AMC 8 · 2011 · #9

Grade 6 rate-ratio
rategraph-reading dimensional-analysis ↑ Prerequisites: rate
📏 Short solution 💡 2 insights 📊 Diagram
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Problem

Carmen takes a long bike ride on a hilly highway. The graph indicates the miles traveled during the time of her ride. What is Carmen's average speed for her entire ride in miles per hour?

Pick an answer.

(A)
2
(B)
2.5
(C)
4
(D)
4.5
(E)
5
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Toolkit + CCSS Solution

Understand

Restated: Carmen rides her bike on a hilly highway. A graph plots her cumulative distance (in miles) against time (in hours). The graph starts at $(0, 0)$ and ends at $(7, 35)$, with bends at $(1, 5)$, $(2, 15)$, and $(4, 20)$. What is her average speed for the entire ride in miles per hour?

Givens: The graph plots miles (vertical) against hours (horizontal); The ride starts at the origin $(0, 0)$; The final point on the graph is $(7, 35)$ — $7$ hours, $35$ miles; Intermediate bends at $(1, 5)$, $(2, 15)$, $(4, 20)$ show changing pace on hills; Answer choices: (A) $2$, (B) $2.5$, (C) $4$, (D) $4.5$, (E) $5$ (mph)

Unknowns: Carmen's average speed for the entire ride, in miles per hour (mph)

Understand

Restated: Carmen rides her bike on a hilly highway. A graph plots her cumulative distance (in miles) against time (in hours). The graph starts at $(0, 0)$ and ends at $(7, 35)$, with bends at $(1, 5)$, $(2, 15)$, and $(4, 20)$. What is her average speed for the entire ride in miles per hour?

Givens: The graph plots miles (vertical) against hours (horizontal); The ride starts at the origin $(0, 0)$; The final point on the graph is $(7, 35)$ — $7$ hours, $35$ miles; Intermediate bends at $(1, 5)$, $(2, 15)$, $(4, 20)$ show changing pace on hills; Answer choices: (A) $2$, (B) $2.5$, (C) $4$, (D) $4.5$, (E) $5$ (mph)

Plan

Primary tool: #8 Analyze the Units

Secondary: #3 Draw a Picture

The question asks for a rate in miles per hour, which screams Tool #8 (Analyze the Units): $\text{mph} = \dfrac{\text{miles}}{\text{hours}}$, so we just need one number for total miles and one number for total hours. Tool #3 (Draw a Picture) is already done for us — the graph is the picture — but we have to read it correctly: only the final point $(7, 35)$ matters for the average, because everything in between is just "how she got there". The intermediate bends are a distractor that tempts students to average the segment speeds, which is wrong.

Execute — Answer: E

#3 Draw a Picture 5.G.A.2 Step 1
  • Read the total time from the horizontal axis.
  • The ride ends at the rightmost point of the graph, which sits at $x = 7$ on the hours axis.
  • So the ride lasted $7$ hours.
$$\text{total time} = 7 \text{ hours}$$

💡 Reading the $x$-coordinate of a point in the first quadrant to recover "how much time" is a Grade 5 coordinate-plane skill.

#3 Draw a Picture 5.G.A.2 Step 2
  • Read the total distance from the vertical axis.
  • The same final point sits at $y = 35$ on the miles axis.
  • So Carmen rode a total of $35$ miles.
$$\text{total distance} = 35 \text{ miles}$$

💡 Reading the $y$-coordinate of the endpoint gives the cumulative distance — the same Grade 5 coordinate-plane idea.

#8 Analyze the Units 6.RP.A.3 Step 3

Apply the definition of average speed: total distance divided by total time, with units already matched (miles and hours), so the answer comes out directly in mph.

$$\text{average speed} = \dfrac{35 \text{ mi}}{7 \text{ hr}} = 5 \text{ mph} \;\Rightarrow\; \textbf{(E)}$$

💡 Computing a unit rate (miles per hour) from a total distance and a total time is core Grade 6 rate reasoning.

[1] #3 5.G.A.2 Read the total time from the horizontal axis. The ride ends at the rightmost poi
[2] #3 5.G.A.2 Read the total distance from the vertical axis. The same final point sits at $y
[3] #8 6.RP.A.3 Apply the definition of average speed: total distance divided by total time, wit

Review

Reasonableness: A casual bike ride on a hilly highway at about $5$ mph is reasonable — closer to a brisk walk than a sport ride, which fits "hilly" terrain. Checking the segments: $0 \to 5$ mi in $1$ hr ($5$ mph), $5 \to 15$ mi in $1$ hr ($10$ mph downhill), $15 \to 20$ mi in $2$ hr ($2.5$ mph uphill), $20 \to 35$ mi in $3$ hr ($5$ mph). The total $5 + 10 + 5 + 15 = 35$ mi over $1 + 1 + 2 + 3 = 7$ hr confirms $35 / 7 = 5$ mph.

Alternative: Tool #6 (Guess and Check) by ruling out the trap. Students who naively average the segment speeds get $\tfrac{5 + 10 + 2.5 + 5}{4} = 5.625$ mph, which isn't even a choice — a warning that "average speed" is not the same as the average of speeds. Only $\dfrac{\text{total distance}}{\text{total time}}$ matches a listed choice.

CCSS standards used (min grade 6)

  • 5.G.A.2 Represent real-world and mathematical problems by graphing points in the first quadrant (Reading the endpoint $(7, 35)$ off the distance-vs-time graph to extract the total time ($7$ hours) and total distance ($35$ miles).)
  • 6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Computing the average speed as the unit rate $35 \div 7 = 5$ miles per hour.)

⭐ Average speed always means total distance $\div$ total time — not the average of the speeds along the way!

⭐ Average speed always means total distance $\div$ total time — not the average of the speeds along the way!