2023 AMC 8 Problems/Problem 11

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

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 $\frac{300,000,000}{5 * 30 * 30} \approx \frac{300,000,000}{5000} = 60,000$ which is $\boxed{\text{(C)}60,000}$

~apex304 and SohumUttamchandani

Solution 2

$292,526,838$ miles is extremely close to $300,000,000$ 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 $\boxed{\textbf{(C) }60,000}$ .

$\dfrac{300,000,000}{6.5\cdot30\cdot24}=\dfrac{10,000,000}{6.5\cdot24}\\ =\dfrac{10,000,000}{13\cdot12}\\ =\dfrac{10,000,000}{156}\\ \approx\dfrac{10,000,000}{150}\\ \approx\dfrac{200,000}{3}\\ \approx60,000$

~MathFun1000

Video Solution (Animated)

https://youtu.be/hwR2VM9tHJ0

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

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2023 AMC 8 Problems/Problem 11

Contents

Problem

Solution 1

Solution 2

Video Solution (Animated)