1998 CEMC Pascal Problems/Problem 1

Problem

The value of $\frac{1 + 3 + 5}{10 + 6 + 2}$ is

$\text{ (A) }\ \frac{1}{6} \qquad\text{ (B) }\ 2 \qquad\text{ (C) }\ \frac{1}{2} \qquad\text{ (D) }\ 1\frac{1}{2} \qquad\text{ (E) }\ 3\frac{1}{10}$

We can add the numbers in the numerator and the denominator of the fraction, and then simplify the fraction:

$\frac{1 + 3 + 5}{10 + 6 + 2} = \frac{4 + 5}{16 + 2} = \frac{9}{18} = \frac{9 \div 9}{18 \div 9} = \boxed {\textbf{(C) } \frac{1}{2}}$

~anabel.disher

We can notice that the denominator is also $10 + 6 + 2 = 2(1 + 3 + 5)$ .

Thus, we have:

$\frac{1 + 3 + 5}{2(1 + 3 + 5)} = \boxed {\textbf{(C) } \frac{1}{2}}$

~anabel.disher