Difference between revisions of "1986 AJHSME Problems/Problem 11"
5849206328x (talk | contribs) (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 ...) |
Aquakitty11 (talk | contribs) |
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==Solution== | ==Solution== | ||
| − | {{ | + | 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== | ==See Also== | ||
| − | [[ | + | {{AJHSME box|year=1986|num-b=10|num-a=12}} |
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:15, 3 July 2013
Problem
If 
means 
, then 
is
Solution
We just plug in and evaluate:
See Also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
