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Arrange the following rational numbers in ascending order:

(i) \(\frac{4}{-9},\frac{-5}{12},\frac{7}{-18},\frac{-2}{3}\)

(ii) \(\frac{-3}{4},\frac{5}{-12},\frac{-7}{16},\frac{9}{-24}\)

(iii) \(\frac{3}{-5},\frac{-7}{10},\frac{-11}{15},\frac{-13}{20}\)

(iv) \(\frac{-4}{7},\frac{-9}{14},\frac{13}{-28},\frac{-23}{42}\)

1 Answer

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(i) \(\frac{4}{-9} = \frac{4\times-1}{-9\times-1} = \frac{-4}{9}\)

And,

\(\frac{4}{-9} = \frac{4\times-1}{-9\times-1} = \frac{-4}{9}\)

Since, the denominators of all the numbers are different therefore we will take LCM of the denominators. 

LCM of 9, 12, 18 and 3 = 36

\(\frac{-4}{9} = \frac{-4\times4}{9\times4} = \frac{-16}{36}\)

\(\frac{-5}{12} = \frac{-5\times3}{12\times3} =\frac{-15}{36}\)

\(\frac{-7}{18} = \frac{-7\times2}{18\times2} = \frac{-14}{36}\)

\(\frac{-2}{3} = \frac{-2\times12}{3\times12}\frac{-24}{36}\)

Clearly,

24 < -16 < -15 < -14

Therefore,

\(\frac{-24}{36}<\frac{-16}{36}<\frac{-15}{36}<\frac{-14}{36}\)

Hence,

\(\frac{-2}{3}<\frac{-4}{-9}<\frac{-5}{36}<\frac{-2}{-18}\)

(ii) \(\frac{5}{-12} = \frac{5\times-1}{-12\times-1} = \frac{-5}{12}\)

And,

\(\frac{9}{-24} = \frac{9\times-1}{-24\times-1} = \frac{-9}{24}\)

Since, the denominators of all the numbers are different therefore we will take LCM of the denominators. 

LCM of 4, 12, 16 and 24 = 48

\(\frac{-3}{4} =\frac{-3\times12}{4\times12} = \frac{-36}{48}\)

\(\frac{-5}{12} =\frac{-5\times4}{12\times4} = \frac{-20}{48}\)

\(\frac{-7}{16} =\frac{-7\times3}{16\times3} = \frac{-21}{48}\)

\(\frac{-9}{24} =\frac{-9\times2}{24\times2} = \frac{-18}{48}\)

Clearly, -36 < -21 < -20 < -18 

Therefore,

\(\frac{-36}{48}<\frac{-21}{48}<\frac{-20}{48}<\frac{-18}{-48}\)

Hence,

\(\frac{-3}{4}<\frac{-7}{16}<\frac{-5}{-12}<\frac{-9}{-24}\)

(iii) \(\frac{3}{-5} = \frac{3\times-1}{-5\times-1} = \frac{-3}{5}\)

Since, the denominators of all the numbers are different therefore we will take LCM of the denominators. 

LCM of 5, 10, 15 and 20 = 60

 \(\frac{-3}{5} =\frac{-3\times12}{5\times12} = \frac{-36}{60}\)

\(\frac{-7}{10} =\frac{-7\times6}{10\times6} = \frac{-42}{60}\)

\(\frac{-11}{15} =\frac{-11\times4}{15\times4} = \frac{-44}{60}\)

\(\frac{-13}{20} =\frac{-13\times3}{20\times3} = \frac{-39}{60}\)

Clearly, -44 < -42 < -39 < -36 

Therefore,

 \(\frac{-44}{60}<\frac{-42}{60}<\frac{-39}{-60}<\frac{-36}{-60}\)

Hence,

\(\frac{-11}{15}<\frac{-7}{10}<\frac{-13}{-20}<\frac{-3}{-5}\)

(iv) \(\frac{13}{-28} = \frac{13\times12}{-28\times-1} =\frac{-13}{28}\)

Since, the denominators of all the numbers are different therefore we will take LCM of the denominators. 

LCM of 7, 14, 28 and 42 = 84

  \(\frac{-4}{7} =\frac{-4\times12}{7\times12} = \frac{-48}{84}\)

\(\frac{-9}{14} =\frac{-9\times6}{14\times6} = \frac{-54}{84}\)

\(\frac{-13}{28} =\frac{-13\times3}{28\times3} = \frac{-39}{84}\)

\(\frac{-23}{42} =\frac{-23\times2}{42\times2} = \frac{-46}{84}\)

Clearly, -54 < -48 < -46 < -39 

Therefore,

\(\frac{-54}{84}<\frac{-48}{84}<\frac{-46}{-84}<\frac{-39}{84}\)

Hence,

\(\frac{-9}{14}<\frac{-4}{7}<\frac{-23}{42}<\frac{-13}{-28}\)

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