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2000 CEMC Pascal Problems/Problem 6

Problem

If $\frac{2}{3}, \frac{23}{30}, \frac{9}{10}, \frac{11}{15}, \text{and} \frac{4}{5}$ are written from smallest to largest then the middle fraction will be

$\text{ (A) }\ \frac{23}{30} \qquad\text{ (B) }\ \frac{4}{5} \qquad\text{ (C) }\ \frac23 \qquad\text{ (D) }\ \frac{9}{10} \qquad\text{ (E) }\ \frac{11}{15}$

Solution 1

To figure out which one is the largest, we can use a common denominator for all of the fractions, and then compare the numerators.

First, we can find the LCM (least common multiple) of the denominators:

$2 = 2$

$30 = 2 \times 3 \times 5$

$10 = 2 \times 5$

$15 = 3 \times 5$

$5 = 5$

Since all of the factors in the other numbers are factors of $30$, we can see that the LCM of the numbers is $30$.

$\frac{2}{3} = \frac{2 \times 10}{3 \times 10} = \frac{20}{30}$

$\frac{23}{30} = \frac{23 \times 1}{30 \times 1} = \frac{23}{30}$

$\frac{9}{10} = \frac{9 \times 3}{10 \times 3} = \frac{27}{30}$

$\frac{11}{15} = \frac{11 \times 2}{15 \times 2} = \frac{22}{30}$

$\frac{4}{5} = \frac{4 \times 6}{5 \times 6} = \frac{24}{30}$

Ordering these fractions from least to greatest, we have:

$\frac{20}{30}, \frac{22}{30}, \frac{23}{30}, \frac{24}{30}, \frac{27}{30}$

The middle fraction is then $\boxed {\textbf {(A) } \frac{23}{30}}$.

~anabel.disher

Solution 2

We can convert the fractions to decimals, and then compare their values.

Using long division, we get:

$\frac{2}{3} = 0.\overline{6}$

$\frac{23}{30} = 0.7\overline{6}$

$\frac{9}{10} = 0.9$

$\frac{11}{15} = 0.7\overline{3}$

$\frac{4}{5} = 0.8$

Ordering these decimals from least to greatest, we have:

$0.\overline{6}, 0.7\overline{3}, 0.7\overline{6}, 0.8, 0.9$

The middle decimal is then $0.7\overline{6}$, which is $\boxed {\textbf {(A) } \frac{23}{30}}$.

~anabel.disher

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