2023 AMC 8 Problems/Problem 11: Difference between revisions

Revision as of 06:23, 2 February 2023

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

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

@@ Line 6: / Line 6: @@
 ==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>\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>
+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.
@@ Line 12: / Line 12: @@
 ==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>
 ~MathFun1000

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