Difference between revisions of "2024 AMC 10A Problems/Problem 2"
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\end{align*} | \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. | 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> | + | Now, use the given <math>4.2</math> length of video and <math>4000</math> special requests, giving you a final answer of <math>\boxed{\textbf{(B) }246}.</math> |
Solution by [[User:Juwushu|juwushu]]. | Solution by [[User:Juwushu|juwushu]]. | ||
| + | |||
| + | Minor edits by ParticlePhysics | ||
==Solution 2== | ==Solution 2== | ||
Revision as of 04:17, 22 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.
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Video Solution
- 5 Video Solution by Central Valley Math Circle
- 6 Video Solution by Math from my desk
- 7 Video Solution (🚀 2 min solve 🚀)
- 8 Video Solution by Daily Dose of Math
- 9 Video Solution by Power Solve
- 10 Video Solution by SpreadTheMathLove
- 11 Video Solution by FrankTutor
- 12 Video Solution by TheBeautyofMath
- 13 Video Solution by Dr. David
- 14 Video Solution by yjtest
- 15 Video solution by TheNeuralMathAcademy
- 16 See Also
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
![\[ T = aL + bG, \]](http://latex.artofproblemsolving.com/b/7/3/b732321f6682cbdbe08e24a43dbb888ddc9a692a.png)
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?
![\[ \textbf{(A) }240\qquad\textbf{(B) }246\qquad\textbf{(C) }252\qquad\textbf{(D) }258\qquad\textbf{(E) }264 \]](http://latex.artofproblemsolving.com/3/8/e/38e9ef91344b10186909d8b7a765c7a3ae06b075.png)
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 
and 
gives you that 
and 
. These values can be plugged back in showing that these values are correct.
Now, use the given 
length of video and 
special requests, giving you a final answer of 
Solution by juwushu.
Minor edits by ParticlePhysics
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: \begin{align*} 69&=15x+8y, \\ 69&=12x+11y. \end{align*}
The solution becomes more obvious in this way, with 
, and since 
, we determine that 
.
The question asks us for 
. Since 
, we have 
.
~
Video Solution
https://youtu.be/l3VrUsZkv8I ~MC
Video Solution by Central Valley Math Circle
~mr_mathman
Video Solution by Math from my desk
https://www.youtube.com/watch?v=ENbD-tbfbhU&t=2s
Video Solution (🚀 2 min solve 🚀)
~Education, the Study of Everything
Video Solution by Daily Dose of Math
~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
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
Video Solution by yjtest
https://www.youtube.com/watch?v=kyDHwV_KEy4
Video solution by TheNeuralMathAcademy
https://www.youtube.com/watch?v=4b_YLnyegtw&t=158s
See Also
| 2024 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by 2023 AMC 10B Problems |
Followed by 2024 AMC 10B Problems | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
