2011 AMC 10B Problems/Problem 15: Difference between revisions

Revision as of 10:06, 13 July 2021

Problem

Let $@$ denote the "averaged with" operation: $a @ b = \frac{a+b}{2}$ . Which of the following distributive laws hold for all numbers $x, y,$ and $z$ ? $\text{I. x @ (y + z) = (x @ y) + (x @ z)}$ $\text{II. x + (y @ z) = (x + y) @ (x + z)}$ $\text{III. x @ (y @ z) = (x @ y) @ (x @ z)}$

$\textbf{(A)}\ \text{I only} \qquad\textbf{(B)}\ \text{II only} \qquad\textbf{(C)}\ \text{III only} \qquad\textbf{(D)}\ \text{I and III only} \qquad\textbf{(E)}\ \text{II and III only}$

Solution

Simplify each operation and see which ones hold true.

$\begin{align*} \text{I.} \qquad x @ (y + z) &= (x @ y) + (x @ z)\\ \frac{x+y+z}{2} &= \frac{x+y}{2} + \frac{x+z}{2}\\ \frac{x+y+z}{2} &\not= \frac{2x+y+z}{2} \end{align*}$

$\begin{align*} \text{II.} \qquad x + (y @ z) &= (x + y) @ (x + z)\\ x+ \frac{y+z}{2} &= \frac{2x+y+z}{2}\\ \frac{2x+y+z}{2} &= \frac{2x+y+z}{2} \end{align*}$

$\begin{align*} \text{III.} \qquad x @ (y @ z) &= (x @ y) @ (x @ z)\\ x @ \frac{y+z}{2} &= \frac{x+y}{2} @ \frac{x+z}{2}\\ \frac{2x+y+z}{4} &= \frac{2x+y+z}{4} \end{align*}$

$\boxed{\textbf{(E)} \text{II and III only}}$

2011 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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 == Solution ==
+Simplify each operation and see which ones hold true.
-Just simplify each operation and see which ones hold true.
 <cmath>\begin{align*}

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2011 AMC 10B Problems/Problem 15: Difference between revisions

Revision as of 10:06, 13 July 2021

Problem

Solution

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