The smallest number in the set $\left\{\frac{1}{2}, \frac{2}{3}, \frac{1}{4}, \frac{5}{6}, \frac{7}{12}\right\}$ is
(A) $\frac{1}{2}$
(B) $\frac{2}{3}$
(C) $\frac{1}{4}$
(D) $\frac{5}{6}$
(E) $\frac{7}{12}$
Solution 1

ANswer: (B)

To determine the smallest number in the set $\left\{\frac{1}{2}, \frac{2}{3}, \frac{1}{4}, \frac{5}{6}, \frac{7}{12}\right\}$, we express each number with a

common denominator of 12 . The set $\left\{\frac{1}{2}, \frac{2}{3}, \frac{1}{4}, \frac{5}{6}, \frac{7}{12}\right\}$ is equivalent to the $\left\{\frac{1 \times 6}{2 \times 6}, \frac{2 \times 4}{3 \times 4}, \frac{1 \times 3}{4 \times 3}, \frac{5 \times 2}{6 \times 2}, \frac{7}{12}\right\}$ or to the set $\left\{\frac{6}{12}, \frac{8}{12}, \frac{3}{12}, \frac{10}{12}, \frac{7}{12}\right\}$.

The smallest number in this set is $\frac{3}{12}$, so $\frac{1}{4}$ is the smallest number in the original set.

## Solution 2

With the exception of $\frac{1}{4}$, each number in the set is greater than or equal to $\frac{1}{2}$.

We can see this by recognizing that the numerator of each fraction is greater than or equal to one half of its denominator.

Thus, $\frac{1}{4}$ is the only number in the list that is less than $\frac{1}{2}$ and so it must be the smallest number in the set.

ANSWER: (C)