1989 AJHSME Problems/Problem 5

Problem

$-15+9\times (6\div 3) =$

$\text{(A)}\ -48 \qquad \text{(B)}\ -12 \qquad \text{(C)}\ -3 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 12$

We use the order of operations here to get

$\begin{align*} -15+9\times (6\div 3) &= -15+9\times 2 \\ &= -15+18 \\ &= 3 \rightarrow \boxed{\text{D}} \end{align*}$

1989 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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