Difference between revisions of "2023 AMC 8 Problems/Problem 11"
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~Star League (https://starleague.us) | ~Star League (https://starleague.us) | ||
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| + | ==Video Solution by Magic Square== | ||
| + | https://youtu.be/-N46BeEKaCQ?t=4695 | ||
==See Also== | ==See Also== | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 10:42, 25 January 2023
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 equivalent to 
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 equivalent to 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 (Animated)
~Star League (https://starleague.us)
Video Solution by Magic Square
https://youtu.be/-N46BeEKaCQ?t=4695
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. 
