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1989 AJHSME Problems/Problem 5

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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

Solution

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*}

See Also

1989 AJHSME (Problems • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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
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Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
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