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2000 CEMC Gauss (Grade 8) Problems/Problem 3

Problem

The value of $\frac{5 + 4 - 3}{5 + 4 + 3}$ is

$\text{ (A) }\ -1 \qquad\text{ (B) }\ \frac{1}{3} \qquad\text{ (C) }\ 2 \qquad\text{ (D) }\ \frac{1}{2} \qquad\text{ (E) }\ -\frac{1}{2}$

Solution

Evaluating the numerator and denominator, we have:

$\frac{5 + 4 - 3}{5 + 4 + 3} = \frac{9 - 3}{9 + 3} = \frac{6}{12}$

Simplifying this gives $\frac{6 \div 6}{12 \div 6} = \boxed {\textbf {(D) } \frac{1}{2}}$.

~anabel.disher

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