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>\boxed{28}</math> feet left when Tarantula A finishes. | 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>\boxed{28}</math> feet left when Tarantula A finishes. | ||
+ | -ShootingStars | ||
== See also == | == See also == |
Revision as of 17:55, 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 |