Difference between revisions of "1997 CEMC Pascal Problems/Problem 1"
(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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<math> \text{ (A) }\ \frac{11}{7} \qquad\text{ (B) }\ 8 \qquad\text{ (C) }\ \frac{7}{2} \qquad\text{ (D) }\ -12 \qquad\text{ (E) }\ 13</math> | <math> \text{ (A) }\ \frac{11}{7} \qquad\text{ (B) }\ 8 \qquad\text{ (C) }\ \frac{7}{2} \qquad\text{ (D) }\ -12 \qquad\text{ (E) }\ 13</math> | ||
==Solution== | ==Solution== | ||
| + | We can simply add the numbers in the [[numerator]] and [[denominator]] of the fraction, and then simplify the fraction: | ||
| + | |||
<math>\frac{4 + 35}{8 - 5} = \frac{39}{3} = \boxed {\textbf {(E) } 13}</math> | <math>\frac{4 + 35}{8 - 5} = \frac{39}{3} = \boxed {\textbf {(E) } 13}</math> | ||
~anabel.disher | ~anabel.disher | ||
Latest revision as of 12:34, 22 June 2025
Problem
equals
Solution
We can simply add the numbers in the numerator and denominator of the fraction, and then simplify the fraction:
~anabel.disher
