Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AMC 8 Problems/Problem 11"
(→‎Solution 3)
 
(20 intermediate revisions by 12 users not shown)
Line 6: Line 6:
  
 
==Solution 1==
 
==Solution 1==
−
Note that <math>6.5</math> months is equivalent to <math>6.5\cdot30\cdot24</math> hours. Therefore, the speed (in miles per hour) is <cmath>.</cmath>
+
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.
 +
 
 
~apex304, SohumUttamchandani, MRENTHUSIASM
 
~apex304, SohumUttamchandani, MRENTHUSIASM
  
 
==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 equivalent to <math>6.5\cdot30\cdot24</math> hours. Now, we can calculate the speed in miles per hour, which we find is about
+
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
 +
 +
==Solution 3==
 +
 +
The rover traveled about <math>292{,}526{,}838 \approx 300{,}000{,}000</math> miles. <math>6.5</math> months is about <math>6\cdot 30 + 15 = 180 + 15 = 195 \approx 200</math> days. <math>200</math> days is about <math>200\cdot 24 \approx 200\cdot 25 = 5{,}000</math> hours. <math>\frac{300{,}000{,}000}{5{,}000} = \boxed{\textbf{(C)}\ 60{,}000}</math>
 +
 +
==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 18: Line 42:
  
 
~Star League (https://starleague.us)
 
~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
 +
 +
==Video Solution by WhyMath==
 +
https://youtu.be/kPhTy-dJp0w
 +
 +
==See Also==
 +
{{AMC8 box|year=2023|num-b=10|num-a=12}}
 +
{{MAA Notice}}

Latest revision as of 17:15, 25 November 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

Solution 3

The rover traveled about $292{,}526{,}838 \approx 300{,}000{,}000$ miles. $6.5$ months is about $6\cdot 30 + 15 = 180 + 15 = 195 \approx 200$ days. $200$ days is about $200\cdot 24 \approx 200\cdot 25 = 5{,}000$ hours. $\frac{300{,}000{,}000}{5{,}000} = \boxed{\textbf{(C)}\ 60{,}000}$

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

Video Solution by WhyMath

https://youtu.be/kPhTy-dJp0w

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_8_Problems/Problem_11&oldid=235586"