2023 AMC 8 Problems/Problem 11: Difference between revisions

Revision as of 01:58, 25 January 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 equivalent to $6.5\cdot30\cdot24$ hours. Therefore, the speed (in miles per hour) is $.$ ~apex304, SohumUttamchandani, MRENTHUSIASM

Solution 2

Note that $292{,}526{,}838 \approx 300{,}000{,}000$ miles. We also know that $6.5$ months is equivalent to $6.5\cdot30\cdot24$ 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 (Animated)

https://youtu.be/hwR2VM9tHJ0

~Star League (https://starleague.us)

@@ Line 1: / Line 1: @@
 ==Problem==
-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?
+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?
-<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>
+<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>
 ==Solution 1==
-Since the answers are so far apart, we just need to compute the # of figures the answer contains. So by approximating all the values for the hourly rate we have <math>\frac{300,000,000}{5 * 30 * 30} \approx \frac{300,000,000}{5000} = 60,000</math> which is <math>\boxed{\text{(C)}60,000}</math>
+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>
+~apex304, SohumUttamchandani, MRENTHUSIASM
-~apex304 and SohumUttamchandani
 ==Solution 2==
-<math>292,526,838</math> miles is extremely close to <math>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 <math>\boxed{\textbf{(C) }60,000}</math>.
+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
+<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}\\
-\approx60,000</cmath>
 ~MathFun1000

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