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2
AJHSME 1991 Problem 2
AJHSME 1991 Problem 2
Source:
June 24, 2011
Problem Statement
16
+
8
4
−
2
=
\frac{16+8}{4-2}=
4
−
2
16
+
8
=
(A)
4
(B)
8
(C)
12
(D)
16
(E)
20
\text{(A)}\ 4 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 20
(A)
4
(B)
8
(C)
12
(D)
16
(E)
20
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