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

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{{duplicate|[[2024 AMC 10A Problems/Problem 2|2024 AMC 10A #2]] and [[2024 AMC 12A Problems/Problem 2|2024 AMC 12A #2]]}}
 
{{duplicate|[[2024 AMC 10A Problems/Problem 2|2024 AMC 10A #2]] and [[2024 AMC 12A Problems/Problem 2|2024 AMC 12A #2]]}}
  
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How to Qualify for the AIME:
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== Problem ==
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https://www.youtube.com/watch?v=ZAYDcjOSTvk
 
  
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Problem 2. https://www.youtube.com/watch?v=ZAYDcjOSTvk
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A model used to estimate the time it will take a content creator to produce a custom video for a subscriber is of the form 
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<cmath>
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\[
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T = aL + bG,
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\] 
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</cmath>
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where \( a \) and \( b \) are constants, \( T \) is the production time in minutes, \( L \) is the length of the video in minutes, and \( G \) is the number of special requests. 
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The model estimates that it will take \( 69 \) minutes to produce a video that is \( 1.5 \) minutes long with \( 800 \) special requests, as well as a video that is \( 1.2 \) minutes long with \( 1100 \) special requests. 
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How many minutes does the model estimate it will take to produce a \( 4.2 \)-minute video with \( 4000 \) special requests? 
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<cmath>
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\[
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\textbf{(A) }240\qquad\textbf{(B) }246\qquad\textbf{(C) }252\qquad\textbf{(D) }258\qquad\textbf{(E) }264
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\]
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</cmath>
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== Solution 1 ==
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Plug in the values into the equation to give you the following two equations:
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\begin{align*}
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69&=1.5a+800b, \\
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69&=1.2a+1100b.
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\end{align*}
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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.
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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>
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Solution by [[User:Juwushu|juwushu]].
  
 
==Solution 2==
 
==Solution 2==
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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>.
  
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~Edited by Rosiefork
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~
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==Video Solution==
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https://youtu.be/l3VrUsZkv8I
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~MC
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==Video Solution by Central Valley Math Circle==
  
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https://youtu.be/UewevLQ9qqE
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~mr_mathman
  
 
== Video Solution by Math from my desk ==
 
== Video Solution by Math from my desk ==
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<i>~Education, the Study of Everything</i>
 
<i>~Education, the Study of Everything</i>
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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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==Video Solution by FrankTutor==
 
==Video Solution by FrankTutor==
 
https://youtu.be/A72QJN_lVj8
 
https://youtu.be/A72QJN_lVj8
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==Video Solution by TheBeautyofMath==
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For AMC 10: https://youtu.be/uKXSZyrIOeU?t=540
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For AMC 12: https://youtu.be/zaswZfIEibA?t=540
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~IceMatrix
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==Video Solution by Dr. David==
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https://youtu.be/mrlTB_0QNyI
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==Video Solution by yjtest==
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https://www.youtube.com/watch?v=kyDHwV_KEy4
  
 
==See also==
 
==See also==

Latest revision as of 15:16, 12 August 2025

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 a content creator to produce a custom video for a subscriber is of the form \[ T = aL + bG, \] where \( a \) and \( b \) are constants, \( T \) is the production time in minutes, \( L \) is the length of the video in minutes, and \( G \) is the number of special requests.

The model estimates that it will take \( 69 \) minutes to produce a video that is \( 1.5 \) minutes long with \( 800 \) special requests, as well as a video that is \( 1.2 \) minutes long with \( 1100 \) special requests.

How many minutes does the model estimate it will take to produce a \( 4.2 \)-minute video with \( 4000 \) special requests? \[ \textbf{(A) }240\qquad\textbf{(B) }246\qquad\textbf{(C) }252\qquad\textbf{(D) }258\qquad\textbf{(E) }264 \]

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 $(40+42)\cdot 3=\boxed{\textbf{(B) }246}$.

~

Video Solution

https://youtu.be/l3VrUsZkv8I ~MC

Video Solution by Central Valley Math Circle

https://youtu.be/UewevLQ9qqE

~mr_mathman

Video Solution by Math from my desk

https://www.youtube.com/watch?v=ENbD-tbfbhU&t=2s

Video Solution (🚀 2 min solve 🚀)

https://youtu.be/OmaG3iG7xFs

~Education, the Study of Everything

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

Video Solution by FrankTutor

https://youtu.be/A72QJN_lVj8

Video Solution by TheBeautyofMath

For AMC 10: https://youtu.be/uKXSZyrIOeU?t=540

For AMC 12: https://youtu.be/zaswZfIEibA?t=540

~IceMatrix

Video Solution by Dr. David

https://youtu.be/mrlTB_0QNyI

Video Solution by yjtest

https://www.youtube.com/watch?v=kyDHwV_KEy4

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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