Question

Arrange the following rational numbers in descending order:
(i) $-2,\frac{-13}{6},\frac{8}{-3},\frac{1}{3}$
(ii) $\frac{-3}{10},\frac{7}{-15},\frac{-11}{20},\frac{14}{-30}$
(iii) $\frac{-5}{6},\frac{-7}{12},\frac{-13}{18},\frac{23}{-24}$
(iv) $\frac{-10}{11},\frac{-19}{22},\frac{-23}{33},\frac{-39}{44}$

Answer 1

(i) We will first write each of the given numbers with positive denominators. We have:

$\frac{8}{-3}=\frac{8\times (-1)}{-3\times (-1)}=\frac{-8}{3}$

Thus, the given numbers are $-2,\frac{-13}{6},\frac{-8}{3}\mathrm{and}\frac{1}{3}$

LCM of 1, 6, 3 and 3 is 6

Now,

$\frac{-2}{1}=\frac{-2\times 6}{1\times 6}=\frac{-12}{6}$

$\frac{-13}{6}=\frac{-13\times 1}{6\times 1}=\frac{-13}{6}$

$\frac{-8}{3}=\frac{-8\times 2}{3\times 2}=\frac{-16}{6}$

and

$\frac{1}{3}=\frac{1\times 2}{3\times 2}=\frac{2}{6}$

Clearly,Thus,

$\frac{2}{6}>\frac{-12}{6}>\frac{-13}{6}>\frac{-16}{6}$

∴​ $\frac{1}{3}>-2>\frac{-13}{6}>\frac{-8}{3}$. i.e $\frac{1}{3}>-2>\frac{-13}{6}>\frac{8}{-3}$


(ii) We will first write each of the given numbers with positive denominators. We have:

$\frac{7}{-15}=\frac{7\times (-1)}{-15\times (-1)}=\frac{-7}{15}$ and $\frac{17}{-30}=\frac{17\times (-1)}{-30\times (-1)}=\frac{-17}{30}$

Thus, the given numbers are $\frac{-3}{10},\frac{-7}{15},\frac{-11}{20}\mathrm{and}\frac{-17}{30}$

LCM of 10, 15, 20 and 30 is 60

Now,

$\frac{-3}{10}=\frac{-3\times 6}{10\times 6}=\frac{-18}{60}$

$\frac{-7}{15}=\frac{-7\times 4}{15\times 4}=\frac{-28}{60}$

$\frac{-11}{20}=\frac{-11\times 3}{20\times 3}=\frac{-33}{60}$

and

$\frac{-17}{30}=\frac{-17\times 2}{30\times 2}=\frac{-34}{60}$

Clearly,

$\frac{-18}{60}>\frac{-28}{60}>\frac{-33}{60}>\frac{-34}{60}$

∴ $\frac{-3}{10}>\frac{-7}{15}>\frac{-11}{20}>\frac{-17}{30}$. i.e $\frac{-3}{10}>\frac{7}{-15}>\frac{-11}{20}>\frac{17}{-30}$

(iii) We will first write each of the given numbers with positive denominators. We have:

$\frac{23}{-24}=\frac{23\times (-1)}{-24\times (-1)}=\frac{-23}{24}$

Thus, the given numbers are $\frac{-5}{6},\frac{-7}{12},\frac{-13}{18}\mathrm{and}\frac{-23}{24}$

LCM of 6, 12, 18 and 24 is 72

Now,

$\frac{-5}{6}=\frac{-5\times 12}{6\times 12}=\frac{-60}{72}$

$\frac{-7}{12}=\frac{-7\times 6}{12\times 6}=\frac{-42}{72}$

$\frac{-13}{18}=\frac{-13\times 4}{18\times 4}=\frac{-52}{72}$

and

$\frac{-23}{24}=\frac{-23\times 3}{24\times 3}=\frac{-69}{72}$

Clearly,

$\frac{-42}{72}>\frac{-52}{72}>\frac{-60}{72}>\frac{-69}{72}$

∴​ $\frac{-7}{12}>\frac{-13}{18}>\frac{-5}{6}>\frac{-23}{24}$. i.e $\frac{-7}{12}>\frac{-13}{18}>\frac{-5}{6}>\frac{23}{-24}$

(iv) The given numbers are $\frac{-10}{11},\frac{-19}{22},\frac{-23}{33}\mathrm{and}\frac{-39}{44}$

LCM of 11, 22, 33 and 44 is 132

Now,

$\frac{-10}{11}=\frac{-10\times 12}{11\times 12}=\frac{-120}{132}$

$\frac{-19}{22}=\frac{-19\times 6}{22\times 6}=\frac{-114}{132}$

$\frac{-23}{33}=\frac{-23\times 4}{33\times 4}=\frac{-92}{132}$

and

$\frac{-39}{44}=\frac{-39\times 3}{44\times 3}=\frac{-117}{132}$

Clearly,

$\frac{-92}{132}>\frac{-114}{132}>\frac{-117}{132}>\frac{-120}{132}$

∴ $\frac{-23}{33}>\frac{-19}{22}>\frac{-39}{44}>\frac{-10}{11}$