AMC 8 · 2005 · #17
Grade 6 rate-ratio
Pick an answer.
AMC 8 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The problem already gives a diagram, so Tool #1 (Draw a Diagram) means using that diagram the right way: add a line from the origin to each dot and compare how steep those lines are. Steeper line = more distance per unit of time = greater average speed. Once each candidate has a line, Tool #3 (Eliminate Possibilities) finishes the job — for a multiple-choice question, we can rank the five candidates by steepness and eliminate anyone whose line is clearly flatter than another's.
Each dot (t, d) is a student's time and distance, so average speed is — the steepness of the line from the origin to the dot.
Grade 6 reads a ratio d : t as a unit rate "distance per unit time." On a graph, that unit rate shows up as how steeply the line climbs.
6.RP.A.3Draw A DiagramDivide d by t for each dot to get its ratio; exact values do not matter, only the ranking. The top ratio is about 3.6.
Grade 6 unit rate: divide distance by time to get "how much distance per one unit of time." The student with the biggest unit rate is the fastest.
6.RP.A.2Draw A DiagramRank the ratios: 3.6 beats every other, so that dot's line from the origin is the steepest and its speed the greatest.
On a multiple-choice problem, once one candidate beats all the others, every other answer is eliminated.
6.RP.A.3Eliminate PossibilitiesRead off the answer: the greatest average speed belongs to Evelyn.
The largest unit rate names the winner.
6.RP.A.3Eliminate PossibilitiesDistance over time is the rate hidden in every dot — and on a distance-time graph, that rate is exactly how steeply a line from the origin climbs to the dot. Steepest line, fastest runner.