AMC 8 · 2010 · #2
Grade 6 arithmeticProblem
If a@b=a+ba×b for a,b positive integers, then what is 5@10?
Pick an answer.
AMC 8 2010 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 custom operation $@$ is defined on positive integers by $a @ b = \dfrac{a \times b}{a + b}$. Compute $5 @ 10$ and match the result to one of the five answer choices.
Givens: The rule $a @ b = \dfrac{a \times b}{a + b}$ for positive integers $a$ and $b$; The specific input pair $a = 5$, $b = 10$; Answer choices: (A) $\tfrac{3}{10}$, (B) $1$, (C) $2$, (D) $\tfrac{10}{3}$, (E) $50$
Unknowns: The value of $5 @ 10$ expressed in lowest terms
Understand
Restated: A custom operation $@$ is defined on positive integers by $a @ b = \dfrac{a \times b}{a + b}$. Compute $5 @ 10$ and match the result to one of the five answer choices.
Givens: The rule $a @ b = \dfrac{a \times b}{a + b}$ for positive integers $a$ and $b$; The specific input pair $a = 5$, $b = 10$; Answer choices: (A) $\tfrac{3}{10}$, (B) $1$, (C) $2$, (D) $\tfrac{10}{3}$, (E) $50$
Plan
Primary tool: #9 Try an Easier / Specific Case
Secondary: #7 Identify Subproblems, #14 Sanity Check
The rule $a @ b = \dfrac{a \times b}{a + b}$ is written with variables, but the question only asks about one specific case: $a = 5$, $b = 10$. Tool #9 (Try a Specific Case) says: don't reason about the general formula — just substitute the given values and compute. Tool #7 (Identify Subproblems) cleanly splits the work into three small pieces — numerator $5 \times 10$, denominator $5 + 10$, then simplify the fraction. Tool #14 (Sanity Check) confirms the result is in lowest terms and matches a choice.
Execute — Answer: D
6.EE.A.2 Step 1 Substitute $a = 5$ and $b = 10$ into the rule, replacing every $a$ with $5$ and every $b$ with $10$.
💡 Plugging concrete numbers into a formula written with letters is the Grade 6 "evaluate expressions at specific values of their variables" skill.
3.OA.C.7 Step 2 Compute the numerator: a single-digit times a multiple of $10$.
💡 Multiplying within $100$ is a Grade 3 fluency fact — multiplying by $10$ just appends a zero.
1.OA.C.6 Step 3 Compute the denominator: simple two-number addition.
💡 Adding within $20$ is a Grade 1 fluency fact.
6.EE.A.2 Step 4 Assemble the fraction from the numerator and denominator computed above.
💡 Putting the pieces back together completes the Tool #7 subproblem strategy.
4.NF.A.1 Step 5 Simplify $\dfrac{50}{15}$ by dividing the numerator and denominator by their common factor $5$.
💡 Dividing top and bottom by the same number to get an equivalent fraction is the Grade 4 "equivalent fractions" standard. The result $\tfrac{10}{3}$ matches choice (D) exactly.
6.EE.A.2 Substitute $a = 5$ and $b = 10$ into the rule, replacing every $a$ with $5$ and 3.OA.C.7 Compute the numerator: a single-digit times a multiple of $10$. 1.OA.C.6 Compute the denominator: simple two-number addition. 6.EE.A.2 Assemble the fraction from the numerator and denominator computed above. 4.NF.A.1 Simplify $\dfrac{50}{15}$ by dividing the numerator and denominator by their com Review
Reasonableness: Quick magnitude check: $5 \times 10 = 50$ and $5 + 10 = 15$, so the answer is roughly $\tfrac{50}{15} \approx 3.33$. Only choice (D) $= \tfrac{10}{3} \approx 3.33$ is in that range — (A) is below $1$, (B) and (C) are too small, (E) is much too big. The fraction $\tfrac{10}{3}$ is also in lowest terms because $\gcd(10, 3) = 1$, which is what an answer choice should look like.
Alternative: Tool #6 (Guess and Check) on the choices: convert each choice to a decimal and compare to $\tfrac{50}{15} = 3.333\ldots$. (A) $0.3$, (B) $1$, (C) $2$, (D) $3.333\ldots$, (E) $50$ — only (D) matches. This independent decimal check confirms the simplified-fraction answer without re-doing the algebra.
CCSS standards used (min grade 6)
1.OA.C.6Add and subtract within $20$ (Computing the denominator $5 + 10 = 15$.)3.OA.C.7Fluently multiply and divide within $100$ (Computing the numerator $5 \times 10 = 50$.)4.NF.A.1Explain why a fraction is equivalent to another by using visual fraction models, with attention to how the number and size of the parts differ (Reducing $\tfrac{50}{15}$ to its lowest-terms equivalent $\tfrac{10}{3}$ by dividing numerator and denominator by the common factor $5$.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Evaluating the variable expression $\dfrac{a \times b}{a + b}$ at the specific values $a = 5$, $b = 10$ — the core move of the problem.)
⭐ This AMC 8 problem only needs Grade 6 expression evaluation — plug in the numbers, then simplify the fraction!
⭐ This AMC 8 problem only needs Grade 6 expression evaluation — plug in the numbers, then simplify the fraction!
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