AMC 8 · 2008 · #13
Grade 6 arithmeticPick an answer.
AMC 8 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
We are not asked for each box's weight — only the total. Tool #11 (Find an Invariant) says to look for a quantity that stays the same no matter which pair you pick. The key invariant here is the count: when you add all three pair weights, every box gets counted exactly twice. Tool #4 (Introduce a Variable) lets us name the box weights x, y, z so we can write that observation as a clean equation. Then the total x + y + z falls out by dividing by 2 — no need to solve for x, y, z separately.
Name the three box weights x, y, z; each pair weighing then gives one equation.
Using letters for unknown weights is the Grade 6 move: "use variables to represent numbers and write expressions when solving a real-world problem."
6.EE.B.6Use Matrix LogicAdd all three equations; on the left each of x, y, z appears exactly twice.
The "each box counted twice" pattern is the invariant. It does not depend on which pair weights you got — only on the fact that every pair was weighed.
6.EE.A.3Work BackwardsCombine like terms: the left becomes 2(x + y + z), and the right adds to 374.
Factoring out the 2 makes the combined weight x + y + z visible as a single block.
6.EE.A.3Work BackwardsDivide both sides by 2: the combined weight is x + y + z = 187 pounds.
One-step equation: divide both sides by the same nonzero number. The answer is the combined weight in pounds.
6.EE.B.7Use Matrix LogicWhen every pair gets weighed, add all the pair totals — each box was counted twice, so half the sum is the answer.
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