2015 UNCO Math Contest II Problems/Problem 4


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?


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.


See also

2015 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions