Difference between revisions of "2024 AMC 10A Problems/Problem 2"
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+ | {{duplicate|[[2024 AMC 10A Problems/Problem 2|2024 AMC 10A #2]] and [[2024 AMC 12A Problems/Problem 2|2024 AMC 12A #2]]}} | ||
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== Problem == | == Problem == | ||
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{{AMC10 box|year=2024|ab=A|num-b=1|num-a=3}} | {{AMC10 box|year=2024|ab=A|num-b=1|num-a=3}} | ||
+ | {{AMC12 box|year=2024|ab=A|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:19, 8 November 2024
- The following problem is from both the 2024 AMC 10A #2 and 2024 AMC 12A #2, so both problems redirect to this page.
Problem
A model used to estimate the time it will take to hike to the top of the mountain on a trail is of the form where and are constants, is the time in minutes, is the length of the trail in miles, and is the altitude gain in feet. The model estimates that it will take minutes to hike to the top if a trail is miles long and ascends feet, as well as if a trail is miles long and ascends feet. How many minutes does the model estimates it will take to hike to the top if the trail is miles long and ascends feet?
Solution 1
Plug in the values into the equation to give you the following two equations: \begin{align*} 69&=1.5a+800b, \\ 69&=1.2a+1100b. \end{align*} Solving for the values and gives you that and . These values can be plugged back in showing that these values are correct. Now, use the given -mile length and -foot change in elevation, giving you a final answer of
Solution by juwushu.
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
See also
2024 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2024 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 1 |
Followed by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.