Question

Arrange the following rational numbers in ascending order:

(i) $\frac{2}{5},\frac{7}{10},\frac{8}{15},\frac{13}{30}$
(ii) $\frac{-3}{4},\frac{5}{-12},\frac{-7}{16},\frac{9}{-24}$
(iii) $\frac{-3}{10},\frac{7}{-15},\frac{-11}{20},\frac{17}{-30}$
(iv) $\frac{2}{3},\frac{3}{4},\frac{5}{-6},\frac{-7}{12}$

Answer 1

$\begin{array}{l}\left(\text{i}\right)\frac{2}{5},\frac{7}{10},\frac{8}{15},\frac{13}{30}\\ \text{L}.\text{C}.\text{M. of 5},\text{1}0,\text{15 and 3}0\mathrm{is}\text{3}0\\ \end{array}$



$\begin{array}{l}\frac{2\times 6}{5\times 6}=\frac{12}{30}\\ \frac{7\times 3}{10\times 3}=\frac{21}{30}\\ \frac{8\times 2}{15\times 2}=\frac{16}{30}\\ \frac{13\times 1}{30\times 1}=\frac{13}{30}\\ \text{Required order:}\frac{2}{5}<\frac{13}{30}<\frac{8}{15}<\frac{7}{10}\end{array}$

$\begin{array}{l}\left(ii\right)\frac{-3}{4},\frac{5}{-12},\frac{-7}{16},\frac{9}{-24}\\ \mathrm{First},\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{convert}\mathrm{each}\mathrm{negative}\mathrm{denominator}\mathrm{into}\mathrm{positive}.\\ \frac{-3}{4},\frac{5\times -1}{-12\times -1},\frac{-7}{16},\frac{9\times -1}{-24\times -1}\\ \begin{array}{l}\frac{-3}{4},\frac{-5}{12},\frac{-7}{16},\frac{-9}{24}\\ \begin{array}{l}\\ \\ \\ \\ \\ \end{array}\end{array}\end{array}$


$\begin{array}{l}\text{L}.\text{C}.\text{M. of 4},\text{12},\text{16 and}\text{24 is 48.}\\ \frac{-3\times 12}{4\times 12}=\frac{-36}{48}\\ \frac{-5\times 4}{12\times 4}=\frac{-20}{48}\\ \frac{-7\times 3}{16\times 3}=\frac{-21}{48}\\ \frac{-9\times 2}{24\times 2}=\frac{-18}{48}\\ \text{Required order:}\frac{-3}{4}<\frac{-7}{16}<\frac{-5}{12}<\frac{-9}{24}\end{array}$
$\begin{array}{l}\left(\text{iii}\right)\frac{-3}{10},\frac{7}{-15},\frac{-11}{20},\frac{17}{-30}\\ \text{First, we need to convert the negative denominators to make them positive.}\\ \frac{-3}{10},\frac{7\times -1}{-15\times -1},\frac{-11}{20},\frac{17\times -1}{-30\times -1}\\ \frac{-3}{10},\frac{-7}{15},\frac{-11}{20},\frac{-17}{30}\end{array}$



$\begin{array}{l}\begin{array}{l}\text{L}.\text{C}.\text{M of 1}0,\text{15},\text{2}0,\text{3}0=60\\ \frac{-3\times 6}{10\times 6}=\frac{-18}{60}\\ \frac{-7\times 4}{15\times 4}=\frac{-28}{60}\\ \frac{-11\times 3}{20\times 3}=\frac{-33}{60}\\ \frac{-17\times 2}{30\times 2}=\frac{-34}{60}\\ \text{Therefore,}\frac{-34}{60}<\frac{-33}{60}<\frac{-28}{60}<\frac{-18}{60}\end{array}\\ \text{i}\text{.e}\text{.}\frac{-17}{30}<\frac{-11}{20}<\frac{-7}{15}<\frac{-3}{10}\\ \left(\text{iv}\right)\frac{2}{3},\frac{3}{4},\frac{5}{-6},\frac{-7}{12}\\ \text{First, we need to convert the negative denominators to positive ones.}\\ \frac{2}{3},\frac{3}{4},\frac{5\times -1}{-6\times -1},\frac{-7}{12}\\ \frac{2}{3},\frac{3}{4},\frac{-5}{6},\frac{-7}{12}\end{array}$



$\begin{array}{l}\begin{array}{l}\text{L}.\text{C}.\text{M of 3},\text{4},\text{6},\text{12}=\text{12}\\ \frac{2\times 4}{3\times 4}=\frac{8}{12}\\ \frac{3\times 3}{4\times 3}=\frac{9}{12}\\ \frac{-5\times 2}{6\times 2}=\frac{-10}{12}\\ \frac{-7\times 1}{12\times 1}=\frac{-7}{12}\end{array}\\ \\ \text{Therefore, the correct order is}\frac{-5}{6}<\frac{-7}{12}<\frac{2}{3}<\frac{3}{4}.\end{array}$