AMC 10 · 2016 · #1
Grade 8 arithmeticPick an answer.
AMC 10 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
This is one compound calculation stacked on top of another, so Tool #7 (Identify Subproblems) is the natural fit: first figure out what a⁻¹ is, then build the numerator, then divide by a. Each piece is a small, safe step. Tool #4 (Introduce a Variable) keeps the meaning of a⁻¹ front and center — a negative exponent is just 1/a — so the symbol does not become a trap.
A negative exponent flips the base into its reciprocal, so a⁻¹ = 1/a, and the reciprocal of 1/2 is 2.
A negative-one exponent just flips the number over, and flipping 1/2 gives 2.
8.EE.A.1Identify SubproblemsSubstitute a⁻¹ = 2 into the top 2a⁻¹ + a⁻¹/2: that is 2·2 + 2/2 = 4 + 1 = 5.
Once a⁻¹ is a plain number, the numerator is ordinary arithmetic you can do left to right.
6.EE.A.2Identify SubproblemsDivide the top by a: 5 ÷ (1/2) = 5·2 = 10, which is choice (D).
Dividing by 1/2 asks "how many halves fit in 5?" — and that is 10.
6.NS.A.1Identify SubproblemsTurn the negative exponent into a reciprocal first, then it is just plug-in-and-compute that you already know.