AMC 8 · 2001 · #12
Grade 6 arithmeticalgebraProblem
If a⊗b=a−ba+b, then (6⊗4)⊗3=
Pick an answer.
AMC 8 2001 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A brand-new operation is defined by the rule $a \otimes b = \dfrac{a+b}{a-b}$. It takes two numbers in and returns one number out. The task is to evaluate $(6 \otimes 4) \otimes 3$, computing the inner $\otimes$ first because of the parentheses.
Givens: Rule: $a \otimes b = \dfrac{a+b}{a-b}$; Expression to evaluate: $(6 \otimes 4) \otimes 3$; Answer choices: (A) $4$, (B) $13$, (C) $15$, (D) $30$, (E) $72$
Unknowns: The numerical value of $(6 \otimes 4) \otimes 3$
Understand
Restated: A brand-new operation is defined by the rule $a \otimes b = \dfrac{a+b}{a-b}$. It takes two numbers in and returns one number out. The task is to evaluate $(6 \otimes 4) \otimes 3$, computing the inner $\otimes$ first because of the parentheses.
Givens: Rule: $a \otimes b = \dfrac{a+b}{a-b}$; Expression to evaluate: $(6 \otimes 4) \otimes 3$; Answer choices: (A) $4$, (B) $13$, (C) $15$, (D) $30$, (E) $72$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #3 Eliminate Possibilities
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 $6 \otimes 4$, then plug that result into the outer $\otimes 3$. Each $\otimes$ 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.
Execute — Answer: A
5.OA.A.1 Step 1 - Split the expression at the parentheses.
- The whole problem becomes two smaller problems: (i) find $6 \otimes 4$, (ii) take that number and apply $\otimes 3$ to it.
- Doing them in this order respects the parentheses.
💡 Reading the parentheses first is the Grade 5 "use parentheses in numerical expressions" rule, and it gives us our two subproblems.
6.EE.A.2 Step 2 - Apply the $\otimes$ recipe with $a = 6$ and $b = 4$.
- The rule says $a \otimes b = \dfrac{a+b}{a-b}$, so substitute the numbers into the letters.
💡 Plugging numbers into letters $a$ and $b$ of a defined formula is exactly Grade 6 "evaluate expressions where letters stand for numbers".
6.NS.B.2 Step 3 Carry out the arithmetic on top and bottom, then divide.
💡 The numerator and denominator are simple whole-number sums and differences, and $10 \div 2 = 5$ is Grade 6 whole-number division.
6.EE.A.2 Step 4 - Now substitute the inner result back: the expression has become $5 \otimes 3$.
- Apply the $\otimes$ recipe again with $a = 5$ and $b = 3$.
💡 We reuse the same Grade 6 substitution move on the outer $\otimes$, and the arithmetic finishes the work.
5.OA.A.1 Step 5 - Match the computed value $4$ against the five answer choices: $4, 13, 15, 30, 72$.
- Only choice (A) hits it; the others can be eliminated.
- (Notice the traps: $15$ in choice (C) comes from doing $(6+4) + (3+2)$ or similar add-everything slips; $72$ in choice (E) is what you get if you accidentally multiply $6 \cdot 4 \cdot 3$, treating $\otimes$ as multiplication.)
💡 Lining the result up against the five choices is the AMC multiple-choice habit and catches the "$\otimes$ is just times" trap labelled (E).
5.OA.A.1 Split the expression at the parentheses. The whole problem becomes two smaller p 6.EE.A.2 Apply the $\otimes$ recipe with $a = 6$ and $b = 4$. The rule says $a \otimes b 6.NS.B.2 Carry out the arithmetic on top and bottom, then divide. 6.EE.A.2 Now substitute the inner result back: the expression has become $5 \otimes 3$. A 5.OA.A.1 Match the computed value $4$ against the five answer choices: $4, 13, 15, 30, 72 Review
Reasonableness: Sanity check by re-deriving the two intermediate values independently. The inner sums $6+4 = 10$ and $6-4 = 2$ give $10/2 = 5$, clean and whole. The outer sums $5+3 = 8$ and $5-3 = 2$ give $8/2 = 4$, 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 $5 \cdot 6$, (E) $72$ is $6 \cdot 4 \cdot 3$ — each one mirrors a familiar slip we sidestepped by reading the recipe carefully.
Alternative: Tool #9 (Solve an Easier Related Problem): rename the symbol with a familiar function. Let $f(a, b) = \dfrac{a+b}{a-b}$. The problem becomes $f(f(6, 4),\, 3)$, a plain composition of one function with itself. Computing $f(6, 4) = 5$ and then $f(5, 3) = 4$ gives the same answer (A). This reframing makes the structure feel like the everyday "feed the output of one machine into the same machine again" idea, which is what defined operations really are.
CCSS standards used (min grade 6)
5.OA.A.1Use parentheses, brackets, or braces in numerical expressions and evaluate (Respecting the parentheses to evaluate the inner $6 \otimes 4$ before the outer $\otimes 3$, and then matching the final value to the multiple-choice list.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Substituting $a = 6,\, b = 4$ into the rule $\dfrac{a+b}{a-b}$, and then $a = 5,\, b = 3$ into the same rule — exactly the Grade 6 "evaluate an expression at given variable values" move.)6.NS.B.2Fluently divide multi-digit numbers using the standard algorithm (Carrying out the whole-number divisions $10 \div 2 = 5$ and $8 \div 2 = 4$ that finish each application of the operation.)
⭐ 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!
⭐ 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!
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