Difference between revisions of "1994 AJHSME Problems/Problem 2"

Latest revision as of 23:12, 4 July 2013

$\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}+\dfrac{55}{10}=$

$\text{(A)}\ 4\dfrac{1}{2} \qquad \text{(B)}\ 6.4 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11$

$1+ 2+ 3 + 4 + 5 + 6 + 7 + 8 + 9 = \dfrac{(9)(10)}{2} = 45$

$\frac{45+55}{10} = \dfrac{100}{10} = \boxed{\text{(D)}\ 10}$

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

@@ Line 6: / Line 6: @@
 ==Solution==
-<math> 1+ 2+ 3 + 4 + 5 + 6 + 7 + 8 + 9 = \dfrac{(9)(10)}{2} = 45 + 55 = 100 </math>
+<math> 1+ 2+ 3 + 4 + 5 + 6 + 7 + 8 + 9 = \dfrac{(9)(10)}{2} = 45 </math>
-<math>\dfrac{100}{10} = \boxed{\text{(D)}\ 10}</math>
+<math>\frac{45+55}{10} = \dfrac{100}{10} = \boxed{\text{(D)}\ 10}</math>
+==See Also==
+{{AJHSME box|year=1994|num-b=1|num-a=3}}
+{{MAA Notice}}