Rational Numbers

Both rational numbers have the same (positive) denominator 7.

$$\begin{aligned}\Rightarrow \frac{4}{7} + \frac{-5}{7} \\[1em] \Rightarrow \frac{4 + (-5)}{7} \\[1em] \Rightarrow \frac{-1}{7}\end{aligned}$$

Hence, $\mathbf{\frac{4}{7} + \frac{-5}{7} = \frac{-1}{7}}$.

First express each rational number with a positive denominator.

$$\begin{aligned}\Rightarrow \frac{7}{-13} = \frac{7 \times (-1)}{-13 \times (-1)} = \frac{-7}{13} \\[1em] \Rightarrow \frac{4}{-13} = \frac{4 \times (-1)}{-13 \times (-1)} = \frac{-4}{13}\end{aligned}$$

Both rational numbers now have the same (positive) denominator 13.

$$\begin{aligned}\Rightarrow \frac{-7}{13} + \frac{-4}{13} \\[1em] \Rightarrow \frac{(-7) + (-4)}{13} \\[1em] \Rightarrow \frac{-11}{13}\end{aligned}$$

Hence, $\mathbf{\frac{7}{-13} + \frac{4}{-13} = \frac{-11}{13}}$.

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

First express rational number with a positive denominator.

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

HCF of 5 and 3 = 1.

Hence, the standard form of$\mathbf{\frac{5}{-3} \text{ is } \frac{-5}{3}}$

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

Here the denominator 36 is already positive.

HCF of 12 and 36 = 12.

Dividing both numerator and denominator by their HCF,

$$\begin{aligned}\Rightarrow \frac{-12}{36} \\[1em] \Rightarrow \frac{-12 \div 12}{36 \div 12} \\[1em] \Rightarrow \frac{-1}{3}\end{aligned}$$

Hence, the standard form of $\mathbf{\frac{-12}{36} \text{ is } \frac{-1}{3}}$

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

In $\frac{2}{7}$, the denominator 7 is already positive.

HCF of 2 and 7 = 1.

So, the numerator and denominator have no common factor other than 1 and $\frac{2}{7}$ is already in standard form.

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

Simplifying,

$$\begin{aligned}\Rightarrow \frac{-4}{9} + 2\frac{12}{13} \\[1em] \Rightarrow \frac{-4}{9} + \frac{38}{13}\\[1em] [\text{ LCM of } 9 \text{ and }13 = 117] \\[1em] \Rightarrow \frac{(-4) \times 13 + 38 \times 9}{117} \\[1em] \Rightarrow \frac{-52 + 342}{117} \\[1em] \Rightarrow \frac{290}{117}\\[1em] \Rightarrow 2\frac{56}{117}\end{aligned}$$

Hence, $\mathbf{\frac{-4}{9} + 2\frac{12}{13} = 2\frac{56}{117}}$

First express the rational number with a positive denominator and convert the mixed fraction into an improper fraction.

Simplifying,

$$\begin{aligned}\Rightarrow \frac{11}{-7} + 8\frac{2}{3} \\[1em] \Rightarrow \frac{-11}{7} + \frac{26}{3} \\[1em] [\text{ LCM of } 7 \text{ and } 3 = 21] \\[1em] \Rightarrow \frac{(-11) \times 3 + 26 \times 7}{21} \\[1em] \Rightarrow \frac{-33 + 182}{21} \\[1em] \Rightarrow \frac{149}{21} \\[1em] \Rightarrow 7\frac{2}{21}\end{aligned}$$

Hence, $\mathbf{\frac{11}{-7} + 8\frac{2}{3} = 7\frac{2}{21}}$

To verify the commutative property of addition, we have to show that $\frac{-4}{3} + \frac{3}{7} = \frac{3}{7} + \frac{-4}{3}$

LCM of 3 and 7 = 21

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{-4}{3} + \frac{3}{7} \\[1em] \Rightarrow \frac{(-4) \times 7 + 3 \times 3}{21} \\[1em] \Rightarrow \frac{-28 + 9}{21} \\[1em] \Rightarrow \frac{-19}{21}\end{aligned}$$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow \frac{3}{7} + \frac{-4}{3} \\[1em] \Rightarrow \frac{3 \times 3 + (-4) \times 7}{21} \\[1em] \Rightarrow \frac{9 - 28}{21} \\[1em] \Rightarrow \frac{-19}{21}\end{aligned}$$

As L.H.S. = R.H.S.

$\Rightarrow \frac{-4}{3} + \frac{3}{7} = \frac{3}{7} + \frac{-4}{3}$

Hence, the commutative property of addition is verified.

First express $\frac{-2}{-5}$ with a positive denominator.

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

To verify the commutative property of addition, we have to show that $\frac{2}{5} + \frac{1}{3} = \frac{1}{3} + \frac{2}{5}$

LCM of 5 and 3 = 15

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{2}{5} + \frac{1}{3} \\[1em] \Rightarrow \frac{2 \times 3 + 1 \times 5}{15} \\[1em] \Rightarrow \frac{6 + 5}{15} \\[1em] \Rightarrow \frac{11}{15}\end{aligned}$$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow \frac{1}{3} + \frac{2}{5} \\[1em] \Rightarrow \frac{1 \times 5 + 2 \times 3}{15} \\[1em] \Rightarrow \frac{5 + 6}{15} \\[1em] \Rightarrow \frac{11}{15}\end{aligned}$$

As L.H.S. = R.H.S.

$\Rightarrow \frac{-2}{-5} + \frac{1}{3} = \frac{1}{3} + \frac{-2}{-5}$

Hence, the commutative property of addition is verified.

To verify the commutative property of addition, we have to show that $\frac{9}{11} + \frac{2}{13} = \frac{2}{13} + \frac{9}{11}$

LCM of 11 and 13 = 143

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{9}{11} + \frac{2}{13} \\[1em] \Rightarrow \frac{9 \times 13 + 2 \times 11}{143} \\[1em] \Rightarrow \frac{117 + 22}{143} \\[1em] \Rightarrow \frac{139}{143}\end{aligned}$$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow \frac{2}{13} + \frac{9}{11} \\[1em] \Rightarrow \frac{2 \times 11 + 9 \times 13}{143} \\[1em] \Rightarrow \frac{22 + 117}{143} \\[1em] \Rightarrow \frac{139}{143}\end{aligned}$$

As L.H.S. = R.H.S.

$\Rightarrow \frac{9}{11} + \frac{2}{13} = \frac{2}{13} + \frac{9}{11}$

Hence, the commutative property of addition is verified.

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

First express the rational number with a positive denominator.

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

Changing the sign of $\frac{-2}{3}$, we get

$$\begin{aligned}\Rightarrow -\left(\frac{-2}{3}\right) \\[1em] \Rightarrow \frac{2}{3}\end{aligned}$$

Hence, the additive inverse of $\mathbf{\frac{2}{-3} \text{ is } \frac{2}{3}}$.

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

First express the rational number with a positive denominator.

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

Changing the sign of $\frac{7}{12}$, we get

$$\begin{aligned}\Rightarrow -\left(\frac{7}{12}\right) \\[1em] \Rightarrow \frac{-7}{12}\end{aligned}$$

Hence, the additive inverse of $\mathbf{\frac{-7}{-12} \text{ is } \frac{-7}{12}}$

Using commutativity and associativity of addition, group together the rational numbers having the same denominator.

$$\begin{aligned}\Rightarrow \frac{4}{5} + \frac{11}{7} + \frac{-7}{5} + \frac{-2}{7} \\[1em] \Rightarrow \left(\frac{4}{5} + \frac{-7}{5}\right) + \left(\frac{11}{7} + \frac{-2}{7}\right) \\[1em] \Rightarrow \frac{4 + (-7)}{5} + \frac{11 + (-2)}{7} \\[1em] \Rightarrow \frac{-3}{5} + \frac{9}{7} \\[1em] [\text{ LCM of } 5 \text{ and } 7 = 35] \\[1em] \Rightarrow \frac{(-3) \times 7 + 9 \times 5}{35} \\[1em] \Rightarrow \frac{-21 + 45}{35} \\[1em] \Rightarrow \frac{24}{35}\end{aligned}$$

Hence, $\mathbf{\frac{4}{5} + \frac{11}{7} + \frac{-7}{5} + \frac{-2}{7} = \frac{24}{35}}$

Using commutativity and associativity of addition, group together the rational numbers whose denominators have a small LCM.

$$\begin{aligned}\Rightarrow \frac{3}{7} + \frac{4}{9} + \frac{-5}{21} + \frac{2}{3} \\[1em] \Rightarrow \left(\frac{3}{7} + \frac{-5}{21}\right) + \left(\frac{4}{9} + \frac{2}{3}\right) \\[1em] \Rightarrow \frac{3 \times 3 + (-5)}{21} + \frac{4 + 2 \times 3}{9} \\[1em] \Rightarrow \frac{9 - 5}{21} + \frac{4 + 6}{9} \\[1em] \Rightarrow \frac{4}{21} + \frac{10}{9} \\[1em] [\text{ LCM of } 21 \text{ and } 9 = 63] \\[1em] \Rightarrow \frac{4 \times 3 + 10 \times 7}{63} \\[1em] \Rightarrow \frac{12 + 70}{63} \\[1em] \Rightarrow \frac{82}{63} \\[1em] \Rightarrow 1\frac{19}{63}\end{aligned}$$

Hence, $\mathbf{\frac{3}{7} + \frac{4}{9} + \frac{-5}{21} + \frac{2}{3} = 1\frac{19}{63}}$

To verify the associative property of addition, we have to show that $a + (b + c) = (a + b) + c$

Solving L.H.S.,

$$\begin{aligned}\Rightarrow b + c = \frac{4}{9} + \frac{-5}{18} \\[1em] \Rightarrow \frac{4 \times 2 + (-5)}{18} \\[1em] \Rightarrow \frac{8 - 5}{18} = \frac{3}{18} = \frac{1}{6} \\[1em] \therefore a + (b + c) = \frac{-11}{27} + \frac{1}{6} \\[1em] [\text{ LCM of } 27 \text{ and } 6 = 54] \\[1em] \Rightarrow \frac{(-11) \times 2 + 1 \times 9}{54} \\[1em] \Rightarrow \frac{-22 + 9}{54} \\[1em] \Rightarrow \frac{-13}{54}\end{aligned}$$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow a + b = \frac{-11}{27} + \frac{4}{9} \\[1em] \Rightarrow \frac{-11 + 4 \times 3}{27} \\[1em] \Rightarrow \frac{-11 + 12}{27} = \frac{1}{27} \\[1em] \therefore (a + b) + c = \frac{1}{27} + \frac{-5}{18} \\[1em] [\text{ LCM of } 27 \text{ and } 18 = 54] \\[1em] \Rightarrow \frac{1 \times 2 + (-5) \times 3}{54} \\[1em] \Rightarrow \frac{2 - 15}{54} \\[1em] \Rightarrow \frac{-13}{54}\end{aligned}$$

As L.H.S. = R.H.S.

$\Rightarrow a + (b + c) = (a + b) + c$

Hence, proved that $\bm{a + (b + c) = (a + b) + c}$.

Subtracting one rational number from another is the same as adding the additive inverse of the rational number that is being subtracted.

$$\begin{aligned}\Rightarrow \frac{-3}{7} - 2\frac{3}{5} \\[1em] \Rightarrow \frac{-3}{7} - \frac{13}{5} \\[1em] \Rightarrow \frac{-3}{7} + \left(\frac{-13}{5}\right) \\[1em] [\text{ LCM of } 7 \text{ and } 5 = 35] \\[1em] \Rightarrow \frac{(-3) \times 5 + (-13) \times 7}{35} \\[1em] \Rightarrow \frac{-15 - 91}{35} \\[1em] \Rightarrow \frac{-106}{35} \\[1em] \Rightarrow -3\frac{1}{35}\end{aligned}$$

Hence, $\mathbf{\frac{-3}{7} - 2\frac{3}{5} = -3\frac{1}{35}}$

Subtracting one rational number from another is the same as adding the additive inverse of the rational number that is being subtracted.

$$\begin{aligned}\Rightarrow 3\frac{5}{8} - \left(\frac{-4}{9}\right) \\[1em] \Rightarrow \frac{29}{8} + \frac{4}{9} \\[1em] [\text{ LCM of } 8 \text{ and } 9 = 72] \\[1em] \Rightarrow \frac{29 \times 9 + 4 \times 8}{72} \\[1em] \Rightarrow \frac{261 + 32}{72} \\[1em] \Rightarrow \frac{293}{72} \\[1em] \Rightarrow 4\frac{5}{72}\end{aligned}$$

Hence, $\mathbf{3\frac{5}{8} - \left(\frac{-4}{9}\right) = 4\frac{5}{72}}$

Subtracting one rational number from another is the same as adding the additive inverse of the rational number that is being subtracted.

$$\begin{aligned}\Rightarrow -4\frac{7}{9} - \left(-3\frac{1}{5}\right) \\[1em] \Rightarrow \frac{-43}{9} - \left(\frac{-16}{5}\right) \\[1em] \Rightarrow \frac{-43}{9} + \frac{16}{5} \\[1em] [\text{ LCM of } 9 \text{ and } 5 = 45] \\[1em] \Rightarrow \frac{(-43) \times 5 + 16 \times 9}{45} \\[1em] \Rightarrow \frac{-215 + 144}{45} \\[1em] \Rightarrow \frac{-71}{45} \\[1em] \Rightarrow -1\frac{26}{45}\end{aligned}$$

Hence, $\mathbf{-4\frac{7}{9} - \left(-3\frac{1}{5}\right) = -1\frac{26}{45}}$

Let the other rational number be $x$.

As the sum of the two rational numbers is $\frac{3}{5}$,

$$\begin{aligned}\Rightarrow \frac{-2}{7} + x = \frac{3}{5} \\[1em] \Rightarrow x = \frac{3}{5} - \left(\frac{-2}{7}\right) \\[1em] \Rightarrow x = \frac{3}{5} + \frac{2}{7} \\[1em] [\text{ LCM of } 5 \text{ and } 7 = 35] \\[1em] \Rightarrow x = \frac{3 \times 7 + 2 \times 5}{35} \\[1em] \Rightarrow x = \frac{21 + 10}{35} \\[1em] \Rightarrow x = \frac{31}{35}\end{aligned}$$

Hence, the other rational number is $\mathbf{\frac{31}{35}}$

Let the required rational number be $x$.

According to the given condition,

$$\begin{aligned}\Rightarrow \frac{-5}{11} + x = \frac{-7}{8} \\[1em] \Rightarrow x = \frac{-7}{8} - \left(\frac{-5}{11}\right) \\[1em] \Rightarrow x = \frac{-7}{8} + \frac{5}{11} \\[1em] [\text{ LCM of } 8 \text{ and } 11 = 88] \\[1em] \Rightarrow x = \frac{(-7) \times 11 + 5 \times 8}{88} \\[1em] \Rightarrow x = \frac{-77 + 40}{88} \\[1em] \Rightarrow x = \frac{-37}{88}\end{aligned}$$

Hence, the required rational number is $\mathbf{\frac{-37}{88}}$

Let the required rational number be $x$.

According to the given condition,

$$\begin{aligned}\Rightarrow -4\frac{3}{5} - x = -3\frac{1}{2} \\[1em] \Rightarrow \frac{-23}{5} - x = \frac{-7}{2} \\[1em] \Rightarrow x = \frac{-23}{5} - \left(\frac{-7}{2}\right) \\[1em] \Rightarrow x = \frac{-23}{5} + \frac{7}{2} \\[1em] [\text{ LCM of } 5 \text{ and } 2 = 10] \\[1em] \Rightarrow x = \frac{(-23) \times 2 + 7 \times 5}{10} \\[1em] \Rightarrow x = \frac{-46 + 35}{10} \\[1em] \Rightarrow x = \frac{-11}{10} = -1\frac{1}{10}\end{aligned}$$

Hence, the required rational number is $\mathbf{\frac{-11}{10}}$

First find the sum of $\frac{-5}{7} \text{ and } \frac{-8}{3}$.

$$\begin{aligned}\Rightarrow \frac{-5}{7} + \frac{-8}{3} \\[1em] [\text{ LCM of } 7 \text{ and } 3 = 21] \\[1em] \Rightarrow \frac{(-5) \times 3 + (-8) \times 7}{21} \\[1em] \Rightarrow \frac{-15 - 56}{21} \\[1em] \Rightarrow \frac{-71}{21}\end{aligned}$$

Now find the sum of $\frac{5}{2} \text{ and } \frac{-11}{12}$.

$$\begin{aligned}\Rightarrow \frac{5}{2} + \frac{-11}{12} \\[1em] [\text{ LCM of } 2 \text{ and } 12 = 12] \\[1em] \Rightarrow \frac{5 \times 6 + (-11)}{12} \\[1em] \Rightarrow \frac{30 - 11}{12} \\[1em] \Rightarrow \frac{19}{12}\end{aligned}$$

Subtracting the first sum from the second sum,

$$\begin{aligned}\Rightarrow \frac{19}{12} - \left(\frac{-71}{21}\right) \\[1em] \Rightarrow \frac{19}{12} + \frac{71}{21} \\[1em] [\text{ LCM of } 12 \text{ and } 21 = 84] \\[1em] \Rightarrow \frac{19 \times 7 + 71 \times 4}{84} \\[1em] \Rightarrow \frac{133 + 284}{84} \\[1em] \Rightarrow \frac{417}{84} = \frac{139}{28} \\[1em] \Rightarrow 4\frac{27}{28}\end{aligned}$$

Hence, the required difference is $\mathbf{4\frac{27}{28}}$

(i) To verify that subtraction of rational numbers is not commutative, we have to show that $x - y \neq y - x$

LCM of 7 and 5 = 35

Solving $x - y$,

$$\begin{aligned}\Rightarrow \frac{-4}{7} - \frac{2}{5} \\[1em] \Rightarrow \frac{(-4) \times 5 - 2 \times 7}{35} \\[1em] \Rightarrow \frac{-20 - 14}{35} \\[1em] \Rightarrow \frac{-34}{35}\end{aligned}$$

Solving $y - x$,

$$\begin{aligned}\Rightarrow \frac{2}{5} - \left(\frac{-4}{7}\right) \\[1em] \Rightarrow \frac{2}{5} + \frac{4}{7} \\[1em] \Rightarrow \frac{2 \times 7 + 4 \times 5}{35} \\[1em] \Rightarrow \frac{14 + 20}{35} \\[1em] \Rightarrow \frac{34}{35}\end{aligned}$$

As $\frac{-34}{35} \neq \frac{34}{35}$

$\Rightarrow x - y \neq y - x$

Hence, proved that $\bm{x - y \neq y - x}$.

(ii) To verify that the additive inverse of a sum equals the sum of the additive inverses, we have to show that $-(x + y) = (-x) + (-y)$

LCM of 7 and 5 = 35

Solving L.H.S.,

$$\begin{aligned}\Rightarrow x + y = \frac{-4}{7} + \frac{2}{5} \\[1em] \Rightarrow \frac{(-4) \times 5 + 2 \times 7}{35} \\[1em] \Rightarrow \frac{-20 + 14}{35} = \frac{-6}{35} \\[1em] \therefore -(x + y) = -\left(\frac{-6}{35}\right) \\[1em] \Rightarrow \frac{6}{35}\end{aligned}$$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow (-x) + (-y) = \frac{4}{7} + \left(\frac{-2}{5}\right) \\[1em] \Rightarrow \frac{4 \times 5 + (-2) \times 7}{35} \\[1em] \Rightarrow \frac{20 - 14}{35} \\[1em] \Rightarrow \frac{6}{35}\end{aligned}$$

As L.H.S. = R.H.S.

$\Rightarrow -(x + y) = (-x) + (-y)$

Hence, proved that $\bm{-(x + y) = (-x) + (-y)}$.

To verify that subtraction of rational numbers is not associative, we have to show that $x - (y - z) \neq (x - y) - z$

Solving L.H.S.,

$$\begin{aligned}\Rightarrow y - z = \frac{-7}{12} - \left(\frac{-2}{3}\right) \\[1em] \Rightarrow \frac{-7}{12} + \frac{2}{3} \\[1em] \Rightarrow \frac{-7 + 2 \times 4}{12} = \frac{-7 + 8}{12} = \frac{1}{12} \\[1em] \therefore x - (y - z) = \frac{4}{9} - \frac{1}{12} \\[1em] [\text{ LCM of } 9 \text{ and } 12 = 36] \\[1em] \Rightarrow \frac{4 \times 4 - 1 \times 3}{36} \\[1em] \Rightarrow \frac{16 - 3}{36} \\[1em] \Rightarrow \frac{13}{36}\end{aligned}$$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow x - y = \frac{4}{9} - \left(\frac{-7}{12}\right) \\[1em] \Rightarrow \frac{4}{9} + \frac{7}{12} \\[1em] \Rightarrow \frac{4 \times 4 + 7 \times 3}{36} = \frac{16 + 21}{36} = \frac{37}{36} \\[1em] \therefore (x - y) - z = \frac{37}{36} - \left(\frac{-2}{3}\right) \\[1em] \Rightarrow \frac{37}{36} + \frac{2}{3} \\[1em] \Rightarrow \frac{37 + 2 \times 12}{36} \\[1em] \Rightarrow \frac{37 + 24}{36} \\[1em] \Rightarrow \frac{61}{36}\end{aligned}$$

As $\frac{13}{36} \neq \frac{61}{36}$

$\Rightarrow x - (y - z) \neq (x - y) - z$

Hence, proved that $\bm{x - (y - z) \neq (x - y) - z}$.

(i) Rational numbers are closed under subtraction, so the difference of two rational numbers is always a rational number.

$$\begin{aligned}\Rightarrow \frac{2}{3} - \frac{4}{5} \\[1em] [\text{ LCM of } 3 \text{ and } 5 = 15] \\[1em] \Rightarrow \frac{2 \times 5 - 4 \times 3}{15} \\[1em] \Rightarrow \frac{10 - 12}{15} \\[1em] \Rightarrow \frac{-2}{15}\end{aligned}$$

As $\frac{-2}{15}$ is a rational number, the given statement is false.

Hence, the given statement is false.

(ii) To find the additive inverse of any rational number, change its sign. The sum of a rational number and its additive inverse is 0.

$$\begin{aligned}\Rightarrow \frac{5}{7} + \left(\frac{-5}{7}\right) \\[1em] \Rightarrow \frac{5 + (-5)}{7} \\[1em] \Rightarrow \frac{0}{7} = 0\end{aligned}$$

Hence, the given statement is true.

(iii) The additive inverse of 0 is obtained by changing its sign, and $-0 = 0$.

As $0 + 0 = 0$, the additive inverse of 0 is 0 itself.

Hence, the given statement is true.

(iv) For any two rational numbers $\frac{p}{q} \text{ and } \frac{r}{s}$, we have $\frac{p}{q} - \frac{r}{s} \neq \frac{r}{s} - \frac{p}{q}$

So, subtraction of rational numbers is not commutative.

Hence, the given statement is false.

(v) For any three rational numbers

$\frac{p}{q}, \frac{r}{s} \text{ and } \frac{t}{u}$, we have $\left(\frac{p}{q} - \frac{r}{s}\right) - \frac{t}{u} \neq \frac{p}{q} - \left(\frac{r}{s} - \frac{t}{u}\right)$

So, subtraction of rational numbers is not associative.

Hence, the given statement is true.

(vi) Although $\frac{p}{q} - 0 = \frac{p}{q}$, we have $0 - \frac{p}{q} = \frac{-p}{q} \neq \frac{p}{q}$

So, subtraction of rational numbers has no identity element.

Hence, the given statement is false.

First express the rational number with a positive denominator, then multiply the numerators together and the denominators together.

$$\begin{aligned}\Rightarrow \frac{6}{-7} \times \frac{14}{30} \\[1em] \Rightarrow \frac{-6}{7} \times \frac{14}{30} \\[1em] \Rightarrow \frac{(-6) \times 14}{7 \times 30} \\[1em] \Rightarrow \frac{-84}{210} \\[1em] \Rightarrow \frac{-2}{5}\end{aligned}$$

Hence, $\mathbf{\frac{6}{-7} \times \frac{14}{30} = \frac{-2}{5}}$

First convert the mixed fractions into improper fractions, then multiply the numerators together and the denominators together.

$$\begin{aligned}\Rightarrow 6\frac{2}{3} \times 1\frac{2}{7} \\[1em] \Rightarrow \frac{20}{3} \times \frac{9}{7} \\[1em] \Rightarrow \frac{20 \times 9}{3 \times 7} \\[1em] \Rightarrow \frac{180}{21} \\[1em] \Rightarrow \frac{60}{7} \\[1em] \Rightarrow 8\frac{4}{7}\end{aligned}$$

Hence, $\mathbf{6\frac{2}{3} \times 1\frac{2}{7} = 8\frac{4}{7}}$

First express the rational number with a positive denominator, then multiply the numerators together and the denominators together.

$$\begin{aligned}\Rightarrow \frac{25}{-9} \times \frac{-3}{10} \\[1em] \Rightarrow \frac{-25}{9} \times \frac{-3}{10} \\[1em] \Rightarrow \frac{(-25) \times (-3)}{9 \times 10} \\[1em] \Rightarrow \frac{75}{90} \\[1em] \Rightarrow \frac{5}{6}\end{aligned}$$

Hence, $\mathbf{\frac{25}{-9} \times \frac{-3}{10} = \frac{5}{6}}$

To verify the commutative property of multiplication, we have to show that $\frac{4}{5} \times \frac{-7}{8} = \frac{-7}{8} \times \frac{4}{5}$

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{4}{5} \times \frac{-7}{8} \\[1em] \Rightarrow \frac{4 \times (-7)}{5 \times 8} \\[1em] \Rightarrow \frac{-28}{40} \\[1em] \Rightarrow \frac{-7}{10}\end{aligned}$$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow \frac{-7}{8} \times \frac{4}{5} \\[1em] \Rightarrow \frac{(-7) \times 4}{8 \times 5} \\[1em] \Rightarrow \frac{-28}{40} \\[1em] \Rightarrow \frac{-7}{10}\end{aligned}$$

As L.H.S. = R.H.S.

$\Rightarrow \frac{4}{5} \times \frac{-7}{8} = \frac{-7}{8} \times \frac{4}{5}$

Hence, the commutative property of multiplication is verified.

First convert the mixed fractions into improper fractions.

$13\frac{1}{3} = \frac{40}{3} \text{ and } 1\frac{1}{8} = \frac{9}{8}$

To verify the commutative property of multiplication, we have to show that $\frac{40}{3} \times \frac{9}{8} = \frac{9}{8} \times \frac{40}{3}$

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{40}{3} \times \frac{9}{8} \\[1em] \Rightarrow \frac{40 \times 9}{3 \times 8} \\[1em] \Rightarrow \frac{360}{24} \\[1em] \Rightarrow 15\end{aligned}$$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow \frac{9}{8} \times \frac{40}{3} \\[1em] \Rightarrow \frac{9 \times 40}{8 \times 3} \\[1em] \Rightarrow \frac{360}{24} \\[1em] \Rightarrow 15\end{aligned}$$

As L.H.S. = R.H.S.

$\Rightarrow 13\frac{1}{3} \times 1\frac{1}{8} = 1\frac{1}{8} \times 13\frac{1}{3}$

Hence, the commutative property of multiplication is verified.

First express each rational number with a positive denominator.

$\frac{-7}{-20} = \frac{7}{20} \text{ and } \frac{5}{-14} = \frac{-5}{14}$

To verify the commutative property of multiplication, we have to show that $\frac{7}{20} \times \frac{-5}{14} = \frac{-5}{14} \times \frac{7}{20}$

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{7}{20} \times \frac{-5}{14} \\[1em] \Rightarrow \frac{7 \times (-5)}{20 \times 14} \\[1em] \Rightarrow \frac{-35}{280} \\[1em] \Rightarrow \frac{-1}{8}\end{aligned}$$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow \frac{-5}{14} \times \frac{7}{20} \\[1em] \Rightarrow \frac{(-5) \times 7}{14 \times 20} \\[1em] \Rightarrow \frac{-35}{280} \\[1em] \Rightarrow \frac{-1}{8}\end{aligned}$$

As L.H.S. = R.H.S.

$\Rightarrow \frac{-7}{-20} \times \frac{5}{-14} = \frac{5}{-14} \times \frac{-7}{-20}$

Hence, the commutative property of multiplication is verified.

Solve the bracket first on each side and then multiply.

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{3}{5} \times \left(\frac{-4}{7} \times \frac{-8}{9}\right) \\[1em] \Rightarrow \frac{3}{5} \times \frac{(-4) \times (-8)}{7 \times 9} \\[1em] \Rightarrow \frac{3}{5} \times \frac{32}{63} \\[1em] \Rightarrow \frac{3 \times 32}{5 \times 63} \\[1em] \Rightarrow \frac{96}{315} \\[1em] \Rightarrow \frac{32}{105}\end{aligned}$$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow \left(\frac{3}{5} \times \frac{-4}{7}\right) \times \frac{-8}{9} \\[1em] \Rightarrow \frac{3 \times (-4)}{5 \times 7} \times \frac{-8}{9} \\[1em] \Rightarrow \frac{-12}{35} \times \frac{-8}{9} \\[1em] \Rightarrow \frac{(-12) \times (-8)}{35 \times 9} \\[1em] \Rightarrow \frac{96}{315} \\[1em] \Rightarrow \frac{32}{105}\end{aligned}$$

As L.H.S. = R.H.S.

$\Rightarrow \frac{3}{5} \times \left(\frac{-4}{7} \times \frac{-8}{9}\right) = \left(\frac{3}{5} \times \frac{-4}{7}\right) \times \frac{-8}{9}$

Hence, verified and the property used is associative property of multiplication of rational numbers.

Solve the bracket first on the left side and then multiply, and on the right side multiply first and then add.

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{5}{9} \times \left(\frac{-3}{2} + \frac{7}{5}\right) \\[1em] [\text{ LCM of } 2 \text{ and } 5 = 10] \\[1em] \Rightarrow \frac{5}{9} \times \frac{(-3) \times 5 + 7 \times 2}{10} \\[1em] \Rightarrow \frac{5}{9} \times \frac{-15 + 14}{10} \\[1em] \Rightarrow \frac{5}{9} \times \frac{-1}{10} \\[1em] \Rightarrow \frac{-5}{90} \\[1em] \Rightarrow \frac{-1}{18}\end{aligned}$$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow \frac{5}{9} \times \frac{-3}{2} + \frac{5}{9} \times \frac{7}{5} \\[1em] \Rightarrow \frac{5 \times (-3)}{9 \times 2} + \frac{5 \times 7}{9 \times 5} \\[1em] \Rightarrow \frac{-15}{18} + \frac{35}{45} \\[1em] \Rightarrow \frac{-5}{6} + \frac{7}{9} \\[1em] [\text{ LCM of } 6 \text{ and } 9 = 18] \\[1em] \Rightarrow \frac{(-5) \times 3 + 7 \times 2}{18} \\[1em] \Rightarrow \frac{-15 + 14}{18} \\[1em] \Rightarrow \frac{-1}{18}\end{aligned}$$

As L.H.S. = R.H.S.

$\Rightarrow \frac{5}{9} \times \left(\frac{-3}{2} + \frac{7}{5}\right) = \frac{5}{9} \times \frac{-3}{2} + \frac{5}{9} \times \frac{7}{5}$

Hence, verified and the property used is distributive property of multiplication over addition of rational numbers.

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

$$\begin{aligned}\Rightarrow 12 = \frac{12}{1} \\[1em] \Rightarrow \text{Multiplicative inverse} = \frac{1}{12}\end{aligned}$$

Hence, the multiplicative inverse of 12 is $\mathbf{\frac{1}{12}}$

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 $\frac{2}{3}$, we get $\frac{3}{2}$

Hence, the multiplicative inverse of $\mathbf{\frac{2}{3} \text{ is } \frac{3}{2}}$

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

$$\begin{aligned}\Rightarrow \text{Multiplicative inverse of } \frac{-4}{7} = \frac{7}{-4} \\[1em] \Rightarrow \frac{-7}{4}\end{aligned}$$

Hence, the multiplicative inverse of $\mathbf{\frac{-4}{7} \text{ is } \frac{-7}{4}}$

First find the product, then interchange its numerator and denominator.

$$\begin{aligned}\Rightarrow \frac{-3}{8} \times \left(\frac{-7}{13}\right) \\[1em] \Rightarrow \frac{(-3) \times (-7)}{8 \times 13} \\[1em] \Rightarrow \frac{21}{104}\end{aligned}$$

Interchanging the numerator and the denominator of $\frac{21}{104}$, we get $\frac{104}{21} = 4\frac{20}{21}$

Hence, the multiplicative inverse of $\mathbf{\frac{-3}{8} \times \left(\frac{-7}{13}\right) \text{ is } 4\frac{20}{21}}$

Rearrange the terms using commutativity of multiplication so that the common factor $\frac{3}{7}$ can be taken out using distributivity.

$$\begin{aligned}\Rightarrow \frac{2}{5} \times \frac{-3}{7} - \frac{1}{14} - \frac{3}{7} \times \frac{3}{5} \\[1em] \Rightarrow \frac{-3}{7} \times \frac{2}{5} - \frac{3}{7} \times \frac{3}{5} - \frac{1}{14} \\[1em] \Rightarrow \frac{3}{7} \times \left(\frac{-2}{5} - \frac{3}{5}\right) - \frac{1}{14} \\[1em] \Rightarrow \frac{3}{7} \times \frac{-5}{5} - \frac{1}{14} \\[1em] \Rightarrow \frac{3}{7} \times (-1) - \frac{1}{14} \\[1em] \Rightarrow \frac{-3}{7} - \frac{1}{14} \\[1em] [\text{ LCM of } 7 \text{ and } 14 = 14] \\[1em] \Rightarrow \frac{(-3) \times 2 - 1}{14} \\[1em] \Rightarrow \frac{-6 - 1}{14} \\[1em] \Rightarrow \frac{-7}{14} \\[1em] \Rightarrow \frac{-1}{2}\end{aligned}$$

Hence, $\mathbf{\frac{2}{5} \times \frac{-3}{7} - \frac{1}{14} - \frac{3}{7} \times \frac{3}{5} = \frac{-1}{2}}$

Rearrange the terms using commutativity of multiplication so that the common factor $\frac{8}{9}$ can be taken out using distributivity.

$$\begin{aligned}\Rightarrow \frac{8}{9} \times \frac{4}{5} + \frac{5}{6} - \frac{9}{5} \times \frac{8}{9} \\[1em] \Rightarrow \frac{8}{9} \times \frac{4}{5} - \frac{8}{9} \times \frac{9}{5} + \frac{5}{6} \\[1em] \Rightarrow \frac{8}{9} \times \left(\frac{4}{5} - \frac{9}{5}\right) + \frac{5}{6} \\[1em] \Rightarrow \frac{8}{9} \times \frac{-5}{5} + \frac{5}{6} \\[1em] \Rightarrow \frac{8}{9} \times (-1) + \frac{5}{6} \\[1em] \Rightarrow \frac{-8}{9} + \frac{5}{6} \\[1em] [\text{ LCM of } 9 \text{ and } 6 = 18] \\[1em] \Rightarrow \frac{(-8) \times 2 + 5 \times 3}{18} \\[1em] \Rightarrow \frac{-16 + 15}{18} \\[1em] \Rightarrow \frac{-1}{18}\end{aligned}$$

Hence, $\mathbf{\frac{8}{9} \times \frac{4}{5} + \frac{5}{6} - \frac{9}{5} \times \frac{8}{9} = \frac{-1}{18}}$

Rearrange and group the rational numbers using commutativity and associativity of multiplication so that the products can be reduced easily.

$$\begin{aligned}\Rightarrow \frac{-3}{7} \times \frac{14}{15} \times \frac{7}{12} \times \left(\frac{-30}{35}\right) \\[1em] \Rightarrow \left(\frac{-3}{7} \times \frac{14}{15}\right) \times \left(\frac{7}{12} \times \frac{-30}{35}\right) \\[1em] \Rightarrow \frac{(-3) \times 14}{7 \times 15} \times \frac{7 \times (-30)}{12 \times 35} \\[1em] \Rightarrow \frac{-42}{105} \times \frac{-210}{420} \\[1em] \Rightarrow \frac{-2}{5} \times \frac{-1}{2} \\[1em] \Rightarrow \frac{(-2) \times (-1)}{5 \times 2} \\[1em] \Rightarrow \frac{2}{10} \\[1em] \Rightarrow \frac{1}{5}\end{aligned}$$

Hence, $\mathbf{\frac{-3}{7} \times \frac{14}{15} \times \frac{7}{12} \times \left(\frac{-30}{35}\right) = \frac{1}{5}}$

Here $r = \frac{-12}{15} = \frac{-4}{5}$

(i) To verify the associative property of multiplication, we have to show that $p \times (q \times r) = (p \times q) \times r$

Solving L.H.S.,

$$\begin{aligned}\Rightarrow q \times r = \frac{3}{4} \times \frac{-4}{5} \\[1em] \Rightarrow \frac{3 \times (-4)}{4 \times 5} = \frac{-12}{20} = \frac{-3}{5} \\[1em] \therefore p \times (q \times r) = \frac{-8}{27} \times \frac{-3}{5} \\[1em] \Rightarrow \frac{(-8) \times (-3)}{27 \times 5} \\[1em] \Rightarrow \frac{24}{135} \\[1em] \Rightarrow \frac{8}{45}\end{aligned}$$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow p \times q = \frac{-8}{27} \times \frac{3}{4} \\[1em] \Rightarrow \frac{(-8) \times 3}{27 \times 4} = \frac{-24}{108} = \frac{-2}{9} \\[1em] \therefore (p \times q) \times r = \frac{-2}{9} \times \frac{-4}{5} \\[1em] \Rightarrow \frac{(-2) \times (-4)}{9 \times 5} \\[1em] \Rightarrow \frac{8}{45}\end{aligned}$$

As L.H.S. = R.H.S.

Hence, proved that $\bm{p \times (q \times r) = (p \times q) \times r}$.

(ii) To verify the distributive property of multiplication over subtraction, we have to show that $p \times (q - r) = p \times q - p \times r$

Solving L.H.S.,

$$\begin{aligned}\Rightarrow q - r = \frac{3}{4} - \left(\frac{-4}{5}\right) \\[1em] \Rightarrow \frac{3}{4} + \frac{4}{5} \\[1em] \Rightarrow \frac{3 \times 5 + 4 \times 4}{20} \\[1em] \Rightarrow \frac{15 + 16}{20} \\[1em] \Rightarrow \frac{31}{20} \\[1em] \therefore p \times (q - r) = \frac{-8}{27} \times \frac{31}{20} \\[1em] \Rightarrow \frac{(-8) \times 31}{27 \times 20} \\[1em] \Rightarrow \frac{-248}{540} \\[1em] \Rightarrow \frac{-62}{135}\end{aligned}$$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow p \times q = \frac{-2}{9} \text{ (found in part (i))} \\[1em] \Rightarrow p \times r = \frac{-8}{27} \times \frac{-4}{5} = \frac{(-8) \times (-4)}{27 \times 5} = \frac{32}{135} \\[1em] \therefore p \times q - p \times r = \frac{-2}{9} - \frac{32}{135} \\[1em] [\text{ LCM of } 9 \text{ and } 135 = 135] \\[1em] \Rightarrow \frac{(-2) \times 15 - 32}{135} \\[1em] \Rightarrow \frac{-30 - 32}{135} \\[1em] \Rightarrow \frac{-62}{135}\end{aligned}$$

As L.H.S. = R.H.S.

Hence, proved that $\bm{p \times (q - r) = p \times q - p \times r}$.

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

Hence, the reciprocal of 0 is not defined.

(ii) As $1 \times 1 = 1 \text{ and } (-1) \times (-1) = 1$, each of these numbers is its own reciprocal.

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

(iii) Since $y$ is the reciprocal of x,

$$\begin{aligned}\Rightarrow y = \frac{1}{x} \\[1em] \Rightarrow y^2 = \frac{1}{x^2} \\[1em] \Rightarrow \text{Reciprocal of } y^2 = x^2\end{aligned}$$

Hence, the reciprocal of y² in terms of x will be x².

(iv) For every non-zero rational number $\frac{p}{q}$, we have $\frac{p}{q} \times \frac{q}{p} = 1$

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

(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.

As $\frac{7}{5}$ is common to both the products, take it out using distributivity.

$$\begin{aligned}\Rightarrow \bigg\lbrace\frac{7}{5} \times \left(\frac{-3}{12}\right)\bigg\rbrace + \bigg\lbrace\frac{7}{5} \times \frac{5}{12}\bigg\rbrace \\[1em] \Rightarrow \frac{7}{5} \times \left(\frac{-3}{12} + \frac{5}{12}\right) \\[1em] \Rightarrow \frac{7}{5} \times \frac{-3 + 5}{12} \\[1em] \Rightarrow \frac{7}{5} \times \frac{2}{12} \\[1em] \Rightarrow \frac{7}{5} \times \frac{1}{6} \\[1em] \Rightarrow \frac{7}{30}\end{aligned}$$

Hence, $\mathbf{\bigg\lbrace\frac{7}{5} \times \left(\frac{-3}{12}\right)\bigg\rbrace + \bigg\lbrace\frac{7}{5} \times \frac{5}{12}\bigg\rbrace = \frac{7}{30}}$

As $\frac{9}{16}$ is common to both the products, take it out using distributivity.

$$\begin{aligned}\Rightarrow \bigg\lbrace\frac{9}{16} \times \frac{4}{12}\bigg\rbrace + \bigg\lbrace\frac{9}{16} \times \left(\frac{-3}{9}\right)\bigg\rbrace \\[1em] \Rightarrow \frac{9}{16} \times \left(\frac{4}{12} + \frac{-3}{9}\right) \\[1em] \Rightarrow \frac{9}{16} \times \left(\frac{1}{3} - \frac{1}{3}\right) \\[1em] \Rightarrow \frac{9}{16} \times 0 \\[1em] \Rightarrow 0\end{aligned}$$

Hence, $\mathbf{\bigg\lbrace\frac{9}{16} \times \frac{4}{12}\bigg\rbrace + \bigg\lbrace\frac{9}{16} \times \left(\frac{-3}{9}\right)\bigg\rbrace = 0}$

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 9 = -9

Multiplicative inverse of 9 = $\frac{1}{9}$

Adding both the inverses,

$$\begin{aligned}\Rightarrow -9 + \frac{1}{9} \\[1em] \Rightarrow \frac{(-9) \times 9 + 1}{9} \\[1em] \Rightarrow \frac{-81 + 1}{9} \\[1em] \Rightarrow \frac{-80}{9} \\[1em] \Rightarrow -8\frac{8}{9}\end{aligned}$$

Hence, the required sum is $\mathbf{-8\frac{8}{9}}$

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 $-\frac{3}{7} = \frac{3}{7}$

Multiplicative inverse of $-\frac{3}{7} = \frac{-7}{3}$

Multiplying both the inverses,

$$\begin{aligned}\Rightarrow \frac{3}{7} \times \frac{-7}{3} \\[1em] \Rightarrow \frac{3 \times (-7)}{7 \times 3} \\[1em] \Rightarrow \frac{-21}{21} \\[1em] \Rightarrow -1\end{aligned}$$

Hence, the required product is -1.

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

$$\begin{aligned}\Rightarrow \frac{-3}{7} \div 4 \\[1em] \Rightarrow \frac{-3}{7} \div \frac{4}{1} \\[1em] \Rightarrow \frac{-3}{7} \times \frac{1}{4} \\[1em] \Rightarrow \frac{(-3) \times 1}{7 \times 4} \\[1em] \Rightarrow \frac{-3}{28}\end{aligned}$$

Hence, $\mathbf{\frac{-3}{7} \div 4 = \frac{-3}{28}}$

First convert the mixed fraction into an improper fraction, then multiply the dividend by the reciprocal of the divisor.

$$\begin{aligned}\Rightarrow 4\frac{5}{8} \div \left(\frac{-4}{9}\right) \\[1em] \Rightarrow \frac{37}{8} \div \left(\frac{-4}{9}\right) \\[1em] \Rightarrow \frac{37}{8} \times \frac{9}{-4} \\[1em] \Rightarrow \frac{37 \times 9}{8 \times (-4)} \\[1em] \Rightarrow \frac{333}{-32} = \frac{-333}{32} \\[1em] \Rightarrow -10\frac{13}{32}\end{aligned}$$

Hence, $\mathbf{4\frac{5}{8} \div \left(\frac{-4}{9}\right) = -10\frac{13}{32}}$

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

$$\begin{aligned}\Rightarrow \frac{-8}{9} \div \frac{-3}{5} \\[1em] \Rightarrow \frac{-8}{9} \times \frac{5}{-3} \\[1em] \Rightarrow \frac{(-8) \times 5}{9 \times (-3)} \\[1em] \Rightarrow \frac{-40}{-27} = \frac{40}{27} \\[1em] \Rightarrow 1\frac{13}{27}\end{aligned}$$

Hence, $\mathbf{\frac{-8}{9} \div \frac{-3}{5} = 1\frac{13}{27}}$

(i) If we exclude 0, the remaining rational numbers are closed under division. Here the divisor $\frac{2}{7}$ is a non-zero rational number.

$$\begin{aligned}\Rightarrow \frac{-9}{13} \div \frac{2}{7} \\[1em] \Rightarrow \frac{-9}{13} \times \frac{7}{2} \\[1em] \Rightarrow \frac{-63}{26}\end{aligned}$$

As $\frac{-63}{26}$ is a rational number, the given statement is true.

Hence, the given statement is true.

(ii) For any three non-zero rational numbers

$\frac{p}{q}, \frac{r}{s} \text{ and } \frac{t}{u}$, we have $\frac{p}{q} \div \left(\frac{r}{s} \div \frac{t}{u}\right) \neq \left(\frac{p}{q} \div \frac{r}{s}\right) \div \frac{t}{u}$

So, division of rational numbers is not associative.

Hence, the given statement is false.

(iii) For any two non-zero rational numbers

$\frac{p}{q} \text{ and } \frac{r}{s}$, we have $\frac{p}{q} \div \frac{r}{s} \neq \frac{r}{s} \div \frac{p}{q}$

So, division of rational numbers is not commutative.

Hence, the given statement is true.

(iv) For any three non-zero rational numbers

$\frac{p}{q}, \frac{r}{s} \text{ and } \frac{t}{u}$, we have $\left(\frac{p}{q} \div \frac{r}{s}\right) \div \frac{t}{u} \neq \frac{p}{q} \div \left(\frac{r}{s} \div \frac{t}{u}\right)$

So, division of rational numbers is not associative.

Hence, the given statement is true.

Let the other rational number be $x$.

As the product of the two rational numbers is $\frac{-11}{12}$,

$$\begin{aligned}\Rightarrow 2\frac{4}{9} \times x = \frac{-11}{12} \\[1em] \Rightarrow \frac{22}{9} \times x = \frac{-11}{12} \\[1em] \Rightarrow x = \frac{-11}{12} \div \frac{22}{9} \\[1em] \Rightarrow x = \frac{-11}{12} \times \frac{9}{22} \\[1em] \Rightarrow x = \frac{(-11) \times 9}{12 \times 22} \\[1em] \Rightarrow x = \frac{-99}{264} \\[1em] \Rightarrow x = \frac{-3}{8}\end{aligned}$$

Hence, the other rational number is $\mathbf{\frac{-3}{8}}$

Let the required rational number be $x$.

According to the given condition,

$$\begin{aligned}\Rightarrow \frac{-7}{12} \times x = \frac{5}{14} \\[1em] \Rightarrow x = \frac{5}{14} \div \frac{-7}{12} \\[1em] \Rightarrow x = \frac{5}{14} \times \frac{12}{-7} \\[1em] \Rightarrow x = \frac{5 \times 12}{14 \times (-7)} \\[1em] \Rightarrow x = \frac{60}{-98} \\[1em] \Rightarrow x = \frac{-30}{49}\end{aligned}$$

Hence, the required rational number is $\mathbf{\frac{-30}{49}}$.

Let the required rational number be $x$.

According to the given condition,

$$\begin{aligned}\Rightarrow -3 \div x = \frac{-9}{13} \\[1em] \Rightarrow \frac{-3}{1} \times \frac{1}{x} = \frac{-9}{13} \\[1em] \Rightarrow \frac{1}{x} = \frac{-9}{13} \div (-3) \\[1em] \Rightarrow \frac{1}{x} = \frac{-9}{13} \times \frac{1}{-3} \\[1em] \Rightarrow \frac{1}{x} = \frac{-9}{-39} = \frac{3}{13} \\[1em] \Rightarrow x = \frac{13}{3} = 4\frac{1}{3}\end{aligned}$$

Hence, the required rational number is $\mathbf{\frac{13}{3}}$.

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

LCM of 8 and 12 = 24

Finding the sum,

$$\begin{aligned}\Rightarrow \frac{-13}{8} + \frac{5}{12} \\[1em] \Rightarrow \frac{(-13) \times 3 + 5 \times 2}{24} \\[1em] \Rightarrow \frac{-39 + 10}{24} \\[1em] \Rightarrow \frac{-29}{24}\end{aligned}$$

Finding the difference,

$$\begin{aligned}\Rightarrow \frac{-13}{8} - \frac{5}{12} \\[1em] \Rightarrow \frac{(-13) \times 3 - 5 \times 2}{24} \\[1em] \Rightarrow \frac{-39 - 10}{24} \\[1em] \Rightarrow \frac{-49}{24}\end{aligned}$$

Dividing the sum by the difference,

$$\begin{aligned}\Rightarrow \frac{-29}{24} \div \frac{-49}{24} \\[1em] \Rightarrow \frac{-29}{24} \times \frac{24}{-49} \\[1em] \Rightarrow \frac{(-29) \times 24}{24 \times (-49)} \\[1em] \Rightarrow \frac{29}{49}\end{aligned}$$

Hence, the required quotient is $\mathbf{\frac{29}{49}}$

First find the sum of $\frac{8}{3} \text{ and } \frac{4}{7} \text{, and the product of } \frac{-3}{7} \text{ and } \frac{14}{9}$.

Finding the sum,

$$\begin{aligned}\Rightarrow \frac{8}{3} + \frac{4}{7} \\[1em] [\text{ LCM of } 3 \text{ and } 7 = 21] \\[1em] \Rightarrow \frac{8 \times 7 + 4 \times 3}{21} \\[1em] \Rightarrow \frac{56 + 12}{21} \\[1em] \Rightarrow \frac{68}{21}\end{aligned}$$

Finding the product,

$$\begin{aligned}\Rightarrow \frac{-3}{7} \times \frac{14}{9} \\[1em] \Rightarrow \frac{(-3) \times 14}{7 \times 9} \\[1em] \Rightarrow \frac{-42}{63} \\[1em] \Rightarrow \frac{-2}{3}\end{aligned}$$

Dividing the sum by the product,

$$\begin{aligned}\Rightarrow \frac{68}{21} \div \frac{-2}{3} \\[1em] \Rightarrow \frac{68}{21} \times \frac{3}{-2} \\[1em] \Rightarrow \frac{68 \times 3}{21 \times (-2)} \\[1em] \Rightarrow \frac{204}{-42} = \frac{-34}{7} \\[1em] \Rightarrow -4\frac{6}{7}\end{aligned}$$

Hence, the required quotient is $\mathbf{-4\frac{6}{7}}$

To verify that division of rational numbers is not associative, we have to show that $(p \div q) \div r \neq p \div (q \div r)$

Solving L.H.S.,

$$\begin{aligned}\Rightarrow p \div q = \frac{-3}{2} \div \frac{4}{5} \\[1em] \Rightarrow \frac{-3}{2} \times \frac{5}{4} = \frac{-15}{8} \\[1em] \therefore (p \div q) \div r = \frac{-15}{8} \div \frac{-7}{12} \\[1em] \Rightarrow \frac{-15}{8} \times \frac{12}{-7} \\[1em] \Rightarrow \frac{-180}{-56} \\[1em] \Rightarrow \frac{45}{14}\end{aligned}$$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow q \div r = \frac{4}{5} \div \frac{-7}{12} \\[1em] \Rightarrow \frac{4}{5} \times \frac{12}{-7} = \frac{48}{-35} = \frac{-48}{35} \\[1em] \therefore p \div (q \div r) = \frac{-3}{2} \div \frac{-48}{35} \\[1em] \Rightarrow \frac{-3}{2} \times \frac{35}{-48} \\[1em] \Rightarrow \frac{-105}{-96} \\[1em] \Rightarrow \frac{35}{32}\end{aligned}$$

As $\frac{45}{14} \neq \frac{35}{32}$

$\Rightarrow (p \div q) \div r \neq p \div (q \div r)$

Hence, proved that $\bm{(p \div q) \div r \neq p \div (q \div r)}$.

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

$$\begin{aligned}\Rightarrow \frac{11}{4} \\[1em] \Rightarrow 2\frac{3}{4} \\[1em] \Rightarrow 2 + \frac{3}{4}\end{aligned}$$

So, $\frac{11}{4}$ lies between 2 and 3.

Divide the line segment between 2 and 3 into four equal parts. The third point of division from 2 represents $\frac{11}{4}$.

Hence, the point marked on the number line represents $\mathbf{\frac{11}{4}}$

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

$$\begin{aligned}\Rightarrow 4\frac{3}{5} = \frac{23}{5} \\[1em] \Rightarrow 4 + \frac{3}{5}\end{aligned}$$

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

Divide the line segment between 4 and 5 into five equal parts. The third point of division from 4 represents $4\frac{3}{5}$.

Hence, the point marked on the number line represents $\mathbf{4\frac{3}{5}}$

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

$$\begin{aligned}\Rightarrow \frac{-9}{7} \\[1em] \Rightarrow -1\frac{2}{7} \\[1em] \Rightarrow -1 + \left(\frac{-2}{7}\right)\end{aligned}$$

So, $\frac{-9}{7}$ lies between -1 and -2.

Divide the line segment between -1 and -2 into seven equal parts. The second point of division to the left of -1 represents $\frac{-9}{7}$.

Hence, the point marked on the number line represents $\mathbf{\frac{-9}{7}}$

First express the rational number with a positive denominator.

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

As $\frac{2}{5}$ is a positive proper fraction, it lies between 0 and 1.

Divide the line segment between 0 and 1 into five equal parts. The second point of division from 0 represents $\frac{2}{5}$.

Hence, the point marked on the number line represents $\mathbf{\frac{2}{5}}$

The line segment between each pair of consecutive integers has been divided into seven equal parts, so each division represents $\frac{1}{7}$.

Counting the divisions from 0, the marked points are the 3rd, 7th, 8th, 12th and 13th points of division.

$$\begin{aligned}\Rightarrow A = \frac{3}{7} \\[1em] \Rightarrow B = \frac{7}{7} = 1 \\[1em] \Rightarrow C = \frac{8}{7} \\[1em] \Rightarrow D = \frac{12}{7} \\[1em] \Rightarrow E = \frac{13}{7}\end{aligned}$$

Hence, $\mathbf{A = \frac{3}{7}, B = 1, C = \frac{8}{7}, D = \frac{12}{7} \text{ and } E = \frac{13}{7}}$

The line segment between each pair of consecutive integers has been divided into eight equal parts, so each division represents $\frac{1}{8}$.

Writing the marked values with denominator 8, we have

$$\begin{aligned}\Rightarrow \frac{-5}{4} = \frac{-10}{8} \\[1em] \Rightarrow -1 = \frac{-8}{8} \\[1em] \Rightarrow \frac{-3}{4} = \frac{-6}{8}\end{aligned}$$

Counting the divisions to the left of 0, the marked points are the 12th, 11th, 7th, 4th and 3rd points of division.

$$\begin{aligned}\Rightarrow T = \frac{-12}{8} = \frac{-3}{2} \\[1em] \Rightarrow S = \frac{-11}{8} \\[1em] \Rightarrow R = \frac{-7}{8} \\[1em] \Rightarrow Q = \frac{-4}{8} = \frac{-1}{2} \\[1em] \Rightarrow P = \frac{-3}{8}\end{aligned}$$

Hence, $\mathbf{T = \frac{-3}{2}, S = \frac{-11}{8}, R = \frac{-7}{8}, Q = \frac{-1}{2} \text{ and } P = \frac{-3}{8}}$

(i) First write each rational number with a positive denominator, then write them with a common denominator and compare the numerators.

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

L.C.M. of 2 and 5 = 10

$$\begin{aligned}\Rightarrow \frac{-1}{2} = \frac{(-1) \times 5}{2 \times 5} = \frac{-5}{10} \\[1em] \Rightarrow \frac{-4}{5} = \frac{(-4) \times 2}{5 \times 2} = \frac{-8}{10}\end{aligned}$$

As -8 < -5, so $\frac{-8}{10} \lt \frac{-5}{10}$

Hence, $\mathbf{\frac{-1}{2} \text{ is greater.}}$

(ii) First write each rational number with a positive denominator.

$\frac{-11}{-17} = \frac{-11 \times (-1)}{-17 \times (-1)} = \frac{11}{17}$

Here $\frac{11}{17}$ is a positive rational number and $\frac{-9}{13}$ is a negative rational number.

Every positive rational number is greater than every negative rational number.

Hence, $\mathbf{\frac{-11}{-17}}$ is greater.

(iii) First write each rational number with a positive denominator, then write them with a common denominator and compare the numerators.

$\frac{13}{-17} = \frac{13 \times (-1)}{-17 \times (-1)} = \frac{-13}{17}$

L.C.M. of 11 and 17 = 187

$$\begin{aligned}\Rightarrow \frac{-5}{11} = \frac{(-5) \times 17}{11 \times 17} = \frac{-85}{187} \\[1em] \Rightarrow \frac{-13}{17} = \frac{(-13) \times 11}{17 \times 11} = \frac{-143}{187}\end{aligned}$$

As -143 < -85, so $\frac{-143}{187}$ < $\frac{-85}{187}$

Hence, $\mathbf{\frac{-5}{11}}$ is greater.

First write each rational number with a positive denominator.

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

So, the given rational numbers are $\frac{-2}{3}, \frac{-1}{4}, \frac{-5}{7}, \frac{2}{11}$

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

$$\begin{aligned}[\text{ LCM of } 3, 4, 7 \text{ and } 11 = 924] \\[1em] \Rightarrow \frac{-2}{3} = \frac{(-2) \times 308}{3 \times 308} = \frac{-616}{924} \\[1em] \Rightarrow \frac{-1}{4} = \frac{(-1) \times 231}{4 \times 231} = \frac{-231}{924} \\[1em] \Rightarrow \frac{-5}{7} = \frac{(-5) \times 132}{7 \times 132} = \frac{-660}{924} \\[1em] \Rightarrow \frac{2}{11} = \frac{2 \times 84}{11 \times 84} = \frac{168}{924}\end{aligned}$$

As -660 < -616 < -231 < 168

so $\frac{-660}{924}$ < $\frac{-616}{924}$ < $\frac{-231}{924}$ < $\frac{168}{924}$

$\Rightarrow \frac{-5}{7}$ < $\frac{-2}{3}$ < $\frac{-1}{4}$ < $\frac{2}{11}$

Hence, the given rational numbers in ascending order are $\mathbf{\frac{-5}{7}, \frac{-2}{3}, \frac{1}{-4}, \frac{2}{11}}$

First write each rational number with a positive denominator.

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

So, the given rational numbers are $\frac{-3}{10}, \frac{-2}{3}, \frac{-4}{5}, 0, \frac{7}{8}$

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

$$\begin{aligned}[\text{ LCM of } 10, 3, 5 \text{ and } 8 = 120] \\[1em] \Rightarrow \frac{-3}{10} = \frac{(-3) \times 12}{10 \times 12} = \frac{-36}{120} \\[1em] \Rightarrow \frac{-2}{3} = \frac{(-2) \times 40}{3 \times 40} = \frac{-80}{120} \\[1em] \Rightarrow \frac{-4}{5} = \frac{(-4) \times 24}{5 \times 24} = \frac{-96}{120} \\[1em] \Rightarrow 0 = \frac{0}{120} \\[1em] \Rightarrow \frac{7}{8} = \frac{7 \times 15}{8 \times 15} = \frac{105}{120}\end{aligned}$$

As 105 > 0 > -36 > -80 > -96

so $\frac{105}{120} \gt \frac{0}{120} \gt \frac{-36}{120} \gt \frac{-80}{120} \gt \frac{-96}{120}$

$\Rightarrow \frac{7}{8} \gt 0 \gt \frac{-3}{10} \gt \frac{-2}{3} \gt \frac{-4}{5}$

Hence, the given rational numbers in descending order are $\mathbf{\frac{7}{8}, 0, \frac{-3}{10}, \frac{-2}{3}, \frac{4}{-5}}$

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

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

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

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

Hence, three rational numbers between -2 and -1 are $\mathbf{\frac{-7}{4}, \frac{-3}{2} \text{ and } \frac{-5}{4}}$

Every rational number lying to the left of -4 on the number line is smaller than -4.

As -9 < -8 < -7 < -6 < -5 < -4, each of these numbers is smaller than -4.

There are infinitely many such rational numbers; for example $\frac{-9}{2}, \frac{-13}{3}, \frac{-17}{4}$ are also smaller than -4.

Hence, five rational numbers smaller than -4 are -5, -6, -7, -8 and -9.

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

$$\begin{aligned}[\text{ LCM of } 7 \text{ and } 3 = 21] \\[1em] \Rightarrow \frac{-3}{7} = \frac{-3 \times 3}{7 \times 3} = \frac{-9}{21} \\[1em] \Rightarrow \frac{2}{3} = \frac{2 \times 7}{3 \times 7} = \frac{14}{21}\end{aligned}$$

We note that -8, -7, -6, …, 12, 13 are the integers between -9 and 14

i.e. -9 < -8 < -7 < -6 < … < 12 < 13 < 14

Hence, twenty rational numbers between $\mathbf{\frac{-3}{7} \text{ and } \frac{2}{3}}$ are

$\mathbf{\frac{-8}{21}, \frac{-7}{21}, \frac{-6}{21}, \frac{-5}{21}, \frac{-4}{21}, \frac{-3}{21}, \frac{-2}{21}, \frac{-1}{21}, \frac{0}{21}, \frac{1}{21}, \frac{2}{21}, \frac{3}{21}, \frac{4}{21}, \frac{5}{21}, \frac{6}{21}, \frac{7}{21}, \frac{8}{21}, \frac{9}{21}, \frac{10}{21}, \frac{11}{21}}$.

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

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

We note that -1, 0, 1, 2, 3, 4 are the integers between -2 and 5

i.e. -2 < -1 < 0 < 1 < 2 < 3 < 4 < 5

Hence, six rational numbers between $\mathbf{\frac{-1}{2} \text{ and } \frac{5}{4}}$ are $\mathbf{\frac{-1}{4}, 0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4} \text{ and } 1}$

Compare each rational number with -1 to see where it lies on the number line.

$\frac{-5}{11}, \frac{-1}{2} \text{ and } \frac{-4}{9}$ are all proper fractions with a negative sign, so each of them lies between -1 and 0 on the number line.

But $\frac{-7}{3} = -2\frac{1}{3}$ is an improper fraction with a negative sign, and it lies between -3 and -2 on the number line.

Hence, $\mathbf{\frac{-7}{3}}$ is different from the other three.

Subtract the total weight of the oranges and the apples from the total weight of the fruits.

Total weight of fruits in the bag = 20 kg

Weight of oranges $= 7\frac{1}{6} \text{ kg} = \frac{43}{6} \text{ kg}$

Weight of apples $= 8\frac{2}{3} \text{ kg} = \frac{26}{3} \text{ kg}$

Finding the weight of oranges and apples together,

$$\begin{aligned}\Rightarrow \frac{43}{6} + \frac{26}{3} \\[1em] [\text{ LCM of } 6 \text{ and } 3 = 6] \\[1em] \Rightarrow \frac{43 + 26 \times 2}{6} \\[1em] \Rightarrow \frac{43 + 52}{6} \\[1em] \Rightarrow \frac{95}{6} \text{ kg}\end{aligned}$$

Finding the weight of grapes,

$$\begin{aligned}\Rightarrow 20 - \frac{95}{6} \\[1em] \Rightarrow \frac{20 \times 6 - 95}{6} \\[1em] \Rightarrow \frac{120 - 95}{6} \\[1em] \Rightarrow \frac{25}{6} = 4\frac{1}{6} \text{ kg}\end{aligned}$$

Hence, the weight of the grapes in the bag is $\mathbf{4\frac{1}{6} \text{ kg}}$

First find the fraction of the population which are children, then find that fraction of the total population.

Total population of the city = 6,63,432

Finding the fraction of the population which are adults,

$$\begin{aligned}\Rightarrow \frac{1}{2} + \frac{1}{3} \\[1em] [\text{ LCM of } 2 \text{ and } 3 = 6] \\[1em] \Rightarrow \frac{3 + 2}{6} \\[1em] \Rightarrow \frac{5}{6}\end{aligned}$$

Finding the fraction of the population which are children,

$$\begin{aligned}\Rightarrow 1 - \frac{5}{6} \\[1em] \Rightarrow \frac{6 - 5}{6} \\[1em] \Rightarrow \frac{1}{6}\end{aligned}$$

Finding the number of children,

$$\begin{aligned}\Rightarrow \frac{1}{6} \times 663432 \\[1em] \Rightarrow 110572\end{aligned}$$

Hence, there are 1,10,572 children in the city.

First find the number of boys, then find how many of them passed and subtract from the total number of boys.

Total number of students who appeared = 400

Number of girls = 130

Number of boys = 400 - 130 = 270

Finding the number of boys who passed,

$\frac{2}{3} \times 270 = 180$

Finding the number of boys who failed,

Number of boys who failed = 270 - 180 = 90

Hence, 90 boys failed in the examination.

We know that distance = speed × time.

Speed of the car $= 40\frac{2}{3} \text{ km/h} = \frac{122}{3} \text{ km/h}$

Time taken $= \frac{9}{10}$ hours

Finding the distance covered,

$$\begin{aligned}\Rightarrow \frac{122}{3} \times \frac{9}{10} \\[1em] \Rightarrow \frac{122 \times 9}{3 \times 10} \\[1em] \Rightarrow \frac{1098}{30} \\[1em] \Rightarrow \frac{183}{5} = 36\frac{3}{5} \text{ km}\end{aligned}$$

Hence, the car will cover a distance of $\mathbf{36\frac{3}{5} \text{ km}}$

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

Side of the square lawn $= 5\frac{7}{9} \text{ m} = \frac{52}{9} \text{ m}$

Finding the area of the square lawn,

$$\begin{aligned}\Rightarrow \frac{52}{9} \times \frac{52}{9} \\[1em] \Rightarrow \frac{52 \times 52}{9 \times 9} \\[1em] \Rightarrow \frac{2704}{81} \\[1em] \Rightarrow 33\frac{31}{81} \text{ sq. m}\end{aligned}$$

Hence, the area of the square lawn is $\mathbf{33\frac{31}{81} \text{ sq. m}}$

We know that perimeter of a rectangle = 2 (length + breadth).

Perimeter of the rectangle $= 15\frac{3}{7} \text{ m} = \frac{108}{7} \text{ m}$

Length of the rectangle $= 4\frac{2}{7} \text{ m} = \frac{30}{7} \text{ m}$

Let the breadth of the rectangle be $b$ metres.

$$\begin{aligned}\Rightarrow \frac{108}{7} = 2\left(\frac{30}{7} + b\right) \\[1em] \Rightarrow \frac{30}{7} + b = \frac{108}{7} \div 2 \\[1em] \Rightarrow \frac{30}{7} + b = \frac{108}{7} \times \frac{1}{2} \\[1em] \Rightarrow \frac{30}{7} + b = \frac{54}{7} \\[1em] \Rightarrow b = \frac{54}{7} - \frac{30}{7} \\[1em] \Rightarrow b = \frac{54 - 30}{7} \\[1em] \Rightarrow b = \frac{24}{7} = 3\frac{3}{7} \text{ m}\end{aligned}$$

Hence, the breadth of the rectangle is $\mathbf{3\frac{3}{7} \text{ m}}$

First find the length of the remaining rope, then divide it by 3.

Total length of the rope $= 325\frac{4}{5} \text{ m} = \frac{1629}{5} \text{ m}$

Length of the piece cut off $= 150\frac{3}{5} \text{ m} = \frac{753}{5} \text{ m}$

Finding the length of the remaining rope,

$$\begin{aligned}\Rightarrow \frac{1629}{5} - \frac{753}{5} \\[1em] \Rightarrow \frac{1629 - 753}{5} \\[1em] \Rightarrow \frac{876}{5} \text{ m}\end{aligned}$$

Finding the length of each part,

$$\begin{aligned}\Rightarrow \frac{876}{5} \div 3 \\[1em] \Rightarrow \frac{876}{5} \times \frac{1}{3} \\[1em] \Rightarrow \frac{876}{15} \\[1em] \Rightarrow \frac{292}{5} = 58\frac{2}{5} \text{ m}\end{aligned}$$

Hence, the length of each part is $\mathbf{58\frac{2}{5} \text{ m}}$

First find the cost of 1 litre of liquid soap by unitary method, then find the cost of 4 litre.

Quantity of liquid soap $= 3\frac{1}{2} \text{ litre} = \frac{7}{2} \text{ litre}$

Cost of $\frac{7}{2} \text{ litre of liquid soap} = ₹270\frac{3}{8} = ₹\frac{2163}{8}$

Finding the cost of 1 litre of liquid soap,

$$\begin{aligned}\Rightarrow \frac{2163}{8} \div \frac{7}{2} \\[1em] \Rightarrow \frac{2163}{8} \times \frac{2}{7} \\[1em] \Rightarrow \frac{4326}{56} \\[1em] \Rightarrow ₹\frac{309}{4}\end{aligned}$$

Finding the cost of 4 litre of liquid soap,

$$\begin{aligned}\Rightarrow \frac{309}{4} \times 4 \\[1em] \Rightarrow ₹309\end{aligned}$$

Hence, the cost of 4 litre of liquid soap is ₹309.

Find the money spent at each stage step by step and subtract it from the amount remaining at that stage.

Monthly income of Ramesh = ₹40,000

Finding the money spent on food,

$$\begin{aligned}\Rightarrow \frac{3}{8} \times 40000 \\[1em] \Rightarrow ₹15000\end{aligned}$$

Remaining money $= ₹40000 - ₹15000 = ₹25000$

Finding the money spent on LIC premium,

$$\begin{aligned}\Rightarrow \frac{1}{5} \times 25000 \\[1em] \Rightarrow ₹5000\end{aligned}$$

Remaining money = $₹25000 - ₹5000 = ₹20000$

Finding the money spent on other expenses,

$$\begin{aligned}\Rightarrow \frac{1}{2} \times 20000 \\[1em] \Rightarrow ₹10000\end{aligned}$$

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

Hence, ₹10,000 is left with Ramesh.

First find the fraction of the bill paid by A and B together, then divide the remaining fraction equally among C, D and E.

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

$$\begin{aligned}\Rightarrow \frac{1}{2} + \frac{1}{5} \\[1em] [\text{ LCM of } 2 \text{ and } 5 = 10] \\[1em] \Rightarrow \frac{5 + 2}{10} \\[1em] \Rightarrow \frac{7}{10}\end{aligned}$$

Finding the remaining fraction of the bill,

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

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

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

Hence, A paid $\mathbf{\frac{1}{2}}$, B paid $\mathbf{\frac{1}{5}}$ and each of C, D and E paid $\mathbf{\frac{1}{10}}$ of the bill.

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

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

$$\begin{aligned}\Rightarrow \frac{2}{5} + \frac{1}{4} \\[1em] [\text{ LCM of } 5 \text{ and } 4 = 20] \\[1em] \Rightarrow \frac{8 + 5}{20} \\[1em] \Rightarrow \frac{13}{20}\end{aligned}$$

Finding the fraction of students who walk to school,

$$\begin{aligned}\Rightarrow 1 - \frac{13}{20} \\[1em] \Rightarrow \frac{20 - 13}{20} \\[1em] \Rightarrow \frac{7}{20}\end{aligned}$$

So, the number of students who walk to school $= \frac{7x}{20}$

According to the given condition,

$$\begin{aligned}\Rightarrow \frac{1}{3} \times \frac{7x}{20} = 224 \\[1em] \Rightarrow \frac{7x}{60} = 224 \\[1em] \Rightarrow x = \frac{224 \times 60}{7} \\[1em] \Rightarrow x = 32 \times 60 \\[1em] \Rightarrow x = 1920\end{aligned}$$

Hence, 1920 students study in that school.

Let the mother's contribution be ₹$x$.

Elder son's contribution $= ₹\frac{3x}{8}$ and younger son's contribution $= ₹\frac{x}{2}$

According to the given condition,

$$\begin{aligned}\Rightarrow x + \frac{3x}{8} + \frac{x}{2} = 60000 \\[1em] \Rightarrow \frac{8x + 3x + 4x}{8} = 60000 \\[1em] \Rightarrow \frac{15x}{8} = 60000 \\[1em] \Rightarrow x = \frac{60000 \times 8}{15} \\[1em] \Rightarrow x = 32000\end{aligned}$$

Finding the elder son's contribution,

$$\begin{aligned}\Rightarrow \frac{3}{8} \times 32000 \\[1em] \Rightarrow ₹12000\end{aligned}$$

Finding the younger son's contribution,

$$\begin{aligned}\Rightarrow \frac{1}{2} \times 32000 \\[1em] \Rightarrow ₹16000\end{aligned}$$

Hence, the mother, the elder son and the younger son contribute ₹32,000, ₹12,000 and ₹16,000 respectively.

Let the number of girls be $x$, so the number of boys $= \frac{2x}{5}$

According to the given condition,

$$\begin{aligned}\Rightarrow x + \frac{2x}{5} = 56 \\[1em] \Rightarrow \frac{5x + 2x}{5} = 56 \\[1em] \Rightarrow \frac{7x}{5} = 56 \\[1em] \Rightarrow x = \frac{56 \times 5}{7} \\[1em] \Rightarrow x = 40\end{aligned}$$

Finding the number of boys,

$$\begin{aligned}\Rightarrow \frac{2}{5} \times 40 \\[1em] \Rightarrow 16\end{aligned}$$

Hence, there are 16 boys and 40 girls in the class.

Let the man originally have ₹$x$.

Money donated to the school $= ₹\frac{x}{10}$

Finding the remaining money,

$$\begin{aligned}\Rightarrow x - \frac{x}{10} \\[1em] \Rightarrow \frac{10x - x}{10} \\[1em] \Rightarrow ₹\frac{9x}{10}\end{aligned}$$

Finding the money donated to the orphanage,

$$\begin{aligned}\Rightarrow \frac{1}{6} \times \frac{9x}{10} \\[1em] \Rightarrow \frac{9x}{60} \\[1em] \Rightarrow ₹\frac{3x}{20}\end{aligned}$$

Finding the money left,

$$\begin{aligned}\Rightarrow \frac{9x}{10} - \frac{3x}{20} \\[1em] \Rightarrow \frac{18x - 3x}{20} \\[1em] \Rightarrow \frac{15x}{20} = ₹\frac{3x}{4}\end{aligned}$$

This money was distributed equally among his three children.

According to the given condition,

$$\begin{aligned}\Rightarrow \frac{3x}{4} \div 3 = 50000 \\[1em] \Rightarrow \frac{3x}{4} \times \frac{1}{3} = 50000 \\[1em] \Rightarrow \frac{x}{4} = 50000 \\[1em] \Rightarrow x = 50000 \times 4 \\[1em] \Rightarrow x = 200000\end{aligned}$$

Hence, the man originally had ₹2,00,000.

Let the required number be $x$.

According to the given condition,

$$\begin{aligned}\Rightarrow \frac{1}{4}x + \frac{1}{3}x = \frac{1}{2}x + 15 \\[1em] [\text{Multiplying both sides by } 12] \\[1em] \Rightarrow 3x + 4x = 6x + 180 \\[1em] \Rightarrow 7x = 6x + 180 \\[1em] \Rightarrow 7x - 6x = 180 \\[1em] \Rightarrow x = 180\end{aligned}$$

Hence, the required number is 180.

Let the given number be $x$.

$$\begin{aligned}\Rightarrow \text{Correct answer} = x \times \frac{4}{5} = \frac{4x}{5} \\[1em] \Rightarrow \text{Answer obtained by the student} = x \div \frac{4}{5} = x \times \frac{5}{4} = \frac{5x}{4}\end{aligned}$$

According to the given condition,

$$\begin{aligned}\Rightarrow \frac{5x}{4} - \frac{4x}{5} = 36 \\[1em] \Rightarrow \frac{25x - 16x}{20} = 36 \\[1em] \Rightarrow \frac{9x}{20} = 36 \\[1em] \Rightarrow x = \frac{36 \times 20}{9} \\[1em] \Rightarrow x = 80\end{aligned}$$

Hence, the given number is 80.

(i) Rational numbers are closed under multiplication, i.e. if $\frac{p}{q} \text{ and } \frac{r}{s}$ are two rational numbers, then $\frac{p}{q} \times \frac{r}{s}$ is also a rational number.

Hence, the product of two rational numbers is a rational number.

(ii) For any two rational numbers $\frac{p}{q} \text{ and } \frac{r}{s}$, we have $\frac{p}{q} - \frac{r}{s} \neq \frac{r}{s} - \frac{p}{q}$

Hence, subtraction of rational numbers is not commutative.

(iii) As $\frac{-7}{4}$ is a negative rational number, it is less than zero.

Every negative rational number lies to the left of zero on the number line.

Hence, the rational number $\mathbf{\frac{-7}{4}}$ lies to the left of zero on the number line.

(iv) For any three non-zero rational numbers, we have $\frac{p}{q} \div \left(\frac{r}{s} \div \frac{t}{u}\right) \neq \left(\frac{p}{q} \div \frac{r}{s}\right) \div \frac{t}{u}$

Hence, division of rational numbers is not associative.

(v) The negative of a rational number is obtained by changing its sign, and the sum of a rational number and its negative is 0.

Hence, negative of a rational number is called its additive inverse.

(vi) For every rational number $\frac{p}{q}$, we have $\frac{p}{q} \times 1 = \frac{p}{q} = 1 \times \frac{p}{q}$

Hence, multiplicative identity of rational numbers is 1.

(vii) For any three rational numbers, we have $\frac{p}{q} \times \left(\frac{r}{s} + \frac{t}{u}\right) = \frac{p}{q} \times \frac{r}{s} + \frac{p}{q} \times \frac{t}{u}$

Hence, multiplication of rational numbers is distributive over addition.

(viii) If we exclude 0, the remaining rational numbers are closed under division, i.e. if $\frac{p}{q} \text{ and } \frac{r}{s}$ are two non-zero rational numbers, then $\frac{p}{q} \div \frac{r}{s}$ is also a rational number.

Hence, division of a rational number by a non-zero rational number is a rational number.

(i) The sum of a rational number and its additive inverse is 0.

$$\begin{aligned}\Rightarrow \frac{5}{9} + \left(\frac{-5}{9}\right) \\[1em] \Rightarrow \frac{5 + (-5)}{9} = \frac{0}{9} = 0\end{aligned}$$

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 $q \neq 0$. For example, $3 = \frac{3}{1} \text{ and } -3 = \frac{-3}{1}$

Hence, the given statement is true.

(iii) There is no rational number which when multiplied by 0 gives 1, so 0 has no multiplicative inverse.

Hence, the given statement is false.

(iv) Rational numbers of the form $\frac{p}{q}$ where $q \neq 1$ are generally not integers. For example, $\frac{-3}{5}$ is a rational number but it is not an integer.

Hence, the given statement is false.

(v) Division of a rational number by zero is not defined, so rational numbers are not closed under division. For example, $\frac{4}{5} \div 0$ is not defined.

Hence, the given statement is false.

(vi) Every terminating decimal number can be written in the form $\frac{p}{q}$ where $q \neq 0$. For example, $0.8 = \frac{4}{5} \text{ and } 0.28 = \frac{7}{25}$

Hence, the given statement is true.

(vii) For every rational number $\frac{p}{q}$, we have $\frac{p}{q} \times 1 = \frac{p}{q}$, so 1 is the multiplicative identity and 0 is the additive identity.

Hence, the given statement is false.

(viii) If $x$ < $y$, then $x$ lies to the left of $y$ on the number line, so $x - y$ is always negative. For example, taking $x = \frac{1}{2} \text{ and } y = \frac{3}{2}$, we get $x - y = -1$

Hence, the given statement is false.

(ix) For any three rational numbers, we have $\left(\frac{p}{q} - \frac{r}{s}\right) - \frac{t}{u} \neq \frac{p}{q} - \left(\frac{r}{s} - \frac{t}{u}\right)$

Hence, the given statement is true.

(x) As $1 \times 1 = 1$, the number 1 is its own reciprocal.

Hence, the given statement is true.

(xi) For every non-zero rational number $\frac{p}{q}$, the rational number $\frac{q}{p}$ is called its multiplicative inverse or reciprocal.

Hence, the given statement is true.

(xii) 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.

First express the rational number with a positive denominator, then change its sign to get the additive inverse.

$$\begin{aligned}\Rightarrow \frac{-2}{-5} = \frac{-2 \times (-1)}{-5 \times (-1)} = \frac{2}{5} \\[1em] \Rightarrow \text{Additive inverse} = -\left(\frac{2}{5}\right) \\[1em] \Rightarrow \frac{-2}{5} = \frac{2}{-5}\end{aligned}$$

Hence, Option 3 is the correct option.

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

$$\begin{aligned}\Rightarrow \text{Multiplicative inverse of } \frac{-3}{7} = \frac{7}{-3} \\[1em] \Rightarrow \frac{-7}{3} \\[1em] [\text{Check : } \frac{-3}{7} \times \frac{-7}{3} = 1]\end{aligned}$$

Hence, Option 2 is the correct option.

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

For every rational number $\frac{p}{q}$, we have $\frac{p}{q} + \left(-\frac{p}{q}\right) = 0$

Hence, Option 2 is the correct option.

Rational numbers are closed under addition, subtraction and multiplication.

However, division of a rational number by zero is not defined. For example, $\frac{4}{5} \div 0$ is not defined.

Hence, Option 4 is the correct option.

Zero divided by any non-zero rational number is always zero.

$$\begin{aligned}\Rightarrow 0 \div \frac{2}{3} \\[1em] \Rightarrow 0 \times \frac{3}{2} \\[1em] \Rightarrow 0\end{aligned}$$

Hence, Option 3 is the correct option.

Divisibility of a rational number by zero is not defined, i.e. $\frac{p}{q} \div 0$ does not exist as 0 has no reciprocal.

Hence, Option 4 is the correct option.

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

If $\frac{p}{q}, \frac{r}{s} \text{ and } \frac{t}{u}$ are any three rational numbers,

Then $\frac{p}{q} + \left(\frac{r}{s} + \frac{t}{u}\right) = \left(\frac{p}{q} + \frac{r}{s}\right) + \frac{t}{u}$

Hence, Option 2 is the correct option.

The reciprocal of a non-zero rational number is obtained by interchanging its numerator and denominator.

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

Hence, Option 2 is the correct option.

First find the additive inverse by changing the sign, then multiply.

$$\begin{aligned}\Rightarrow \text{Additive inverse of } \frac{-2}{5} = \frac{2}{5} \\[1em] \Rightarrow \frac{-2}{5} \times \frac{2}{5} \\[1em] \Rightarrow \frac{(-2) \times 2}{5 \times 5} \\[1em] \Rightarrow \frac{-4}{25}\end{aligned}$$

Hence, Option 3 is the correct option.

First find the reciprocal by interchanging the numerator and denominator, then add.

$$\begin{aligned}\Rightarrow \text{Reciprocal of } \frac{4}{7} = \frac{7}{4} \\[1em] \Rightarrow \frac{4}{7} + \frac{7}{4} \\[1em] [\text{ LCM of } 7 \text{ and } 4 = 28] \\[1em] \Rightarrow \frac{4 \times 4 + 7 \times 7}{28} \\[1em] \Rightarrow \frac{16 + 49}{28} \\[1em] \Rightarrow \frac{65}{28}\end{aligned}$$

Hence, Option 2 is the correct option.

As the sum of the two rational numbers is 0, the other rational number is the additive inverse of $\frac{-4}{5}$.

$$\begin{aligned}\Rightarrow \text{Additive inverse of } \frac{-4}{5} = \frac{4}{5} \\[1em] [\text{Check : } \frac{-4}{5} + \frac{4}{5} = 0]\end{aligned}$$

Hence, Option 2 is the correct option.

As the product of the two rational numbers is 1, the other rational number is the reciprocal of $\frac{10}{3}$.

$$\begin{aligned}\Rightarrow \text{Reciprocal of } \frac{10}{3} = \frac{3}{10} \\[1em] [\text{Check : } \frac{10}{3} \times \frac{3}{10} = 1]\end{aligned}$$

Hence, Option 1 is the correct option.

The line segment between -1 and 0 has been divided into seven equal parts, so each division represents $\frac{1}{7}$.

The point P is the fourth point of division to the right of -1.

$$\begin{aligned}\Rightarrow P = -1 + \frac{4}{7} \\[1em] \Rightarrow \frac{-7 + 4}{7} \\[1em] \Rightarrow \frac{-3}{7}\end{aligned}$$

Hence, Option 2 is the correct option.

Let the required number be $x$.

According to the given condition,

$$\begin{aligned}\Rightarrow \frac{-5}{3} - x = \frac{-2}{7} \\[1em] \Rightarrow x = \frac{-5}{3} - \left(\frac{-2}{7}\right) \\[1em] \Rightarrow x = \frac{-5}{3} + \frac{2}{7} \\[1em] [\text{ LCM of } 3 \text{ and } 7 = 21] \\[1em] \Rightarrow x = \frac{(-5) \times 7 + 2 \times 3}{21} \\[1em] \Rightarrow x = \frac{-35 + 6}{21} \\[1em] \Rightarrow x = \frac{-29}{21}\end{aligned}$$

Hence, Option 3 is the correct option.

The reciprocal of a non-zero rational number is obtained by interchanging its numerator and denominator, so the sign of the rational number remains unchanged.

For example, the reciprocal of $\frac{-3}{5} \text{ is } \frac{5}{-3} = \frac{-5}{3}$, which is negative.

Hence, Option 2 is the correct option.

Between any two different rational numbers, we can always insert their mean, and repeating this process we can insert as many rational numbers as we want.

Hence, Option 4 is the correct option.

If $x \text{ and } y$ are rational numbers such that $x$ < $y$, then the mean of $x \text{ and } y$ always lies between them, i.e. $x$ < $\frac{x + y}{2}$ < $y$

Hence, Option 1 is the correct option.

First express the rational number with a positive denominator, then change its sign.

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

As the additive inverse is $\frac{-5}{3}$ and not $\frac{3}{5}$, Statement I is false.

For every rational number $\frac{p}{q}$, we have $\frac{p}{q} + \left(-\frac{p}{q}\right) = 0$

So, Statement II is true.

Hence, Option 2 is the correct option.

The rational number 0 has no multiplicative inverse, since there is no rational number which when multiplied by 0 gives 1.

So, Statement I is false.

For a non-zero rational number $\frac{p}{q}$, the number $\frac{q}{p}$ is called its multiplicative inverse or reciprocal.

So, Statement II is true.

Hence, Option 2 is the correct option.

For any three rational numbers, we have $\frac{p}{q} + \left(\frac{r}{s} + \frac{t}{u}\right) = \left(\frac{p}{q} + \frac{r}{s}\right) + \frac{t}{u}$

So, Statement I is true.

For any three rational numbers, we have $\frac{p}{q} - \left(\frac{r}{s} - \frac{t}{u}\right) \neq \left(\frac{p}{q} - \frac{r}{s}\right) - \frac{t}{u}$

So, Statement II is false.

Hence, Option 1 is the correct option.

When a rational number is multiplied by zero, the product is always zero, i.e. $\frac{p}{q} \times 0 = 0 = 0 \times \frac{p}{q}$

So, Statement I is true.

Divisibility of a rational number by zero is not defined, i.e. $\frac{p}{q} \div 0$ does not exist.

So, Statement II is true.

Hence, Option 3 is the correct option.

Addition and multiplication are commutative for all these systems of numbers, while subtraction and division are not commutative for any of them.

For example, $5 - 3 \neq 3 - 5 \text{ and } 6 \div 2 \neq 2 \div 6$

Hence, the completed table is

Numbers	Commutative for addition	Commutative for subtraction	Commutative for multiplication	Commutative for division
Rational numbers	Yes	No	Yes	No
Integers	Yes	No	Yes	No
Whole numbers	Yes	No	Yes	No
Natural numbers	Yes	No	Yes	No

Addition and multiplication are associative for all these systems of numbers, while subtraction and division are not associative for any of them.

For example, $(9 - 5) - 2 \neq 9 - (5 - 2) \text{ and } (12 \div 6) \div 2 \neq 12 \div (6 \div 2)$

Hence, the completed table is

Numbers	Associative for addition	Associative for subtraction	Associative for multiplication	Associative for division
Rational numbers	Yes	No	Yes	No
Integers	Yes	No	Yes	No
Whole numbers	Yes	No	Yes	No
Natural numbers	Yes	No	Yes	No

First convert the mixed fractions into improper fractions and then add.

$$\begin{aligned}\Rightarrow 4\frac{2}{5} + 3\frac{7}{8} \\[1em] \Rightarrow \frac{22}{5} + \frac{31}{8} \\[1em] [\text{ LCM of } 5 \text{ and } 8 = 40] \\[1em] \Rightarrow \frac{22 \times 8 + 31 \times 5}{40} \\[1em] \Rightarrow \frac{176 + 155}{40} \\[1em] \Rightarrow \frac{331}{40} = 8\frac{11}{40}\end{aligned}$$

Hence, $\mathbf{4\frac{2}{5} + 3\frac{7}{8} = 8\frac{11}{40}}$

First convert the mixed fractions into improper fractions and then subtract.

$$\begin{aligned}\Rightarrow 7\frac{3}{4} - 4\frac{3}{5} \\[1em] \Rightarrow \frac{31}{4} - \frac{23}{5} \\[1em] [\text{ LCM of } 4 \text{ and } 5 = 20] \\[1em] \Rightarrow \frac{31 \times 5 - 23 \times 4}{20} \\[1em] \Rightarrow \frac{155 - 92}{20} \\[1em] \Rightarrow \frac{63}{20} = 3\frac{3}{20}\end{aligned}$$

Hence, $\mathbf{7\frac{3}{4} - 4\frac{3}{5} = 3\frac{3}{20}}$

Subtracting a negative rational number is the same as adding its additive inverse.

$$\begin{aligned}\Rightarrow 6\frac{8}{9} - \left(\frac{-7}{3}\right) \\[1em] \Rightarrow \frac{62}{9} + \frac{7}{3} \\[1em] [\text{ LCM of } 9 \text{ and } 3 = 9] \\[1em] \Rightarrow \frac{62 + 7 \times 3}{9} \\[1em] \Rightarrow \frac{62 + 21}{9} \\[1em] \Rightarrow \frac{83}{9} = 9\frac{2}{9}\end{aligned}$$

Hence, $\mathbf{6\frac{8}{9} - \left(\frac{-7}{3}\right) = 9\frac{2}{9}}$

First convert the mixed fractions into improper fractions and then multiply.

$$\begin{aligned}\Rightarrow \left(-16\frac{2}{3}\right) \times 14\frac{1}{2} \\[1em] \Rightarrow \frac{-50}{3} \times \frac{29}{2} \\[1em] \Rightarrow \frac{(-50) \times 29}{3 \times 2} \\[1em] \Rightarrow \frac{-1450}{6} \\[1em] \Rightarrow \frac{-725}{3} = -241\frac{2}{3}\end{aligned}$$

Hence, $\mathbf{\left(-16\frac{2}{3}\right) \times 14\frac{1}{2} = -241\frac{2}{3}}$

First convert the mixed fraction into an improper fraction, then multiply by the reciprocal of the divisor.

$$\begin{aligned}\Rightarrow \frac{-7}{8} \div 15\frac{3}{4} \\[1em] \Rightarrow \frac{-7}{8} \div \frac{63}{4} \\[1em] \Rightarrow \frac{-7}{8} \times \frac{4}{63} \\[1em] \Rightarrow \frac{(-7) \times 4}{8 \times 63} \\[1em] \Rightarrow \frac{-28}{504} = \frac{-1}{18}\end{aligned}$$

Hence, $\mathbf{\frac{-7}{8} \div 15\frac{3}{4} = \frac{-1}{18}}$

As 1 is the multiplicative identity of rational numbers, the product remains unchanged.

$$\begin{aligned}\Rightarrow \frac{-12}{13} \times 1 \\[1em] \Rightarrow \frac{-12}{13}\end{aligned}$$

Hence, $\mathbf{\frac{-12}{13} \times 1 = \frac{-12}{13}}$

When a rational number is multiplied by zero, the product is always zero.

$$\begin{aligned}\Rightarrow \frac{4}{9} \times 0 \\[1em] \Rightarrow 0\end{aligned}$$

Hence, $\mathbf{\frac{4}{9} \times 0 = 0}$

Zero divided by any non-zero rational number is always zero.

$$\begin{aligned}\Rightarrow 0 \div \left(-4\frac{7}{10}\right) \\[1em] \Rightarrow 0 \div \left(\frac{-47}{10}\right) \\[1em] \Rightarrow 0 \times \frac{10}{-47} \\[1em] \Rightarrow 0\end{aligned}$$

Hence, $\mathbf{0 \div \left(-4\frac{7}{10}\right) = 0}$

Let the required number be $x$.

According to the given condition,

$$\begin{aligned}\Rightarrow \frac{-4}{11} + x = \frac{-3}{8} \\[1em] \Rightarrow x = \frac{-3}{8} - \left(\frac{-4}{11}\right) \\[1em] \Rightarrow x = \frac{-3}{8} + \frac{4}{11} \\[1em] [\text{ LCM of } 8 \text{ and } 11 = 88] \\[1em] \Rightarrow x = \frac{(-3) \times 11 + 4 \times 8}{88} \\[1em] \Rightarrow x = \frac{-33 + 32}{88} \\[1em] \Rightarrow x = \frac{-1}{88}\end{aligned}$$

Hence, the required number is $\mathbf{\frac{-1}{88}}$

First find the sum of the two given rational numbers.

$$\begin{aligned}\Rightarrow \frac{3}{14} + \frac{-4}{7} \\[1em] [\text{ LCM of } 14 \text{ and } 7 = 14] \\[1em] \Rightarrow \frac{3 + (-4) \times 2}{14} \\[1em] \Rightarrow \frac{3 - 8}{14} \\[1em] \Rightarrow \frac{-5}{14}\end{aligned}$$

Let the required rational number be $x$.

According to the given condition,

$$\begin{aligned}\Rightarrow \frac{-5}{14} - x = \frac{13}{21} \\[1em] \Rightarrow x = \frac{-5}{14} - \frac{13}{21} \\[1em] [\text{ LCM of } 14 \text{ and } 21 = 42] \\[1em] \Rightarrow x = \frac{(-5) \times 3 - 13 \times 2}{42} \\[1em] \Rightarrow x = \frac{-15 - 26}{42} \\[1em] \Rightarrow x = \frac{-41}{42}\end{aligned}$$

Hence, the required rational number is $\mathbf{\frac{-41}{42}}$

Let the other rational number be $x$.

As the product of the two rational numbers is $\frac{25}{42}$,

$$\begin{aligned}\Rightarrow -2\frac{6}{7} \times x = \frac{25}{42} \\[1em] \Rightarrow \frac{-20}{7} \times x = \frac{25}{42} \\[1em] \Rightarrow x = \frac{25}{42} \div \frac{-20}{7} \\[1em] \Rightarrow x = \frac{25}{42} \times \frac{7}{-20} \\[1em] \Rightarrow x = \frac{25 \times 7}{42 \times (-20)} \\[1em] \Rightarrow x = \frac{175}{-840} \\[1em] \Rightarrow x = \frac{-5}{24}\end{aligned}$$

Hence, the other rational number is $\mathbf{\frac{-5}{24}}$

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

Finding the sum,

$$\begin{aligned}\Rightarrow \frac{4}{13} + \frac{-3}{2} \\[1em] [\text{ LCM of } 13 \text{ and } 2 = 26] \\[1em] \Rightarrow \frac{4 \times 2 + (-3) \times 13}{26} \\[1em] \Rightarrow \frac{8 - 39}{26} \\[1em] \Rightarrow \frac{-31}{26}\end{aligned}$$

Finding the product,

$$\begin{aligned}\Rightarrow \frac{4}{13} \times \frac{-3}{2} \\[1em] \Rightarrow \frac{4 \times (-3)}{13 \times 2} \\[1em] \Rightarrow \frac{-12}{26} = \frac{-6}{13}\end{aligned}$$

Dividing the sum by the product,

$$\begin{aligned}\Rightarrow \frac{-31}{26} \div \frac{-6}{13} \\[1em] \Rightarrow \frac{-31}{26} \times \frac{13}{-6} \\[1em] \Rightarrow \frac{(-31) \times 13}{26 \times (-6)} \\[1em] \Rightarrow \frac{-403}{-156} \\[1em] \Rightarrow \frac{31}{12} = 2\frac{7}{12}\end{aligned}$$

Hence, the required quotient is $\mathbf{2\frac{7}{12}}$

Rearrange the terms using commutativity of multiplication so that the common factor $\frac{4}{5}$ can be taken out using distributivity.

$$\begin{aligned}\Rightarrow \frac{3}{13} \times \frac{4}{5} - \frac{7}{5} - \frac{4}{5} \times \frac{5}{13} \\[1em] \Rightarrow \frac{4}{5} \times \frac{3}{13} - \frac{4}{5} \times \frac{5}{13} - \frac{7}{5} \\[1em] \Rightarrow \frac{4}{5} \times \left(\frac{3}{13} - \frac{5}{13}\right) - \frac{7}{5} \\[1em] \Rightarrow \frac{4}{5} \times \frac{-2}{13} - \frac{7}{5} \\[1em] \Rightarrow \frac{-8}{65} - \frac{7}{5} \\[1em] [\text{ LCM of } 65 \text{ and } 5 = 65] \\[1em] \Rightarrow \frac{-8 - 7 \times 13}{65} \\[1em] \Rightarrow \frac{-8 - 91}{65} \\[1em] \Rightarrow \frac{-99}{65} = -1\frac{34}{65}\end{aligned}$$

Hence, $\mathbf{\frac{3}{13} \times \frac{4}{5} - \frac{7}{5} - \frac{4}{5} \times \frac{5}{13} = -1\frac{34}{65}}$

Using commutativity and associativity of addition, group together the rational numbers having the same denominator.

$$\begin{aligned}\Rightarrow \frac{5}{7} + \frac{2}{11} + \frac{-8}{7} + \frac{6}{11} \\[1em] \Rightarrow \left(\frac{5}{7} + \frac{-8}{7}\right) + \left(\frac{2}{11} + \frac{6}{11}\right) \\[1em] \Rightarrow \frac{5 + (-8)}{7} + \frac{2 + 6}{11} \\[1em] \Rightarrow \frac{-3}{7} + \frac{8}{11} \\[1em] [\text{ LCM of } 7 \text{ and } 11 = 77] \\[1em] \Rightarrow \frac{(-3) \times 11 + 8 \times 7}{77} \\[1em] \Rightarrow \frac{-33 + 56}{77} \\[1em] \Rightarrow \frac{23}{77}\end{aligned}$$

Hence, $\mathbf{\frac{5}{7} + \frac{2}{11} + \frac{-8}{7} + \frac{6}{11} = \frac{23}{77}}$

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

$$\begin{aligned}\Rightarrow -13\frac{7}{8} = \frac{-111}{8} \\[1em] \Rightarrow \text{Additive inverse} = -\left(\frac{-111}{8}\right) \\[1em] \Rightarrow \frac{111}{8} = 13\frac{7}{8}\end{aligned}$$

Hence, the additive inverse of $\mathbf{-13\frac{7}{8} \text{ is } 13\frac{7}{8}}$

First reduce the fractional part to its lowest form, then change the sign.

$$\begin{aligned}\Rightarrow 4\frac{3}{6} = 4\frac{1}{2} = \frac{9}{2} \\[1em] \Rightarrow \text{Additive inverse} = -\left(\frac{9}{2}\right) \\[1em] \Rightarrow \frac{-9}{2} = -4\frac{1}{2}\end{aligned}$$

Hence, the additive inverse of $\mathbf{4\frac{3}{6} \text{ is } -4\frac{1}{2}}$

First reduce the rational number to its lowest form, then interchange its numerator and denominator.

$$\begin{aligned}\Rightarrow \frac{-23}{46} = \frac{-23 \div 23}{46 \div 23} = \frac{-1}{2} \\[1em] \Rightarrow \text{Multiplicative inverse} = \frac{2}{-1} = -2 \\[1em] [\text{Check : } \frac{-1}{2} \times (-2) = 1]\end{aligned}$$

Hence, the multiplicative inverse of $\mathbf{\frac{-23}{46} \text{ is } -2}$

There is no rational number which when multiplied by 0 gives 1, so 0 has no multiplicative inverse (reciprocal).

Hence, the multiplicative inverse of 0 does not exist.

As $\frac{-3}{11}$ is a negative proper fraction, it lies between -1 and 0.

Divide the line segment between -1 and 0 into eleven equal parts. The third point of division to the left of 0 represents $\frac{-3}{11}$.

Hence, the point marked on the number line represents $\mathbf{\frac{-3}{11}}$

As $\frac{5}{7}$ is a positive proper fraction, it lies between 0 and 1.

Divide the line segment between 0 and 1 into seven equal parts. The fifth point of division from 0 represents $\frac{5}{7}$.

Hence, the point marked on the number line represents $\mathbf{\frac{5}{7}}$

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

$$\begin{aligned}[\text{ LCM of } 7 \text{ and } 5 = 35] \\[1em] \Rightarrow \frac{-3}{7} = \frac{-3 \times 5}{7 \times 5} = \frac{-15}{35} \\[1em] \Rightarrow \frac{2}{5} = \frac{2 \times 7}{5 \times 7} = \frac{14}{35}\end{aligned}$$

We note that -14, -13, -12, …, 12, 13 are the integers between -15 and 14.

Choosing any five of these integers as numerators over the common denominator 35, say -1, 0, 1, 12 and 13, we get the rational numbers $\frac{-1}{35}, \frac{0}{35}, \frac{1}{35}, \frac{12}{35} \text{ and } \frac{13}{35}$.

Hence, five rational numbers between $\mathbf{\frac{-3}{7} \text{ and } \frac{2}{5}}$ are $\mathbf{\frac{-1}{35}, 0, \frac{1}{35}, \frac{12}{35} \text{ and } \frac{13}{35}}$

First find the fraction of the cake which was nuts, then find that fraction of the total weight.

Total weight of the wedding cake = 8 kg

Finding the fraction of the cake which was flour, sugar and cream,

$$\begin{aligned}\Rightarrow \frac{2}{5} + \frac{5}{16} + \frac{1}{4} \\[1em] [\text{ LCM of } 5, 16 \text{ and } 4 = 80] \\[1em] \Rightarrow \frac{2 \times 16 + 5 \times 5 + 1 \times 20}{80} \\[1em] \Rightarrow \frac{32 + 25 + 20}{80} \\[1em] \Rightarrow \frac{77}{80}\end{aligned}$$

Finding the fraction of the cake which was nuts,

$$\begin{aligned}\Rightarrow 1 - \frac{77}{80} \\[1em] \Rightarrow \frac{80 - 77}{80} \\[1em] \Rightarrow \frac{3}{80}\end{aligned}$$

Finding the weight of nuts,

$$\begin{aligned}\Rightarrow \frac{3}{80} \times 8 \\[1em] \Rightarrow \frac{24}{80} \\[1em] \Rightarrow \frac{3}{10} \text{ kg} \\[1em] \Rightarrow \frac{3}{10} \times 1000 \text{ g} = 300 \text{ g}\end{aligned}$$

Hence, the weight of nuts is $\mathbf{300 \text{ g}}$

To verify the associative property of addition, we have to show that $p + (q + r) = (p + q) + r$

Solving L.H.S.,

$$\begin{aligned}\Rightarrow q + r = \frac{2}{3} + \left(\frac{-8}{11}\right) \\[1em] = \frac{2 \times 11 + (-8) \times 3}{33} \\[1em] = \frac{22 - 24}{33} \\[1em] = \frac{-2}{33}\end{aligned}$$

Calculating, $p + (q + r)$ now :

$$\begin{aligned}\Rightarrow p + (q + r) = \frac{-4}{9} + \frac{-2}{33} \\[1em] = \frac{(-4) \times 11 + (-2) \times 3}{99} \\[1em] = \frac{-44 - 6}{99} \\[1em] = \frac{-50}{99}\end{aligned}$$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow p + q = \frac{-4}{9} + \frac{2}{3} \\[1em] = \frac{-4 + 2 \times 3}{9} \\[1em] = \frac{-4 + 6}{9} \\[1em] = \frac{2}{9}\end{aligned}$$

Calculating, $(p + q) + r$ now :

$$\begin{aligned}\Rightarrow (p + q) + r = \frac{2}{9} + \left(\frac{-8}{11}\right) \\[1em] = \frac{2 \times 11 + (-8) \times 9}{99} \\[1em] = \frac{22 - 72}{99} \\[1em] = \frac{-50}{99}\end{aligned}$$

As L.H.S. = R.H.S.

$\Rightarrow p + (q + r) = (p + q) + r$

Hence, proved that $\bm{p + (q + r) = (p + q) + r}$.

To verify the commutative property of multiplication, we have to show that $p \times q = q \times p$

Solving L.H.S.,

$$\begin{aligned}\Rightarrow p \times q = \frac{-4}{9} \times \frac{2}{3} \\[1em] = \frac{(-4) \times 2}{9 \times 3} \\[1em] = \frac{-8}{27}\end{aligned}$$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow q \times p = \frac{2}{3} \times \frac{-4}{9} \\[1em] = \frac{2 \times (-4)}{3 \times 9} \\[1em] = \frac{-8}{27}\end{aligned}$$

As L.H.S. = R.H.S.

$\Rightarrow p \times q = q \times p$

Hence, proved that $\bm{p \times q = q \times p}$.

To verify the distributive property of multiplication over addition, we have to show that $p \times (q + r) = p \times q + p \times r$

Solving L.H.S.,

Here $q + r = \frac{-2}{33}$ (found in part (i))

$$\begin{aligned}\Rightarrow p \times (q + r) = \frac{-4}{9} \times \frac{-2}{33} \\[1em] = \frac{(-4) \times (-2)}{9 \times 33} \\[1em] = \frac{8}{297}\end{aligned}$$

Solving R.H.S.,

Here $p \times q = \frac{-8}{27}$ (found in part (ii))

$$\begin{aligned}\Rightarrow p \times r = \frac{-4}{9} \times \left(\frac{-8}{11}\right) \\[1em] = \frac{(-4) \times (-8)}{9 \times 11} \\[1em] = \frac{32}{99}\end{aligned}$$

Calculating, $p \times q + p \times r$ now :

$$\begin{aligned}\Rightarrow p \times q + p \times r = \frac{-8}{27} + \frac{32}{99} \\[1em] [\text{ LCM of } 27 \text{ and } 99 = 297] \\[1em] = \frac{(-8) \times 11 + 32 \times 3}{297} \\[1em] = \frac{-88 + 96}{297} \\[1em] = \frac{8}{297}\end{aligned}$$

As L.H.S. = R.H.S.

$\Rightarrow p \times (q + r) = p \times q + p \times r$

Hence, proved that $\bm{p \times (q + r) = p \times q + p \times r}$.

To verify the given result, we have to show that $(p + q) \div r = p \div r + q \div r$

Solving L.H.S.,

Here $p + q = \frac{2}{9}$ (found in part (i))

$$\begin{aligned}\Rightarrow (p + q) \div r = \frac{2}{9} \div \left(\frac{-8}{11}\right) \\[1em] = \frac{2}{9} \times \frac{11}{-8} \\[1em] = \frac{22}{-72} \\[1em] = \frac{-11}{36}\end{aligned}$$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow p \div r = \frac{-4}{9} \div \left(\frac{-8}{11}\right) \\[1em] = \frac{-4}{9} \times \frac{11}{-8} \\[1em] = \frac{-44}{-72} \\[1em] = \frac{11}{18}\end{aligned}$$

Calculating, $q \div r$ now :

$$\begin{aligned}\Rightarrow q \div r = \frac{2}{3} \div \left(\frac{-8}{11}\right) \\[1em] = \frac{2}{3} \times \frac{11}{-8} \\[1em] = \frac{22}{-24} \\[1em] = \frac{-11}{12}\end{aligned}$$

Calculating, $p \div r + q \div r$ now :

$$\begin{aligned}\Rightarrow p \div r + q \div r = \frac{11}{18} + \left(\frac{-11}{12}\right) \\[1em] [\text{ LCM of } 18 \text{ and } 12 = 36] \\[1em] = \frac{11 \times 2 + (-11) \times 3}{36} \\[1em] = \frac{22 - 33}{36} \\[1em] = \frac{-11}{36}\end{aligned}$$

As L.H.S. = R.H.S.

$\Rightarrow (p + q) \div r = p \div r + q \div r$

Hence, proved that $\bm{(p + q) \div r = p \div r + q \div r}$.