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 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?
\[ \textbf{(A) }240\qquad\textbf{(B) }246\qquad\textbf{(C) }252\qquad\textbf{(D) }258\qquad\textbf{(E) }264 \]
Contents
- 1 Solution 1
- 2 Solution 2
- 3 Video Solution
- 4 Video Solution by Central Valley Math Circle
- 5 Video Solution by Math from my desk
- 6 Video Solution (🚀 2 min solve 🚀)
- 7 Video Solution by Daily Dose of Math
- 8 Video Solution by Power Solve
- 9 Video Solution by SpreadTheMathLove
- 10 Video Solution by FrankTutor
- 11 Video Solution by TheBeautyofMath
- 12 Video Solution by Dr. David
- 13 Video Solution by yjtest
- 14 See also
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 
-mile length and 
-foot change in elevation, giving you a final answer of 
Solution by juwushu.
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
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. 
