know distance and time and can find the rate.

**2d = [(d/3 + (d/6)]r**
**2d/[(d/3 + (d/6)] = r**
**2/[(1/3 + (1/6)] = r**
**12/3 = 4 = r**
Therefore, the man's average rate up and down the hill was 4 miles per
hour.

The man travels on level ground to the hill at 4 miles per hour, goes
up and down the hill at 4 miles per hour and then travels back home on level
ground at 4 miles per hour. Thus, each segment of his trip covers the same
distance since he is traveling at the same rate on all three segments.

However, he spends 2/3 of his time on level ground and 1/3 of his time
on the hill. Now, if we let **d** be the total distance the man travels,
we have an equation that looks like this:

**d = (2/3)(5 hours)(4 miles/hour) + (1/3)(5 hours)(4 miles/hour)**
**d = 40/3 + 20/3**
**d = 60/3**

or

d = 20 miles
Therefore, the man traveled a total distance of **20 miles** during
his journey.

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©1998 by Luke Rapley