Math Help −> Puzzles −> Academic decathlon 2004 logic quiz −> Pass the baton −> Answer 

Answer:

The best way to solve this is to look at the last runner first, and work backwards.  The last runner goes 1/4 of the remaining distance plus 1/4 mile.  If you let x stand for the distance the last runner travels, then

x = (1/4) x + 1/4
(3/4) x = 1/4
x = 1/3 -- the last runner runs 1/3 of a mile.

Now, let's look at the middle runner.  He runs 1/3 of the remaining distance plus 1/3 mile.  Let y stand for the distance from the middle runner's starting point to the end of the race.  This distance is 1/3 of itself plus the extra 1/3 mile the middle runner runs, plus the 1/3 mile the third runner runs.

y = (1/3) y + 1/3 + 1/3
(2/3) y = 2/3
y = 1 -- the distance run by the last two runners.

Now, let's look at the first runner.  He runs 1/2 of the total distance plus 1/2 mile.  Let z stand for the total distance.  This distance is 1/2 of itself plus the extra 1/2 mile run by the first runner, plus the 1 mile run by the last two runners.  So we have

z = (1/2) z + 1/2 + 1
(1/2) z = 3/2
z = 3

The total distance is three miles.

Let's check this answer...

The first runner runs half of 3 miles, or a mile and a half, then he runs an extra half-mile, for a total of two miles, leaving 1 mile left in the race.

The second runner runs 1/3 of that remaining mile, plus another 1/3 mile, leaving 1/3 mile for the last runner.

The third runner runs 1/4 of the remaining distance, which is 1/12 of a mile, leaving 1/4 mile, which is exactly the extra distance run by the third runner.

So it all checks out.

If you wanted to use "old fashioned algebra" to solve this problem, you could do it, but it gets very messy.  Here's how it works.

Let "x" be the total distance of the race.