Difference between revisions of "2015 UNCO Math Contest II Problems/Problem 4"
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== Solution == | == 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. | ||
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== See also == | == See also == |
Latest revision as of 17:01, 18 December 2023
Problem
Tarantulas and start together at the same time and race straight along a foot path, each running at a constant speed the whole distance. When reaches the end, still has feet more to run. When reaches the end, has feet more to run. How many more feet does Tarantula have to run when Tarantula reaches the end?
Solution
Convert the speeds to fractions that relate between the various speeds. Tarantula A runs feet in the time Tarantula B runs feet, which implies that Tarantula B's speed is th of Tarantula A's speed. Similarly, Tarantula C's speed is or th of Tarantula B's speed. Multiplying by yields . Multiplying this by feet (the total distance), we find that when Tarantula A finishes, Tarantula C is feet in, so Tarantula C has feet left when Tarantula A finishes.
-ShootingStars
See also
2015 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
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 |