Difference between revisions of "1991 AJHSME Problems/Problem 2"

Revision as of 15:59, 7 August 2009

$\frac{16+8}{4-2}=$

$\text{(A)}\ 4 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 20$

$\begin{align*} \frac{16+8}{4-2} &= \frac{24}{2} \\ &= 12\rightarrow \boxed{\text{C}}. \end{align*}$

1991 AJHSME (Problems • Answer Key • Resources)
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