Difference between revisions of "2023 AMC 8 Problems/Problem 11"
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<cmath>\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}.</cmath> | <cmath>\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}.</cmath> | ||
~MathFun1000 | ~MathFun1000 | ||
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+ | == Video Solution (HOW TO THINK CREATIVELY!!!)== | ||
+ | https://youtu.be/AKJ4XmfYEsA | ||
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+ | ~Education the Study of everything | ||
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==Video Solution by SpreadTheMathLove== | ==Video Solution by SpreadTheMathLove== |
Revision as of 12:35, 11 June 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 approximately hours. Therefore, the speed (in miles per hour) is 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 ~MathFun1000
Video Solution (HOW TO THINK CREATIVELY!!!)
~Education the Study of everything
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
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.