AMC 8 · 2002 · #6
Grade 6 rate-ratioPick an answer.
AMC 8 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The story has two clearly separate phases, so Tool #7 (Identify Subproblems) splits the picture into Phase 1 (filling, before the bath is full) and Phase 2 (overflowing, after the bath is full). In each phase the net rate is constant, so each phase is a straight line; the slope just changes at the moment the bath fills up. Tool #1 (Draw a Diagram) then sketches the expected shape — a positive-slope ray from the origin followed by a horizontal segment — and we match that two-piece silhouette against the five options.
Filling phase: inflow 20 ml/min minus drain 18 ml/min gives a net rise of 2 ml/min.
Combining two rates by subtraction is the Grade 6 "unit rate" move: the net effect is a single constant rate.
6.RP.A.3Identify SubproblemsA steady 2 ml/min rise adds equal volume each minute, so the graph is a straight line with positive slope from the origin.
Grade 6 "two-variable equations": a constant rate → a linear relationship → a straight-line graph through (0,0).
6.EE.C.9Draw A DiagramOverflow phase: once full, the spare 2 ml/min spills over the edge, so the volume inside stays pinned at capacity.
When inflow and total outflow balance, the volume stops changing — on the graph, that is a flat horizontal segment.
6.RP.A.3Identify SubproblemsStitch the phases: a rising ray, then a flat top. Only graph A shows a positive slope followed by a horizontal line.
Choosing a graph is a Grade 6 "match the story to the picture" task once the algebra of each phase is in hand.
6.EE.C.9Draw A DiagramSplit the story into two phases — filling (constant positive rate, so a straight line going up) and overflowing (volume stuck at capacity, so a horizontal line) — and only graph (A) shows that two-piece shape.