Rational Numbers

To get the denominator 7, divide both the numerator and the denominator of the given rational number by the same number.

Note that 112 ÷ 7 = 16;

$\frac{-64}{112} = \frac{-64 \div 16}{112 \div 16} = \frac{-4}{7}$

Hence, $\mathbf{\frac{-64}{112} = \frac{-4}{7}}.$

First reduce the given rational number to its lowest terms, then multiply both the numerator and the denominator by the same number to get the denominator 25.

HCF of 48 and 60 = 12.

$$\begin{aligned}\Rightarrow \frac{-48}{60} = \frac{-48 \div 12}{60 \div 12} = \frac{-4}{5} \\[1em] [5 \times 5 = 25] \\[1em] \Rightarrow \frac{-4}{5} = \frac{-4 \times 5}{5 \times 5} \\[1em] \Rightarrow \frac{-20}{25}\end{aligned}$$

Hence, $\mathbf{\frac{-48}{60} = \frac{-20}{25}}$.

A rational number $\frac{p}{q}$ is in standard form if $q \gt 0$ and $p, q$ have no common divisor other than 1.

Here the denominator 30 is already positive.

The greatest common divisor of 12 and 30 is 6.

Now,

$\frac{-12}{30} =\frac{-12 \div 6}{30 \div 6} = \frac{-2}{5}$

Hence, the standard form of $\mathbf{\frac{-12}{30} \text{ is } \frac{-2}{5}}$.

A rational number $\frac{p}{q}$ is in standard form if $q \gt 0$ and $p, q$ have no common divisor other than 1.

Here the denominator 49 is already positive.

The greatest common divisor of 14 and 49 is 7.

Now,

$\frac{-14}{49} = \frac{-14 \div 7}{49 \div 7} = \frac{-2}{7}$

Hence, the standard form of $\mathbf{\frac{-14}{49} \text{ is } \frac{-2}{7}}$.

A rational number $\frac{p}{q}$ is in standard form if $q \gt 0$ and $p, q$ have no common divisor other than 1.

Here the denominator -64 is negative, so first express the rational number with a positive denominator.

$\frac{24}{-64} = \frac{24 \times (-1)}{-64 \times (-1)} = \frac{-24}{64}$

The greatest common divisor of 24 and 64 is 8.

Now,

$\frac{-24}{64} = \frac{-24 \div 8}{64 \div 8} = \frac{-3}{8}$

Hence, the standard form of $\mathbf{\frac{24}{-64} \text{ is } \frac{-3}{8}}$.

A rational number $\frac{p}{q}$ is in standard form if $q \gt 0$ and $p, q$ have no common divisor other than 1.

Here the denominator -63 is negative, so first express the rational number with a positive denominator.

$\frac{-36}{-63} = \frac{-36 \times (-1)}{-63 \times (-1)} = \frac{36}{63}$

The greatest common divisor of 36 and 63 is 9.

Now,

$\frac{36}{63} = \frac{36 \div 9}{63 \div 9} = \frac{4}{7}$

Hence, the standard form of $\mathbf{\frac{-36}{-63} \text{ is } \frac{4}{7}}$.

Write both the rational numbers with the LCM of their denominators as the common denominator and compare the numerators.

L.C.M. of 32 and 24 = 96

$$\begin{aligned}\Rightarrow \frac{15}{32} = \frac{15 \times 3}{32 \times 3} = \frac{45}{96} \\[1em] \Rightarrow \frac{17}{24} = \frac{17 \times 4}{24 \times 4} = \frac{68}{96}\end{aligned}$$

As 45 < 68, so $\frac{45}{96} \lt \frac{68}{96}$.

Hence, $\mathbf{\frac{15}{32} \lt \frac{17}{24}}$.

Write both the rational numbers with the LCM of their denominators as the common denominator and compare the numerators.

L.C.M. of 11 and 18 = 198

$$\begin{aligned}\Rightarrow \frac{10}{11} = \frac{10 \times 18}{11 \times 18} = \frac{180}{198} \\[1em] \Rightarrow \frac{17}{18} = \frac{17 \times 11}{18 \times 11} = \frac{187}{198}\end{aligned}$$

As 180 < 187, so $\frac{180}{198} \lt \frac{187}{198}$

Hence, $\mathbf{\frac{10}{11} \lt \frac{17}{18}}$.

Write both the rational numbers with the LCM of their denominators as the common denominator and compare the numerators.

L.C.M. of 12 and 4 = 12

$$\begin{aligned}\Rightarrow \frac{-5}{12} = \frac{-5}{12} \\[1em] \Rightarrow \frac{-3}{4} = \frac{(-3) \times 3}{4 \times 3} = \frac{-9}{12}\end{aligned}$$

As -9 < -5, so $\frac{-9}{12} \lt \frac{-5}{12}$.

Hence, $\mathbf{\frac{-5}{12} \gt \frac{-3}{4}}$.

Write both the rational numbers with the LCM of their denominators as the common denominator and compare the numerators.

First express $\frac{9}{-20}$ with a positive denominator.

$\frac{9}{-20} = \frac{9 \times (-1)}{-20 \times (-1)} = \frac{-9}{20}$

L.C.M. of 24 and 20 = 120

$$\begin{aligned}\Rightarrow \frac{-7}{24} = \frac{(-7) \times 5}{24 \times 5} = \frac{-35}{120} \\[1em] \Rightarrow \frac{-9}{20} = \frac{(-9) \times 6}{20 \times 6} = \frac{-54}{120}\end{aligned}$$

As -54 < -35, so $\frac{-54}{120} \lt \frac{-35}{120}$.

Hence, $\mathbf{\frac{-7}{24} \gt \frac{9}{-20}}$.

All the denominators are already positive, so write each rational number with the LCM of the denominators as the common denominator and arrange the numerators in an increasing order.

LCM of 6, 9, 12 and 18 = 36

$$\begin{aligned}\Rightarrow \frac{5}{6} = \frac{5 \times 6}{6 \times 6} = \frac{30}{36} \\[1em] \Rightarrow \frac{7}{9} = \frac{7 \times 4}{9 \times 4} = \frac{28}{36} \\[1em] \Rightarrow \frac{11}{12} = \frac{11 \times 3}{12 \times 3} = \frac{33}{36} \\[1em] \Rightarrow \frac{13}{18} = \frac{13 \times 2}{18 \times 2} = \frac{26}{36}\end{aligned}$$

As 26 < 28 < 30 < 33

So, $\frac{26}{36} \lt \frac{28}{36} \lt \frac{30}{36} \lt \frac{33}{36}$

$\Rightarrow \frac{13}{18} \lt \frac{7}{9} \lt \frac{5}{6} \lt \frac{11}{12}$

Hence, the given rational numbers in ascending order are $\mathbf{\frac{13}{18}, \frac{7}{9}, \frac{5}{6} \text{ and } \frac{11}{12}}$.

First write each rational number with a positive denominator, then write each of them with the LCM of the denominators as the common denominator and arrange the numerators in increasing order.

$$\begin{aligned}\Rightarrow \frac{5}{-7} = \frac{5 \times (-1)}{-7 \times (-1)} = \frac{-5}{7} \\[1em] \Rightarrow \frac{7}{-12} = \frac{7 \times (-1)}{-12 \times (-1)} = \frac{-7}{12}\end{aligned}$$

So, the given rational numbers are $\frac{-5}{7}, \frac{-9}{14}, \frac{-5}{6} \text{ and } \frac{-7}{12}$.

Now write each rational number with the LCM of the denominators as the common denominator.

LCM of 7, 14, 6 and 12 = 84

$$\begin{aligned}\Rightarrow \frac{-5}{7} = \frac{(-5) \times 12}{7 \times 12} = \frac{-60}{84} \\[1em] \Rightarrow \frac{-9}{14} = \frac{(-9) \times 6}{14 \times 6} = \frac{-54}{84} \\[1em] \Rightarrow \frac{-5}{6} = \frac{(-5) \times 14}{6 \times 14} = \frac{-70}{84} \\[1em] \Rightarrow \frac{-7}{12} = \frac{(-7) \times 7}{12 \times 7} = \frac{-49}{84}\end{aligned}$$

As -70 < -60 < -54 < -49,

So $\frac{-70}{84} \lt \frac{-60}{84} \lt \frac{-54}{84} \lt \frac{-49}{84}$

$\Rightarrow \frac{-5}{6} \lt \frac{-5}{7} \lt \frac{-9}{14} \lt \frac{-7}{12}$

Hence, the given rational numbers in ascending order are $\mathbf{\frac{-5}{6}, \frac{5}{-7}, \frac{-9}{14} \text{ and } \frac{7}{-12}}$.

First write each rational number with a positive denominator, then write each of them with the LCM of the denominators as the common denominator and arrange the numerators in increasing order.

$$\begin{aligned}\Rightarrow -2 = \frac{-2}{1} \\[1em] \Rightarrow \frac{8}{-3} = \frac{8 \times (-1)}{-3 \times (-1)} = \frac{-8}{3}\end{aligned}$$

So, the given rational numbers are $\frac{-2}{1}, \frac{1}{3}, \frac{-13}{6} \text{ and } \frac{-8}{3}$.

Now write each rational number with the LCM of the denominators as the common denominator.

LCM of 1, 3, 6 and 3 = 6

$$\begin{aligned}\Rightarrow \frac{-2}{1} = \frac{(-2) \times 6}{1 \times 6} = \frac{-12}{6} \\[1em] \Rightarrow \frac{1}{3} = \frac{1 \times 2}{3 \times 2} = \frac{2}{6} \\[1em] \Rightarrow \frac{-13}{6} = \frac{-13}{6} \\[1em] \Rightarrow \frac{-8}{3} = \frac{(-8) \times 2}{3 \times 2} = \frac{-16}{6}\end{aligned}$$

As -16 < -13 < -12 < 2,

so $\frac{-16}{6} \lt \frac{-13}{6} \lt \frac{-12}{6} \lt \frac{2}{6}$

$\Rightarrow \frac{-8}{3} \lt \frac{-13}{6} \lt -2 \lt \frac{1}{3}$

Hence, the given rational numbers in ascending order are $\mathbf{\frac{8}{-3}, \frac{-13}{6}, -2 \text{ and } \frac{1}{3}}$.

First write each rational number with a positive denominator, then write each of them with the LCM of the denominators as the common denominator and arrange the numerators in increasing order.

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

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

Now write each rational number with the LCM of the denominators as the common denominator.

LCM of 28, 42, 7 and 14 = 84

$$\begin{aligned}\Rightarrow \frac{-13}{28} = \frac{(-13) \times 3}{28 \times 3} = \frac{-39}{84} \\[1em] \Rightarrow \frac{-23}{42} = \frac{(-23) \times 2}{42 \times 2} = \frac{-46}{84} \\[1em] \Rightarrow \frac{-4}{7} = \frac{(-4) \times 12}{7 \times 12} = \frac{-48}{84} \\[1em] \Rightarrow \frac{-9}{14} = \frac{(-9) \times 6}{14 \times 6} = \frac{-54}{84}\end{aligned}$$

As -54 < -48 < -46 < -39

so $\frac{-54}{84} \lt \frac{-48}{84} \lt \frac{-46}{84} \lt \frac{-39}{84}$

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

Hence, the given rational numbers in ascending order are $\mathbf{\frac{-9}{14}, \frac{-4}{7}, \frac{-23}{42} \text{ and } \frac{13}{-28}}$.

Draw a number line and mark a point O on it, representing the integer 0. Each point of division represents an integer and each length between two successive integers is a unit length.

As $\frac{5}{6}$ is a positive proper fraction, it lies between 0 and 1.

Divide the unit length between 0 and 1 into 6 equal segments by making 5 equally spaced marks. Starting from 0, count forward 5 segments to the right. Label this specific position as point A, representing $\frac{5}{6}$ on the number line, as shown below :

Hence, the point A on the number line represents $\mathbf{\frac{5}{6}}$.

Draw a number line and mark a point O on it, representing the integer 0. Each point of division represents an integer and each length between two successive integers is a unit length.

Convert the improper fraction into a mixed fraction to find the two integers between which it lies.

$\frac{14}{3} = 4\frac{2}{3}$

So, $\frac{14}{3}$ lies between 4 and 5.

Divide the unit length between 4 and 5 into 3 equal segments by making 2 equally spaced marks. Starting from the integer 4, count forward (to the right) 2 segments. Label this specific position as point B, representing $4\frac{2}{3}$ i.e. $\frac{14}{3}$, as shown below :

Hence, the point B on the number line represents $\mathbf{\frac{14}{3}}$.

Draw a number line and mark a point O on it, representing the integer 0. Each point of division represents an integer and each length between two successive integers is a unit length.

As $\frac{-3}{7}$ is a negative proper fraction, it lies between -1 and 0.

Divide the unit length between 0 and -1 into 7 equal segments by making 6 equally spaced marks. Starting from 0, count backward (to the left) 3 segments. Label this specific position as point C, representing $\frac{-3}{7}$, as shown below :

Hence, the point C on the number line represents $\mathbf{\frac{-3}{7}}$.

Draw a number line and mark a point O on it, representing the integer 0. Each point of division represents an integer and each length between two successive integers is a unit length.

Convert the improper fraction into a mixed fraction to find the two integers between which it lies.

$\frac{-17}{5} = -3\frac{2}{5}$

So, $\frac{-17}{5}$ lies between -3 and -4.

Divide the unit length between -3 and -4 into 5 equal segments by making 4 equally spaced marks. Starting from the integer -3, count backward (to the left) 2 segments. Label this specific position as point D, representing $-3\frac{2}{5}$ i.e. $\frac{-17}{5}$, as shown below :

Hence, the point D on the number line represents $\mathbf{\frac{-17}{5}}$.

Draw a number line and mark a point O on it, representing the integer 0. Each point of division represents an integer and each length between two successive integers is a unit length.

The given number is already a mixed fraction, so we can directly find the two integers between which it lies.

$-2\frac{2}{7} = \frac{-16}{7}$

So, $-2\frac{2}{7}$ lies between -2 and -3.

Divide the unit length between -2 and -3 into 7 equal segments by making 6 equally spaced marks. Starting from the integer -2, count backward (to the left) 2 segments. Label this specific position as point E, representing $-2\frac{2}{7}$ i.e. $\frac{-16}{7}$, as shown below :

Hence, the point E on the number line represents $\mathbf{-2\frac{2}{7}}$.

(i) Every whole number $p$ can be written as $\frac{p}{1}$, which is of the form $\frac{p}{q}$ where $p, q$ are integers and $q ≠ 0$. For example, $0 = \frac{0}{1} \text{ and } 7 = \frac{7}{1}$

Hence, the given statement is true.

(ii) Every integer $p$ can be written as $\frac{p}{1}$, which is of the form $\frac{p}{q}$ where $p, q$ are integers and $q ≠ 0$. For example, $-5 = \frac{-5}{1} \text{ and } 5 = \frac{5}{1}$.

Hence, the given statement is true.

(iii) An equivalent rational number is obtained by multiplying (or dividing) both the numerator and the denominator by the same non-zero number, not by adding the same number to both.

$$\begin{aligned}\Rightarrow \frac{2 + 4}{3 + 4} = \frac{6}{7} \\[1em] \Rightarrow \frac{2}{3} ≠ \frac{6}{7} \quad [\text{as } 2 \times 7 ≠ 3 \times 6]\end{aligned}$$

Hence, the given statement is false.

(iv) Write both the rational numbers with the LCM of their denominators as the common denominator.

L.C.M. of 6 and 5 = 30

$$\begin{aligned}\Rightarrow \frac{-5}{6} = \frac{(-5) \times 5}{6 \times 5} = \frac{-25}{30} \\[1em] \Rightarrow \frac{-6}{5} = \frac{(-6) \times 6}{5 \times 6} = \frac{-36}{30}\end{aligned}$$

As -36 < -25, so $\frac{-6}{5} \lt \frac{-5}{6}$.

Hence, the given statement is false.

(v) A rational number is positive if its numerator and denominator are either both positive or both negative.

$\frac{-3}{-4} = \frac{-3 \times (-1)}{-4 \times (-1)} = \frac{3}{4}$, which is a positive rational number.

Hence, the given statement is false.

(vi) 0 is a whole number and it can also be written as $\frac{0}{1}$, which is of the form $\frac{p}{q}$ where $q ≠ 0$, so 0 is a rational number as well.

Hence, the given statement is false.

Draw a number line and mark a point O on it, representing the integer 0. Mark the integers to the right (1, 2, 3) and to the left (-1, -2, -3).

Convert each of the given improper fractions into a mixed fraction to find the two integers between which it lies.

$$\begin{aligned}\Rightarrow \frac{13}{5} = 2\frac{3}{5} \\[1em] \Rightarrow \frac{-13}{5} = -2\frac{3}{5}\end{aligned}$$

So, $\frac{13}{5}$ lies between 2 and 3, and $\frac{-13}{5}$ lies between -2 and -3.

Divide the unit length between 2 and 3, as well as the unit length between -2 and -3, into 5 equal segments by making 4 equally spaced marks in each interval.

For $\frac{13}{5}$ : Starting from the integer 2, count forward (to the right) 3 segments. Label this specific position as point A, representing $2\frac{3}{5}$ i.e. $\frac{13}{5}$.

For $\frac{-13}{5}$ : Starting from the integer -2, count backward (to the left) 3 segments. Label this specific position as point B, representing $-2\frac{3}{5}$ i.e. $\frac{-13}{5}$, as shown below :

Hence, the points A and B on the number line represent $\mathbf{\frac{13}{5} \text{ and } \frac{-13}{5}}$.

To find the additive inverse of any rational number, change its sign.

Changing the sign of $\frac{9}{13}$, we get $\frac{-9}{13}$.

[Check : $\frac{9}{13} + \frac{-9}{13} = \frac{0}{13} = 0$]

Hence, the additive inverse of $\mathbf{\frac{9}{13} \text{ is } \frac{-9}{13}}$.

To find the additive inverse of any rational number, change its sign.

Changing the sign of $\frac{16}{7}$, we get $\frac{-16}{7}$.

[Check : $\frac{16}{7} + \frac{-16}{7} = \frac{0}{7} = 0$]

Hence, the additive inverse of $\mathbf{\frac{16}{7} \text{ is } \frac{-16}{7}}$.

To find the additive inverse of any rational number, change its sign.

Changing the sign of $\frac{-3}{23}$, we get $\frac{3}{23}$.

[Check : $\frac{-3}{23} + \frac{3}{23} = \frac{0}{23} = 0$]

Hence, the additive inverse of $\mathbf{\frac{-3}{23} \text{ is } \frac{3}{23}}$.

To find the additive inverse of any rational number, change its sign.

First express the rational number with a positive denominator.

$\frac{8}{-11} = \frac{8 \times (-1)}{-11 \times (-1)} = \frac{-8}{11}$

Changing the sign of $\frac{-8}{11}$, we get $\frac{8}{11}$.

[Check : $\frac{-8}{11} + \frac{8}{11} = \frac{0}{11} = 0$]

Hence, the additive inverse of $\mathbf{\frac{8}{-11} \text{ is } \frac{8}{11}}$.

To find the additive inverse of any rational number, change its sign.

Changing the sign of $\frac{-22}{15}$, we get $\frac{22}{15}$.

[Check: $\frac{-22}{15} + \frac{22}{15} = \frac{0}{15} = 0$]

Hence, the additive inverse of $\mathbf{\frac{-22}{15} \text{ is } \frac{22}{15}}$.

To find the additive inverse of any rational number, change its sign.

First express the rational number with a positive denominator.

$\frac{-11}{-9} = \frac{-11 \times (-1)}{-9 \times (-1)} = \frac{11}{9}$

Changing the sign of $\frac{11}{9}$, we get $\frac{-11}{9}$.

[Check : $\frac{11}{9} + \frac{-11}{9} = \frac{0}{9} = 0$]

Hence, the additive inverse of $\mathbf{\frac{-11}{-9} \text{ is } \frac{-11}{9}}$.

Both rational numbers have the same (positive) denominator 17, so add the numerators keeping the denominator the same.

$$\begin{aligned}\frac{-7}{17} + \frac{6}{17} = \frac{(-7) + 6}{17} \\[1em] = \frac{-1}{17}\end{aligned}$$

Hence, $\mathbf{\frac{-7}{17} + \frac{6}{17} = \frac{-1}{17}}$.

First express each rational number with a positive denominator.

$\frac{7}{-12} = \frac{7 \times (-1)}{-12 \times (-1)} = \frac{-7}{12}$

Both rational numbers now have the same (positive) denominator 12, so add the numerators keeping the denominator the same.

$$\begin{aligned}\frac{-5}{12} + \frac{-7}{12} = \frac{(-5) + (-7)}{12} \\[1em] = \frac{-12}{12} \\[1em] = -1\end{aligned}$$

Hence, $\mathbf{\frac{-5}{12} + \frac{7}{-12} = -1}$.

The denominators are different, so express the rational numbers with the LCM of their denominators as the common denominator and then add.

LCM of 15 and 12 = 60

$$\begin{aligned}\frac{8}{15} + \frac{5}{12} = \frac{8 \times 4 + 5 \times 5}{60} \\[1em] = \frac{32 + 25}{60} \\[1em] = \frac{57}{60} \\[1em] = \frac{19}{20}\end{aligned}$$

Hence, $\mathbf{\frac{8}{15} + \frac{5}{12} = \frac{19}{20}}$.

First express each rational number with a positive denominator.

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

The denominators are different, so express the rational numbers with the LCM of their denominators as the common denominator and then add.

LCM of 18 and 12 = 36

$$\begin{aligned}\frac{-11}{18} + \frac{-5}{12} = \frac{(-11) \times 2 + (-5) \times 3}{36} \\[1em] = \frac{-22 - 15}{36} \\[1em] = \frac{-37}{36} \\[1em] = -1\frac{1}{36}\end{aligned}$$

Hence, $\mathbf{\frac{-11}{18} + \frac{5}{-12} = -1\frac{1}{36}}$.

All the denominators are positive but different, so express the rational numbers with the LCM of their denominators as the common denominator and then add.

LCM of 6, 4, 8 and 3 = 24

$$\begin{aligned}\frac{-11}{6} + \frac{-3}{4} + \frac{5}{8} + \frac{-7}{3} = \frac{(-11) \times 4 + (-3) \times 6 + 5 \times 3 + (-7) \times 8}{24} \\[1em] = \frac{-44 - 18 + 15 - 56}{24} \\[1em] = \frac{15 - 118}{24} \\[1em] = \frac{-103}{24} \\[1em] = -4\frac{7}{24}\end{aligned}$$

Hence, $\mathbf{\frac{-11}{6} + \frac{-3}{4} + \frac{5}{8} + \frac{-7}{3} = -4\frac{7}{24}}$.

First express each rational number with a positive denominator.

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

The denominators are different, so express the rational numbers with the LCM of their denominators as the common denominator and then add.

LCM of 7, 3, 21 and 9 = 63

$$\begin{aligned}\frac{4}{7} + \frac{-2}{3} + \frac{5}{21} + \frac{-8}{9} = \frac{4 \times 9 + (-2) \times 21 + 5 \times 3 + (-8) \times 7}{63} \\[1em] = \frac{36 - 42 + 15 - 56}{63} \\[1em] = \frac{51 - 98}{63} \\[1em] = \frac{-47}{63}\end{aligned}$$

Hence, $\mathbf{\frac{4}{7} + \frac{2}{-3} + \frac{5}{21} + \frac{-8}{9} = \frac{-47}{63}}$.

To subtract one rational number from another, add the additive inverse of the rational number to be subtracted.

Subtracting $\frac{2}{3}$ from $\frac{5}{6}$,

LCM of 6 and 3 = 6

$$\begin{aligned}\frac{5}{6} - \frac{2}{3} = \frac{5 - 2 \times 2}{6} \\[1em] = \frac{5 - 4}{6} \\[1em] = \frac{1}{6}\end{aligned}$$

Hence, the required difference is $\mathbf{\frac{1}{6}}$.

To subtract one rational number from another, add the additive inverse of the rational number to be subtracted.

Subtracting $\frac{-2}{5}$ from $\frac{-5}{7}$,

LCM of 7 and 5 = 35

$$\begin{aligned}\frac{-5}{7} - \left(\frac{-2}{5}\right) = \frac{-5}{7} + \frac{2}{5} \\[1em] = \frac{(-5) \times 5 + 2 \times 7}{35} \\[1em] = \frac{-25 + 14}{35} \\[1em] = \frac{-11}{35}\end{aligned}$$

Hence, the required difference is $\mathbf{\frac{-11}{35}}$.

To subtract one rational number from another, add the additive inverse of the rational number to be subtracted.

Subtracting $\frac{4}{9}$ from $\frac{-7}{8}$,

LCM of 8 and 9 = 72

$$\begin{aligned}\frac{-7}{8} - \frac{4}{9} = \frac{(-7) \times 9 - 4 \times 8}{72} \\[1em] = \frac{-63 - 32}{72} \\[1em] = \frac{-95}{72} \\[1em] = -1\frac{23}{72}\end{aligned}$$

Hence, the required difference is $\mathbf{-1\frac{23}{72}}$.

To subtract one rational number from another, add the additive inverse of the rational number to be subtracted.

Subtracting $\frac{-11}{6}$ from $\frac{8}{3}$,

LCM of 3 and 6 = 6

$$\begin{aligned}\frac{8}{3} - \left(\frac{-11}{6}\right) = \frac{8}{3} + \frac{11}{6} \\[1em] = \frac{8 \times 2 + 11}{6} \\[1em] = \frac{16 + 11}{6} \\[1em] = \frac{27}{6} = \frac{9}{2} \\[1em] = 4\frac{1}{2}\end{aligned}$$

Hence, the required difference is $\mathbf{4\frac{1}{2}}$.

Let the other rational number be $x$.

As the sum of the two rational numbers is $\frac{-4}{9}$,

$$\begin{aligned}\Rightarrow \frac{13}{6} + x = \frac{-4}{9} \\[1em] \Rightarrow x = \frac{-4}{9} - \frac{13}{6} \\[1em] [\text{ LCM of } 9 \text{ and } 6 = 18] \\[1em] \Rightarrow x = \frac{(-4) \times 2 - 13 \times 3}{18} \\[1em] \Rightarrow x = \frac{-8 - 39}{18} \\[1em] \Rightarrow x = \frac{-47}{18} = -2\frac{11}{18}\end{aligned}$$

Hence, the other rational number is $\mathbf{\frac{-47}{18}}$.

Let the required number be $x$.

According to the given condition,

$$\begin{aligned}\Rightarrow \frac{-2}{3} + x = \frac{-1}{7} \\[1em] \Rightarrow x = \frac{-1}{7} - \left(\frac{-2}{3}\right) \\[1em] \Rightarrow x = \frac{-1}{7} + \frac{2}{3} \\[1em] [\text{ LCM of } 7 \text{ and } 3 = 21] \\[1em] \Rightarrow x = \frac{(-1) \times 3 + 2 \times 7}{21} \\[1em] \Rightarrow x = \frac{-3 + 14}{21} \\[1em] \Rightarrow x = \frac{11}{21}\end{aligned}$$

Hence, the required number is $\mathbf{\frac{11}{21}}$.

Let the required number be $x$.

According to the given condition,

$$\begin{aligned}\Rightarrow -2 - x = \frac{7}{11} \\[1em] \Rightarrow x = -2 - \frac{7}{11} \\[1em] \Rightarrow x = \frac{(-2) \times 11 - 7}{11} \\[1em] \Rightarrow x = \frac{-22 - 7}{11} \\[1em] \Rightarrow x = \frac{-29}{11} = -2\frac{7}{11}\end{aligned}$$

Hence, the required number is $\mathbf{\frac{-29}{11}}$.

Let the required number be $x$.

According to the given condition,

$$\begin{aligned}\Rightarrow -1 + x = \frac{5}{7} \\[1em] \Rightarrow x = \frac{5}{7} - (-1) \\[1em] \Rightarrow x = \frac{5}{7} + 1 \\[1em] \Rightarrow x = \frac{5 + 7}{7} \\[1em] \Rightarrow x = \frac{12}{7} = 1\frac{5}{7}\end{aligned}$$

Hence, the required number is $\mathbf{\frac{12}{7}}$.

Let the required number be $x$.

According to the given condition,

$$\begin{aligned}\Rightarrow \frac{-2}{3} - x = \frac{-1}{6} \\[1em] \Rightarrow x = \frac{-2}{3} - \left(\frac{-1}{6}\right) \\[1em] \Rightarrow x = \frac{-2}{3} + \frac{1}{6} \\[1em] [\text{ LCM of } 3 \text{ and } 6 = 6] \\[1em] \Rightarrow x = \frac{(-2) \times 2 + 1}{6} \\[1em] \Rightarrow x = \frac{-4 + 1}{6} \\[1em] \Rightarrow x = \frac{-3}{6} = \frac{-1}{2}\end{aligned}$$

Hence, the required number is $\mathbf{\frac{-1}{2}}$.

(i) Rational numbers are closed under subtraction, i.e. if $\frac{a}{b} \text{ and } \frac{c}{d}$ are two rational numbers, then $\frac{a}{b} - \frac{c}{d}$ is also a rational number.

Hence, the given statement is true.

(ii) Solving L.H.S.,

$\frac{2}{3} - \frac{3}{5} = \frac{2 \times 5 - 3 \times 3}{15} = \frac{10 - 9}{15} = \frac{1}{15}$

Solving R.H.S.,

$\frac{3}{5} - \frac{2}{3} = \frac{3 \times 3 - 2 \times 5}{15} = \frac{9 - 10}{15} = \frac{-1}{15}$

Clearly, L.H.S. ≠ R.H.S.

Hence, the given statement is false.

(iii) For every rational number $\frac{a}{b}$, we have $\frac{a}{b} + 0 = \frac{a}{b} = 0 + \frac{a}{b}$, so 0 is the additive identity and 1 is the multiplicative identity.

Hence, the given statement is false.

(iv) For any two rational numbers $\frac{a}{b} \text{ and } \frac{c}{d}$, we have $\frac{a}{b} - \frac{c}{d} ≠ \frac{c}{d} - \frac{a}{b}$. For example, $\frac{2}{3} - \frac{3}{5} = \frac{1}{15}$ but $\frac{3}{5} - \frac{2}{3} = \frac{-1}{15}$.

Hence, the given statement is false.

(v) The additive inverse of the rational number 0 is 0 itself, as 0 + 0 = 0.

Hence, the given statement is true.

(vi) Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{19}{-5} + \frac{-3}{11} \\[1em] \Rightarrow \frac{-19}{5} + \frac{-3}{11} \\[1em] \Rightarrow \frac{(-19) \times 11 + (-3) \times 5}{55} = \frac{-209 - 15}{55} = \frac{-224}{55}\end{aligned}$$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow \frac{19}{5} + \frac{3}{11} \\[1em] \Rightarrow \frac{19 \times 11 + 3 \times 5}{55} = \frac{209 + 15}{55} = \frac{224}{55}\end{aligned}$$

Clearly, L.H.S. ≠ R.H.S.

Hence, the given statement is false.

The product of two rational numbers $\frac{a}{b} \text{ and } \frac{c}{d}$ is given by $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$.

Multiplying,

$$\begin{aligned}\frac{4}{9} \times \frac{7}{12} = \frac{4 \times 7}{9 \times 12} \\[1em] = \frac{28}{108} \\[1em] = \frac{7}{27}\end{aligned}$$

Hence, $\mathbf{\frac{4}{9} \times \frac{7}{12} = \frac{7}{27}}$.

The product of two rational numbers $\frac{a}{b} \text{ and } \frac{c}{d}$ is given by $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$.

Multiplying,

$$\begin{aligned}-9 \times \frac{7}{18} = \frac{-9}{1} \times \frac{7}{18} \\[1em] = \frac{(-9) \times 7}{1 \times 18} \\[1em] = \frac{-63}{18} = \frac{-7}{2} \\[1em] = -3\frac{1}{2}\end{aligned}$$

Hence, $\mathbf{-9 \times \frac{7}{18} = -3\frac{1}{2}}$.

The product of two rational numbers $\frac{a}{b} \text{ and } \frac{c}{d}$ is given by $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$.

Multiplying,

$$\begin{aligned}\frac{-3}{16} \times \frac{8}{-15} = \frac{(-3) \times 8}{16 \times (-15)} \\[1em] = \frac{-24}{-240} \\[1em] = \frac{1}{10}\end{aligned}$$

Hence, $\mathbf{\frac{-3}{16} \times \frac{8}{-15} = \frac{1}{10}}$.

The product of two rational numbers $\frac{a}{b} \text{ and } \frac{c}{d}$ is given by $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$.

Multiplying,

$$\begin{aligned}\frac{6}{7} \times \frac{-21}{12} = \frac{6 \times (-21)}{7 \times 12} \\[1em] = \frac{-126}{84} = \frac{-3}{2} \\[1em] = -1\frac{1}{2}\end{aligned}$$

Hence, $\mathbf{\frac{6}{7} \times \frac{-21}{12} = -1\frac{1}{2}}$.

The product of two rational numbers $\frac{a}{b} \text{ and } \frac{c}{d}$ is given by $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$.

Multiplying,

$$\begin{aligned}\frac{5}{-18} \times \frac{-9}{20} = \frac{5 \times (-9)}{(-18) \times 20} \\[1em] = \frac{-45}{-360} \\[1em] = \frac{1}{8}\end{aligned}$$

Hence, $\mathbf{\frac{5}{-18} \times \frac{-9}{20} = \frac{1}{8}}$.

The product of two rational numbers $\frac{a}{b} \text{ and } \frac{c}{d}$ is given by $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$.

Multiplying,

$$\begin{aligned}\frac{-13}{15} \times \frac{-25}{26} = \frac{(-13) \times (-25)}{15 \times 26} \\[1em] = \frac{325}{390} \\[1em] = \frac{5}{6}\end{aligned}$$

Hence, $\mathbf{\frac{-13}{15} \times \frac{-25}{26} = \frac{5}{6}}$.

The product of two rational numbers $\frac{a}{b} \text{ and } \frac{c}{d}$ is given by $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$.

Multiplying,

$$\begin{aligned}\frac{7}{24} \times (-48) = \frac{7}{24} \times \frac{-48}{1} \\[1em] = \frac{7 \times (-48)}{24 \times 1} \\[1em] = \frac{-336}{24} \\[1em] = -14\end{aligned}$$

Hence, $\mathbf{\frac{7}{24} \times (-48) = -14}$.

The product of two rational numbers $\frac{a}{b} \text{ and } \frac{c}{d}$ is given by $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$.

Multiplying,

$$\begin{aligned}\frac{-13}{5} \times (-10) = \frac{-13}{5} \times \frac{-10}{1} \\[1em] = \frac{(-13) \times (-10)}{5 \times 1} \\[1em] = \frac{130}{5} \\[1em] = 26\end{aligned}$$

Hence, $\mathbf{\frac{-13}{5} \times (-10) = 26}$.

To find the multiplicative inverse (or reciprocal) of a non-zero rational number, interchange its numerator and denominator.

Interchanging the numerator and the denominator of $\frac{-17}{12}$, we get $\frac{12}{-17} = \frac{-12}{17}$

$[\text{Check : } \frac{-17}{12} \times \frac{-12}{17} = 1]$

Hence, the multiplicative inverse of $\mathbf{\frac{-17}{12} \text{ is } \frac{-12}{17}}$.

To find the multiplicative inverse (or reciprocal) of a non-zero rational number, interchange its numerator and denominator.

Writing -16 in the form $\frac{p}{q}$ and interchanging the numerator and the denominator,

$$\begin{aligned}\Rightarrow -16 = \frac{-16}{1} \\[1em] \Rightarrow \text{Multiplicative inverse} = \frac{1}{-16} = \frac{-1}{16}\end{aligned}$$

Hence, the multiplicative inverse of -16 is $\mathbf{\frac{-1}{16}}$.

To find the multiplicative inverse (or reciprocal) of a non-zero rational number, interchange its numerator and denominator.

Here $\frac{0}{2} = 0$, and there is no rational number which when multiplied by 0 gives 1.

Hence, the multiplicative inverse of $\mathbf{\frac{0}{2}}$ does not exist.

To find the multiplicative inverse (or reciprocal) of a non-zero rational number, interchange its numerator and denominator.

First express the rational number with a positive denominator.

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

Interchanging the numerator and the denominator of $\frac{3}{5}$, we get $\frac{5}{3}$.

$[\text{Check : } \frac{3}{5} \times \frac{5}{3} = 1]$

Hence, the multiplicative inverse of $\mathbf{\frac{-3}{-5} \text{ is } \frac{5}{3}}$.

To find the multiplicative inverse (or reciprocal) of a non-zero rational number, interchange its numerator and denominator.

First express the rational number with a positive denominator.

$\frac{2}{-5} = \frac{2 \times (-1)}{-5 \times (-1)} = \frac{-2}{5}$

Interchanging the numerator and the denominator of $\frac{-2}{5}$, we get $\frac{5}{-2} = \frac{-5}{2}$

$[\text{Check : } \frac{-2}{5} \times \frac{-5}{2} = 1]$

Hence, the multiplicative inverse of $\mathbf{\frac{2}{-5} \text{ is } \frac{-5}{2}}$.

To divide one rational number by another, multiply the dividend by the reciprocal of the divisor.

Dividing,

$$\begin{aligned}\frac{17}{8} \div \frac{51}{4} = \frac{17}{8} \times \frac{4}{51} \\[1em] = \frac{17 \times 4}{8 \times 51} \\[1em] = \frac{68}{408} \\[1em] = \frac{1}{6}\end{aligned}$$

Hence, $\mathbf{\frac{17}{8} \div \frac{51}{4} = \frac{1}{6}}$.

To divide one rational number by another, multiply the dividend by the reciprocal of the divisor.

Dividing,

$$\begin{aligned}\frac{-16}{35} \div \frac{15}{14} = \frac{-16}{35} \times \frac{14}{15} \\[1em] = \frac{(-16) \times 14}{35 \times 15} \\[1em] = \frac{-224}{525} \\[1em] = \frac{-32}{75}\end{aligned}$$

Hence, $\mathbf{\frac{-16}{35} \div \frac{15}{14} = \frac{-32}{75}}$.

To divide one rational number by another, multiply the dividend by the reciprocal of the divisor.

Dividing,

$$\begin{aligned}\frac{-12}{7} \div (-16) = \frac{-12}{7} \div \frac{-16}{1} \\[1em] = \frac{-12}{7} \times \frac{1}{-16} \\[1em] = \frac{(-12) \times 1}{7 \times (-16)} \\[1em] = \frac{-12}{-112} = \frac{3}{28}\end{aligned}$$

Hence, $\mathbf{\frac{-12}{7} \div (-16) = \frac{3}{28}}$.

To divide one rational number by another, multiply the dividend by the reciprocal of the divisor.

Dividing,

$$\begin{aligned}-9 \div \left(\frac{-5}{18}\right) = \frac{-9}{1} \times \frac{18}{-5} \\[1em] = \frac{(-9) \times 18}{1 \times (-5)} \\[1em] = \frac{-162}{-5} = \frac{162}{5} \\[1em] = 32\frac{2}{5}\end{aligned}$$

Hence, $\mathbf{-9 \div \left(\frac{-5}{18}\right) = 32\frac{2}{5}}$.

Here the order of the two rational numbers has been changed but the product remains the same.

If $\frac{a}{b} \text{ and } \frac{c}{d}$ are any two rational numbers, then $\frac{a}{b} \times \frac{c}{d} = \frac{c}{d} \times \frac{a}{b}$

Hence, the given statement illustrates the commutative law of multiplication.

Here the grouping of the three rational numbers has been changed but the product remains the same.

If $\frac{a}{b}, \frac{c}{d} \text{ and } \frac{e}{f}$ are any three rational numbers, then

$\left(\frac{a}{b} \times \frac{c}{d}\right) \times \frac{e}{f} = \frac{a}{b} \times \left(\frac{c}{d} \times \frac{e}{f}\right)$

Hence, the given statement illustrates the associative law of multiplication.

Here a rational number has been multiplied by the sum of two rational numbers, which is the same as the sum of the two separate products.

If $\frac{a}{b}, \frac{c}{d} \text{ and } \frac{e}{f}$ are any three rational numbers, then

$\frac{a}{b} \times \left(\frac{c}{d} + \frac{e}{f}\right) = \left(\frac{a}{b} \times \frac{c}{d}\right) + \left(\frac{a}{b} \times \frac{e}{f}\right)$

Hence, the given statement illustrates the distributive law of multiplication over addition.

Here the rational number remains unchanged on being multiplied by 1.

For every rational number $\frac{a}{b}$, we have $\frac{a}{b} \times 1 = 1 \times \frac{a}{b} = \frac{a}{b}$.

Hence, the given statement illustrates the existence of multiplicative identity, 1 being the multiplicative identity for rational numbers.

Here the product of the rational number and its reciprocal is 1.

For every non-zero rational number $\frac{a}{b}$, we have $\frac{a}{b} \times \frac{b}{a} = \frac{b}{a} \times \frac{a}{b} = 1$

Hence, the given statement illustrates the existence of multiplicative inverse (or reciprocal).

Here the product of the rational number and 0 is 0.

For every rational number $\frac{a}{b}$, we have $\frac{a}{b} \times 0 = 0 \times \frac{a}{b} = 0$

Hence, the given statement illustrates the multiplicative property of zero.

Solve the L.H.S. and the R.H.S. separately by multiplying the dividend with the reciprocal of the divisor, and then compare them.

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{-8}{9} \div \frac{-4}{3} \\[1em] \Rightarrow \frac{-8}{9} \times \frac{3}{-4} \\[1em] \Rightarrow \frac{(-8) \times 3}{9 \times (-4)} = \frac{-24}{-36} = \frac{2}{3}\end{aligned}$$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow \frac{-4}{3} \div \frac{-8}{9} \\[1em] \Rightarrow \frac{-4}{3} \times \frac{9}{-8} \\[1em] \Rightarrow \frac{(-4) \times 9}{3 \times (-8)} = \frac{-36}{-24} = \frac{3}{2}\end{aligned}$$

As L.H.S. $≠$ R.H.S., division of rational numbers is not commutative.

Hence, the given statement is false.

Solve the L.H.S. and the R.H.S. separately by multiplying the dividend with the reciprocal of the divisor, and then compare them.

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{-7}{24} \div \frac{3}{-16} \\[1em] \Rightarrow \frac{-7}{24} \times \frac{-16}{3} \\[1em] \Rightarrow \frac{(-7) \times (-16)}{24 \times 3} = \frac{112}{72} = \frac{14}{9}\end{aligned}$$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow \frac{3}{-16} \div \frac{-7}{24} \\[1em] \Rightarrow \frac{3}{-16} \times \frac{24}{-7} \\[1em] \Rightarrow \frac{3 \times 24}{(-16) \times (-7)} = \frac{72}{112} = \frac{9}{14}\end{aligned}$$

As L.H.S. $≠$ R.H.S., division of rational numbers is not commutative.

Hence, the given statement is false.

Solve the L.H.S. and the R.H.S. separately by first simplifying the bracket, and then compare them.

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{-3}{5} \div \frac{-12}{35} = \frac{-3}{5} \times \frac{35}{-12} \\[1em] \Rightarrow \frac{(-3) \times 35}{5 \times (-12)} = \frac{-105}{-60} = \frac{7}{4} \\[1em] \therefore \left(\frac{-3}{5} \div \frac{-12}{35}\right) \div \frac{1}{4} = \frac{7}{4} \div \frac{1}{4} \\[1em] \Rightarrow \frac{7}{4} \times \frac{4}{1} = 7\end{aligned}$$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow \frac{-12}{35} \div \frac{1}{4} = \frac{-12}{35} \times \frac{4}{1} = \frac{-48}{35} \\[1em] \therefore \frac{-3}{5} \div \left(\frac{-12}{35} \div \frac{1}{4}\right) = \frac{-3}{5} \div \frac{-48}{35} \\[1em] \Rightarrow \frac{-3}{5} \times \frac{35}{-48} = \frac{-105}{-240} = \frac{7}{16}\end{aligned}$$

As L.H.S. $≠$ R.H.S., division of rational numbers is not associative.

Hence, the given statement is false.

(i) There is no rational number which when multiplied by 0 gives 1.

Hence, 0 is the rational number that does not have a multiplicative inverse.

(ii) As 1 × 1 = 1 and (-1) × (-1) = 1, each of these numbers is its own reciprocal.

Hence, 1 and -1 are the rational numbers which are equal to their own reciprocals.

(iii) For every non-zero rational number $\frac{a}{b}$, we have $\frac{a}{b} \times \frac{b}{a} = 1$.

Hence, the product of a rational number and its reciprocal is 1.

(iv) The reciprocal is obtained by interchanging the numerator and the denominator, so the sign of the rational number does not change. For example, the reciprocal of $\frac{3}{5} \text{ is } \frac{5}{3}$.

Hence, the reciprocal of a positive rational number is a positive rational number.

(v) The reciprocal is obtained by interchanging the numerator and the denominator, so the sign of the rational number does not change. For example, the reciprocal of $\frac{-3}{5} \text{ is } \frac{5}{-3} = \frac{-5}{3}$.

Hence, the reciprocal of a negative rational number is a negative rational number.

(vi) As the product of any rational number and 0 is 0 and never 1, no rational number has 0 as its reciprocal.

Hence, 0 is not the reciprocal of any number.

(vii) Interchanging the numerator and the denominator of $\frac{1}{a}$, we get $\frac{a}{1} = a$.

Check : $\frac{1}{a} \times a = 1$

Hence, the multiplicative inverse of a rational number $\bm{\frac{1}{a}}, a ≠ 0$ is $\bm{a}$.

Let the other rational number be $x$.

As the product of the two rational numbers is -7,

$$\begin{aligned}\Rightarrow \frac{-8}{11} \times x = -7 \\[1em] \Rightarrow x = -7 \div \frac{-8}{11} \\[1em] \Rightarrow x = \frac{-7}{1} \times \frac{11}{-8} \\[1em] \Rightarrow x = \frac{(-7) \times 11}{1 \times (-8)} \\[1em] \Rightarrow x = \frac{-77}{-8} = \frac{77}{8} \\[1em] \Rightarrow x = 9\frac{5}{8}\end{aligned}$$

Hence, the other rational number is $\mathbf{\frac{77}{8}}$.

Let the other rational number be $x$.

As the product of the two rational numbers is $\frac{-16}{9}$,

$$\begin{aligned}\Rightarrow \frac{-4}{3} \times x = \frac{-16}{9} \\[1em] \Rightarrow x = \frac{-16}{9} \div \frac{-4}{3} \\[1em] \Rightarrow x = \frac{-16}{9} \times \frac{3}{-4} \\[1em] \Rightarrow x = \frac{(-16) \times 3}{9 \times (-4)} \\[1em] \Rightarrow x = \frac{-48}{-36} = \frac{4}{3} \\[1em] \Rightarrow x = 1\frac{1}{3}\end{aligned}$$

Hence, the other rational number is $\mathbf{\frac{4}{3}}$.

Let the required rational number be $x$.

According to the given condition,

$$\begin{aligned}\Rightarrow \frac{1}{26} \div x = \frac{-8}{39} \\[1em] \Rightarrow x = \frac{1}{26} \div \frac{-8}{39} \\[1em] \Rightarrow x = \frac{1}{26} \times \frac{39}{-8} \\[1em] \Rightarrow x = \frac{1 \times 39}{26 \times (-8)} \\[1em] \Rightarrow x = \frac{39}{-208} = \frac{-3}{16}\end{aligned}$$

Hence, the required rational number is $\mathbf{\frac{-3}{16}}$.

First find the sum of $\frac{13}{5} \text{ and } \frac{-12}{7}$ and the product of $\frac{-31}{7} \text{ and } \frac{1}{-2}$.

Note that LCM of 5 and 7 is 35.

Finding the sum,

$$\begin{aligned}\frac{13}{5} + \frac{-12}{7} = \frac{13 \times 7 + (-12) \times 5}{35} \\[1em] = \frac{91 - 60}{35} \\[1em] = \frac{31}{35}\end{aligned}$$

Finding the product,

$$\begin{aligned}\frac{-31}{7} \times \frac{1}{-2} = \frac{(-31) \times 1}{7 \times (-2)} \\[1em] = \frac{-31}{-14} = \frac{31}{14}\end{aligned}$$

Dividing the sum by the product,

$$\begin{aligned}\frac{31}{35} \div \frac{31}{14} = \frac{31}{35} \times \frac{14}{31} \\[1em] = \frac{31 \times 14}{35 \times 31} \\[1em] = \frac{14}{35} = \frac{2}{5}\end{aligned}$$

Hence, the required quotient is $\mathbf{\frac{2}{5}}$.

First find the sum and the difference of the given rational numbers.

LCM of 12 and 3 = 12

Finding the sum,

$$\begin{aligned}\frac{65}{12} + \frac{8}{3} = \frac{65 + 8 \times 4}{12} \\[1em] = \frac{65 + 32}{12} \\[1em] = \frac{97}{12}\end{aligned}$$

Finding the difference,

$$\begin{aligned}\frac{65}{12} - \frac{8}{3} = \frac{65 - 8 \times 4}{12} \\[1em] = \frac{65 - 32}{12} \\[1em] = \frac{33}{12} = \frac{11}{4}\end{aligned}$$

Dividing the sum by the difference,

$$\begin{aligned}\frac{97}{12} \div \frac{11}{4} = \frac{97}{12} \times \frac{4}{11} \\[1em] = \frac{97 \times 4}{12 \times 11} \\[1em] = \frac{388}{132} = \frac{97}{33} \\[1em] = 2\frac{31}{33}\end{aligned}$$

Hence, the required quotient is $\mathbf{2\frac{31}{33}}$.

We know that total cost = cost of 1 metre × length of cloth.

Length of the cloth $= 3\frac{2}{5} \text{ m} = \frac{17}{5} \text{ m}$

Cost of 1 metre of cloth $= ₹ 63\frac{3}{4} = ₹ \frac{255}{4}$

Finding the total cost,

$$\begin{aligned}\frac{255}{4} \times \frac{17}{5} = \frac{255 \times 17}{4 \times 5} \\[1em] = \frac{4335}{20} \\[1em] = \frac{867}{4} = 216\frac{3}{4}\end{aligned}$$

Hence, the cost of the cloth is $\mathbf{₹ 216\frac{3}{4} \text{ or } ₹ 216.75}$.

We know that area of a rectangle = length × breadth.

Length of the rectangular part $= 36\frac{3}{5} \text{ m} = \frac{183}{5} \text{ m}$

Breadth of the rectangular part $= 16\frac{2}{5} \text{ m} = \frac{82}{5} \text{ m}$

Finding the area,

$$\begin{aligned}\frac{183}{5} \times \frac{82}{5} = \frac{183 \times 82}{5 \times 5} \\[1em] = \frac{15006}{25} \\[1em] = 600\frac{6}{25} \text{ sq. m}\end{aligned}$$

Hence, the area of the rectangular part is $\mathbf{600\frac{6}{25} \text{ sq. m}}$.

We know that area of a square = side × side.

Side of the square plot $= 8\frac{1}{2} \text{ m} = \frac{17}{2} \text{ m}$

Finding the area,

$$\begin{aligned}\frac{17}{2} \times \frac{17}{2} = \frac{17 \times 17}{2 \times 2} \\[1em] = \frac{289}{4} \\[1em] = 72\frac{1}{4} \text{ sq. m}\end{aligned}$$

Hence, the area of the square plot is $\mathbf{72\frac{1}{4} \text{ sq. m}}$.

To find the length of each piece, divide the total length of the cord by the number of pieces.

Total length of the cord $= 71\frac{1}{2} \text{ m} = \frac{143}{2} \text{ m}$

Number of pieces = 26

Finding the length of each piece,

$$\begin{aligned}\frac{143}{2} \div 26 = \frac{143}{2} \times \frac{1}{26} \\[1em] = \frac{143 \times 1}{2 \times 26} \\[1em] = \frac{143}{52} = \frac{11}{4} \\[1em] = 2\frac{3}{4} \text{ m}\end{aligned}$$

Hence, the length of each piece is $\mathbf{2\frac{3}{4} \text{ m}}$.

We know that area of a rectangle = length × breadth, so length = area ÷ breadth.

Area of the room $= 65\frac{1}{4} \text{ sq. m} = \frac{261}{4} \text{ sq. m}$

Breadth of the room $= 5\frac{1}{16} \text{ m} = \frac{81}{16} \text{ m}$

Finding the length,

$$\begin{aligned}\frac{261}{4} \div \frac{81}{16} = \frac{261}{4} \times \frac{16}{81} \\[1em] = \frac{261 \times 16}{4 \times 81} \\[1em] = \frac{4176}{324} = \frac{116}{9} \\[1em] = 12\frac{8}{9} \text{ m}\end{aligned}$$

Hence, the length of the room is $\mathbf{12\frac{8}{9} \text{ m}}$.

If $x$ and $y$ are two rational numbers, then $\frac{1}{2}(x + y)$ is a rational number lying between them.

LCM of 10 and 17 = 170

Finding the required rational number,

$$\begin{aligned}\frac{1}{2} \times \left(\frac{7}{10} + \frac{10}{17}\right) = \frac{1}{2} \times \left(\frac{7 \times 17 + 10 \times 10}{170}\right) \\[1em] = \frac{1}{2} \times \frac{119 + 100}{170} \\[1em] = \frac{1}{2} \times \frac{219}{170} \\[1em] = \frac{219}{340}\end{aligned}$$

Hence, $\mathbf{\frac{219}{340}}$ is a rational number lying between $\mathbf{\frac{7}{10} \text{ and } \frac{10}{17}}$.

If $x$ and $y$ are two rational numbers, then $\frac{1}{2}(x + y)$ is a rational number lying between them.

First convert the mixed fraction into an improper fraction.

$1\frac{3}{8} = \frac{11}{8}$

Finding the required rational number,

$$\begin{aligned}\frac{1}{2} \times \left(\frac{11}{8} + 2\right) = \frac{1}{2} \times \frac{11 + 2 \times 8}{8} \\[1em] = \frac{1}{2} \times \frac{27}{8} \\[1em] = \frac{27}{16} = 1\frac{11}{16}\end{aligned}$$

Hence, $\mathbf{\frac{27}{16}}$ is a rational number lying between $\mathbf{1\frac{3}{8} \text{ and } 2}$.

If $x$ and $y$ are two rational numbers, then $\frac{1}{2}(x + y)$ is a rational number lying between them.

Finding the required rational number,

LCM of 5 and 7 = 35

$$\begin{aligned}\frac{1}{2} \times \left(\frac{-3}{5} + \frac{-4}{7}\right) = \frac{1}{2} \times \left(\frac{(-3) \times 7 + (-4) \times 5}{35}\right) \\[1em] = \frac{1}{2} \times \frac{-21 - 20}{35} \\[1em] = \frac{1}{2} \times \frac{-41}{35} \\[1em] = \frac{-41}{70}\end{aligned}$$

Hence, $\mathbf{\frac{-41}{70}}$ is a rational number lying between $\mathbf{\frac{-3}{5} \text{ and } \frac{-4}{7}}$.

If $x$ and $y$ are two rational numbers, then $\frac{1}{2}(x + y)$ is a rational number lying between them.

Finding the required rational number,

$$\begin{aligned}\frac{1}{2} \times \left(-2 + \frac{-17}{21}\right) = \frac{1}{2} \times \frac{(-2) \times 21 - 17}{21} \\[1em] = \frac{1}{2} \times \frac{-42 - 17}{21} \\[1em] = \frac{1}{2} \times \frac{-59}{21} \\[1em] = \frac{-59}{42} = -1\frac{17}{42}\end{aligned}$$

Hence, $\mathbf{\frac{-59}{42}}$ is a rational number lying between $\mathbf{-2 \text{ and } \frac{-17}{21}}$.

Write both the numbers with a common denominator, then multiply the numerator and the denominator of each by one more than the number of rational numbers required.

$$\begin{aligned}\Rightarrow 4 = \frac{12}{3} \\[1em] \Rightarrow 4\frac{2}{3} = \frac{14}{3}\end{aligned}$$

To insert 3 rational numbers, multiply both the numerator and the denominator of each rational number by 3 + 1 i.e. 4.

$$\begin{aligned}\Rightarrow \frac{12}{3} = \frac{12 \times 4}{3 \times 4} = \frac{48}{12} \\[1em] \Rightarrow \frac{14}{3} = \frac{14 \times 4}{3 \times 4} = \frac{56}{12}\end{aligned}$$

As 48 < 49 < 50 < 51 < 56,

$\therefore \frac{48}{12} \lt \frac{49}{12} \lt \frac{50}{12} \lt \frac{51}{12} \lt \frac{56}{12}$

Hence, three rational numbers between 4 and $\mathbf{4\frac{2}{3}}$ are $\mathbf{\frac{49}{12}, \frac{50}{12} \text{ and } \frac{51}{12}}$.

Write both the rational numbers with a common denominator, then multiply the numerator and the denominator of each by one more than the number of rational numbers required.

$$\begin{aligned}[\text{ LCM of } 2 \text{ and } 4 = 4] \\[1em] \Rightarrow \frac{-1}{2} = \frac{-1 \times 2}{2 \times 2} = \frac{-2}{4} \\[1em] \Rightarrow \frac{-1}{4} = \frac{-1}{4}\end{aligned}$$

To insert 3 rational numbers, multiply both the numerator and the denominator of each rational number by 3 + 1 i.e. 4.

$$\begin{aligned}\Rightarrow \frac{-2}{4} = \frac{-2 \times 4}{4 \times 4} = \frac{-8}{16} \\[1em] \Rightarrow \frac{-1}{4} = \frac{-1 \times 4}{4 \times 4} = \frac{-4}{16}\end{aligned}$$

As -8 < -7 < -6 < -5 < -4,

$\therefore \frac{-8}{16} \lt \frac{-7}{16} \lt \frac{-6}{16} \lt \frac{-5}{16} \lt \frac{-4}{16}$

Hence, three rational numbers between $\mathbf{\frac{-1}{2} \text{ and } \frac{-1}{4}}$ are $\mathbf{\frac{-7}{16}, \frac{-6}{16} \text{ and } \frac{-5}{16}}$.

Write both the numbers with a common denominator, then multiply the numerator and the denominator of each by one more than the number of rational numbers required.

To insert 3 rational numbers between 2 and 3, multiply both the numerator and the denominator of each number by 3 + 1 i.e. 4.

$$\begin{aligned}\Rightarrow 2 = \frac{2 \times 4}{1 \times 4} = \frac{8}{4} \\[1em] \Rightarrow 3 = \frac{3 \times 4}{1 \times 4} = \frac{12}{4}\end{aligned}$$

As 8 < 9 < 10 < 11 < 12,

$\therefore \frac{8}{4} \lt \frac{9}{4} \lt \frac{10}{4} \lt \frac{11}{4} \lt \frac{12}{4}$

Hence, three rational numbers between 2 and 3 are $\mathbf{\frac{9}{4}, \frac{10}{4} \text{ and } \frac{11}{4}}$.

Write both the rational numbers with a common denominator, then multiply the numerator and the denominator of each by one more than the number of rational numbers required.

$$\begin{aligned}[\text{ LCM of } 5 \text{ and } 3 = 15] \\[1em] \Rightarrow \frac{3}{5} = \frac{3 \times 3}{5 \times 3} = \frac{9}{15} \\[1em] \Rightarrow \frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15}\end{aligned}$$

To insert 5 rational numbers, multiply both the numerator and the denominator of each rational number by 5 + 1 i.e. 6.

$$\begin{aligned}\Rightarrow \frac{9}{15} = \frac{9 \times 6}{15 \times 6} = \frac{54}{90} \\[1em] \Rightarrow \frac{10}{15} = \frac{10 \times 6}{15 \times 6} = \frac{60}{90}\end{aligned}$$

As 54 < 55 < 56 < 57 < 58 < 59 < 60

$\therefore \frac{54}{90} \lt \frac{55}{90} \lt \frac{56}{90} \lt \frac{57}{90} \lt \frac{58}{90} \lt \frac{59}{90} \lt \frac{60}{90}$

Hence, five rational numbers between $\mathbf{\frac{3}{5} \text{ and } \frac{2}{3}}$ are $\mathbf{\frac{55}{90}, \frac{56}{90}, \frac{57}{90}, \frac{58}{90} \text{ and } \frac{59}{90}}$.

Write both the numbers with a common denominator, then multiply the numerator and the denominator of each by one more than the number of rational numbers required.

First convert the mixed fraction into an improper fraction and write both the numbers with a common denominator.

$$\begin{aligned}\Rightarrow -1\frac{1}{2} = \frac{-3}{2} \\[1em] \Rightarrow -2 = \frac{-4}{2}\end{aligned}$$

To insert 5 rational numbers, multiply both the numerator and the denominator of each rational number by 5 + 1 i.e. 6.

$$\begin{aligned}\Rightarrow \frac{-4}{2} = \frac{-4 \times 6}{2 \times 6} = \frac{-24}{12} \\[1em] \Rightarrow \frac{-3}{2} = \frac{-3 \times 6}{2 \times 6} = \frac{-18}{12}\end{aligned}$$

As -24 < -23 < -22 < -21 < -20 < -19 < -18,

$\therefore \frac{-24}{12} \lt \frac{-23}{12} \lt \frac{-22}{12} \lt \frac{-21}{12} \lt \frac{-20}{12} \lt \frac{-19}{12} \lt \frac{-18}{12}$

Hence, five rational numbers between -2 and $\mathbf{-1\frac{1}{2}}$ are

$\mathbf{\frac{-23}{12}, \frac{-22}{12}, \frac{-21}{12}, \frac{-20}{12} \text{ and } \frac{-19}{12}}$.

Write both the numbers with a common denominator, then multiply the numerator and the denominator of each by one more than the number of rational numbers required.

To insert 5 rational numbers between -3 and -2, multiply both the numerator and the denominator of each number by 5 + 1 i.e. 6.

$$\begin{aligned}\Rightarrow -3 = \frac{-3 \times 6}{1 \times 6} = \frac{-18}{6} \\[1em] \Rightarrow -2 = \frac{-2 \times 6}{1 \times 6} = \frac{-12}{6}\end{aligned}$$

As -18 < -17 < -16 < -15 < -14 < -13 < -12

$\therefore \frac{-18}{6} \lt \frac{-17}{6} \lt \frac{-16}{6} \lt \frac{-15}{6} \lt \frac{-14}{6} \lt \frac{-13}{6} \lt \frac{-12}{6}$

Hence, five rational numbers between -3 and -2 are

$\mathbf{\frac{-17}{6}, \frac{-16}{6}, \frac{-15}{6}, \frac{-14}{6} \text{ and } \frac{-13}{6}}$.

The given rational numbers have different denominators, so first write them with a common denominator.

$$\begin{aligned}[\text{ LCM of } 4 \text{ and } 6 = 12] \\[1em] \Rightarrow \frac{-3}{4} = \frac{-3 \times 3}{4 \times 3} = \frac{-9}{12} \\[1em] \Rightarrow \frac{5}{6} = \frac{5 \times 2}{6 \times 2} = \frac{10}{12}\end{aligned}$$

We note that -8, -7, -6, …, 8, 9 are the integers between -9 and 10

i.e. -9 < -8 < -7 < -6 < -5 < -4 < -3 < -2 < -1 < 0 < 1 < … < 9 < 10

Choosing any ten of these integers as numerators over the common denominator 12, we get the required rational numbers.

Hence, ten rational numbers between $\mathbf{\frac{-3}{4} \text{ and } \frac{5}{6}}$ are

$\mathbf{\frac{-8}{12}, \frac{-7}{12}, \frac{-6}{12}, \frac{-5}{12}, \frac{-4}{12}, \frac{-3}{12}, \frac{-2}{12}, \frac{-1}{12}, \frac{0}{12} \text{ and } \frac{1}{12}}$.

To insert 12 rational numbers, write both the numbers with the denominator 12 + 1 i.e. 13.

$$\begin{aligned}\Rightarrow -1 = \frac{-1 \times 13}{1 \times 13} = \frac{-13}{13} \\[1em] \Rightarrow 2 = \frac{2 \times 13}{1 \times 13} = \frac{26}{13}\end{aligned}$$

We note that -12, -11, -10, …, 24, 25 are the integers between -13 and 26

i.e. -13 < -12 < -11 < -10 < … < -2 < -1 < 0 < … < 25 < 26

Choosing any twelve of these integers as numerators over the common denominator 13, we get the required rational numbers.

Hence, twelve rational numbers between -1 and 2 are

$\mathbf{\frac{-12}{13}, \frac{-11}{13}, \frac{-10}{13}, \frac{-9}{13}, \frac{-8}{13}, \frac{-7}{13}, \frac{-6}{13}, \frac{-5}{13}, \frac{-4}{13}, \frac{-3}{13}, \frac{-2}{13} \text{ and } \frac{-1}{13}}$.

The numbers which can be expressed in the form $\frac{p}{q}$, where $p \text{ and } q$ are integers and $q ≠ 0$, are called rational numbers.

The numerator $p$ may also be 0, as $\frac{0}{q} = 0$ is a rational number, but the denominator $q$ can never be 0 since division by 0 is not defined.

Hence, Option 3 is the correct option.

To find the additive inverse of any rational number, change its sign.

Additive inverse of $\frac{-5}{12} = \frac{5}{12}$

Hence, Option 1 is the correct option.

To find the multiplicative inverse of a non-zero rational number, interchange its numerator and denominator.

Multiplicative inverse of $\frac{-5}{12} = \frac{12}{-5} = \frac{-12}{5}$

Hence, Option 3 is the correct option.

To find the additive inverse, change the sign of the number, and to find the multiplicative inverse, interchange its numerator and denominator.

Additive inverse of $\frac{-1}{4} = \frac{1}{4}$

Multiplicative inverse of $\frac{-1}{4} = \frac{4}{-1} = -4$

Multiplying both the inverses,

$$\begin{aligned}\Rightarrow \frac{1}{4} \times (-4) \\[1em] \Rightarrow \frac{1 \times (-4)}{4} \\[1em] \Rightarrow \frac{-4}{4} = -1\end{aligned}$$

Hence, Option 2 is the correct option.

The reciprocal is obtained by interchanging the numerator and the denominator, so the sign of the rational number does not change.

For example, the reciprocal of $\frac{-4}{7} \text{ is } \frac{7}{-4} = \frac{-7}{4}$, which is again a negative rational number.

Hence, Option 2 is the correct option.

A number is equal to its own reciprocal only when its product with itself is 1.

⇒ 1 × 1 = 1

⇒ (-1) × (-1) = 1

Also, 0 has no reciprocal at all.

Hence, Option 3 is the correct option.

If $\frac{a}{b}, \frac{c}{d} \text{ and } \frac{e}{f}$ are any three rational numbers, then the associative law of multiplication states that $\frac{a}{b} \times \left(\frac{c}{d} \times \frac{e}{f}\right) = \left(\frac{a}{b} \times \frac{c}{d}\right) \times \frac{e}{f}$

In Option 1, only the grouping of the same three rational numbers, taken in the same order, has been changed.

In Options 2 and 4, the order of the rational numbers has been changed, which shows the commutative law.

In Option 3, the order of the rational numbers has been changed and $\frac{6}{7}$ on the left side has been replaced by $\frac{-6}{7}$ on the right side, so the two sides are not equal at all.

Hence, Option 1 is the correct option.

By the distributive law of multiplication over addition, for any three rational numbers $\frac{a}{b}, \frac{c}{d} \text{ and } \frac{e}{f}$, we have

$\frac{a}{b} \times \left(\frac{c}{d} + \frac{e}{f}\right) = \left(\frac{a}{b} \times \frac{c}{d}\right) + \left(\frac{a}{b} \times \frac{e}{f}\right)$

So, $\frac{-1}{4}$ has to be multiplied by each of the two rational numbers inside the bracket separately.

Expanding the given expression by this law,

$\frac{-1}{4} \times \left[\frac{2}{5} + \left(\frac{-5}{6}\right)\right] = \left[\frac{-1}{4} \times \frac{2}{5}\right] + \left[\frac{-1}{4} \times \left(\frac{-5}{6}\right)\right]$

Hence, Option 1 is the correct option.

Let the required rational number be $x$.

According to the given condition,

$$\begin{aligned}\Rightarrow \frac{-5}{24} \times x = 10 \\[1em] \Rightarrow x = 10 \div \frac{-5}{24} \\[1em] \Rightarrow x = \frac{10}{1} \times \frac{24}{-5} \\[1em] \Rightarrow x = \frac{10 \times 24}{-5} \\[1em] \Rightarrow x = \frac{240}{-5} = -48\end{aligned}$$

Hence, Option 4 is the correct option.

Let the required number be $x$.

According to the given condition,

$$\begin{aligned}\Rightarrow \frac{-3}{4} + x = \frac{7}{6} \\[1em] \Rightarrow x = \frac{7}{6} - \left(\frac{-3}{4}\right) \\[1em] \Rightarrow x = \frac{7}{6} + \frac{3}{4} \\[1em] [\text{ LCM of } 6 \text{ and } 4 = 12] \\[1em] \Rightarrow x = \frac{7 \times 2 + 3 \times 3}{12} \\[1em] \Rightarrow x = \frac{14 + 9}{12} = \frac{23}{12}\end{aligned}$$

Hence, Option 4 is the correct option.

Let the required number be $x$.

According to the given condition,

$$\begin{aligned}\Rightarrow \frac{-7}{8} - x = \frac{5}{12} \\[1em] \Rightarrow x = \frac{-7}{8} - \frac{5}{12} \\[1em] [\text{ LCM of } 8 \text{ and } 12 = 24] \\[1em] \Rightarrow x = \frac{(-7) \times 3 - 5 \times 2}{24} \\[1em] \Rightarrow x = \frac{-21 - 10}{24} = \frac{-31}{24}\end{aligned}$$

Hence, Option 3 is the correct option.

Let the other rational number be $x$.

As the sum of the two rational numbers is -3,

$$\begin{aligned}\Rightarrow \frac{-7}{5} + x = -3 \\[1em] \Rightarrow x = -3 - \left(\frac{-7}{5}\right) \\[1em] \Rightarrow x = -3 + \frac{7}{5} \\[1em] \Rightarrow x = \frac{(-3) \times 5 + 7}{5} \\[1em] \Rightarrow x = \frac{-15 + 7}{5} = \frac{-8}{5}\end{aligned}$$

Hence, Option 1 is the correct option.

Let the other rational number be $x$.

As the product of the two rational numbers is $\frac{-16}{35}$,

$$\begin{aligned}\Rightarrow \frac{-15}{14} \times x = \frac{-16}{35} \\[1em] \Rightarrow x = \frac{-16}{35} \div \frac{-15}{14} \\[1em] \Rightarrow x = \frac{-16}{35} \times \frac{14}{-15} \\[1em] \Rightarrow x = \frac{(-16) \times 14}{35 \times (-15)} \\[1em] \Rightarrow x = \frac{-224}{-525} = \frac{32}{75}\end{aligned}$$

Hence, Option 4 is the correct option.

A rational number $\frac{p}{q}$ is in standard form if $q \gt 0 \text{ and } p, q$ have no common divisor other than 1.

In Option 1, 12 and 26 have the common divisor 2, so $\frac{-12}{26} = \frac{-6}{13}$

In Option 2, 49 and 70 have the common divisor 7, so $\frac{-49}{70} = \frac{-7}{10}$

In Option 3, the denominator 16 is positive and the HCF of 9 and 16 is 1.

In Option 4, the denominator is negative, so $\frac{28}{-105} = \frac{-28}{105} = \frac{-4}{15}$

Hence, Option 3 is the correct option.

Multiplying the two rational numbers,

$$\begin{aligned}\frac{-9}{16} \times \frac{8}{15} = \frac{(-9) \times 8}{16 \times 15} \\[1em] = \frac{-72}{240} \\[1em] = \frac{-3}{10}\end{aligned}$$

Hence, Option 1 is the correct option.

To divide one rational number by another, multiply the dividend by the reciprocal of the divisor.

$$\begin{aligned}\frac{-5}{9} \div \frac{2}{3} = \frac{-5}{9} \times \frac{3}{2} \\[1em] = \frac{(-5) \times 3}{9 \times 2} \\[1em] = \frac{-15}{18} = \frac{-5}{6}\end{aligned}$$

Hence, Option 2 is the correct option.

Let the required rational number be $x$.

According to the given condition,

$$\begin{aligned}\Rightarrow \frac{4}{9} \div x = \frac{-8}{15} \\[1em] \Rightarrow x = \frac{4}{9} \div \frac{-8}{15} \\[1em] \Rightarrow x = \frac{4}{9} \times \frac{15}{-8} \\[1em] \Rightarrow x = \frac{4 \times 15}{9 \times (-8)} \\[1em] \Rightarrow x = \frac{60}{-72} = \frac{-5}{6}\end{aligned}$$

Hence, Option 4 is the correct option.

If $x$ and $y$ are two rational numbers, then $\frac{1}{2}(x + y)$ is a rational number lying between them.

This process of inserting the mean can be repeated as many times as we want.

Hence, Option 4 is the correct option.

Finding the average of the two given rational numbers,

$$\begin{aligned}\Rightarrow \frac{1}{2} \times \left(\frac{1}{5} + \frac{1}{2}\right) \\[1em] [\text{ LCM of } 5 \text{ and } 2 = 10] \\[1em] \Rightarrow \frac{1}{2} \times \frac{2 + 5}{10} \\[1em] \Rightarrow \frac{1}{2} \times \frac{7}{10} = \frac{7}{20}\end{aligned}$$

As $\frac{1}{5} = \frac{4}{20} \text{ and } \frac{1}{2} = \frac{10}{20}$, we see that $\frac{4}{20} \lt \frac{7}{20} \lt \frac{10}{20}$

Hence, Option 4 is the correct option.

Write the given rational numbers with the LCM of their denominators as the common denominator and compare them with the given options.

$$\begin{aligned}[\text{ LCM of } 3 \text{ and } 5 = 30] \\[1em] \Rightarrow \frac{1}{3} = \frac{1 \times 10}{3 \times 10} = \frac{10}{30} \\[1em] \Rightarrow \frac{4}{5} = \frac{4 \times 6}{5 \times 6} = \frac{24}{30}\end{aligned}$$

As 10 < 17 < 24, so $\frac{10}{30} \lt \frac{17}{30} \lt \frac{24}{30}$

Hence, Option 3 is the correct option.

First write each rational number with a positive denominator.

$$\begin{aligned}\Rightarrow \frac{5}{-8} = \frac{5 \times (-1)}{-8 \times (-1)} = \frac{-5}{8} \\[1em] \Rightarrow \frac{2}{-3} = \frac{2 \times (-1)}{-3 \times (-1)} = \frac{-2}{3}\end{aligned}$$

Now write each rational number with the LCM of the denominators as the common denominator.

$$\begin{aligned}[\text{ LCM of } 10, 8 \text{ and } 3 = 120] \\[1em] \Rightarrow \frac{-7}{10} = \frac{(-7) \times 12}{10 \times 12} = \frac{-84}{120} \\[1em] \Rightarrow \frac{-5}{8} = \frac{(-5) \times 15}{8 \times 15} = \frac{-75}{120} \\[1em] \Rightarrow \frac{-2}{3} = \frac{(-2) \times 40}{3 \times 40} = \frac{-80}{120}\end{aligned}$$

As -84 < -80 < -75, so $\frac{-84}{120} \lt \frac{-80}{120} \lt \frac{-75}{120}$.

$\Rightarrow \frac{-7}{10} \lt \frac{2}{-3} \lt \frac{5}{-8}$

Hence, Option 2 is the correct option.

(i) Here the order of the two rational numbers has been changed but the sum remains the same.

If $\frac{a}{b} \text{ and } \frac{c}{d}$ are any two rational numbers, then $\frac{a}{b} + \frac{c}{d} = \frac{c}{d} + \frac{a}{b}$.

Hence, the given statement depicts the commutative property of addition of rational numbers.

(ii) For every non-zero rational number $\frac{a}{b}$, the rational number $\frac{b}{a}$ satisfies $\frac{a}{b} \times \frac{b}{a} = 1$

Hence, the multiplicative inverse of a rational number is also called its reciprocal.

(iii) The multiplicative inverse is obtained by interchanging the numerator and the denominator, so the sign of the rational number does not change. For example, the multiplicative inverse of $\frac{-2}{9} \text{ is } \frac{9}{-2} = \frac{-9}{2}$

Hence, the multiplicative inverse of a negative rational number is a negative rational number.

(iv) As 1 × 1 = 1 and (-1) × (-1) = 1, each of these numbers is its own reciprocal.

Hence, the numbers 1 and -1 are their own reciprocals.

(v) For every non-zero rational number $\frac{a}{b}$, we have $\frac{a}{b} \div \frac{a}{b} = 1$

$\left(\frac{-10}{11}\right) \div \left(\frac{-10}{11}\right) = \frac{-10}{11} \times \frac{11}{-10} = 1$

Hence, $\mathbf{\left(-\frac{10}{11}\right) \div \left(-\frac{10}{11}\right) = 1}$.

(vi) The product of a non-zero rational number and its reciprocal is 1, so the required number is the reciprocal of $-\frac{7}{9}$.

$$\begin{aligned}\Rightarrow \text{Reciprocal of } \frac{-7}{9} = \frac{9}{-7} = \frac{-9}{7} \\[1em] \Rightarrow \frac{-7}{9} \times \frac{-9}{7} = 1\end{aligned}$$

Hence, $\mathbf{\left(-\frac{7}{9}\right) \times \left(\frac{-9}{7}\right) = 1}$.

(vii) All three rational numbers are negative and have the same numerator -5, so write them with a common denominator and compare.

$$\begin{aligned}[\text{ LCM of } 11, 13 \text{ and } 7 = 1001] \\[1em] \Rightarrow \frac{-5}{11} = \frac{(-5) \times 91}{11 \times 91} = \frac{-455}{1001} \\[1em] \Rightarrow \frac{-5}{13} = \frac{(-5) \times 77}{13 \times 77} = \frac{-385}{1001} \\[1em] \Rightarrow \frac{-5}{7} = \frac{(-5) \times 143}{7 \times 143} = \frac{-715}{1001}\end{aligned}$$

As -715 < -455 < -385, so $\frac{-5}{7} \lt \frac{-5}{11} \lt \frac{-5}{13}$.

Hence, the greatest rational number is $\mathbf{\frac{-5}{13}}$.

(i) There is no rational number which when multiplied by 0 gives 1. So the rational number 0 has no reciprocal.

Hence, the given statement is false.

(ii) For any two rational numbers $\frac{a}{b} \text{ and } \frac{c}{d}$, we have $\frac{a}{b} - \frac{c}{d} ≠ \frac{c}{d} - \frac{a}{b}$.

For example, $\frac{1}{2} - \frac{1}{3} = \frac{1}{6}$, but $\frac{1}{3} - \frac{1}{2} = \frac{-1}{6}$.

Hence, the given statement is false.

(iii) The additive inverse is obtained by changing the sign of the rational number. So the additive inverse of $\frac{1}{4} \text{ is } \frac{-1}{4}$ and not -4.

$[\text{Check : } \frac{1}{4} + \frac{-1}{4} = 0 \text{ but } \frac{1}{4} + (-4) = \frac{-15}{4}]$

Hence, the given statement is false.

(iv) A rational number is of the form $\frac{p}{q}$ where $q ≠ 0$. Here the denominator is 0 and division by 0 is not defined.

Hence, the given statement is false.

(v) Every negative rational number is smaller than 0, and there is no smallest rational number as we can always find a rational number smaller than any given one. For example, -1 < 0 and -2 < -1 and so on.

Hence, the given statement is false.

(vi) First express the rational number with a positive denominator.

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

To find the multiplicative inverse of a non-zero rational number, interchange its numerator and denominator.

Interchanging the numerator and the denominator of $\frac{4}{9}$, we get $\frac{9}{4}$, which is the same as the multiplicative inverse given in the statement.

$[\text{Check : } \frac{4}{9} \times \frac{9}{4} = 1]$

Hence, the given statement is true.

(vii) Between any two different rational numbers we can always insert their mean, and this process can be repeated as many times as we want.

Hence, the given statement is true.

For any two rational numbers a and b, each of a + b, a - b and a × b is again a rational number, so rational numbers are closed under addition, subtraction and multiplication.

So, the Reason (R) is true.

But the Assertion (A) claims that rational numbers are not closed under subtraction, which is wrong. For example, $\frac{2}{3} - \frac{1}{4} = \frac{5}{12}$, which is a rational number.

So, the Assertion (A) is false.

Hence, Option 4 is the correct option.

The negative (additive inverse) of a rational number is obtained by changing its sign.

⇒ -5 + 5 = 0

⇒ 5 + (-5) = 0

So, the negative of -5 is 5 and the negative of 5 is -5, i.e. the Assertion (A) is true.

The Reason (R) states that the negative of a negative rational number is negative, which is wrong, as the negative of a negative rational number is a positive rational number.

So, the Reason (R) is false.

Hence, Option 3 is the correct option.

For any three rational numbers x, y and z, we have x × (y + z) = x × y + x × z, which is the distributive law of multiplication over addition.

So, the Assertion (A) is true.

Also, for any three rational numbers, (x + y) + z = x + (y + z) and (x × y) × z = x × (y × z), so rational numbers are associative under addition as well as under multiplication.

So, the Reason (R) is also true, but it states the associative law, whereas the Assertion (A) is based on the distributive law.

So, the Reason (R) is not the correct explanation of the Assertion (A).

Hence, Option 2 is the correct option.

Subtract each expense one by one from the amount remaining at that stage.

Monthly income of Ramesh = ₹ 40,000

Finding the money spent on food,

$\frac{3}{8} \times 40000 = 3 \times 5000 = ₹ 15000$

Money left after buying food = ₹ 40000 - ₹ 15000 = ₹ 25000

Finding the money spent on LIC premium,

$\frac{1}{5} \times 25000 = ₹ 5000$

Money left after paying the LIC premium = ₹ 25000 - ₹ 5000 = ₹ 20000

Finding the money spent on other expenses,

$\frac{1}{2} \times 20000 = ₹ 10000$

Money left with him = ₹ 20000 - ₹ 10000 = ₹ 10000

Hence, Option 2 is the correct option.

First find the fraction of the bill left after A and B have paid, then divide it equally among C, D and E.

Finding the fraction of the bill paid by A and B together,

LCM of 2 and 5 is 10.

$$\begin{aligned}\frac{1}{2} + \frac{1}{5} = \frac{5 + 2}{10} \\[1em] = \frac{7}{10}\end{aligned}$$

Finding the fraction of the bill left,

$$\begin{aligned}1 - \frac{7}{10} = \frac{10 - 7}{10} \\[1em] = \frac{3}{10}\end{aligned}$$

Finding the fraction of the bill paid by each of C, D and E,

$$\begin{aligned}\frac{3}{10} \div 3 = \frac{3}{10} \times \frac{1}{3} \\[1em] = \frac{1}{10}\end{aligned}$$

Hence, Option 4 is the correct option.

First find the fraction of the students who walk on their own, then use it to find the total number of students.

Finding the fraction of the students who come by car or by bus,

LCM of 5 and 4 = 20

$$\begin{aligned}\frac{2}{5} + \frac{1}{4} = \frac{2 \times 4 + 1 \times 5}{20} \\[1em] = \frac{8 + 5}{20} = \frac{13}{20}\end{aligned}$$

Finding the fraction of the students who walk to school,

$1 - \frac{13}{20} = \frac{20 - 13}{20} = \frac{7}{20}$

Finding the fraction of the students who walk on their own,

$\frac{1}{3} \times \frac{7}{20} = \frac{7}{60}$

Let the total number of students in the school be $x$.

$$\begin{aligned}\Rightarrow \frac{7}{60} \times x = 224 \\[1em] \Rightarrow x = 224 \times \frac{60}{7} \\[1em] \Rightarrow x = 32 \times 60 \\[1em] \Rightarrow x = 1920\end{aligned}$$

Hence, Option 3 is the correct option.

First find the sum of the three given rational numbers.

LCM of 7, 21 and 14 = 42

$$\begin{aligned}\frac{3}{7} + \left(\frac{-8}{21}\right) + \left(\frac{-3}{14}\right) = \frac{3 \times 6 + (-8) \times 2 + (-3) \times 3}{42} \\[1em] = \frac{18 - 16 - 9}{42} \\[1em] = \frac{-7}{42} = \frac{-1}{6}\end{aligned}$$

Let the required rational number be $x$.

According to the given condition,

LCM of 14 and 6 = 42

$$\begin{aligned}\Rightarrow \frac{-1}{6} + x = \frac{-5}{14} \\[1em] \Rightarrow x = \frac{-5}{14} - \left(\frac{-1}{6}\right) \\[1em] \Rightarrow x = \frac{-5}{14} + \frac{1}{6} \\[1em] \Rightarrow x = \frac{(-5) \times 3 + 1 \times 7}{42} \\[1em] \Rightarrow x = \frac{-15 + 7}{42} \\[1em] \Rightarrow x = \frac{-8}{42} = \frac{-4}{21}\end{aligned}$$

Hence, Option 1 is the correct option.

Examine each expression separately, keeping in mind that $p$ and $q$ are non-zero and $p ≠ q$.

(i) As $p ≠ q$, the numerator $p - q$ is non-zero and $q - p = -(p - q)$

$\Rightarrow \frac{p - q}{q - p} = \frac{p - q}{-(p - q)} = -1$

So, this expression is always equal to -1 and can never be equal to 1.

(ii) Here the numerator and the denominator are the same, as $p + q = q + p$ by the commutative law of addition.

$\Rightarrow \frac{p + q}{q + p} = 1$

So, this expression is equal to 1.

(iii) Here the numerator and the denominator are the same, as $pq = qp$ by the commutative law of multiplication.

$\Rightarrow \frac{pq}{qp} = 1$

So, this expression is equal to 1.

(iv) Dividing,

$$\begin{aligned}\Rightarrow \frac{p}{q} \div \frac{q}{p} \\[1em] \Rightarrow \frac{p}{q} \times \frac{p}{q} \\[1em] \Rightarrow \frac{p^2}{q^2}\end{aligned}$$

This is equal to 1 whenever $p^2 = q^2$, which happens for $p = -q$. For example, taking $p = 2$ and $q = -2$, we get $\frac{p^2}{q^2} = \frac{4}{4} = 1$

So, this expression can be equal to 1.

Thus, only expression (i) can never be equal to 1.

Hence, Option 1 is the correct option.

Rational numbers are closed under division only when the divisor is a non-zero rational number, as division by 0 is not defined.

As a ÷ b does not satisfy the closure property, the divisor b must be 0.

⇒ b = 0

⇒ a + b = 3

⇒ a + 0 = 3

⇒ a = 3

Finding the values of ab and a - b,

⇒ ab = 3 × 0 = 0

⇒ a - b = 3 - 0 = 3

Hence, Option 3 is the correct option.