AMC 8 · 2001 · #19
Grade 6 rate-ratioProblem
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.
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
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)
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
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.
💡 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.
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 $s_M$.
💡 Doubling the speed must halve the time when the distance is locked in. Same area, double height, so width halves.
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.
💡 These two visual checks — height doubles, width halves — are all we need to score each graph.
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.
💡 Four options break a rule; the survivor is the answer. Eliminating is faster than re-deriving for each graph.
6.RP.A.3 Name the invariant. Both cars cover the same distance $d$. On a speed-time graph 6.RP.A.3 Use the speed condition to pin down the time ratio. Substitute $s_N = 2 s_M$ int 6.RP.A.1 Translate the two conditions into what the solid line for N must look like next 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.1Understand 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.3Use 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.6Use 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).
⭐ 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).