AMC 8 · 2023 · #11

Grade 6 rate-ratio

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

rateunit-conversionestimation dimensional-analysisestimation ↑ Prerequisites: rateunit-conversion

📏 Medium solution 💡 3 insights

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?

Pick an answer.

(A)

$6{,}000$

(B)

$12{,}000$

(C)

$60{,}000$

(D)

$120{,}000$

(E)

$600{,}000$

AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

Toolkit + CCSS Solution

Understand

Restated: NASA's Perseverance Rover flew $292{,}526{,}838$ miles to Mars in about $6.5$ months. Of the five answer choices, which one is closest to the rover's average speed in miles per hour?

Givens: Total distance traveled: $292{,}526{,}838$ miles; Total travel time: about $6.5$ months; Answer choices: (A) $6{,}000$, (B) $12{,}000$, (C) $60{,}000$, (D) $120{,}000$, (E) $600{,}000$

Unknowns: Average speed in miles per hour (the choice closest to it)

Understand

Restated: NASA's Perseverance Rover flew $292{,}526{,}838$ miles to Mars in about $6.5$ months. Of the five answer choices, which one is closest to the rover's average speed in miles per hour?

Givens: Total distance traveled: $292{,}526{,}838$ miles; Total travel time: about $6.5$ months; Answer choices: (A) $6{,}000$, (B) $12{,}000$, (C) $60{,}000$, (D) $120{,}000$, (E) $600{,}000$

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #8 Analyze the Units, #7 Identify Subproblems

The numbers are huge ($292{,}526{,}838$ miles) and the answer choices are spread far apart (each is $2{\times}$ or $10{\times}$ the previous), so we don't need an exact answer. Tool #9 (Easier Related Problem) says: replace ugly numbers with nearby round numbers and compute the easier version. Tool #8 (Analyze the Units) keeps us honest about miles vs. months vs. hours so we don't divide the wrong things. Tool #7 (Subproblems) splits the work into two small jobs — convert $6.5$ months to hours, then divide distance by hours.

Execute — Answer: C

#9 Solve an Easier Related Problem 3.NBT.A.1 Step 1

Round the distance to a friendly number.
Since $292{,}526{,}838$ is very close to $300{,}000{,}000$, replace it with $3\times 10^8$ miles.
The choices are far apart, so this rounding will not flip the answer (Tool #9).

$$292{,}526{,}838\;\text{mi}\;\approx\;300{,}000{,}000\;\text{mi}$$

💡 Rounding a whole number to a nearby place value (here, hundred-millions) is an extension of the Grade 3 "round to nearest 10 or 100" skill — the rule is the same, the place is just larger.

#8 Analyze the Units 4.MD.A.1 Step 2

Convert $6.5$ months to hours so the units of the answer come out as miles/hour (Tool #8).
Use $1\text{ month}\approx 30\text{ days}$ and $1\text{ day}=24\text{ hours}$.
This is the first subproblem (Tool #7): a pure unit-conversion job, ignore distance for now.

$$6.5\;\text{mo}\times 30\,\dfrac{\text{day}}{\text{mo}}\times 24\,\dfrac{\text{hr}}{\text{day}}=6.5\times 720=4680\;\text{hr}$$

💡 Converting a larger unit (months) into a smaller unit (hours) by multiplying conversion factors is the Grade 4 measurement standard.

#9 Solve an Easier Related Problem 3.NBT.A.1 Step 3

Round the hour total too.
$4680$ is close to $5000$, and $5000$ divides into $3\times 10^8$ very cleanly.
Tool #9 again: we want a number we can divide in our head.

$$4680\;\text{hr}\;\approx\;5000\;\text{hr}$$

💡 Rounding $4680$ to the nearest thousand is the same place-value-rounding idea introduced in Grade 3 (rounding to nearest 10/100), just one place higher.

#9 Solve an Easier Related Problem 6.RP.A.2 Step 4

Divide the rounded distance by the rounded time — this is the second subproblem (Tool #7).
The unit cancellation $\dfrac{\text{mi}}{\text{hr}}$ confirms we are computing speed in miles per hour (Tool #8).

$$\dfrac{300{,}000{,}000\;\text{mi}}{5000\;\text{hr}}=\dfrac{300{,}000{,}000}{5000}\;\dfrac{\text{mi}}{\text{hr}}=60{,}000\;\text{mi/hr}$$

💡 Speed in miles per hour IS a unit rate (miles per $1$ hour), which is the Grade 6 ratio-and-rate concept.

#9 Solve an Easier Related Problem 6.RP.A.3 Step 5

Compare $60{,}000$ to the five choices.
It matches choice (C) exactly, and the next choices on either side ($12{,}000$ and $120{,}000$) are off by a factor of about $5$ or more, so the rounding error can't move the answer.
The answer is (C).

$$60{,}000\;\text{mi/hr}\;\Rightarrow\;\textbf{(C)}\ 60{,}000$$

💡 Picking the choice closest to our computed unit rate is the Grade 6 "use rate reasoning to solve real-world problems" standard.

[1] #9 3.NBT.A.1 Round the distance to a friendly number. Since $292{,}526{,}838$ is very close t

[2] #8 4.MD.A.1 Convert $6.5$ months to hours so the units of the answer come out as miles/hour

[3] #9 3.NBT.A.1 Round the hour total too. $4680$ is close to $5000$, and $5000$ divides into $3\

[4] #9 6.RP.A.2 Divide the rounded distance by the rounded time — this is the second subproblem

[5] #9 6.RP.A.3 Compare $60{,}000$ to the five choices. It matches choice (C) exactly, and the n

Review

Reasonableness: Sanity-check the magnitude. About $300{,}000{,}000$ miles over about $6.5\times 30=195$ days is roughly $1{,}500{,}000$ miles per day. Dividing by $24$ gives about $62{,}500$ miles per hour, very close to our $60{,}000$. The two neighboring choices, $12{,}000$ (a factor of $5$ smaller) and $120{,}000$ (a factor of $2$ bigger), are both far enough away that no reasonable rounding flips the answer. (C) is consistent.

Alternative: Use Tool #3 (Eliminate Possibilities). For each choice, multiply by $4680\;\text{hr}$ and see which product is closest to $3\times 10^8\;\text{mi}$: (A) $6{,}000\times 4680\approx 2.8\times 10^7$ — too small; (B) $12{,}000\times 4680\approx 5.6\times 10^7$ — still too small; (C) $60{,}000\times 4680\approx 2.8\times 10^8$ — matches; (D) $120{,}000\times 4680\approx 5.6\times 10^8$ — too big; (E) $600{,}000\times 4680\approx 2.8\times 10^9$ — way too big. Only (C) is in the right neighborhood.

CCSS standards used (min grade 6)

3.NBT.A.1 Round whole numbers to the nearest 10 or 100 (Rounding $292{,}526{,}838\to 3\times 10^8$ miles and $4680\to 5000$ hours so the division is easy (place-value rounding, extended to larger places — the closest match in our standards DB).)
4.MD.A.1 Know relative sizes of measurement units and convert larger to smaller units (Converting $6.5$ months into hours via $30\;\text{days/month}\times 24\;\text{hours/day}$.)
6.RP.A.2 Understand the concept of a unit rate and use rate language (Interpreting "miles per hour" as a unit rate (miles per ONE hour), so dividing total miles by total hours gives the speed.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Applying the speed = distance / time rate relationship to a real-world space-travel scenario and picking the closest answer choice.)

⭐ This AMC 8 problem only needs Grade 6 unit-rate (miles-per-hour) reasoning you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

More like this