2011 AMC 10B Problem 15

Attempt Problem 15 of the 2011 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 10B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

15.

Let @@ denote the “averaged with” operation: a@b=a+b2.a @ b = \frac{a+b}{2}. Which of the following distributive laws hold for all numbers x,x, y,y, and z?z?
I. x@(y+z)=(x@y)+(x@z)x @ (y + z) = (x @ y) + (x @ z)
II. x+(y@z)=(x+y)@(x+z)x + (y @ z) = (x + y) @ (x + z)
III. x@(y@z)=(x@y)@(x@z)x @ (y @ z) = (x @ y) @ (x @ z)
I\mathrm{I} only
II\mathrm{II} only
III\mathrm{III} only
I\mathrm{I} and III\mathrm{III} only
II\mathrm{II} and III\mathrm{III} only
Answer: E
Concepts:custom operationdistributive property
Small hint:
Expand each side using a@b=a+b2a@b=\frac{a+b}{2}
Big hint:
Test each of I, II, and III symbolically
Solution:
In statement I, the left-hand side equals x+y+z2\dfrac{x+y+z}2 and the right-hand side equals x+y+z2,x+\dfrac{y+z}2, so they are not always equal.
In statement II, both sides equal x+y+z2,x+\dfrac{y+z}2, so it holds.
In statement III, both sides equal 2x+y+z4,\dfrac{2x+y+z}4, so it also holds.
Thus, the correct answer is E .
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