AMC 10 · 2011 · #3
Grade 6 arithmeticSuppose [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}?
Pick an answer.
AMC 10 2011 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: Two made-up symbols are defined: $[a\ b]$ means the average of $a$ and $b$, and $\{a\ b\ c\}$ means the average of $a$, $b$, and $c$. Evaluate the nested expression $\{\{1\ 1\ 0\}\ [0\ 1]\ 0\}$ and match it to a choice.
Givens: $[a\ b] = \dfrac{a+b}{2}$ (average of two numbers); $\{a\ b\ c\} = \dfrac{a+b+c}{3}$ (average of three numbers); The expression to evaluate: $\{\{1\ 1\ 0\}\ [0\ 1]\ 0\}$; Answer choices: (A) $\frac{2}{9}$, (B) $\frac{5}{18}$, (C) $\frac{1}{3}$, (D) $\frac{7}{18}$, (E) $\frac{2}{3}$
Unknowns: The single fraction the whole nested expression equals
Understand
Restated: Two made-up symbols are defined: $[a\ b]$ means the average of $a$ and $b$, and $\{a\ b\ c\}$ means the average of $a$, $b$, and $c$. Evaluate the nested expression $\{\{1\ 1\ 0\}\ [0\ 1]\ 0\}$ and match it to a choice.
Givens: $[a\ b] = \dfrac{a+b}{2}$ (average of two numbers); $\{a\ b\ c\} = \dfrac{a+b+c}{3}$ (average of three numbers); The expression to evaluate: $\{\{1\ 1\ 0\}\ [0\ 1]\ 0\}$; Answer choices: (A) $\frac{2}{9}$, (B) $\frac{5}{18}$, (C) $\frac{1}{3}$, (D) $\frac{7}{18}$, (E) $\frac{2}{3}$
Plan
Primary tool: #7 Identify Subproblems
The scary part is the notation, not the arithmetic. Tool #7 says: don't try to do it all at once — break the compound expression into small, independent pieces. Here the two inner brackets $\{1\ 1\ 0\}$ and $[0\ 1]$ are self-contained mini-problems. Solve each one to a plain fraction first, drop those two numbers into the outer $\{\ \ \}$, and the last step is just one more average. Working inside-out keeps every stage a simple 'add and divide'.
Execute — Answer: D
6.SP.B.5 Step 1 Average the inner three
- Start with the innermost group $\{1\ 1\ 0\}$.
- It means the average of $1$, $1$, and $0$: add them and divide by $3$.
💡 An average of three numbers is just their total shared equally three ways.
6.SP.B.5 Step 2 Average the inner pair
Next handle $[0\ 1]$, the average of the two numbers $0$ and $1$: add them and divide by $2$.
💡 The average of two numbers sits exactly halfway between them.
5.NF.A.1 Step 3 Add the outer three
- Now the outer expression is $\{\tfrac{2}{3}\ \tfrac{1}{2}\ 0\}$.
- First add its three numbers.
- Rewrite $\tfrac{2}{3}$ and $\tfrac{1}{2}$ with the common denominator $6$.
💡 Fractions can only be added once their pieces are the same size, so switch both to sixths.
6.NS.A.1 Step 4 Divide the total by three
- The outer average divides that sum by $3$.
- Dividing by $3$ is the same as multiplying by $\tfrac{1}{3}$, which multiplies the denominator by $3$.
💡 Splitting a fraction into three equal parts triples how many pieces the whole is cut into, so the denominator grows from $6$ to $18$.
6.SP.B.5 Start with the innermost group $\{1\ 1\ 0\}$. It means the average of $1$, $1$, 6.SP.B.5 Next handle $[0\ 1]$, the average of the two numbers $0$ and $1$: add them and d 5.NF.A.1 Now the outer expression is $\{\tfrac{2}{3}\ \tfrac{1}{2}\ 0\}$. First add its t 6.NS.A.1 The outer average divides that sum by $3$. Dividing by $3$ is the same as multip Review
Reasonableness: The three outer numbers are $\tfrac{2}{3}\approx 0.67$, $\tfrac{1}{2}=0.5$, and $0$, so their average must land between $0$ and $\tfrac{2}{3}$ and near the middle value. $\tfrac{7}{18}\approx 0.39$ fits that window perfectly. Choice (E) $\tfrac{2}{3}$ is too big (that is a single input, not the average), and (A) $\tfrac{2}{9}\approx 0.22$ is too small, so (D) is the consistent answer.
Alternative: Convert everything to eighteenths from the start: $\tfrac{2}{3}=\tfrac{12}{18}$, $\tfrac{1}{2}=\tfrac{9}{18}$, $0=\tfrac{0}{18}$. Their sum is $\tfrac{21}{18}$, and dividing by $3$ gives $\tfrac{7}{18}$ directly — same answer, with the common denominator chosen up front.
CCSS standards used (min grade 6)
6.SP.B.5Summarize numerical data sets by reporting number of observations and measures (Reading each bracket symbol as a mean and computing $\{1\ 1\ 0\}=\tfrac{2}{3}$ and $[0\ 1]=\tfrac{1}{2}$ as sum-divided-by-count.)5.NF.A.1Add and subtract fractions with unlike denominators (Adding the outer three values $\tfrac{2}{3}+\tfrac{1}{2}+0$ by rewriting them over the common denominator $6$ to get $\tfrac{7}{6}$.)6.NS.A.1Interpret and compute quotients of fractions and solve word problems (Dividing the sum $\tfrac{7}{6}$ by $3$ to finish the outer average as $\tfrac{7}{18}$.)
⭐ When a problem invents strange brackets, just translate them into a word you know — here it's 'average' — then work from the inside out, one small piece at a time.
⭐ When a problem invents strange brackets, just translate them into a word you know — here it's 'average' — then work from the inside out, one small piece at a time.
More like this
Same archetype — closest grade level first.