Difference between revisions of "2023 AMC 8 Problems/Problem 11"
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==Solution 1== | ==Solution 1== | ||
− | Note that <math>6.5</math> months is | + | Note that <math>6.5</math> months is approximately <math>6.5\cdot30\cdot24</math> hours. Therefore, the speed (in miles per hour) is <cmath>\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}.</cmath> |
As the answer choices are far apart from each other, we can ensure that the approximation is correct. | As the answer choices are far apart from each other, we can ensure that the approximation is correct. | ||
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==Solution 2== | ==Solution 2== | ||
− | Note that <math>292{,}526{,}838 \approx 300{,}000{,}000</math> miles. We also know that <math>6.5</math> months is | + | Note that <math>292{,}526{,}838 \approx 300{,}000{,}000</math> miles. We also know that <math>6.5</math> months is approximately <math>6.5\cdot30\cdot24</math> hours. Now, we can calculate the speed in miles per hour, which we find is about |
<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 | ||
+ | |||
+ | ==Remark== | ||
+ | This problem is a great example of a situation where rounding all the numbers to ones we can easily work with is an excellent first step. In any competitive timed test, especially the AMC, if you see problems that require you to choose answers that are the closest, this is a sign to you that you can speed through this problem and move on quickly. | ||
+ | |||
+ | ~Nivaar | ||
+ | |||
+ | ==Video Solution (HOW TO THINK CREATIVELY!!!)== | ||
+ | https://youtu.be/AKJ4XmfYEsA | ||
+ | |||
+ | ~Education the Study of everything | ||
+ | |||
+ | ==Video Solution by Math-X (Smart and Simple)== | ||
+ | https://youtu.be/Ku_c1YHnLt0?si=qbC0eobIhyCNKboM&t=1647 | ||
+ | |||
+ | ~Math-X | ||
+ | |||
+ | ==Video Solution by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=AJqTqVLEFnI | ||
==Video Solution (Animated)== | ==Video Solution (Animated)== | ||
Line 23: | Line 41: | ||
==Video Solution by Magic Square== | ==Video Solution by Magic Square== | ||
https://youtu.be/-N46BeEKaCQ?t=4695 | https://youtu.be/-N46BeEKaCQ?t=4695 | ||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/DBqko2xATxs&t=967 | ||
+ | |||
+ | ==Video Solution by harungurcan== | ||
+ | https://www.youtube.com/watch?v=oIGy79w1H8o&t=707s | ||
+ | |||
+ | ~harungurcan | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2023|num-b=10|num-a=12}} | {{AMC8 box|year=2023|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 03:51, 24 January 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Remark
- 5 Video Solution (HOW TO THINK CREATIVELY!!!)
- 6 Video Solution by Math-X (Smart and Simple)
- 7 Video Solution by SpreadTheMathLove
- 8 Video Solution (Animated)
- 9 Video Solution by Magic Square
- 10 Video Solution by Interstigation
- 11 Video Solution by harungurcan
- 12 See Also
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
Remark
This problem is a great example of a situation where rounding all the numbers to ones we can easily work with is an excellent first step. In any competitive timed test, especially the AMC, if you see problems that require you to choose answers that are the closest, this is a sign to you that you can speed through this problem and move on quickly.
~Nivaar
Video Solution (HOW TO THINK CREATIVELY!!!)
~Education the Study of everything
Video Solution by Math-X (Smart and Simple)
https://youtu.be/Ku_c1YHnLt0?si=qbC0eobIhyCNKboM&t=1647
~Math-X
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/DBqko2xATxs&t=967
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.