AMC 10 · 2003 · #4
Grade 6 rate-ratioIt takes Anna 30 minutes to walk uphill 1 km from her home to school, but it takes her only 10 minutes to walk from school to her home along the same route. What is her average speed, in km/hr, for the round trip?
Pick an answer.
AMC 10 2003 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: Anna walks $1$ km uphill from home to school in $30$ minutes, then walks the same $1$ km back downhill in $10$ minutes. Find her average speed, in km/hr, for the whole round trip.
Givens: Uphill leg: $1$ km in $30$ minutes; Downhill leg (same route): $1$ km in $10$ minutes; Answer choices: (A) $3$, (B) $3.125$, (C) $3.5$, (D) $4$, (E) $4.5$
Unknowns: The average speed for the round trip, measured in km/hr
Understand
Restated: Anna walks $1$ km uphill from home to school in $30$ minutes, then walks the same $1$ km back downhill in $10$ minutes. Find her average speed, in km/hr, for the whole round trip.
Givens: Uphill leg: $1$ km in $30$ minutes; Downhill leg (same route): $1$ km in $10$ minutes; Answer choices: (A) $3$, (B) $3.125$, (C) $3.5$, (D) $4$, (E) $4.5$
Plan
Primary tool: #8 Analyze the Units
Secondary: #7 Identify Subproblems
The word "speed" with a target unit of km/hr is a signal for Tool #8 (Analyze the Units): average speed is defined as $\frac{\text{total distance in km}}{\text{total time in hr}}$, so the units themselves tell us exactly what to collect — a total distance and a total time — and warn us to convert minutes to hours before dividing. Tool #7 (Identify Subproblems) then splits the job cleanly: first add up the distance, separately add up the time, and only at the end combine them. The units framing also guards against the classic trap of averaging the two speeds ($2$ and $6$) to get $4$, which mixes rates instead of dividing one total by another.
Execute — Answer: A
6.RP.A.3 Step 1 Read the target unit
- The answer must come out in km/hr, so average speed is total distance (km) over total time (hr).
- This is one division at the end, not an average of the uphill and downhill speeds.
💡 The unit "km per hr" literally reads as kilometres divided by hours, so it names the two totals you must gather.
5.MD.A.1 Step 2 Add distance and time
- Subproblem 1 — total distance: $1$ km out plus $1$ km back is $2$ km.
- Subproblem 2 — total time: $30 + 10 = 40$ minutes, which must be changed into hours because the answer is per hour.
- Since $60$ minutes make an hour, $40$ minutes is $\frac{40}{60} = \frac{2}{3}$ of an hour.
💡 Keep the two piles separate — distance is just $1+1$, and time only becomes usable once it is in hours to match the km/hr goal.
6.NS.A.1 Step 3 Divide totals for the speed
- Divide the total distance by the total time.
- Dividing by $\frac{2}{3}$ means multiplying by its reciprocal $\frac{3}{2}$, which gives $3$.
- So Anna's average speed is $3$ km/hr, choice (A).
💡 Splitting $2$ km over $\frac{2}{3}$ of an hour is the same as asking how far she goes in a full hour, and $2 \div \frac{2}{3}$ scales it up to $3$.
6.RP.A.3 The answer must come out in km/hr, so average speed is total distance (km) over 5.MD.A.1 Subproblem 1 — total distance: $1$ km out plus $1$ km back is $2$ km. Subproblem 6.NS.A.1 Divide the total distance by the total time. Dividing by $\frac{2}{3}$ means mul Review
Reasonableness: The two actual speeds are $2$ km/hr uphill ($1$ km in half an hour) and $6$ km/hr downhill ($1$ km in a sixth of an hour), so the round-trip average must land between $2$ and $6$ — and $3$ does. It sits closer to the slow speed than to $4$, which is correct because she spends much more time crawling uphill than coasting downhill. This also exposes the trap answer (D) $4$: that is the plain average $\frac{2+6}{2}$, which wrongly ignores that the slow leg eats up most of the time.
Alternative: Tool #5 (Look for a Pattern) via the harmonic mean: when equal distances are covered at speeds $x$ and $y$, the average speed is $\frac{2xy}{x+y}$. With $x=2$ and $y=6$, that is $\frac{2\cdot 2\cdot 6}{2+6} = \frac{24}{8} = 3$ km/hr, matching (A). The total-distance-over-total-time method is more direct here because both distances are already known, but the harmonic-mean formula confirms the same value.
CCSS standards used (min grade 6)
6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Interpreting average speed as the single rate total distance $\div$ total time, rather than averaging the two leg speeds.)5.MD.A.1Convert among different-sized standard measurement units within a given system (Converting the total travel time of $40$ minutes into $\frac{2}{3}$ of an hour so the speed comes out in km/hr.)6.NS.A.1Interpret and compute quotients of fractions and solve word problems (Computing $2 \div \frac{2}{3} = 2 \times \frac{3}{2} = 3$ to get the average speed.)
⭐ Average speed is all your distance divided by all your time — never just the average of the two speeds, because the slow part eats up more of the clock.
⭐ Average speed is all your distance divided by all your time — never just the average of the two speeds, because the slow part eats up more of the clock.
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