AMC 8 · 2001 · #5
Grade 6 rate-ratioarithmeticProblem
On a dark and stormy night Snoopy suddenly saw a flash of lightning. Ten seconds later he heard the sound of thunder. The speed of sound is 1088 feet per second and one mile is 5280 feet. Estimate, to the nearest half-mile, how far Snoopy was from the flash of lightning.
Pick an answer.
AMC 8 2001 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.
Toolkit + CCSS Solution
Understand
Restated: Snoopy sees a lightning flash and hears the thunder $10$ seconds later. Sound travels at $1088$ feet per second and one mile is $5280$ feet. Estimate, to the nearest half-mile, how far the lightning was from Snoopy.
Givens: Delay between flash and thunder: $10$ seconds; Speed of sound: $1088$ feet per second; Conversion: $1$ mile $= 5280$ feet; Light reaches Snoopy almost instantly, so the $10$-second delay is the sound's travel time; Answer choices: (A) $1$, (B) $1\tfrac{1}{2}$, (C) $2$, (D) $2\tfrac{1}{2}$, (E) $3$ (miles)
Unknowns: The distance from Snoopy to the lightning, rounded to the nearest half-mile
Understand
Restated: Snoopy sees a lightning flash and hears the thunder $10$ seconds later. Sound travels at $1088$ feet per second and one mile is $5280$ feet. Estimate, to the nearest half-mile, how far the lightning was from Snoopy.
Givens: Delay between flash and thunder: $10$ seconds; Speed of sound: $1088$ feet per second; Conversion: $1$ mile $= 5280$ feet; Light reaches Snoopy almost instantly, so the $10$-second delay is the sound's travel time; Answer choices: (A) $1$, (B) $1\tfrac{1}{2}$, (C) $2$, (D) $2\tfrac{1}{2}$, (E) $3$ (miles)
Plan
Primary tool: #8 Analyze the Units
Secondary: #7 Identify Subproblems
The numbers come dressed in three different units — seconds, feet per second, feet per mile — but the answer must be in miles. Tool #8 (Analyze the Units) gives the recipe: multiply $\tfrac{\text{feet}}{\text{second}}$ by seconds to cancel "second" and land on feet, then divide by $\tfrac{\text{feet}}{\text{mile}}$ to cancel "feet" and land on miles. Tool #7 (Identify Subproblems) splits that recipe into three clean steps — (a) feet of sound travel, (b) feet $\to$ miles, (c) round to the nearest half-mile — so each line does one job.
Execute — Answer: C
6.RP.A.3 Step 1 - Subproblem 1: distance the sound traveled, in feet.
- Multiply the speed by the time.
- The unit $\tfrac{\text{feet}}{\text{second}} \times \text{seconds}$ cancels "second" and leaves "feet."
💡 Rate $\times$ time is the Grade 6 rate-reasoning move; multiplying by $10$ just shifts the digits one place.
5.MD.A.1 Step 2 - Subproblem 2: convert feet to miles.
- Divide by $5280 \tfrac{\text{ft}}{\text{mile}}$.
- Dividing "feet" by "feet per mile" cancels "feet" and leaves "miles."
💡 Converting feet to miles by the given factor $5280$ is the Grade 5 "convert measurement units" standard.
5.NBT.A.4 Step 3 - Subproblem 3: round $2.06$ miles to the nearest half-mile.
- The nearby half-mile marks are $2$ and $2\tfrac{1}{2}$.
- Compare the gaps and pick the smaller one.
💡 Rounding to the nearest half is the same idea as Grade 5 decimal rounding — pick whichever benchmark is closer.
6.RP.A.3 Subproblem 1: distance the sound traveled, in feet. Multiply the speed by the ti 5.MD.A.1 Subproblem 2: convert feet to miles. Divide by $5280 \tfrac{\text{ft}}{\text{mil 5.NBT.A.4 Subproblem 3: round $2.06$ miles to the nearest half-mile. The nearby half-mile Review
Reasonableness: Quick mental check using the well-known rule of thumb "sound travels about $1$ mile every $5$ seconds." In $10$ seconds, the sound covers roughly $\tfrac{10}{5} = 2$ miles — exactly the answer (C). The exact speed $1088$ ft/s gives $1088 \times 5 = 5440$ ft per $5$ seconds, which is just $160$ ft over a mile, so the rule of thumb slightly underestimates and our $2.06$ miles is the small correction. Either way, $2$ is the nearest half-mile.
Alternative: Tool #3 (Eliminate Possibilities) on the choices. Convert each choice to feet and compare with the actual $10{,}880$ ft of sound travel: (A) $1$ mi $= 5280$ ft (too small), (B) $1.5$ mi $= 7920$ ft (still too small), (C) $2$ mi $= 10{,}560$ ft (off by only $320$ ft), (D) $2.5$ mi $= 13{,}200$ ft (off by $2320$ ft), (E) $3$ mi $= 15{,}840$ ft (way too big). Choice (C) is by far the closest, confirming the answer.
CCSS standards used (min grade 6)
6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Multiplying the rate $1088 \tfrac{\text{ft}}{\text{s}}$ by $10$ s to get $10{,}880$ ft of sound travel.)5.MD.A.1Convert among different-sized standard measurement units within a given measurement system (Using the conversion $1$ mile $= 5280$ ft to turn $10{,}880$ ft into $\tfrac{10{,}880}{5280} \approx 2.06$ miles.)5.NBT.A.4Use place value understanding to round decimals to any place (Rounding $2.06$ to the nearest half-mile by comparing the distances to $2$ and $2.5$.)
⭐ Let the units guide you: seconds times feet-per-second gives feet, then dividing by feet-per-mile gives miles. Sound covers $10{,}880$ ft in $10$ s, which is about $2.06$ mi — closest to $2$ miles, answer (C).
⭐ Let the units guide you: seconds times feet-per-second gives feet, then dividing by feet-per-mile gives miles. Sound covers $10{,}880$ ft in $10$ s, which is about $2.06$ mi — closest to $2$ miles, answer (C).
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