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: 69=1.5a+800b,69=1.2a+1100b. 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: 69=15x+8y,69=12x+11y.

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 Daily Dose of Math

https://youtu.be/W0NMzXaULx4

~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

See also

2024 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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