Competition · AMC preparation · step 4 of 4
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 = 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.
Name the invariant
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 BackwardsCompare the two times
With 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.
When two cars cover the same distance and one drives twice as fast, that faster car spends exactly half the time on the road.
▸ Why?
Distance is speed multiplied by travel time, and since both cars cover the same distance, the faster car's speed-times-time must equal the slower car's speed-times-time.
▸ Why?
At a steady speed every equal unit of time adds the same fixed distance, so the whole trip is that one speed counted once for each unit of time — speed times time.
▸ Why?
Both cars travel the very same trip length, so their two distance amounts name one and the same number and can be set equal to each other.
▸ Why?
Substitute the doubled speed into that equality: the slower speed times the slower time equals twice the slower speed times the faster time, so the shared slower speed sits on both sides and can be cancelled.
▸ Why?
Dividing both sides by the shared slower speed undoes that multiplication and clears it away, leaving the slower time equal to two of the faster times.
▸ Why?
With the slower time equal to two of the faster times, the faster time is that duration cut into two equal pieces — half the slower time.
▸ Why?
Halving reverses doubling, so if one time is twice another, that other time is half of it.
Describe the graph shape
So 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 PossibilitiesEliminate the wrong graphs
Scoring 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).
- Name the invariant
- Compare the two times
- Describe the graph shape
- Eliminate the wrong graphs
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