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5
1989 AJHSME Problem 5
1989 AJHSME Problem 5
Source:
June 25, 2011
Problem Statement
−
15
+
9
×
(
6
÷
3
)
=
-15+9\times (6\div 3) =
−
15
+
9
×
(
6
÷
3
)
=
(A)
−
48
(B)
−
12
(C)
−
3
(D)
3
(E)
12
\text{(A)}\ -48 \qquad \text{(B)}\ -12 \qquad \text{(C)}\ -3 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 12
(A)
−
48
(B)
−
12
(C)
−
3
(D)
3
(E)
12
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