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Difference between revisions of "2011 AMC 10B Problems/Problem 15"

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== Solution ==
 
== Solution ==
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Simplify each operation and see which ones hold true.
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Just simplify each operation and see which ones hold true.
 
  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
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\text{III.} \qquad x @ (y @ z) &= (x @ y) @ (x @ z)\\
 
\text{III.} \qquad x @ (y @ z) &= (x @ y) @ (x @ z)\\
 
x @ \frac{y+z}{2} &= \frac{x+y}{2} @ \frac{x+z}{2}\\
 
x @ \frac{y+z}{2} &= \frac{x+y}{2} @ \frac{x+z}{2}\\
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\frac{2x+y+z}{4} &\not= \frac{2x+y+z}{2}
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\frac{2x+y+z}{4} &= \frac{2x+y+z}{4}
 
\end{align*}</cmath>
 
\end{align*}</cmath>
  
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<math>\boxed{\textbf{(B)} \text{II only}}</math>
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<math>\boxed{\textbf{(E)} \text{II and III only}}</math>
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==Solution 2==
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Alternatively, substitute arbitrary values for <math>x</math>, <math>y</math>, and <math>z</math> where <math>x\neq y\neq z</math>. For example, <math>x=1</math>, <math>y=2</math>, and <math>z=3</math> give that only II and III work. <math>\boxed{\textbf{(E)} \text{II and III only}}</math>
  
 
== See Also==
 
== See Also==

Latest revision as of 00:48, 26 June 2022

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*}

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

Solution 2

Alternatively, substitute arbitrary values for $x$, $y$, and $z$ where $x\neq y\neq z$. For example, $x=1$, $y=2$, and $z=3$ give that only II and III work. $\boxed{\textbf{(E)} \text{II and III only}}$

See Also

2011 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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