# 2015 UNCO Math Contest II Problems/Problem 4

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## Problem

Tarantulas $A, B,$ and $C$ start together at the same time and race straight along a $100$ foot path, each running at a constant speed the whole distance. When $A$ reaches the end, $B$ still has $10$ feet more to run. When $B$ reaches the end, $C$ has $20$ feet more to run. How many more feet does Tarantula $C$ have to run when Tarantula $A$ reaches the end?

## Solution

Convert the speeds to fractions that relate between the various speeds. Tarantula A runs $100$ feet in the time Tarantula B runs $90$ feet, which implies that Tarantula B's speed is $9/10$th of Tarantula A's speed. Similarly, Tarantula C's speed is $8/10$ or $4/5$th of Tarantula B's speed. Multiplying $9/10$ by $4/5$ yields $36/50$. Multiplying this by $100$ feet (the total distance), we find that when Tarantula A finishes, Tarantula C is $72$ feet in, so Tarantula C has $100-72=$ $\boxed{28}$ feet left when Tarantula A finishes.

-ShootingStars

## See also

 2015 UNCO Math Contest II (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 All UNCO Math Contest Problems and Solutions