AMC 10 · 2015 · #2
Grade 7 geometry-2dPick an answer.
AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Two unknowns (how many triangles, how many squares) are linked by two facts (a tile count and an edge count), so Tool #4 (Introduce a Variable) names one unknown and writes the other in terms of it. Tool #13 (Convert to Algebra) turns the two sentences — '25 tiles' and '84 edges' — into a single equation that can be solved. Tool #14 (Extreme Principle) gives a fast check: imagine every tile is a triangle, then see how many edges are missing.
Name the unknown
Let s be the squares; the rest, 25 - s, are triangles.
If two amounts must add to a fixed total, naming one of them automatically gives the other by subtraction.
6.EE.B.6Use Matrix LogicWrite the edge count as an expression
Squares contribute 4s edges, triangles contribute 3(25 - s) edges; add them for the total.
Count each shape's edges separately, then add — the total is just the parts combined.
6.EE.A.2Convert To AlgebraSet up and simplify the equation
Set 4s + 3(25 - s) = 84 and simplify to s + 75 = 84.
Multiplying out and merging like terms shrinks a messy expression down to one clean unknown.
7.EE.A.1Use Matrix LogicSolve for the number of squares
Subtract 75 from both sides: s = 9 square tiles — choice (D).
Once the equation reads 'something plus 75 equals 84', the something is just 84 - 75.
7.EE.B.4Use Matrix LogicWhen two amounts add to a fixed total, name just one with a letter, write the rest in terms of it, and one equation finishes the job.
- Name the unknown
- Write the edge count as an expression
- Set up and simplify the equation
- Solve for the number of squares