AMC 8 · 2001 · #12

Grade 6 arithmeticalgebra

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

function-evaluationfraction-arithmeticorder-of-operationsformula-substitution identify-subproblems ↑ Prerequisites: order-of-operationsfraction-arithmetic

📏 Short solution 💡 2 insights

Problem

A brand-new operation is defined by the rule a ⊗ b =

\frac{a+b}{a-b}

. It takes two numbers in and returns one number out. The task is to evaluate (6 ⊗ 4) ⊗ 3, computing the inner ⊗ first because of the parentheses.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2001 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The expression has a clean inside-then-outside shape thanks to the parentheses, so Tool #7 (Identify Subproblems) is the natural fit: first compute the inner 6 ⊗ 4, then plug that result into the outer ⊗ 3. Each ⊗ is just a tiny recipe — read the letters a and b, line them up with the actual numbers on either side, and substitute. Tool #3 (Eliminate Possibilities) is the AMC multiple-choice safety net: after computing, check that the value matches one of the five offered choices and that the trap distractors can be ruled out.

1STEP 1

Read the parentheses first: split (6 ⊗ 4) ⊗ 3 into an inner job and an outer job.

(6 ⊗ 4) ⊗ 3 = [ 6 ⊗ 4 ] ⊗ 3

2STEP 2

Apply the ⊗ rule to the inner pair with a = 6, b = 4: 6 ⊗ 4 = $\frac{6+4}{6-4}$ .

6 ⊗ 4 =

\frac{6+4}{6-4}

3STEP 3

Do the arithmetic: $\frac{6+4}{6-4}$ = $\frac{10}{2}$ = 5, so the inner ⊗ equals 5.

\frac{6+4}{6-4}

\frac{10}{2}

= 5

4STEP 4

Substitute back — now 5 ⊗ 3 = $\frac{5+3}{5-3}$ = $\frac{8}{2}$ = 4.

5 ⊗ 3 =

\frac{5+3}{5-3}

\frac{8}{2}

= 4

5STEP 5

Only choice (A) equals 4; (E) 72 is the ⊗-as-multiply trap 6·4·3, and (C) 15 is just adding everything.

4 → (A)

Answer

Sanity check by re-deriving the two intermediate values independently. The inner sums 6+4 = 10 and 6-4 = 2 give

\frac{10}{2}

= 5, clean and whole. The outer sums 5+3 = 8 and 5-3 = 2 give

\frac{8}{2}

= 4, again clean and whole. Both divisions land on whole numbers, which is a strong signal we substituted correctly — a mis-step in the rule almost always leaves a fraction or a negative. Also, the trap distractors all map to specific mistakes: (C) 15 is the sum of everything, (D) 30 is 5 · 6, (E) 72 is 6 · 4 · 3 — each one mirrors a familiar slip we sidestepped by reading the recipe carefully.

💡Key takeaway

This AMC 8 problem only needs Grade 6 expression evaluation — plug numbers into the letters of a formula, then do it once more — that you already know!

More like this