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2011 AMC 10A Problem 3

2011 AMC 10A Problem 3AMC 10 Step by Step

Name: Date: September 13, 2026

Functions · difficulty 1 of 5 · about 2 min · position P1-10

Suppose $[a\ b]$ denotes the average of $a$ and $b$ , and $\{a\ b\ c\}$ denotes the average of $a, b$ , and $c$ . What is $\{\{1\ 1\ 0\}\ [0\ 1]\ 0\}$ ?

A)
$\frac{2}{9}$
B)
$\frac{5}{18}$
C)
$\frac{1}{3}$
D)
$\frac{7}{18}$
E)
$\frac{2}{3}$

Answer

Key insight

Evaluate the innermost brackets first: {1 1 0} = 2/3 and [0 1] = 1/2, then average those two with 0.

Solution

Work from the inside out.

The inner triple: $\{1\ 1\ 0\} = \dfrac{1 + 1 + 0}{3} = \dfrac{2}{3}$ .

The inner pair: $[0\ 1] = \dfrac{0 + 1}{2} = \dfrac{1}{2}$ .

The outer expression is the average of $\frac{2}{3}$ , $\frac{1}{2}$ and $0$ :
$\frac{\frac{2}{3} + \frac{1}{2} + 0}{3} = \frac{\frac{7}{6}}{3} = \frac{7}{18}.$

The answer is $\boxed{\textbf{(D)}\ \frac{7}{18}}$ .

Why this works

A newly defined symbol is just a function with a stated rule; nested symbols are evaluated like nested parentheses, innermost first. Keeping the count in each average straight (two entries versus three) is the whole task.

The trap

Forgetting to divide by 3 in the outer average (adding 2/3 + 1/2 = 7/6 and stopping), or averaging only the two nonzero entries.

Common mistakes

Forgetting to divide by 3 in the outer average (adding 2/3 + 1/2 = 7/6 and stopping), or averaging only the two nonzero entries.
Reading $\{1\ 1\ 0\}$ as the average of two numbers, or dropping the trailing $0$ from the outer average and getting $\frac{7}{12}$ .

Techniques

Set up the equation/formula and compute; no special trick needed