AMC 8 · 2011 · #13
Grade 6 geometry-2d
Pick an answer.
AMC 8 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The figure is the whole point: two side-15 squares slide horizontally until their union is a 15 × 25 rectangle. Tool #5 (Draw a Picture) — or just reading the given Asymptote diagram carefully — pins down that the overlap is a rectangle whose height equals the square's side (15) and whose width w is what we have to find. Tool #11 (Look for Structure) gives the clean inclusion-exclusion idea along the long edge: the two squares together span length 15 + 15 = 30, but they actually cover only 25, so the missing 5 is exactly the doubly-covered overlap width.
Multiply the rectangle's dimensions 15 × 25 to get its area 375.
Area of a rectangle as length times width is a Grade 3 standard.
3.MD.C.7Look For A PatternBoth squares fill the full height 15, so the overlap is a rectangle 15 tall and w wide.
Drawing or reading the picture shows the overlap is a vertical strip with the same height as the squares.
3.MD.C.7Look For A PatternTwo squares side by side span 15+15=30, but AQRD is only 25 long, so the overlap width is 5.
When two segments of total length 30 fit into a span of 25, the surplus 5 is exactly the overlap — Grade 4 multi-step reasoning.
4.OA.A.3Work BackwardsMultiply the overlap's width 5 by its height 15 to get area 75.
Same rectangle-area idea as Step 1, applied to the smaller rectangle.
3.MD.C.7Look For A PatternDivide 75 by 375 to get , which is 20% — choice (C).
Writing a part-to-whole ratio as a percent is Grade 6 ratio reasoning.
6.RP.A.3Work BackwardsThis AMC 8 problem only needs Grade 6 ratio reasoning: find the overlap width, then turn the part-to-whole ratio into a percent.
More like this
Same archetype — closest grade level first.
