Navigation 
 Home 
 Search 
 Site map 

 Contact Graeme 
 Home 
 Email 
 Twitter

 Skip Navigation LinksMath Help > Math Puzzles > 2004 Logic Quiz > Pass the Baton > Pass the Baton

Academic Decathlon 2004 Logic Quiz

Pass the Baton  (400 points)

In a relay race the first runner of the team hands on the baton after having raced half the distance plus 1/2 mile.  The second runner runs 1/3 of the remaining distance plus 1/3 mile.  The third runner reaches the goal after having raced 1/4 of the remaining distance plus 1/4 mile.  How long was the course?

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.

x = (1/2) x + 1/2  the distance run by the first runner
 + (1/3)(x - ((1/2)x+1/2)) + 1/3  the distance run by the second runner
 + (1/4)(x - ((1/2)x+1/2) - ((1/3)(x - ((1/2)x+1/2)) + 1/3)) + 1/4  the distance run by the third runner
x = (1/2) x + 1/2  distributing everything...
 + (1/3)x - (1/6)x - 1/6 + 1/3
 + (1/4)x - (1/8)x - 1/8 - (1/12)x + (1/24)x + 1/24 - 1/12 + 1/4
 multiplying through by 24...
24x = 12 x + 12
 + 8x - 4x - 4 + 8
 + 6x - 3x - 3 - 2x + x + 1 - 2 + 6
 gathering like terms...
24x = 12x + 8x - 4x + 6x - 3x - 2x + x
 + 12 - 4 + 8 - 3 + 1 - 2 + 6
24x = 18x + 18
6x = 18
x = 3

So the method works, but the problem with it is that distributing all those minus-signs is a very error-prone process, so it is unlikely that a single person using this method will solve the problem within the 50 minutes for the whole test (I certainly didn't!).

 

Related pages in this website

See the NEXT puzzle in the Academic Decathlon 2004 Logic Quiz

 

The webmaster and author of this Math Help site is Graeme McRae.