Difference between revisions of "2023 AMC 8 Problems/Problem 11"

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==Problem==
  
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NASA’s Perseverance Rover was launched on July <math>30,</math> <math>2020.</math> After traveling <math>292{,}526{,}838</math> miles, it landed on Mars in Jezero Crater about <math>6.5</math> months later. Which of the following is closest to the Rover’s average interplanetary speed in miles per hour?
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<math>\textbf{(A)}\ 6{,}000 \qquad \textbf{(B)}\ 12{,}000 \qquad \textbf{(C)}\ 60{,}000 \qquad \textbf{(D)}\ 120{,}000 \qquad \textbf{(E)}\ 600{,}000</math>
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==Solution 1==
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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>
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As the answer choices are far apart from each other, we can ensure that the approximation is correct.
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~apex304, SohumUttamchandani, MRENTHUSIASM
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==Solution 2==
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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
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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>
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~MathFun1000
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==Remark==
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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.
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~Nivaar
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==Video Solution (HOW TO THINK CREATIVELY!!!)==
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https://youtu.be/AKJ4XmfYEsA
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~Education the Study of everything
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==Video Solution by Math-X (Smart and Simple)==
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https://youtu.be/Ku_c1YHnLt0?si=qbC0eobIhyCNKboM&t=1647
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~Math-X
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==Video Solution by SpreadTheMathLove==
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https://www.youtube.com/watch?v=AJqTqVLEFnI
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==Video Solution (Animated)==
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https://youtu.be/hwR2VM9tHJ0
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~Star League (https://starleague.us)
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==Video Solution by Magic Square==
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https://youtu.be/-N46BeEKaCQ?t=4695
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==Video Solution by Interstigation==
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https://youtu.be/DBqko2xATxs&t=967
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==Video Solution by harungurcan==
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https://www.youtube.com/watch?v=oIGy79w1H8o&t=707s
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~harungurcan
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==Video Solution by Dr. David==
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https://youtu.be/JQUlnX_h__E
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==See Also==
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{{AMC8 box|year=2023|num-b=10|num-a=12}}
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{{MAA Notice}}

Revision as of 23:08, 27 October 2024

Problem

NASA’s Perseverance Rover was launched on July $30,$ $2020.$ After traveling $292{,}526{,}838$ miles, it landed on Mars in Jezero Crater about $6.5$ months later. Which of the following is closest to the Rover’s average interplanetary speed in miles per hour?

$\textbf{(A)}\ 6{,}000 \qquad \textbf{(B)}\ 12{,}000 \qquad \textbf{(C)}\ 60{,}000 \qquad \textbf{(D)}\ 120{,}000 \qquad \textbf{(E)}\ 600{,}000$

Solution 1

Note that $6.5$ months is approximately $6.5\cdot30\cdot24$ hours. Therefore, the speed (in miles per hour) is \[\frac{292{,}526{,}838}{6.5\cdot30\cdot24} \approx \frac{300{,}000{,}000}{6.5\cdot30\cdot24} = \frac{10{,}000{,}000}{6.5\cdot24} \approx \frac{10{,}000{,}000}{6.4\cdot25} = \frac{10{,}000{,}000}{160} = 62500 \approx \boxed{\textbf{(C)}\ 60{,}000}.\] 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 $292{,}526{,}838 \approx 300{,}000{,}000$ miles. We also know that $6.5$ months is approximately $6.5\cdot30\cdot24$ hours. Now, we can calculate the speed in miles per hour, which we find is about \[\dfrac{300{,}000{,}000}{6.5\cdot30\cdot24}=\dfrac{10{,}000{,}000}{6.5\cdot24}=\dfrac{10{,}000{,}000}{13\cdot12}=\dfrac{10{,}000{,}000}{156}\approx\dfrac{10{,}000{,}000}{150}\approx\dfrac{200{,}000}{3}\approx\boxed{\textbf{(C)}\ 60{,}000}.\] ~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)

https://youtu.be/hwR2VM9tHJ0

~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

Video Solution by Dr. David

https://youtu.be/JQUlnX_h__E

See Also

2023 AMC 8 (ProblemsAnswer KeyResources)
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

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