Difference between revisions of "1991 AJHSME Problems/Problem 2"
5849206328x (talk | contribs) (Created page with '==Problem== <math>\frac{16+8}{4-2}=</math> <math>\text{(A)}\ 4 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 20</math> ==Solution== <cma…') |
|||
Line 16: | Line 16: | ||
{{AJHSME box|year=1991|num-b=1|num-a=3}} | {{AJHSME box|year=1991|num-b=1|num-a=3}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 23:06, 4 July 2013
Problem
Solution
See Also
1991 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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.