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Difference between revisions of "1973 AHSME Problems/Problem 5"

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However, since <math>x</math> is a variable, but <math>i</math> must be a constant, fixed number, there is no identity element for the binary operation of averaging. Therefore, Statement V is false.
 
However, since <math>x</math> is a variable, but <math>i</math> must be a constant, fixed number, there is no identity element for the binary operation of averaging. Therefore, Statement V is false.
  
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Statements II and IV are true, so the answer is \boxed{\textbf{(D)}\ II and IV only}$.
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Statements II and IV are true, so the answer is <math>\boxed{\textbf{(D)} \text{II and IV only}}</math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME 35p box|year=1973|num-b=4|num-a=6}}
 
{{AHSME 35p box|year=1973|num-b=4|num-a=6}}

Revision as of 16:26, 28 August 2018

Problem

Of the following five statements, I to V, about the binary operation of averaging (arithmetic mean),

$\text{I. Averaging is associative }$

$\text{II. Averaging is commutative }$

$\text{III. Averaging distributes over addition }$

$\text{IV. Addition distributes over averaging }$

$\text{V. Averaging has an identity element }$

those which are always true are

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

$\textbf{(D)}\ \text{II and IV only}\qquad\textbf{(E)}\ \text{II and V only}$

Solution

We can consider each statement independently and see which ones are true. The average of two numbers $x$ and $y$ is $avg(x, y) = \frac{x + y}{2}$.

Statement I \[avg(avg(x, y), z) = avg(x, avg(y, z))\] \[\frac{\frac{x + y}{2} + z}{2} = \frac{x + \frac{y + z}{2}}{2}\] \[\frac{x + y + 2z}{4} = \frac{2x + y + z}{4}\] The above statement is not true for all $(x, y, z)$, so Statement I is false.


Statement II \[avg(x, y) = avg(y, x)\] \[\frac{x + y}{2} = \frac{y + x}{2}\] Since addition is commutative, the above is true for all $(x,y)$, so Statement II is true.


Statement III \[avg(x, y + z) = avg(x, y) + avg(x, z)\] \[\frac{x + y + z}{2} = \frac{x + y}{2} + \frac{x + z}{2}\] \[\frac{x + y + z}{2} = \frac{2x + y + z}{2}\] The above statement is not true for all $(x, y, z)$, so Statement III is false.


Statement IV \[x + avg(y, z) = avg(x + y, x + z)\] \[x + \frac{y + z}{2} = \frac{2x + y + z}{2}\] \[\frac{2x + y + z}{2} = \frac{2x + y + z}{2}\] The left side and the right side are equivalent expressions, so Statement IV is true.


Statement V Let $i$ be the identity element. If such $i$ exists, then the following is true: \[avg(x, i) = x\] \[\frac{x + i}{2} = x\] \[x + i = 2x\] \[i = x\] However, since $x$ is a variable, but $i$ must be a constant, fixed number, there is no identity element for the binary operation of averaging. Therefore, Statement V is false.

Statements II and IV are true, so the answer is $\boxed{\textbf{(D)} \text{II and IV only}}$.

See Also

1973 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions
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