Difference between revisions of "2024 AMC 10A Problems/Problem 2"
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==Solution 2== | ==Solution 2== | ||
− | Alternatively, observe that using <math> | + | Alternatively, observe that using <math>a=10x</math> and <math>b=\frac{y}{100}</math> makes the numbers much more closer to each other in terms of magnitude. |
Plugging in the new variables: | Plugging in the new variables: | ||
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The question asks us for <math>4.2a+4000b=42x+40y</math>. Since <math>x=y</math>, we have <math>(40+42)\cdot 3=\boxed{\textbf{(B) }246}</math>. | The question asks us for <math>4.2a+4000b=42x+40y</math>. Since <math>x=y</math>, we have <math>(40+42)\cdot 3=\boxed{\textbf{(B) }246}</math>. | ||
+ | |||
+ | ~Edited by Rosiefork | ||
== Video Solution by Pi Academy == | == Video Solution by Pi Academy == |
Revision as of 10:46, 10 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
[hide]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:
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:
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 Pi Academy
https://youtu.be/GPoTfGAf8bc?si=JYDhLVzfHUbXa3DW
~Thesmartgreekmathdude
Video Solution by Power Solve
https://youtu.be/j-37jvqzhrg?si=2zTY21MFpVd22dcR&t=100
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.