Difference between revisions of "2000 CEMC Gauss (Grade 8) Problems/Problem 3"

Latest revision as of 13:39, 20 October 2025

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

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

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

2000 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources)
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CEMC Gauss (Grade 8)

Revision as of 15:00, 11 September 2025 (view source) Anabel.disher (talk \| contribs) (Created page with "==Problem== The value of <math>\frac{5 + 4 - 3}{5 + 4 + 3}</math> is <math> \text{ (A) }\ -1 \qquad\text{ (B) }\ \frac{1}{3} \qquad\text{ (C) }\ 2 \qquad\text{ (D) }\ \frac...")		Latest revision as of 13:39, 20 October 2025 (view source) Anabel.disher (talk \| contribs)
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	~anabel.disher		~anabel.disher
		+	{{CEMC box\|year=2000\|competition=Gauss (Grade 8)\|num-b=2\|num-a=4}}