Difference between revisions of "1986 AJHSME Problems/Problem 11"

 
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==Solution==
 
==Solution==
  
You might have realized that <math>A*B</math> gets you the average of A and B. If you're good at averages, you'll immediately realize that the average of <math>3</math> and <math>5</math> is <math>4</math>, and the average of <math>4</math> and <math>8</math> is <math>6</math>.
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We just plug in and evaluate:
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<cmath>\begin{align*}
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(3*5)*8 &= \left( \frac{3+5}{2}\right) *8 \\
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&= 4*8 \\
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&= \frac{4+8}{2} \\
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&= 6 \\
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\end{align*}</cmath>
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<math>\boxed{\text{A}}</math>
  
 
==See Also==
 
==See Also==
  
[[1986 AJHSME Problems]]
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{{AJHSME box|year=1986|num-b=10|num-a=12}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 21:15, 3 July 2013

Problem

If $\text{A}*\text{B}$ means $\frac{\text{A}+\text{B}}{2}$, then $(3*5)*8$ is

$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 16\qquad \text{(E)}\ 30$

Solution

We just plug in and evaluate: \begin{align*} (3*5)*8 &= \left( \frac{3+5}{2}\right) *8 \\ &= 4*8 \\ &= \frac{4+8}{2} \\ &= 6 \\ \end{align*}

$\boxed{\text{A}}$

See Also

1986 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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 AJHSME/AMC 8 Problems and Solutions

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