Difference between revisions of "2015 UNCO Math Contest II Problems/Problem 4"
(→Solution) |
m (→Solution) |
||
| (One intermediate revision by the same user not shown) | |||
| Line 7: | Line 7: | ||
== 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>100-72=</math> <math>\boxed{28}</math> feet left when Tarantula A finishes. |
| + | |||
-ShootingStars | -ShootingStars | ||
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 | ||
