Difference between revisions of "2024 AMC 10A Problems/Problem 2"
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| + | ==Video Solution by Number Craft== | ||
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| + | https://youtu.be/k1rTBtiDWqY | ||
== Video Solution by Daily Dose of Math == | == Video Solution by Daily Dose of Math == | ||
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https://www.youtube.com/watch?v=6SQ74nt3ynw | https://www.youtube.com/watch?v=6SQ74nt3ynw | ||
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| − | https://youtu.be/ | + | https://youtu.be/A72QJN_lVj8 |
==See also== | ==See also== | ||
Latest revision as of 21:49, 30 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.
Contents
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.
Solution 2
Alternatively, observe that using 
and 
makes the numbers much more closer to each other in terms of magnitude.
Plugging in the new variables: \begin{align*} 69&=15x+8y, \\ 69&=12x+11y. \end{align*}
The solution becomes more obvious in this way, with 
, and since 
, we determine that 
.
The question asks us for 
. Since 
, we have 
.
~Edited by Rosiefork
Video Solution by Number Craft
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
Video Solution by Power Solve
https://youtu.be/j-37jvqzhrg?si=2zTY21MFpVd22dcR&t=100
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=6SQ74nt3ynw
Video Solution by FrankTutor
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. 
