Arrange the following rational number in ascending order $\frac{3}{7}, \frac{4}{5}, \frac{7}{9}, \frac{1}{2}$

\frac{4}{5}, \frac{7}{5}, \frac{3}{9}, \frac{1}{2}

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\frac{3}{7}, \frac{1}{2}, \frac{7}{9}, \frac{4}{5}

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\frac{4}{5}, \frac{7}{9}, \frac{1}{2}, \frac{3}{7}

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\frac{1}{2}, \frac{3}{7}, \frac{7}{9}, \frac{4}{5}

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Solution

The correct option is B $\frac{3}{7}, \frac{1}{2}, \frac{7}{9}, \frac{4}{5}$
The LCM of 2, 5, 7 and 9 is 630.
$\frac{3}{7} = \frac{270}{630}$
$\frac{4}{5} = \frac{504}{630}$
$\frac{7}{9} = \frac{490}{630}$
$\frac{1}{2} = \frac{315}{630}$
The ascending order will be $\frac{3}{7}, \frac{1}{2}, \frac{7}{9}, \frac{4}{5}$

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Similar questions

(1) \frac{- 3}{5}, \frac{7}{- 10}, \frac{- 5}{6}

(2) \frac{2}{3}, \frac{3}{4}, \frac{5}{6}, \frac{- 7}{12}

\frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{7}{9}, \frac{1}{8}, \frac{7}{12}

On arranging the following fractions $\frac{4}{9}, \frac{7}{9}, \frac{1}{9}, \frac{3}{9}$ in ascending order we get:

3 x^{2} - 8 x - 3 = 0, x = 9, \frac{1}{2}, 3

\frac{3}{5}

\frac{2}{5}

\frac{1}{5}

\frac{7}{6}

\frac{1}{7}

\frac{3}{8}

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