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

(New page: ==Problem== If <math>\text{A}*\text{B}</math> means <math>\frac{\text{A}+\text{B}}{2}</math>, then <math>(3*5)*8</math> is <math>\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 ...)
 
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==Solution==
 
==Solution==
  
{{Solution}}
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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>.
  
 
==See Also==
 
==See Also==
  
 
[[1986 AJHSME Problems]]
 
[[1986 AJHSME Problems]]

Revision as of 10:54, 24 January 2009

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

You might have realized that $A*B$ gets you the average of A and B. If you're good at averages, you'll immediately realize that the average of $3$ and $5$ is $4$, and the average of $4$ and $8$ is $6$.

See Also

1986 AJHSME Problems

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