RS Aggarwal solution class 8 chapter 1 Rational Numbers Exercise 1A

Exercise 1A

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Q1 | Ex-1A | Class 8 | RS AGGARWAL | chapter 1 | Rational Numbers | myhelper

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Question 1:

Express $\frac{-3}{5}$ as a rational number with denominator
(i) 20
(ii) −30
(iii) 35
(iv) −40

Answer 1:

If $\frac{a}{b}$ is a fraction and $m$ is a non-zero integer, then $\frac{a}{b}=\frac{a\times m}{b\times m}$.

Now,

(i) $\frac{-3}{5}=\frac{-3\times 4}{5\times 4}=\frac{-12}{20}$

(ii) $\frac{-3}{5}=\frac{-3\times -6}{5\times -6}=\frac{18}{-30}$

(iii)$\frac{-3}{5}=\frac{-3\times 7}{5\times 7}=\frac{-21}{35}$

(iv)$\frac{-3}{5}=\frac{-3\times -8}{5\times -8}=\frac{24}{-40}$

Q2 | Ex-1A | Class 8 | RS AGGARWAL | chapter 1 | Rational Numbers | myhelper

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Question 2:

Express $\frac{-42}{98}$ as a rational number with denominator 7.

Answer 2:

If $\frac{a}{b}$ is a rational number and $m$ is a common divisor of $a \mathrm{and} b$, then $\frac{a}{b}=\frac{a÷m}{b÷m}$.

∴ $\frac{-42}{98}=\frac{-42÷14}{98÷14}=\frac{-3}{7}$

Q3 | Ex-1A | Class 8 | RS AGGARWAL | chapter 1 | Rational Numbers | myhelper

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Question 3:

Express $\frac{-48}{60}$ as a rational number with denominator 5.

Answer 3:

If $\frac{a}{b}$ is a rational integer and $m$ is a common divisor of $a \mathrm{and} b$, then $\frac{a}{b}=\frac{a÷m}{b÷m}$.

∴​ $\frac{-48}{60}=\frac{-48÷12}{60÷12}=\frac{-4}{5}$

Q4 | Ex-1A | Class 8 | RS AGGARWAL | chapter 1 | Rational Numbers | myhelper

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Question 4:

Express each of the following rational numbers in standard form:
(i) $\frac{-12}{30}$
(ii) $\frac{-14}{49}$
(iii) $\frac{24}{-64}$
(iv) $\frac{-36}{-63}$

Answer 4:

A rational number $\frac{a}{b}$ is said to be in the standard form if $a$ and $b$ have no common divisor other than unity and $b>0$.
Thus,

(i) The greatest common divisor of 12 and 30 is 6.
   
     ∴ $\frac{-12}{30}=\frac{-12÷6}{30÷6}=\frac{-2}{5}$ (In the standard form)

(ii)The greatest common divisor of 14 and 49 is 7.    
     
    ∴ $\frac{-14}{49}=\frac{-14÷7}{49÷7}=\frac{-2}{7}$ (In the standard form)

(iii) $\frac{24}{-64}=\frac{24\times (-1)}{-64\times -1}=\frac{-24}{64}$
   
     The greatest common divisor of 24 and 64 is 8.          
    
     ∴ $\frac{-24}{64}=\frac{-24÷8}{64÷8}=\frac{-3}{8}$ (In the standard form)

(iv) $\frac{-36}{-63}=\frac{-36\times (-1)}{-63\times -1}=\frac{36}{63}$
 
     The greatest common divisor of 36 and 63 is 9.     
  
      ∴ $\frac{36}{63}=\frac{36÷9}{63÷9}=\frac{4}{7}$ (In the standard form)

Q5 | Ex-1A | Class 8 | RS AGGARWAL | chapter 1 | Rational Numbers | myhelper

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Question 5:

Which of the two rational numbers is greater in the given pair?
(i) $\frac{3}{8} \mathrm{or} 0$
(ii) $\frac{-2}{9} \mathrm{or} 0$
(iii) $\frac{-3}{4} \mathrm{or} \frac{1}{4}$
(iv) $\frac{-5}{7} \mathrm{or} \frac{-4}{7}$
(v) $\frac{2}{3} \mathrm{or} \frac{3}{4}$
(vi) $\frac{-1}{2} \mathrm{or} -1$

Answer 5:

We know:
(i) Every positive rational number is greater than 0.
(ii) Every negative rational number is less than 0.

Thus, we have:

(i)$\frac{3}{8}$ is a positive rational number.
    ∴ $\frac{3}{8}>0$

(ii)$\frac{-2}{9}$ is a negative rational number.
    ∴ $\frac{-2}{9}<0$

(iii) $\frac{-3}{4}$ is a negative rational number.
    ∴ $\frac{-3}{4}<0$
    Also,
    $\frac{1}{4}$ is a positive rational number.
    ∴ $\frac{1}{4}>0$
    Combining the two inequalities, we get:
   $\frac{-3}{4}<\frac{1}{4}$

(iv)Both $\frac{-5}{7}$ and $\frac{-4}{7}$ have the same denominator, that is, 7.
    So, we can directly compare the numerators.

    ∴ $\frac{-5}{7}<\frac{-4}{7}$

(v)The two rational numbers are $\frac{2}{3}$ and $\frac{3}{4}$.
    The LCM of the denominators 3 and 4 is 12.
    Now,
   $\frac{2}{3}=\frac{2\times 4}{3\times 4}=\frac{8}{12}$
    Also, 
   $\frac{3}{4}=\frac{3\times 3}{4\times 3}=\frac{9}{12}$
    Further
   $\frac{8}{12}<\frac{9}{12}$

    ∴$\frac{2}{3}<\frac{3}{4}$

(vi)The two rational numbers are $\frac{-1}{2}$ and $-1$.
    We can write $-1=\frac{-1}{1}$.
    The LCM of the denominators 2 and 1 is 2.
    Now,
    $\frac{-1}{2}=\frac{-1\times 1}{2\times 1}=\frac{-1}{2}$
    Also,
    $\frac{-1}{1}=\frac{-1\times 2}{1\times 2}=\frac{-2}{2}$
    ∵ $\frac{-2}{1}<\frac{-1}{1}$
    ∴ $-1<\frac{-1}{2}$

Q6 | Ex-1A | Class 8 | RS AGGARWAL | chapter 1 | Rational Numbers | myhelper

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Question 6:

Which of the two rational numbers is greater in the given pair?
(i) $\frac{-4}{3} \mathrm{or} \frac{-8}{7}$

Sol :

The two rational numbers are $\frac{-4}{3}\mathrm{and}\frac{-8}{7}$.

The LCM of the denominators 3 and 7 is 21.

Now,
 
$\frac{-4}{3}=\frac{-4\times 7}{3\times 7}=\frac{-28}{21}$

Also,

$\frac{-8}{7}=\frac{-8\times 3}{7\times 3}=\frac{-24}{21}$

Further,
 
$\frac{-28}{21}<\frac{-24}{21}$

∴ $\frac{-4}{3}<\frac{-8}{7}$

(ii) $\frac{7}{-9} \mathrm{or} \frac{-5}{\mathrm{8 }}$

Sol :

The two rational numbers are $\frac{7}{-9}\mathrm{and}\frac{-5}{8}$.

The first fraction can be expressed as $\frac{7}{-9}=\frac{7\times -1}{-9\times -1}=\frac{-7}{9}$.

The LCM of the denominators 9 and 8 is 72.

Now, 

$\frac{-7}{9}=\frac{-7\times 8}{9\times 8}=\frac{-56}{72}$

Also,

$\frac{-5}{8}=\frac{-5\times 9}{8\times 9}=\frac{-45}{72}$

Further,
 
$\frac{-56}{72}<\frac{-45}{72}$

∴​ $\frac{7}{-9}<\frac{-5}{8}$

(iii) $\frac{-1}{3} \mathrm{or} \frac{4}{-5}$​

Sol :

The two rational numbers are $\frac{-1}{3}\mathrm{and}\frac{4}{-5}$.

$\frac{4}{-5}=\frac{4\times -1}{-5\times -1}=\frac{-4}{5}$

The LCM of the denominators 3 and 5 is 15.

Now, 

$\frac{-1}{3}=\frac{-1\times 5}{3\times 5}=\frac{-5}{15}$

Also,

$\frac{-4}{5}=\frac{-4\times 3}{5\times 3}=\frac{-12}{15}$

Further,
 
$\frac{-12}{15}<\frac{-5}{15}$

∴ $\frac{4}{-5}<\frac{-1}{3}$

(iv) $\frac{9}{-13} \mathrm{or} \frac{7}{-12}$

Sol :

The two rational numbers are $\frac{9}{-13}\mathrm{and}\frac{7}{-12}$.

$\mathrm{Now}, \frac{ 9}{-13}=\frac{9\times -1}{-13\times -1}=\frac{-9}{13} \mathrm{and} \frac{7}{-12}=\frac{7\times -1}{-12\times -1}=\frac{-7}{12}$

The LCM of the denominators 13 and 12 is 156.

Now, 

$\frac{-9}{13}=\frac{-9\times 12}{13\times 12}=\frac{-108}{156}$

Also,

$\frac{-7}{12}=\frac{-7\times 13}{12\times 13}=\frac{-91}{156}$

Further,
 
$\frac{-108}{156}<\frac{-91}{156}$

∴ $\frac{9}{-13}<\frac{7}{-12}$

(v) $\frac{4}{-5} \mathrm{or} \frac{-7}{10}$

Sol :

The two rational numbers are $\frac{4}{-5} \mathrm{and} \frac{-7}{10}$.

∴​ $\frac{4}{-5}=\frac{4\times -1}{-5\times -1}=\frac{-4}{5}$

The LCM of the denominators 5 and 10 is 10.

Now,
 
$\frac{-4}{5}=\frac{-4\times 2}{5\times 2}=\frac{-8}{10}$

Also,

$\frac{-7}{10}=\frac{-7\times 1}{10\times 1}=\frac{-7}{10}$

Further,
 
$\frac{-8}{10}<\frac{-7}{10}$

∴ $\frac{-4}{5}<\frac{-7}{10}, or, \frac{4}{-5}<\frac{-7}{10}$

(vi) $\frac{-12}{5} \mathrm{or} -3$

Sol :

The two rational numbers are $\frac{-12}{5}\mathrm{and} -3.$
$-3 \mathrm{can} \mathrm{be} \mathrm{written} \mathrm{as} \frac{-3}{1}$.

The LCM of the denominators is 5.

Now,
 
$\frac{-3}{1}=\frac{-3\times 5}{1\times 5}=\frac{-15}{5}$

Because $\frac{-15}{5}<\frac{-12}{5}$, we can conclude that $-3<\frac{-12}{5}$.

Q7 | Ex-1A | Class 8 | RS AGGARWAL | chapter 1 | Rational Numbers | myhelper (METHOD-1)

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Q7 | Ex-1A | Class 8 | RS AGGARWAL | chapter 1 | Rational Numbers | myhelper (METHOD-2)

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Question 7:

Fill in the blanks with the correct symbol out of >, = and <:
(i) $\frac{-3}{7}.....\frac{6}{-13}$
(ii) $\frac{5}{-13}.....\frac{-35}{91}$
(iii) $-2 .....\frac{-13}{5}$
(iv) $\frac{-2}{3}.....\frac{5}{-8}$
(v) $0 .....\frac{-3}{-5}$
(vi) $\frac{-8}{9}.....\frac{-9}{10}$

Answer 7:

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

One number = $\frac{-3}{7}$ 
Other number =$\frac{6}{-13}=\frac{6\times (-1)}{-13\times (-1)}=\frac{-6}{13}$

 LCM of 7 and 13 = 91

 ∴ $\frac{-3}{7}=\frac{-3\times 13}{7\times 13}=\frac{-39}{91}$ 

And,

$\frac{-6}{13}=\frac{-6\times 7}{13\times 7}=\frac{-42}{91}$$\frac{-6}{13}=\frac{-6\times 7}{13\times 7}=\frac{-42}{91}$$\frac{-6}{13}=\frac{-6\times 7}{13\times 7}=\frac{-42}{91}$

Clearly,

$-39>-41$

∴ ​$\frac{-39}{91 }>\frac{-42}{91}$

Thus,

$\frac{-3}{7}>\frac{6}{-13}$

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

One number = $\frac{5}{-13}=\frac{5\times (-1)}{-13\times (-1)}=\frac{-5}{13}$ 

Other number =$\frac{-35}{91}$

 LCM of 13 and 91 = 91

 ∴ $\frac{-5}{13}=\frac{-5\times 7}{13\times 7}=\frac{-35}{91}$ and $\frac{-35}{91}$

Clearly,
 
$-35=-35$

 ∴ $\frac{-35}{91 }=\frac{-35}{91}$

Thus,
 
$\frac{-5}{13}=\frac{-35}{91}$ 
 

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

One number = $-2$
 
We can write -2 as$\frac{-2}{1}$.
Other number =$\frac{-13}{5}$

 LCM of 1 and 5 = 5

 ∴​ $\frac{-2}{1}=\frac{-2\times 5}{1\times 5}=\frac{-10}{5}$ and $\frac{-13}{5}=\frac{-13\times 1}{5\times 1}=\frac{-13}{5}$

Clearly,

$-10>-13$

  ∴ $\frac{-10}{5}>\frac{-13}{5}$

Thus,
 
$\frac{-2}{1}>\frac{-13}{5}$ 
 
$-2>\frac{-13}{5}$

(iv) We will write each of the given numbers with positive denominators.

One number = $\frac{-2}{3}$ 
Other number =$\frac{5}{-8}=\frac{5\times (-1)}{-8\times (-1)}=\frac{-5}{8}$

 LCM of 3 and 8 = 24

∴ ​$\frac{-2}{3}=\frac{-2\times 8}{3\times 8}=\frac{-16}{24}$ and $\frac{-5}{8}=\frac{-5\times 3}{8\times 3}=\frac{-15}{24}$

Clearly,

$-16<-15$
 
∴ $\frac{-16}{24}<\frac{-15}{24}$

Thus,
 
$\frac{-2}{3}<\frac{-5}{8}$ 
 
$\frac{-2}{3}<\frac{5}{-8}$

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

$\frac{3}{5}$ is a positive number.

Because every positive rational number is greater than 0, $\frac{3}{5}>0\Rightarrow 0<\frac{3}{5}$.

(vi) We will write each of the given numbers with positive denominators.

One number = $\frac{-8}{9}$

Other number = $\frac{-9}{10}$

 LCM of 9 and 10 = 90

∴​$\frac{-8}{9}=\frac{-8\times 10}{9\times 10}=\frac{-80}{90}$ and $\frac{-9}{10}=\frac{-9\times 9}{10\times 9}=\frac{-81}{90}$

Clearly,

$-81<-80$

∴​$\frac{-81}{90}<\frac{-80}{90}$

Thus,
 
$\frac{-9}{10}<\frac{-8}{9}$ 

Q8 | Ex-1A | Class 8 | RS AGGARWAL | chapter 1 | Rational Numbers | myhelper

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Question 8:

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

Answer 8:

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

We have:

$\frac{4}{-9}=\frac{4\times (-1)}{-9\times (-1)}=\frac{-4}{9}$ and$\frac{7}{-18}=\frac{7\times (-1)}{-18\times (-1)}=\frac{-7}{18}$

Thus, the given numbers are $\frac{-4}{9}, \frac{-5}{12}, \frac{-7}{18} \mathrm{and} \frac{-2}{3}.$

LCM of 9, 12, 18 and 3 is 36.

Now, 

$\frac{-4}{9}=\frac{-4\times 4}{9\times 4}=\frac{-16}{36}$

$\frac{-5}{12}=\frac{-5\times 3}{12\times 3}=\frac{-15}{36}$

$\frac{-7}{18}=\frac{-7\times 2}{18\times 2}=\frac{-14}{36}$

$\frac{-2}{3}=\frac{-2\times 12}{3\times 12}=\frac{-24}{36}$

Clearly, 

$\frac{-24}{36}<\frac{-16}{36}<\frac{-15}{36}<\frac{-14}{36}$

∴ ​$\frac{-2}{3}<\frac{-4}{9}<\frac{-5}{12}<\frac{-7}{18}$ 
 
That is

$\frac{-2}{3}<\frac{4}{-9}<\frac{-5}{12}<\frac{7}{-18}$

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

We have:

$\frac{5}{-12}=\frac{5\times (-1)}{-12\times (-1)}=\frac{-5}{12}$ and$\frac{9}{-24}=\frac{9\times (-1)}{-24\times (-1)}=\frac{-9}{24}$

Thus, the given numbers are $\frac{-3}{4}, \frac{-5}{12}, \frac{-7}{16} \mathrm{and} \frac{-9}{24}.$

LCM of  4, 12, 16 and 24 is 48.

Now,
 
$\frac{-3}{4}=\frac{-3\times 12}{4\times 12}=\frac{-36}{48}$

$\frac{-5}{12}=\frac{-5\times 4}{12\times 4}=\frac{-20}{48}$

$\frac{-7}{16}=\frac{-7\times 3}{16\times 3}=\frac{-21}{48}$

$\frac{-9}{24}=\frac{-9\times 2}{24\times 2}=\frac{-18}{48}$

Clearly, 

$\frac{-36}{48}<\frac{-21}{48}<\frac{-20}{48}<\frac{-18}{48}$

∴​ $\frac{-3}{4}<\frac{-7}{16}<\frac{-5}{12}<\frac{-9}{24}$ 

That is

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

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

We have:

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

Thus, the given numbers are $\frac{-3}{5}, \frac{-7}{10}, \frac{-11}{15} \mathrm{and} \frac{-13}{20}.$

LCM of 5, 10, 15 and 20 is 60.

Now, 

$\frac{-3}{5}=\frac{-3\times 12}{5\times 12}=\frac{-36}{60}$

$\frac{-7}{10}=\frac{-7\times 6}{10\times 6}=\frac{-42}{60}$

$\frac{-11}{15}=\frac{-11\times 4}{15\times 4}=\frac{-44}{60}$

$\frac{-13}{20}=\frac{-13\times 3}{20\times 3}=\frac{-39}{60}$

Clearly,
 
$\frac{-44}{60}<\frac{-42}{60}<\frac{-39}{60}<\frac{-36}{60}$

∴​ $\frac{-11}{15}<\frac{-7}{10}<\frac{-13}{20}<\frac{-3}{5}$.

That is 

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

(iv) We will write each of the given numbers with positive denominators.

We have:

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

Thus, the given numbers are $\frac{-4}{7}, \frac{-9}{14}, \frac{-13}{28} \mathrm{and} \frac{-23}{42}.$

LCM of 7, 14, 28 and 42 is 84.

Now, 

$\frac{-4}{7}=\frac{-4\times 12}{7\times 12}=\frac{-48}{84}$

$\frac{-9}{14}=\frac{-9\times 6}{14\times 6}=\frac{-54}{84}$

$\frac{-13}{28}=\frac{-13\times 3}{28\times 3}=\frac{-39}{84}$

$\frac{-23}{42}=\frac{-23\times 2}{42\times 2}=\frac{-46}{84}$

Clearly, 

$\frac{-54}{84}<\frac{-48}{84}<\frac{-46}{84}<\frac{-39}{84}$

∴​ $\frac{-9}{14}<\frac{-4}{7}<\frac{-23}{42}<\frac{-13}{28}$.
 
That is

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

Q9 | Ex-1A | Class 8 | RS AGGARWAL | chapter 1 | Rational Numbers | myhelper

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Question 9:

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 9:

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

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

Q10 | Ex-1A | Class 8 | RS AGGARWAL | chapter 1 | Rational Numbers | myhelper

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Question 10:

Which of the following statements are true and which are false?
(i) Every whole number is a rational number.
(ii) Every integer is a rational number.
(iii) 0 is a whole number but it is not a rational number.

Answer 10:

1. True
A whole number can be expressed as $\frac{a}{b}, \mathrm{with} b=1 \mathrm{and} a\ge 0$. Thus, every whole number is rational.

2. True
Every integer is a rational number because any integer can be expressed as $\frac{a}{b}, \mathrm{with} b=1 \mathrm{and} 0>a\ge 0$. Thus, every integer is a rational number.

3. False
$0=\frac{a}{b}, \mathrm{for} a=0 \mathrm{and} b\ne 0.$ Thus, 0 is a rational and whole number.