AMC 8 · 2001 · #19

Grade 6 rate-ratio

Archetype Arithmetic Expression Decomposition └ Sub-archetype Successive Multiplicative Steps

rateratio-proportiongraph-reading identify-subproblems ↑ Prerequisites: rateratio-proportion

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Car M traveled at a constant speed for a given time. This is shown by the dashed line. Car N traveled at twice the speed for the same distance. If Car M and Car N's speed and time are shown as solid line, which graph illustrates this?

Pick an answer.

(A)

(speed-time graph) N at twice M's speed, both run for the same time

(B)

(speed-time graph) N at twice M's speed, but N runs longer than M

(C)

(speed-time graph) N's speed is lower than M's, same time

(D)

(speed-time graph) N at twice M's speed, N runs for half the time M does

(E)

(speed-time graph) N's speed is lower than M's, same time

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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: Car M travels at a constant speed for some time, shown as a dashed line on a speed-vs-time graph. Car N covers the same distance as M, but at twice M's speed. On each option's speed-time graph, M is the dashed line and N is the solid line. Which graph correctly shows N?

Givens: The graphs plot speed on the vertical axis and time on the horizontal axis; Car M's speed is shown by the dashed horizontal segment; its length on the time axis is M's travel time; Car N travels at twice M's speed: $s_N = 2 s_M$; Cars M and N cover the same distance; Answer choices: five speed-time graphs labeled (A)-(E)

Unknowns: Which of the five graphs (A)-(E) is consistent with both conditions on N

Understand

Plan

Primary tool: #11 Find an Invariant

Secondary: #3 Eliminate Possibilities

The unchanging quantity here is the distance — Tool #11 (Find an Invariant). On a speed-time graph, distance shows up as the area of the rectangle under each car's segment, so the two rectangles must have equal area. Once you fix the speed ratio $s_N = 2 s_M$, the equal-area condition forces $t_N = \tfrac{1}{2} t_M$. With both N-conditions (height doubles, width halves) pinned down, Tool #3 (Eliminate Possibilities) sweeps the five graphs: any graph that violates either the height or the width condition is out, and only one survives.

Execute — Answer: D

#11 Find an Invariant 6.RP.A.3 Step 1

Name the invariant.
Both cars cover the same distance $d$.
On a speed-time graph, a constant-speed segment of speed $s$ lasting time $t$ encloses a rectangle of area $s \times t$, and that area is the distance traveled.

$$d = s_M \cdot t_M = s_N \cdot t_N$$

💡 Distance is the hidden quantity that both cars share. Reading it as the area under a speed-time segment turns the problem into a rectangle comparison.

#11 Find an Invariant 6.RP.A.3 Step 2

Use the speed condition to pin down the time ratio.
Substitute $s_N = 2 s_M$ into the equal-distance equation and cancel $s_M$.

$$s_M \cdot t_M = (2 s_M) \cdot t_N \;\Rightarrow\; t_N = \dfrac{t_M}{2}$$

💡 Doubling the speed must halve the time when the distance is locked in. Same area, double height, so width halves.

#3 Eliminate Possibilities 6.RP.A.1 Step 3

Translate the two conditions into what the solid line for N must look like next to the dashed line for M: twice as tall, half as wide.

$$\text{height}(N) = 2 \cdot \text{height}(M), \quad \text{width}(N) = \tfrac{1}{2} \cdot \text{width}(M)$$

💡 These two visual checks — height doubles, width halves — are all we need to score each graph.

#3 Eliminate Possibilities 6.RP.A.3 Step 4

Apply the checks to each option and eliminate.
(A) height doubles, but widths are equal (time not halved) — out.
(B) height doubles, but N is wider than M (N runs longer) — out.
(C) N sits below M (speed lower, not doubled) — out.
(D) N is twice as tall and half as wide — passes both checks.
(E) N sits below M — out.

$$\text{(A) fail width}, \;\text{(B) fail width}, \;\text{(C) fail height}, \;\text{(D) pass}, \;\text{(E) fail height} \;\Rightarrow\; \textbf{(D)}$$

💡 Four options break a rule; the survivor is the answer. Eliminating is faster than re-deriving for each graph.

[1] #11 6.RP.A.3 Name the invariant. Both cars cover the same distance $d$. On a speed-time graph

[2] #11 6.RP.A.3 Use the speed condition to pin down the time ratio. Substitute $s_N = 2 s_M$ int

[3] #3 6.RP.A.1 Translate the two conditions into what the solid line for N must look like next

[4] #3 6.RP.A.3 Apply the checks to each option and eliminate. (A) height doubles, but widths ar

Review

Reasonableness: Plug in numbers to confirm. Suppose M drives $60$ mph for $2$ hours: distance $= 60 \times 2 = 120$ miles. Then N drives at $120$ mph and must cover the same $120$ miles, so N's time is $\tfrac{120}{120} = 1$ hour. On the graph, N's line should be twice as high as M's ($120$ vs $60$) and half as long along the time axis ($1$ vs $2$). Graph (D) is the only one that shows both, so the answer is consistent.

Alternative: Tool #1 (Draw a Diagram): sketch two rectangles with the same area but one twice as tall as the other. Anyone who has played with $2 \times 6$ vs $4 \times 3$ tiles knows the taller rectangle must be the narrower one. Then match that picture — tall-and-narrow N next to short-and-wide M — to the graph that shows it: only (D).

CCSS standards used (min grade 6)

6.RP.A.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship (Reading $s_N = 2 s_M$ as a $2{:}1$ height ratio on the speed axis, and translating the equal-distance condition into the $1{:}2$ width ratio on the time axis.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Using $\text{distance} = \text{speed} \times \text{time}$ to derive $t_N = \tfrac{1}{2} t_M$ from $s_N = 2 s_M$ when the two distances are equal, then checking each graph against this rate relationship.)
6.EE.B.6 Use variables to represent numbers and write expressions when solving real-world problems (Naming the speeds and times $s_M, s_N, t_M, t_N$ and the shared distance $d$ so the same-distance condition can be written as one equation and solved for the time ratio.)

⭐ On a speed-time graph, distance is the area of the rectangle under each segment. Same distance with double the speed means half the time — twice as tall, half as wide. That single visual check picks (D).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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