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1991 AJHSME Problems/Problem 2

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Problem

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

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

Solution

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

See Also

1991 AJHSME (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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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