Difference between revisions of "2024 AMC 10A Problems/Problem 2"

(Solution)
 
(19 intermediate revisions by 14 users not shown)
Line 1: Line 1:
Spell "red"
+
{{duplicate|[[2024 AMC 10A Problems/Problem 2|2024 AMC 10A #2]] and [[2024 AMC 12A Problems/Problem 2|2024 AMC 12A #2]]}}
  
<math>\textbf{(A) } \text{R E D} \qquad\textbf{(B) } \text{L S T E R} \qquad\textbf{(C) } \text{Q E D} \qquad\textbf{(D) } \text{L M N O P} \qquad\textbf{(E) } 42 </math>
+
== 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 <math>T=aL+bG,</math> where <math>a</math> and <math>b</math> are constants, <math>T</math> is the time in minutes, <math>L</math> is the length of the trail in miles, and <math>G</math> is the altitude gain in feet. The model estimates that it will take <math>69</math> minutes to hike to the top if a trail is <math>1.5</math> miles long and ascends <math>800</math> feet, as well as if a trail is <math>1.2</math> miles long and ascends <math>1100</math> feet. How many minutes does the model estimates it will take to hike to the top if the trail is <math>4.2</math> miles long and ascends <math>4000</math> feet?
  
==Solution==
+
<math>\textbf{(A) }240\qquad\textbf{(B) }246\qquad\textbf{(C) }252\qquad\textbf{(D) }258\qquad\textbf{(E) }264</math>
Looking at the answer choices, we find that the correct answer is <math>\textbf{(A)}</math>.
 
  
==Video Solution==
+
== Solution 1 ==
https://youtu.be/sHQgxWPJSKc?si=bTNk8EXGJw5pThy7
+
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 <math>a</math> and <math>b</math> gives you that <math>a=30</math> and <math>b=\frac{3}{100}</math>. These values can be plugged back in showing that these values are correct.
 +
Now, use the given <math>4.2</math>-mile length and <math>4000</math>-foot change in elevation, giving you a final answer of <math>\boxed{\textbf{(B) }246}.</math>
 +
 
 +
Solution by [[User:Juwushu|juwushu]].
 +
 
 +
==Solution 2==
 +
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:
 +
\begin{align*}
 +
69&=15x+8y, \\
 +
69&=12x+11y.
 +
\end{align*}
 +
 
 +
The solution becomes more obvious in this way, with <math>15+8=12+11=23</math>, and since <math>23\cdot 3=69</math>, we determine that <math>x=y=3</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 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==
 +
{{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}}

Latest revision as of 17:29, 14 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 $T=aL+bG,$ 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 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png