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2014 CEMC Gauss (Grade 8) Problems/Problem 9

Problem

Consider the set of fractions $\{\frac{3}{7}, \frac{3}{2}, \frac{6}{7}, \frac{3}{5}\}$. Ordered from smallest to largest, the set is

$\text{ (A) }\ \{\frac{3}{7}, \frac{3}{5}, \frac{6}{7}, \frac{3}{2}\} \qquad\text{ (B) }\ \{\frac{3}{2}, \frac{3}{5}, \frac{6}{7}, \frac{3}{7}\} \qquad\text{ (C) }\ \{\frac{3}{2}, \frac{3}{5}, \frac{3}{7}, \frac{6}{7}\} \qquad\text{ (D) }\ \{\frac{3}{5}, \frac{3}{7}, \frac{6}{7}, \frac{3}{2}\}\qquad\text{ (E) }\ \{\frac{3}{7}, \frac{3}{5}, \frac{3}{2}, \frac{6}{7}\}$

Solution 1

We can give everything a common denominator of $70$. This gives:

$\frac{3}{7} = \frac{3 \times 10}{7 \times 10} = \frac{30}{70}$

$\frac{3}{2} = \frac{3 \times 35}{2 \times 35} = \frac{105}{70}$

$\frac{6}{7} = \frac{6 \times 10}{7 \times 10} = \frac{60}{70}$

$\frac{3}{5} = \frac{3 \times 14}{5 \times 14} = \frac{42}{70}$

Ordering these from least to greatest by comparing their numerators (since their denominators are now all $70$), we get $\boxed {\textbf{ (A) }\ \{\frac{3}{7}, \frac{3}{5}, \frac{6}{7}, \frac{3}{2}\}}$.

~anabel.disher

Solution 2

$\frac{3}{2} > 1$ because its numerator is more than its denominator, but all of the other numbers are less than $1$ because their denominators are all lower than their numerators. This means it must be the highest.

$\frac{3}{7} < \frac{3}{5}$ because the denominator of $\frac{3}{7}$ is greater than the denominator of $\frac{3}{5}$.

$\frac{3}{5} = \frac{3 \times 7}{5 \times 7} = \frac{21}{35}$, and $\frac{6}{7} = \frac{6 \times 5}{7 \times 5} = \frac{30}{35}$, which has a higher numerator but the same denominator than the other fraction.

Thus, the answer is $\boxed {\textbf{ (A) }\ \{\frac{3}{7}, \frac{3}{5}, \frac{6}{7}, \frac{3}{2}\}}$.

~anabel.disher

Solution 3 (answer choices)

Since $\frac{3}{2} > 1$ while the others are less than $1$, we know that the answer must be A or D. However, $\frac{3}{7} < \frac{3}{5}$ due to $\frac{3}{7}$ having a higher denominator, so the answer must be $\boxed {\textbf{ (A) }\ \{\frac{3}{7}, \frac{3}{5}, \frac{6}{7}, \frac{3}{2}\}}$.

~anabel.disher

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