2011 AMC 10A Problem 3

Below is the professionally curated solution for Problem 3 of the 2011 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 10A solutions, or check the answer key.

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Concepts:custom operationmeanfraction

Difficulty rating: 870

3.

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

29\dfrac{2}{9}

518\dfrac{5}{18}

13\dfrac{1}{3}

718\dfrac{7}{18}

23\dfrac{2}{3}

Solution:

We have that {1 1 0}=1+1+03=23. \{1 \ 1 \ 0\} = \dfrac{1 + 1 + 0}{3} = \dfrac{2}{3}.

We also get that [0 1]=1+02=12. [0 \ 1] = \dfrac{1 + 0}{2} = \dfrac{1}{2}.

Finally, {23 12 0}=23+12+03=718. \left\{\dfrac{2}{3} \ \dfrac{1}{2} \ 0\right\} = \dfrac{\dfrac{2}{3} + \dfrac{1}{2} + 0}{3} = \dfrac{7}{18}.

Thus, D is the correct answer.

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