Competition · AMC preparation · step 4 of 4
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.
Find the net filling rate
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 SubproblemsTurn the rate into a line
A 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 DiagramSee what overflow does
Overflow 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 SubproblemsJoin the two phases
Stitch 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.
The volume-vs-time graph climbs as a straight line from the origin while the bath fills, then holds flat as a horizontal line once the bath is full.
▸ Why?
While the bath is filling it starts empty and gains the same fixed amount of water every minute, so its volume plots as a straight line rising from the origin.
▸ Why?
Each minute 20 ml flows in and 18 ml drains out; the drain removes part of what just came in, leaving 20 - 18 = 2 ml behind every minute.
▸ Why?
Adding the same 2 ml minute after minute makes the volume after t minutes equal to t groups of 2 ml, so the plotted points step up by equal amounts and fall on one straight line.
▸ Why?
Once the bath is full its volume stops changing, so the graph levels off into a horizontal line.
▸ Why?
At full, the 20 ml arriving each minute leaves again in two parts — 18 ml down the drain and 2 ml over the rim — and those parts together make up the whole 20 ml.
▸ Why?
Since exactly as much water leaves as enters each minute, the net change is zero, and adding zero leaves the volume exactly where it already was.
Split 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.
- Find the net filling rate
- Turn the rate into a line
- See what overflow does
- Join the two phases
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