# 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

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 (Animated)

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

## (Creative Thinking) Video Solution

~Education the Study of everything

## Video Solution by harungurcan

~harungurcan

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 