# 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

## Remark

This problem is a great example of a situation where rounding all the numbers to ones we can easily work with is an excellent first step. In any competitive timed test, especially the AMC, if you see problems that require you to choose answers that are the closest, this is a sign to you that you can speed through this problem and move on quickly.

~Nivaar

## Video Solution (HOW TO THINK CREATIVELY!!!)

~Education the Study of everything

~Math-X

## Video Solution (Animated)

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

~harungurcan