AMC 8 · 2001 · #19
Grade 6 rate-ratioPick an answer.
AMC 8 2001 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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 = 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.
Both cars cover the same distance d, and on a speed-time graph that distance is the area of the rectangle under each constant-speed segment.
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.3Work BackwardsWith the distance fixed, doubling the speed forces N to take half the time M does.
Doubling the speed must halve the time when the distance is locked in. Same area, double height, so width halves.
6.RP.A.3Work BackwardsSo next to the dashed M, the solid line for N must be twice as tall and half as wide.
These two visual checks — height doubles, width halves — are all we need to score each graph.
6.RP.A.1Eliminate PossibilitiesScoring each option, only one is both twice as tall and half as wide; A/B fail the width and C/E fail the height → (D).
Four options break a rule; the survivor is the answer. Eliminating is faster than re-deriving for each graph.
6.RP.A.3Eliminate PossibilitiesOn 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).