2023 AMC 8 Problems/Problem 11
Contents
Problem
NASA’s Perseverance Rover was launched on July 
After traveling 
miles, it landed on Mars in Jezero Crater about 
months later. Which of the following is closest to the Rover’s average interplanetary speed in miles per hour?
Solution 1
Note that months is approximately 
hours. Therefore, the speed (in miles per hour) is ![\[\frac{292{,}526{,}838}{6.5\cdot30\cdot24} \approx \frac{300{,}000{,}000}{6.5\cdot30\cdot24} = \frac{10{,}000{,}000}{6.5\cdot24} \approx \frac{10{,}000{,}000}{6.4\cdot25} = \frac{10{,}000{,}000}{160} = 62500 \approx \boxed{\textbf{(C)}\ 60{,}000}.\]](http://latex.artofproblemsolving.com/3/9/d/39d951a0814c9962084b6966c14bddc467365df6.png)
As the answer choices are far apart from each other, we can ensure that the approximation is correct.
~apex304, SohumUttamchandani, MRENTHUSIASM
Solution 2
Note that 
miles. We also know that months is approximately hours. Now, we can calculate the speed in miles per hour, which we find is about
![\[\dfrac{300{,}000{,}000}{6.5\cdot30\cdot24}=\dfrac{10{,}000{,}000}{6.5\cdot24}=\dfrac{10{,}000{,}000}{13\cdot12}=\dfrac{10{,}000{,}000}{156}\approx\dfrac{10{,}000{,}000}{150}\approx\dfrac{200{,}000}{3}\approx\boxed{\textbf{(C)}\ 60{,}000}.\]](http://latex.artofproblemsolving.com/7/c/d/7cd20ea8b618574baf611e0fb4c72ac4060d2a98.png)
~MathFun1000
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=AJqTqVLEFnI
Video Solution (Animated)
~Star League (https://starleague.us)
Video Solution by Magic Square
https://youtu.be/-N46BeEKaCQ?t=4695
Video Solution by Interstigation
https://youtu.be/1bA7fD7Lg54?t=727
(Creative Thinking) Video Solution
~Education the Study of everything
Video Solution by harungurcan
https://www.youtube.com/watch?v=oIGy79w1H8o&t=707s
~harungurcan
See Also
| 2023 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
