2023 AMC 8

Complete problem set with solutions and individual problem pages

Problem 11 Medium

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?

  • A.

    6{,}000

  • B.

    12{,}000

  • C.

    60{,}000

  • D.

    120{,}000

  • E.

    600{,}000

Answer:C

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.

 

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}.

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