2024 AMC 10A Problems/Problem 2

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 $T=aL+bG,$ where $a$ and $b$ are constants, $T$ is the time in minutes, $L$ is the length of the trail in miles, and $G$ is the altitude gain in feet. The model estimates that it will take $69$ minutes to hike to the top if a trail is $1.5$ miles long and ascends $800$ feet, as well as if a trail is $1.2$ miles long and ascends $1100$ feet. How many minutes does the model estimates it will take to hike to the top if the trail is $4.2$ miles long and ascends $4000$ feet?

$\textbf{(A) }240\qquad\textbf{(B) }246\qquad\textbf{(C) }252\qquad\textbf{(D) }258\qquad\textbf{(E) }264$

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 $a$ and $b$ gives you that $a=30$ and $b=\frac{3}{100}$ . These values can be plugged back in showing that these values are correct. Now, use the given $4.2$ -mile length and $4000$ -foot change in elevation, giving you a final answer of $\boxed{\textbf{(B) }246}.$

Solution by juwushu.

Solution 2

Alternatively, observe that using $a=10x$ and $b=\frac{y}{100}$ 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 $15+8=12+11=23$ , and since $23\cdot 3=69$ , we determine that $x=y=3$ .

The question asks us for $4.2a+4000b=42x+40y$ . Since $x=y$ , we have $(40+42)\cdot 3=\boxed{\textbf{(B) }246}$ .

~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

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.

Page

Toolbox

Search