Difference between revisions of "1994 AJHSME Problems/Problem 2"
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<math>\text{(A)}\ 4\dfrac{1}{2} \qquad \text{(B)}\ 6.4 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11</math> | <math>\text{(A)}\ 4\dfrac{1}{2} \qquad \text{(B)}\ 6.4 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11</math> | ||
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| + | ==Solution== | ||
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| + | <math> 1+ 2+ 3 + 4 + 5 + 6 + 7 + 8 + 9 = \dfrac{(9)(10)}{2} = 45 </math> | ||
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| + | <math>\frac{45+55}{10} = \dfrac{100}{10} = \boxed{\text{(D)}\ 10}</math> | ||
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| + | ==See Also== | ||
| + | {{AJHSME box|year=1994|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 00:12, 5 July 2013
Problem
Solution
See Also
| 1994 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. 
