View source for 2015 UNCO Math Contest II Problems/Problem 4

== Problem ==

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


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

-ShootingStars

== See also ==
{{UNCO Math Contest box|year=2015|n=II|num-b=3|num-a=5}}

[[Category:Introductory Algebra Problems]]