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1994 AMC 8
2
1994 AJHSME Problem 2
1994 AJHSME Problem 2
Source:
July 9, 2011
Problem Statement
1
10
+
2
10
+
3
10
+
4
10
+
5
10
+
6
10
+
7
10
+
8
10
+
9
10
+
55
10
=
\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}=
10
1
+
10
2
+
10
3
+
10
4
+
10
5
+
10
6
+
10
7
+
10
8
+
10
9
+
10
55
=
(A)
4
1
2
(B)
6.4
(C)
9
(D)
10
(E)
11
\text{(A)}\ 4\dfrac{1}{2} \qquad \text{(B)}\ 6.4 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11
(A)
4
2
1
(B)
6.4
(C)
9
(D)
10
(E)
11
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