AMC 10 · 2002 · #2
Grade 6 arithmeticGiven that a, b, and c are non-zero real numbers, define (a,b,c)=ba+cb+ac, find (2,12,9).
Pick an answer.
AMC 10 2002 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 new operation is defined by $(a, b, c) = \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a}$ for non-zero real numbers $a$, $b$, $c$. Find the value of $(2, 12, 9)$.
Givens: The rule $(a, b, c) = \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a}$; $a$, $b$, $c$ are non-zero real numbers; The specific input is $(2, 12, 9)$, so $a = 2$, $b = 12$, $c = 9$; Answer choices: (A) $4$, (B) $5$, (C) $6$, (D) $7$, (E) $8$
Unknowns: The single number that $(2, 12, 9)$ equals
Understand
Restated: A new operation is defined by $(a, b, c) = \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a}$ for non-zero real numbers $a$, $b$, $c$. Find the value of $(2, 12, 9)$.
Givens: The rule $(a, b, c) = \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a}$; $a$, $b$, $c$ are non-zero real numbers; The specific input is $(2, 12, 9)$, so $a = 2$, $b = 12$, $c = 9$; Answer choices: (A) $4$, (B) $5$, (C) $6$, (D) $7$, (E) $8$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #3 Eliminate Possibilities
The strange-looking symbol $(a, b, c)$ is just a recipe: it tells you to build three separate fractions and add them. Tool #7 (Identify Subproblems) handles this cleanly — substitute the numbers to get the three fractions, simplify each one on its own, then add them over a common denominator. Tool #3 (Eliminate Possibilities) is the safety net: since every answer choice is a whole number, a quick size estimate of the three pieces pins the total to a single choice and catches any slip in the fraction work.
Execute — Answer: C
6.EE.A.2 Step 1 Substitute the numbers into the rule
- The definition says the first letter goes on top of the first fraction, the second on top of the second, and the third on top of the third.
- With $a = 2$, $b = 12$, $c = 9$: $\dfrac{a}{b} = \dfrac{2}{12}$, $\dfrac{b}{c} = \dfrac{12}{9}$, and $\dfrac{c}{a} = \dfrac{9}{2}$.
- So $(2, 12, 9) = \dfrac{2}{12} + \dfrac{12}{9} + \dfrac{9}{2}$.
💡 A made-up operation is just a fill-in-the-blank recipe: drop each number into the slot its letter names.
5.NF.B.3 Step 2 Simplify each fraction
- Reduce each fraction first so the numbers stay small.
- $\dfrac{2}{12} = \dfrac{1}{6}$ (divide top and bottom by $2$) and $\dfrac{12}{9} = \dfrac{4}{3}$ (divide top and bottom by $3$).
- The last fraction $\dfrac{9}{2}$ is already in lowest terms.
💡 Shrinking each fraction to lowest terms keeps the denominators small so the final addition is easy.
5.NF.A.1 Step 3 Add over a common denominator
- The denominators are $6$, $3$, and $2$, which all divide into $6$.
- Rewrite each fraction with denominator $6$: $\dfrac{1}{6} = \dfrac{1}{6}$, $\dfrac{4}{3} = \dfrac{8}{6}$, and $\dfrac{9}{2} = \dfrac{27}{6}$.
- Adding the numerators gives $\dfrac{1 + 8 + 27}{6} = \dfrac{36}{6} = 6$.
- So $(2, 12, 9) = 6$, which is choice (C).
💡 Once every fraction shares one denominator, adding them is just adding the top numbers.
6.EE.A.2 The definition says the first letter goes on top of the first fraction, the seco 5.NF.B.3 Reduce each fraction first so the numbers stay small. $\dfrac{2}{12} = \dfrac{1} 5.NF.A.1 The denominators are $6$, $3$, and $2$, which all divide into $6$. Rewrite each Review
Reasonableness: A quick size estimate confirms the total. The three pieces are about $0.17$, $1.33$, and $4.5$, and $0.17 + 1.33 + 4.5 \approx 6$. Because every answer choice is a whole number, the total must land exactly on $6$ — the neighbours $5$ and $7$ are too far from the estimate. This also guards the order trap: if the fractions had been built in the wrong order the sum would not have come out to a clean integer near this estimate.
Alternative: Skip simplifying and put all three fractions straight over the common denominator $36$: $\dfrac{2}{12} = \dfrac{6}{36}$, $\dfrac{12}{9} = \dfrac{48}{36}$, $\dfrac{9}{2} = \dfrac{162}{36}$. Then $\dfrac{6 + 48 + 162}{36} = \dfrac{216}{36} = 6$, the same answer (C).
CCSS standards used (min grade 6)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Reading the definition $(a, b, c) = \tfrac{a}{b} + \tfrac{b}{c} + \tfrac{c}{a}$ and substituting $a = 2$, $b = 12$, $c = 9$ in the correct order.)5.NF.B.3Interpret a fraction as division of the numerator by the denominator (Reducing $\tfrac{2}{12}$ to $\tfrac{1}{6}$ and $\tfrac{12}{9}$ to $\tfrac{4}{3}$.)5.NF.A.1Add and subtract fractions with unlike denominators (Rewriting the three fractions over the common denominator $6$ and adding them to get $\tfrac{36}{6} = 6$.)
⭐ A made-up operation is just a recipe: put each number where its letter goes, then do the ordinary fraction arithmetic carefully.
⭐ A made-up operation is just a recipe: put each number where its letter goes, then do the ordinary fraction arithmetic carefully.
More like this
Same archetype — closest grade level first.