Consider the set of fractions $\left\{\frac{3}{7}, \frac{3}{2}, \frac{6}{7}, \frac{3}{5}\right\}$. Ordered from smallest to largest, the set is
(A) $\left\{\frac{3}{7}, \frac{3}{5}, \frac{6}{7}, \frac{3}{2}\right\}$
(B) $\left\{\frac{3}{2}, \frac{3}{5}, \frac{6}{7}, \frac{3}{7}\right\}$
(C) $\left\{\frac{3}{2}, \frac{3}{5}, \frac{3}{7}, \frac{6}{7}\right\}$
(D) $\left\{\frac{3}{5}, \frac{3}{7}, \frac{6}{7}, \frac{3}{2}\right\}$
(E) $\left\{\frac{3}{7}, \frac{3}{5}, \frac{3}{2}, \frac{6}{7}\right\}$
Solution 1

To order the fractions $\left\{\frac{3}{7}, \frac{3}{2}, \frac{6}{7}, \frac{3}{5}\right\}$ from smallest to largest, we first express each fraction with a common denominator of $7 \times 2 \times 5=70$.

The set $\left\{\frac{3}{7}, \frac{3}{2}, \frac{6}{7}, \frac{3}{5}\right\}$ is equivalent to the $\operatorname{set}\left\{\frac{3 \times 10}{7 \times 10}, \frac{3 \times 35}{2 \times 35}, \frac{6 \times 10}{7 \times 10}, \frac{3 \times 14}{5 \times 14}\right\}$ or to the set $\left\{\frac{30}{70}, \frac{105}{70}, \frac{60}{70}, \frac{42}{70}\right\}$. Ordered from smallest to largest, the set is $\left\{\frac{30}{70}, \frac{42}{70}, \frac{60}{70}, \frac{105}{70}\right\}$, so ordered from smallest to largest the original set is $\left\{\frac{3}{7}, \frac{3}{5}, \frac{6}{7}, \frac{3}{2}\right\}$.

## Solution 2

Since the numerators of the given fractions are each either 3 or 6 , we rewrite each fraction with a common numerator of 6 .

The set $\left\{\frac{3}{7}, \frac{3}{2}, \frac{6}{7}, \frac{3}{5}\right\}$ is equivalent to the $\left\{\frac{3 \times 2}{7 \times 2}, \frac{3 \times 2}{2 \times 2}, \frac{6}{7}, \frac{3 \times 2}{5 \times 2}\right\}$ or to the $\left\{\frac{6}{14}, \frac{6}{4}, \frac{6}{7}, \frac{6}{10}\right\}$.

Since the numerators are equal, then the larger the denominator, the smaller the fraction.

Ordered from smallest to largest, the set is $\left\{\frac{6}{14}, \frac{6}{10}, \frac{6}{7}, \frac{6}{4}\right\}$, so ordered from smallest to largest the original set is $\left\{\frac{3}{7}, \frac{3}{5}, \frac{6}{7}, \frac{3}{2}\right\}$.

ANSWER: (A)

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