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1997 CEMC Pascal Problems/Problem 1

Revision as of 12:33, 22 June 2025 by Anabel.disher (talk | contribs) (Created page with "==Problem== <math>\frac{4 + 35}{8 - 5}</math> equals <math> \text{ (A) }\ \frac{11}{7} \qquad\text{ (B) }\ 8 \qquad\text{ (C) }\ \frac{7}{2} \qquad\text{ (D) }\ -12 \qquad\te...")
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Problem

$\frac{4 + 35}{8 - 5}$ equals

$\text{ (A) }\ \frac{11}{7} \qquad\text{ (B) }\ 8 \qquad\text{ (C) }\ \frac{7}{2} \qquad\text{ (D) }\ -12 \qquad\text{ (E) }\ 13$

Solution

$\frac{4 + 35}{8 - 5} = \frac{39}{3} = \boxed {\textbf {(E) } 13}$

~anabel.disher

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