, 2013

Solution 1

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 set $\left\{ \frac{1\times6}{2\times6}, \frac{2\times4}{3\times4}, \frac{1\times3}{4\times3}, \frac{5\times2}{6\times2}, \frac{7}{12} \right\}$ 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.

Thirty-six hundredths equals $\frac{36}{100}$ or $0.36$.