Difference between revisions of "1989 AJHSME Problems/Problem 5"

(New page: ==Problem== <math>-15+9\times (6\div 3) =</math> <math>\text{(A)}\ -48 \qquad \text{(B)}\ -12 \qquad \text{(C)}\ -3 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 12</math> ==Solution== We us...)
 
 
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{{AJHSME box|year=1989|num-b=4|num-a=6}}
 
{{AJHSME box|year=1989|num-b=4|num-a=6}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 22:57, 4 July 2013

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 (ProblemsAnswer KeyResources)
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
All AJHSME/AMC 8 Problems and Solutions

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