ML Aggarwal Solution Class 9 Chapter 1 Rational and Irrational Numbers Exercise 1.1

 Exercise 1.1

Question 1

Sol :

Given rational numbers are $\frac{2}{9}$ and $\frac{3}{8}$

L.C.M of denominators 9 and 8 is 72

Equivalent fractions of $\frac{2}{9}$ and $\frac{3}{8}$ with denominator 72

$\frac{2}{9}=\frac{2\times 8}{9\times 8}=\frac{16}{72}$

$\frac{3}{8}=\frac{3\times 9}{8\times 9}=\frac{27}{72}$

Since, 16<27,  $\frac{16}{72}<\frac{27}{72}$

$\frac{2}{9}<\frac{3}{8}$

A rational number between $\frac{2}{9}$ and $\frac{3}{8}$ is

$=\frac{\frac{2}{9}+\frac{3}{8}}{2}$

$=\frac{\frac{2 \times 8+3 \times 9}{72}}{2}$

$=\frac{16+27}{72 \times 2}$

$=\frac{43}{144}$

Descending order of the numbers is $\frac{3}{8}, \frac{43}{144}, \frac{2}{9}$

$\frac{2}{9} < \frac{43}{144}<\frac{3}{8}$


Question 2

Sol :

Method I

L.C.M of 3 and 4 is 12

Rational number between $\frac{1}{3}$ and $\frac{1}{4}$ is $\frac{\frac{1}{3}+\frac{1}{4}}{2}$

$=\frac{\frac{4+3}{12}}{2}$

$=\frac{7}{12 \times 2}$

$=\frac{7}{24}$


$\frac{1}{3}=\frac{1 \times 4}{3 \times 4}=\frac{4}{12}$

$\frac{1}{4}=\frac{1 \times 3}{4 \times 3}=\frac{3}{12}$

Since, 4>3 

$\frac{4}{12}>\frac{3}{12}$

$\frac{1}{3}>\frac{1}{4}$

'$\frac{7}{24}$' in between $\frac{1}{3}$ and $\frac{1}{4}$So, $ \frac{1}{3}>\frac{7}{24}>\frac{1}{4}$...(i)

Rational number between $\frac{1}{3}$ and $\frac{7}{24}$ is $\frac{\frac{1}{3}+\frac{7}{24}}{2}$

$=\frac{\frac{1 \times 8+7 \times 1}{24}}{2}$

$=\frac{15}{2 \times 24}$

$=\frac{15}{48}$

$\frac{15}{48}$ is in between $\frac{1}{3}$ and $\frac{7}{24}$ So, $\frac{1}{3}>\frac{15}{48}>\frac{7}{24}$...(2)

From (1) and (2) ascending order of number (increasing order) is $\frac{1}{4}<\frac{7}{24}<\frac{15}{48}<\frac{1}{3}$


Question 3

Sol :

Given rational numbers are $-\frac{1}{3}$ and $-\frac{1}{2}$

L.C.M of 3 and 2 is 6

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

Since , 2<3

-2>-3

$\frac{-2}{6}>\frac{-3}{6}$

So, $\frac{-1}{3}>\frac{-1}{2}$

Rational number between $\frac{-1}{3}$ and $\frac{-1}{2}$ is $\frac{\frac{-1}{3}+\left(\frac{-1}{2}\right)}{2}$

$\frac{\frac{-1 \times 2+(-1)(3)}{6}}{2}$

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

$=\frac{-5}{12}$

∴$\frac{-1}{3}>\frac{-5}{12}>\frac{-1}{2}$...(i)


Question 4

Sol :

L.C.M of 3 and 5 is 15

$\frac{1}{3}=\frac{1 \times 5}{3 \times 5}=\frac{5}{15}$, $\frac{4}{5}=\frac{4 \times 3}{5 \times 3}=\frac{12}{15}$

Since, 5<12

$\frac{5}{15}<\frac{12}{15}$

$\frac{1}{3}<\frac{4}{5}$

Rational number between $\frac{1}{3}$ and $\frac{4}{5}$ is

$\frac{\frac{1}{3}+\frac{4}{5}}{2}$

$=\frac{\frac{1 \times 5+4 \times 3}{3 \times 5}}{2}$

$=\frac{\frac{5+12}{15}}{2}$

$=\frac{17}{30}$



Rational number between $\frac{1}{3}$ and $\frac{17}{30}$ is 

$\frac{\frac{1}{3}+\frac{17}{30}}{2}$

$=\frac{\frac{1 \times 10+17}{30}}{2}$

$=\frac{27}{60}$




Rational number between '$\frac{17}{30}$' and '$\frac{4}{5}$' is 

$\frac{\frac{17}{30}+\frac{4}{5}}{2}$

$=\frac{\frac{17+4 \times 6}{30}}{2}$

$=\frac{41}{60}$




Rational numbers between $\frac{1}{3}$ and $\frac{4}{5}$ are




Descending order(decreasing order) of numbers are
$\frac{4}{5}, \frac{41}{60}, \frac{17}{30}, \frac{27}{60}, \frac{1}{3}$


Question 5

Sol :

A rational number between 4 and 4.5 is $\frac{4+4 \cdot 5}{2}=\frac{8 \cdot 5}{2}$=4.25

A rational number between 4 and 4.25 is $=\frac{4+4 \cdot 25}{2}=\frac{8 \cdot 25}{2}$

=4.125

A rational number between 4 and 4.125 is $=\frac{4+4 \cdot 125}{2}=\frac{8 \cdot 125}{2}$

=4.0625

Three rational number between 4 and 4.5 is 4.0625, 4.125, 4.25


Question 6

Sol :

We need to insert six rational number between 3 and 4. SO, we multiply both numerator and denominator of rational numbers with 6+1 i.e. 7
So,$\frac{3}{1}=\frac{3 \times 7}{1 \times 7}=\frac{21}{7}$
$\frac{4}{1}=\frac{4 \times 7}{1 \times 7}=\frac{28}{7}$

We have 21<22<23<24<25<26<27
$\frac{21}{2}<\frac{22}{7}<\frac{23}{7}<\frac{24}{7}<\frac{25}{7}<\frac{26}{7}<\frac{27}{7}<\frac{28}{7}$

Therefore, six rational numbers between 3 and 4 are
$\frac{22}{7}, \frac{23}{7}, \frac{24}{7}, \frac{25}{7}, \frac{26}{7}, \frac{27}{7}$

Question 7

Sol :

We need to insert five rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$. So, we multiply both numerator and denominator with 5+1 i.e. 6

So, $\frac{3}{5}=\frac{3 \times 6}{5 \times 6}=\frac{18}{30}$

$\frac{4}{5}=\frac{4 \times 6}{5 \times 6}=\frac{24}{30}$

We have 18<19<20<21<22<23<24

$\frac{18}{30}<\frac{19}{30}<\frac{20}{30}<\frac{21}{30}<\frac{22}{30}<\frac{23}{30}<\frac{24}{30}$

Therefore, five rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$

$\frac{19}{30}, \frac{20}{30}, \frac{21}{30}, \frac{22}{30}, \frac{23}{30}$


Question 8

Sol :

L.C.M of 5 and 7 is 35

$\frac{-2}{5}=\frac{-2 \times 7}{5 \times 7}=\frac{-14}{35}$ 

$ \frac{1}{7}=\frac{1 \times 5}{7 \times 5}=\frac{5}{35}$

We need to inset ten rational numbers between $\frac{-2}{5}\left(=\frac{-14}{35}\right)$ and $\frac{1}{7}\left(=\frac{5}{35}\right)$.

So, we can select any ten numbers between -14 and 5 as numerator and 35 as denominator

$\frac{-13}{35}, \frac{-12}{35}, \frac{-11}{35}, \frac{-10}{35}, \frac{-9}{35}, \frac{-8}{35}, \frac{-7}{35}, \frac{-6}{35},\frac{-5}{35}, \frac{-4}{35}$ are ten rational numbers which are in between $\frac{-2}{5}$ and $\frac{1}{7}$


Question 9

Sol :

L C M of 2 and 3 is 6

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

$\frac{2}{3}=\frac{2 \times 2}{3 \times 2}=\frac{4}{6}$

We need to inset six rational numbers. So, multiplying both numerator and denominator by 6+1 i.e. 7 

$\frac{3}{6}=\frac{3 \times 7}{6 \times 7}=\frac{21}{42}$

$\frac{4}{6}=\frac{4 \times 7}{6 \times 7}=\frac{28}{42}$

Since, 21<22<23<24<25<26<27<28

$\frac{21}{42}<\frac{22}{42}<\frac{23}{42}<\frac{24}{42}<\frac{25}{42}<\frac{26}{42}<\frac{27}{42}<\frac{28}{42}$

Therefore, Six numbers between $\frac{1}{2}$ and $\frac{2}{3}$ are

$\frac{22}{42}, \frac{23}{42}, \frac{24}{42}, \frac{25 .}{42}, \frac{26}{42}, \frac{27}{42}$

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