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2011 AMC 10B Problem 15

2011 AMC 10B Problem 15AMC 10 Step by Step

Name: Date: September 13, 2026

Functions · difficulty 2 of 5 · about 3 min · position P11-15

Let $@$ denote the "averaged with" operation: $a @ b = (a+b)/2$ . Which of the following distributive laws hold for all numbers $x, y,$ and $z$ ? $\text{I. x @ (y + z) = (x @ y) + (x @ z)}$ $\text{II. x + (y @ z) = (x + y) @ (x + z)}$ $\text{III. x @ (y @ z) = (x @ y) @ (x @ z)}$

A)
$\text{I only}$
B)
$\text{II only}$
C)
$\text{III only}$
D)
$\text{I and III only}$
E)
$\text{II and III only}$

Answer

Key insight

Expand each side by the definition: I fails because its right side carries an extra x/2, while II and III both simplify to matching expressions.

Solution

Write out both sides of each law using $a @ b = \frac{a + b}{2}$ .

I. Left: $x @ (y + z) = \frac{x + y + z}{2}$ . Right: $\frac{x + y}{2} + \frac{x + z}{2} = \frac{2x + y + z}{2}$ . These differ by $\frac{x}{2}$ , so I fails (for example $x = 2$ , $y = z = 0$ gives $1 \ne 2$ ).

II. Left: $x + \frac{y + z}{2} = \frac{2x + y + z}{2}$ . Right: $\frac{(x + y) + (x + z)}{2} = \frac{2x + y + z}{2}$ . Equal, so II holds.

III. Left: $x @ \frac{y + z}{2} = \frac{x + \frac{y+z}{2}}{2} = \frac{2x + y + z}{4}$ . Right: $\frac{x+y}{2} @ \frac{x+z}{2} = \frac{\frac{x+y}{2} + \frac{x+z}{2}}{2} = \frac{2x + y + z}{4}$ . Equal, so III holds.

The answer is $\boxed{\textbf{(E)}\ \text{II and III only}}$ .

Why this works

Averaging is a weighted sum, so it commutes with adding a constant (II) and with averaging again (III), but it is not linear in the sense of distributing over $+$ : averaging $x$ with a sum is not the same as summing two averages, which double-counts $x$ . Translating a made-up operation back to ordinary algebra is always the first step.

Alternative approach

Plug in $x = 2$ , $y = 0$ , $z = 0$ : I gives $1$ versus $2$ (fails); II gives $2$ versus $2$ ; III gives $\frac{1}{2}$ versus $\frac{1}{2}$ . A single numeric test kills I and is consistent with (E); a second test such as $x = 1$ , $y = 2$ , $z = 4$ confirms II and III.

The trap

Assuming an operation that looks like multiplication distributes over addition, accepting I, or making an algebra slip in the four-term expression of III.

Common mistakes

Assuming an operation that looks like multiplication distributes over addition, accepting I, or making an algebra slip in the four-term expression of III.
Testing only with $x = y = z$ , where every law trivially holds, and concluding all three are valid.

Techniques

Set up the equation/formula and compute; no special trick needed · Test small/specific values or special cases to find or verify the answer