AMC 10 · 2011 · #3
Grade 6 arithmeticPick an answer.
AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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'.
Average the inner three
The innermost group {1 1 0} is the average of 1, 1, and 0: add them, divide by 3, and get 2/3.
An average of three numbers is just their total shared equally three ways.
6.SP.B.5Identify SubproblemsAverage the inner pair
Next, [0 1] is the average of 0 and 1: add them, divide by 2, and get 1/2.
The average of two numbers sits exactly halfway between them.
6.SP.B.5Identify SubproblemsAdd the outer three
The outer expression is now {2/3 1/2 0}; rewrite the fractions over the common denominator 6 and add to get 7/6.
Fractions can only be added once their pieces are the same size, so switch both to sixths.
5.NF.A.1Identify SubproblemsDivide the total by three
The outer average splits that sum three ways, and dividing by 3 triples the denominator: 7/6 becomes 7/18.
Splitting a fraction into three equal parts triples how many pieces the whole is cut into, so the denominator grows from 6 to 18.
Splitting a fraction into three equal parts triples how many pieces the whole is cut into.
▸ Why?
A share of a share is a smaller share, cut from the same whole.
▸ Why?
Scaling the bottom while the top stays fixed still names an amount measured against the same whole.
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.
- Average the inner three
- Average the inner pair
- Add the outer three
- Divide the total by three