AMC 8 · 2010 · #2

Grade 6 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation

function-evaluationfraction-arithmeticformula-substitution identify-subproblems ↑ Prerequisites: fraction-arithmeticmulti-digit-arithmetic

📏 Short solution 💡 2 insights

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Problem

A custom operation @ is defined on positive integers by a @ b =

\frac{a × b}{a + b}

. Compute 5 @ 10 and match the result to one of the five answer choices.

Pick an answer.

(A)

$\frac{3}{10}$

(B)

(C)

(D)

$\frac{10}{3}$

(E)

AMC 8 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Try an Easier / Specific Case

The rule a @ b = $\frac{a × b}{a + b}$ 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 × 10, denominator 5 + 10, then simplify the fraction. Tool #14 (Sanity Check) confirms the result is in lowest terms and matches a choice.

1STEP 1

Substitute a = 5 and b = 10 into the rule.

5 @ 10 =

\frac{5 × 10}{5 + 10}

2STEP 2

Compute the numerator: 5 × 10 gives 50.

5 × 10 = 50

3STEP 3

Compute the denominator: 5 + 10 gives 15.

5 + 10 = 15

4STEP 4

Assemble the pieces into the fraction $\frac{50}{15}$ .

\frac{5 × 10}{5 + 10}

\frac{50}{15}

5STEP 5

Divide top and bottom by 5 to simplify $\frac{50}{15}$ to $\frac{10}{3}$ , choice (D).

\frac{50}{15}

\frac{50 ÷ 5}{15 ÷ 5}

\frac{10}{3}

→ (D)

Answer

\frac{10}{3}

Quick magnitude check: 5 × 10 = 50 and 5 + 10 = 15, so the answer is roughly

\frac{50}{15}

≈ 3.33. Only choice (D) =

\frac{10}{3}

≈ 3.33 is in that range — (A) is below 1, (B) and (C) are too small, (E) is much too big. The fraction

\frac{10}{3}

is also in lowest terms because gcd(10, 3) = 1, which is what an answer choice should look like.

💡Key takeaway

This AMC 8 problem only needs Grade 6 expression evaluation — plug in the numbers, then simplify the fraction!

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