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

Latest revision as of 20:15, 3 July 2013

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

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

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

$\boxed{\text{A}}$

1986 AJHSME (Problems • Answer Key • Resources)
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

@@ Line 7: / Line 7: @@
 ==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>.
+We just plug in and evaluate:
+<cmath>\begin{align*}
+(3*5)*8 &= \left( \frac{3+5}{2}\right) *8 \\
+&= 4*8 \\
+&= \frac{4+8}{2} \\
+&= 6 \\
+\end{align*}</cmath>
+<math>\boxed{\text{A}}</math>
 ==See Also==
-[[1986 AJHSME Problems]]
+{{AJHSME box|year=1986|num-b=10|num-a=12}}
+[[Category:Introductory Algebra Problems]]
+{{MAA Notice}}