Competition · AMC preparation · step 4 of 4
AMC 8 · 2000 · #4
Grade 6 arithmetic
Pick an answer.
AMC 8 2000 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Five graphs, only one correct: this is the textbook setup for Tool #3 (Eliminate Possibilities). Read each graph's 1960 point against the target 5% first — that single quick check kills most of the graphs immediately. Tool #1 (Draw a Diagram) supports the read by treating each gridline as 10% so the points can be located precisely. No arithmetic is needed beyond reading values off the axis.
List the four target points
List the four targets the answer must hit: (1960, 5%), (1970, 8%), (1980, 15%), (1990, 30%).
Grade 5 coordinate-plane reading: each data point is a (x, y) pair where x is the year and y is the percent.
5.G.A.2Draw A DiagramCheck the 1960 value
Easiest filter first: check the 1960 point against 5%. Graph (A) starts at 10% — too high — so (A) is out.
One wrong point is enough. Reading the leftmost point on the line is the quickest check.
A line graph can be ruled out the moment one of its plotted points sits at a different height than that year's percent requires.
▸ Why?
The correct graph has to reproduce the whole data set — all four year-percent pairs at once — so a single point at the wrong height already shows a different data set and cannot be the graph we want.
▸ Why?
The complete correct picture is made of its four points together; change even one point and the parts no longer rebuild the same whole.
▸ Why?
For any one year, deciding whether its point is right means comparing the value the point shows against the single percent that year was given.
▸ Why?
The year's point matches only if it is paired with exactly the one percent listed for that year, so a point pushed to another height is paired with a different value and no longer matches.
▸ Why?
You read the percent a point shows from its height, because the axis is evenly scaled — each gridline step stands for a fixed 10 percent — so a point above or below the target height reads as more or fewer percent.
Check the 1970 value
Now test the 1970 point of 8% (just under the first gridline). Graphs (B) and (D) sit on 10%, so (B) and (D) are out.
8% should sit a little below the first gridline; a point sitting on the gridline (10%) is visibly wrong.
6.SP.B.4Eliminate PossibilitiesCheck the 1980 value
Two left: (C) and (E). At 1980 the value must be 15%. Graph (C) jumps to about 25%, so (C) is out and (E) survives.
15% is the midpoint between gridlines 10 and 20. (C) overshoots; (E) lands right.
6.SP.B.4Eliminate PossibilitiesConfirm the surviving graph
Confirm (E): its four points read 5%, 8%, 15%, 30% — every year matches, so the answer is (E).
After elimination, the survivor must still be verified end-to-end — never skip the confirmation.
5.G.A.2Eliminate PossibilitiesWhen the question asks "which graph?", do not redraw everything — pick one data point, read it off each option, and eliminate. The first mismatched point kills the graph. Here, checking 1960 → 5% and 1970 → 8% already leaves only (E).
- List the four target points
- Check the 1960 value
- Check the 1970 value
- Check the 1980 value
- Confirm the surviving graph
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