Rational Numbers

Largest number of two digits = 99

Smallest number of four digits = 1000

Hence, the required rational number = $\frac{99}{1000}$.

(i) Numerator of $\frac{-125}{127}$ is -125.

(ii) Numerator of $\frac{37}{-137}$ is 37.

(iii) Numerator of $\frac{-85}{93}$ is -85.

(iv) 2 can be written as $\frac{2}{1}$, so its numerator is 2.

(v) 0 can be written as $\frac{0}{1}$, so its numerator is 0.

(i) Denominator of $\frac{7}{-15}$ is -15.

(ii) Denominator of $\frac{-18}{29}$ is 29.

(iii) Denominator of $\frac{-3}{4}$ is 4.

(iv) -7 can be written as $\frac{-7}{1}$, so its denominator is 1.

(v) 0 can be written as $\frac{0}{1}, \frac{0}{2}, \frac{0}{3}, \frac{0}{4}, \ldots$

Zero does not have any unique denominator, so its denominator can be any non-zero number.

Numerator = (-5) × (-4) = 20

Denominator = (28 - 27) × (8 - 5) = 1 × 3 = 3

Hence, the required rational number = $\frac{20}{3}$.

(i) $\frac{-15}{1}$ in integer form is -15.

(ii) $\frac{23}{-1}$ in integer form is -23.

(iii) Given,

$$\begin{aligned}\Rightarrow 18 = \frac{18}{a} \\[1em] \Rightarrow 18 \times a = 18 \times 1 \\[1em] \Rightarrow a = \frac{18}{18} \\[1em] \Rightarrow a = 1\end{aligned}$$

Hence, a = 1

(iv) Given,

$$\begin{aligned}\Rightarrow -57 = \frac{57}{a}\\[1em] \Rightarrow -57 \times a = 57 \times 1 \\[1em] \Rightarrow a = \frac{57}{-57} \\[1em] \Rightarrow a = -1\end{aligned}$$

Hence, a = -1

A rational number is positive if its numerator and denominator are both of the same sign, and negative if they are of opposite signs.

Positive rational numbers: $\frac{-3}{-5}, \frac{3}{5} \text{ and } \frac{-13}{-3}$

Negative rational numbers: $\frac{-3}{5}, \frac{3}{-5}, \frac{15}{-8} \text{ and } \frac{-15}{8}$

(0 is neither positive nor negative).

(i) Three rational numbers equivalent to $\frac{3}{5}$ are :

$$\begin{aligned}\frac{3 \times 2}{5 \times 2} = \frac{6}{10}, \\[1em] \frac{3 \times 3}{5 \times 3} = \frac{9}{15}, \\[1em] \frac{3 \times 4}{5 \times 4} = \frac{12}{20}\end{aligned}$$

Hence, $\frac{6}{10}, \frac{9}{15} \text{ and } \frac{12}{20}$ are three rational numbers equivalent to $\frac{3}{5}$.

(ii) Three rational numbers equivalent to $\frac{4}{-7}$ are :

$$\begin{aligned}\frac{4 \times 2}{-7 \times 2} = \frac{8}{-14}, \\[1em] \frac{4 \times 3}{-7 \times 3} = \frac{12}{-21}, \\[1em] \frac{4 \times 4}{-7 \times 4} = \frac{16}{-28}\end{aligned}$$

Hence, $\frac{8}{-14}, \frac{12}{-21} \text{ and } \frac{16}{-28}$ are three rational numbers equivalent to $\frac{4}{-7}$.

(iii) Three rational numbers equivalent to $\frac{-5}{9}$ are :

$$\begin{aligned}\frac{-5 \times 2}{9 \times 2} = \frac{-10}{18}, \\[1em] \frac{-5 \times 3}{9 \times 3} = \frac{-15}{27}, \\[1em] \frac{-5 \times 4}{9 \times 4} = \frac{-20}{36}\end{aligned}$$

Hence, $\frac{-10}{18}, \frac{-15}{27} \text{ and } \frac{-20}{36}$ are three rational numbers equivalent to $\frac{-5}{9}$.

(iv) Three rational numbers equivalent to $\frac{8}{-15}$ are :

$$\begin{aligned}\frac{8 \times 2}{-15 \times 2} = \frac{16}{-30}, \\[1em] \frac{8 \times 3}{-15 \times 3} = \frac{24}{-45}, \\[1em] \frac{8 \times 4}{-15 \times 4} = \frac{32}{-60}\end{aligned}$$

Hence, $\frac{16}{-30}, \frac{24}{-45} \text{ and } \frac{32}{-60}$ are three rational numbers equivalent to $\frac{8}{-15}$.

A number $\frac{p}{q}$ is a rational number only if q ≠ 0.

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

(ii) 0 = $\frac{0}{1}$, which is a rational number.

(iii) $\frac{0}{4} = 0$, which is a rational number.

(iv) $\frac{8}{0}$ has denominator 0, so it is not a rational number.

(v) $\frac{0}{0}$ has denominator 0, so it is not a rational number.

Hence, $\frac{8}{0}\text{ and }\frac{0}{0}$ are not rational numbers.

(i) $5 = \frac{5 \times 7}{1 \times 7} = \frac{35}{7}$

(ii) $-8 = \frac{-8 \times 7}{1 \times 7} = \frac{-56}{7}$

(iii) $0 = \frac{0 \times 7}{1 \times 7} = \frac{0}{7}$

(iv) $-16 = \frac{-16 \times 7}{1 \times 7} = \frac{-112}{7}$

(v) $7 = \frac{7 \times 7}{1 \times 7} = \frac{49}{7}$

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

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

(iii) $\frac{3}{5} = \frac{3 \times 9}{5 \times 9} = \frac{27}{45}$

(iv) $\frac{3}{5} = \frac{3 \times 5}{5 \times 5} = \frac{15}{25}$

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

(i) $\frac{4}{7} = \frac{4 \times 3}{7 \times 3} = \frac{12}{21}$

(ii) $\frac{4}{7} = \frac{4 \times (-3)}{7 \times (-3)} = \frac{-12}{-21}$

(iii) $\frac{4}{7} = \frac{4 \times (-4)}{7 \times (-4)} = \frac{-16}{-28}$

(iv) $\frac{4}{7} = \frac{4 \times (-5)}{7 \times (-5)} = \frac{-20}{-35}$

(v) $\frac{4}{7} = \frac{4 \times 5}{7 \times 5} = \frac{20}{35}$

(i) Given,

$$\begin{aligned}\Rightarrow -\frac{2}{3} = \frac{6}{x} \\[1em] \Rightarrow -2 \times x = 3 \times 6 \\[1em] \Rightarrow -2x = 18 \\[1em] \Rightarrow x = \frac{18}{-2} \\[1em] \Rightarrow x = -9\end{aligned}$$

Hence, x = -9

(ii) Given,

$$\begin{aligned}\Rightarrow \frac{7}{-4} = \frac{x}{8}\\[1em] \Rightarrow -4 \times x = 7 \times 8\\[1em] \Rightarrow -4x = 56 \\[1em] \Rightarrow x = \frac{56}{-4} \\[1em] \Rightarrow x = -14\end{aligned}$$

Hence, x = -14

(iii) Given,

$$\begin{aligned}\Rightarrow \frac{3}{7} = \frac{x}{-35}\\[1em] \Rightarrow 7 \times x = 3 \times (-35) \\[1em] \Rightarrow 7x = -105 \\[1em] \Rightarrow x = \frac{-105}{7} \\[1em] \Rightarrow x = -15\end{aligned}$$

Hence, x = -15

(iv) Given,

$$\begin{aligned}\Rightarrow \frac{-48}{x} = 6\\[1em] \Rightarrow 6x = -48 \\[1em] \Rightarrow x = \frac{-48}{6} \\[1em] \Rightarrow x = -8\end{aligned}$$

Hence, x = -8

(v) Given,

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

Hence, x = 12

(vi) Given,

$$\begin{aligned}\Rightarrow \frac{-27}{x} = 9\\[1em] \Rightarrow 9x = -27 \\[1em] \Rightarrow x = \frac{-27}{9} \\[1em] \Rightarrow x = -3\end{aligned}$$

Hence, x = -3

(i) By Prime Factorization,

$$\begin{array}{l|r} 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

HCF of 12 and 15 = 3.

$\frac{12}{15} = \frac{12 ÷ 3}{15 ÷ 3} = \frac{4}{5}$

Hence, $\frac{12}{15} = \frac{4}{5}$

(ii) By Prime Factorization,

$$\begin{array}{l|r} 2 & 120 \\ \hline 2 & 60 \\ \hline 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 144 \\ \hline 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$

HCF of 120 and 144 = 2 × 2 × 2 × 3 = 24.

$\frac{-120}{144} = \frac{-120 ÷ 24}{144 ÷ 24} = \frac{-5}{6}$

Hence, $\frac{-120}{144} = \frac{-5}{6}$

(iii) $\frac{-48}{-72} = \frac{48}{72}$

By Prime Factorization,

$$\begin{array}{l|r} 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$

HCF of 48 and 72 = 2 × 2 × 2 × 3 = 24.

$\frac{48}{72} = \frac{48 ÷ 24}{72 ÷ 24} = \frac{2}{3}$

Hence, $\frac{-48}{-72} = \frac{2}{3}$

(iv) By Prime Factorization,

$$\begin{array}{l|r} 2 & 14 \\ \hline 7 & 7 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 56 \\ \hline 2 & 28 \\ \hline 2 & 14 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$$

HCF of 14 and 56 = 2 × 7 = 14.

$\frac{14}{-56} = \frac{14 ÷ 14}{-56 ÷ 14} = \frac{1}{-4} = -\frac{1}{4}$

Hence, $\frac{14}{-56} = -\frac{1}{4}$

A rational number is in standard form if its denominator is positive and the rational number is in its lowest terms.

(i) $\frac{-7}{-8} = \frac{-7 \times (-1)}{-8 \times (-1)} = \frac{7}{8}$

Hence, standard form of $\frac{-7}{-8}\text{ is }\frac{7}{8}$.

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

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

(iii) $\frac{-7}{-20} = \frac{-7 \times (-1)}{-20 \times (-1)} = \frac{7}{20}$

Hence, standard form of $\frac{-7}{-20}\text{ is }\frac{7}{20}$.

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

Hence, standard form of $\frac{4}{-9}\text{ is }\frac{-4}{9}$.

(i) Since the denominator of each rational number is 4, divide each unit length between 0 and 1, and between 0 and -1, into four equal parts.

To represent $\frac{3}{4}$, move 3 parts to the right of 0; to represent $-\frac{1}{4}$, move 1 part to the left of 0.

(ii) Since the denominator of each rational number is 5, divide each unit length into five equal parts.

To represent $\frac{2}{5}$, move 2 parts to the right of 0; to represent $\frac{-3}{5}$, move 3 parts to the left of 0.

(iii) $-\frac{2}{3} = -\frac{4}{6}$, so both numbers have denominator 6. Divide each unit length into six equal parts.

To represent $\frac{5}{6}$, move 5 parts to the right of 0; to represent $-\frac{2}{3} = -\frac{4}{6}$, move 4 parts to the left of 0.

(iv) Since the denominator of each rational number is 5, divide each unit length into five equal parts.

To represent $\frac{2}{5}$, move 2 parts to the right of 0; to represent $-\frac{4}{5}$, move 4 parts to the left of 0.

(v) Since the denominator of each rational number is 4, divide each unit length into four equal parts.

To represent $\frac{1}{4}$, move 1 part to the right of 0; to represent $-\frac{5}{4}$, move 5 parts to the left of 0.

(i) On cross-multiplying $\frac{3}{5}$ and $\frac{5}{7}$:

3 × 7 = 21 and 5 × 5 = 25

Since 21 < 25,

Hence, $\frac{3}{5} \lt \frac{5}{7}$

(ii) $\frac{-7}{2}$ is a negative rational number and $\frac{5}{2}$ is a positive rational number.

Since every negative rational number is less than every positive rational number,

Hence, $\frac{-7}{2} \lt \frac{5}{2}$

(iii) -3 is a negative number and $2\frac{3}{4}$ is a positive number.

Since every negative number is less than every positive number,

Hence, $-3 \lt 2\frac{3}{4}$

(iv) $-1\frac{1}{2}$ is a negative number and 0 is neither positive nor negative.

Since every negative number is less than 0,

Hence, $-1\frac{1}{2} \lt 0$

(v) 0 is neither positive nor negative and $\frac{3}{4}$ is a positive number.

Since 0 is less than every positive number,

Hence, $0 \lt \frac{3}{4}$

(vi) 3 is a positive number and -1 is a negative number.

Since every positive number is greater than every negative number,

Hence, 3 > -1

(i) $-\frac{1}{4}$ is a negative number and 0 is neither positive nor negative.

Since every negative number is less than 0,

Hence, $-\frac{1}{4} \lt 0$

(ii) $\frac{1}{4}$ is a positive number and 0 is neither positive nor negative.

Since 0 is less than every positive number,

Hence, $\frac{1}{4} \gt 0$

(iii) $-\frac{3}{8}$ is a negative number and $\frac{2}{5}$ is a positive number.

Since every negative number is less than every positive number,

Hence, $-\frac{3}{8} \lt \frac{2}{5}$

(iv) $\frac{7}{-12} = \frac{-7}{12}$

By Division Method,

$$\begin{array}{l|r} 2 & 8, 12 \\ \hline 2 & 4, 6 \\ \hline 2 & 2, 3 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}$$

LCM of 8 and 12 = 2 × 2 × 2 × 3 = 24

$\frac{-5 \times 3}{8 \times 3} = \frac{-15}{24} \text{ and } \frac{-7 \times 2}{12 \times 2} = \frac{-14}{24}$

Since -15 < -14,

Hence, $\frac{-5}{8} \lt \frac{7}{-12}$

(v) $\frac{5}{-9} = \frac{-5}{9} \text{ and } \frac{-5}{-9} = \frac{5}{9}$

$\frac{-5}{9}$ is negative and $\frac{5}{9}$ is positive.

Since every negative number is less than every positive number,

Hence, $\frac{5}{-9} \lt \frac{-5}{-9}$

(vi) $\frac{5}{-6} = \frac{-5}{6}$

By prime factorization,

$$\begin{array}{l|r} 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$

LCM of 8 and 6 is 2 × 2 × 2 × 3 = 24

$\frac{-7 \times 3}{8 \times 3} = \frac{-21}{24} \text{ and } \frac{-5 \times 4}{6 \times 4} = \frac{-20}{24}$

Since -21 < -20,

Hence, $\frac{-7}{8} \lt \frac{5}{-6}$

(vii) $\frac{-3}{-8} = \frac{3}{8}$

On cross-multiplying $\frac{2}{7} \text{ and } \frac{3}{8}$:

$2 \times 8 = 16 \text{ and } 7 \times 3 = 21$

Since 16 < 21,

Hence, $\frac{2}{7} \lt \frac{-3}{-8}$

(i) Writing each number with a positive denominator:

$\frac{-11}{-30} = \frac{11}{30} \text{ and } \frac{5}{-15} = \frac{-1}{3}$

So, the numbers are $\frac{7}{10}, \frac{11}{30} \text{ and } \frac{-1}{3}$.

By Division Method,

$$\begin{array}{l|r} 2 & 10, 30, 3 \\ \hline 3 & 5, 15, 3 \\ \hline 5 & 5, 5, 1 \\ \hline & 1, 1, 1 \end{array}$$

LCM of 10, 30 and 3 is 2 × 3 × 5 = 30

$$\begin{aligned}\frac{7 \times 3}{10 \times 3} = \frac{21}{30}, \\[1em] \frac{11 \times 1}{30 \times 1} = \frac{11}{30}, \\[1em] \frac{-1 \times 10}{3 \times 10} = \frac{-10}{30}\end{aligned}$$

Since -10 < 11 < 21,

$\frac{-10}{30} \lt \frac{11}{30} \lt \frac{21}{30}$

Hence, $\frac{5}{-15} \lt \frac{-11}{-30} \lt \frac{7}{10}$.

(ii) Writing each number with a positive denominator:

$\frac{4}{-9} = \frac{-4}{9} \text{ and } \frac{2}{-3} = \frac{-2}{3}$

So, the numbers are $\frac{-4}{9}, \frac{-5}{12} \text{ and } \frac{-2}{3}$.

By Division Method,

$$\begin{array}{l|r} 2 & 9, 12, 3 \\ \hline 2 & 9, 6, 3 \\ \hline 3 & 9, 3, 3 \\ \hline 3 & 3, 1, 1 \\ \hline & 1, 1, 1 \end{array}$$

LCM of 9, 12 and 3 is 2 × 2 × 3 × 3 = 36

$$\begin{aligned}\frac{-4 \times 4}{9 \times 4} = \frac{-16}{36}, \\[1em] \frac{-5 \times 3}{12 \times 3} = \frac{-15}{36}, \\[1em] \frac{-2 \times 12}{3 \times 12} = \frac{-24}{36}\end{aligned}$$

Since -24 < -16 < -15,

$\frac{-24}{36} \lt \frac{-16}{36} \lt \frac{-15}{36}$

Hence, $\frac{2}{-3} \lt \frac{4}{-9} \lt \frac{-5}{12}$.

(i) Writing each number with a positive denominator:

$\frac{13}{-16} = \frac{-13}{16}$

So, the numbers are $\frac{5}{8}, \frac{-13}{16} \text{ and } \frac{-7}{12}$.

By Division Method,

$$\begin{array}{l|r} 2 & 8, 16, 12 \\ \hline 2 & 4, 8, 6 \\ \hline 2 & 2, 4, 3 \\ \hline 2 & 1, 2, 3 \\ \hline 3 & 1, 1, 3 \\ \hline & 1, 1, 1 \end{array}$$

LCM of 8, 16 and 12 is 2 × 2 × 2 × 2 × 3 = 48

$$\begin{aligned}\frac{5 \times 6}{8 \times 6} = \frac{30}{48}, \\[1em] \frac{-13 \times 3}{16 \times 3} = \frac{-39}{48}, \\[1em] \frac{-7 \times 4}{12 \times 4} = \frac{-28}{48}\end{aligned}$$

Since 30 > -28 > -39,

$\frac{30}{48} \gt \frac{-28}{48} \gt \frac{-39}{48}$

Hence, $\frac{5}{8} \gt \frac{-7}{12} \gt \frac{13}{-16}$.

(ii) Writing each number with a positive denominator:

$\frac{3}{-10} = \frac{-3}{10} \text{ and } \frac{8}{-20} = \frac{-2}{5}$

So, the numbers are $\frac{-3}{10}, \frac{-13}{30} \text{ and } \frac{-2}{5}$.

By Division Method,

$$\begin{array}{l|r} 2 & 10, 30, 5 \\ \hline 3 & 5, 15, 5 \\ \hline 5 & 5, 5, 5 \\ \hline & 1, 1, 1 \end{array}$$

LCM of 10, 30 and 5 is 2 × 3 × 5 = 30

$$\begin{aligned}\frac{-3 \times 3}{10 \times 3} = \frac{-9}{30}, \\[1em] \frac{-13 \times 1}{30 \times 1} = \frac{-13}{30}, \\[1em] \frac{-2 \times 6}{5 \times 6} = \frac{-12}{30}\end{aligned}$$

Since -9 > -12 > -13,

$\frac{-9}{30} \gt \frac{-12}{30} \gt \frac{-13}{30}$

Hence, $\frac{3}{-10} \gt \frac{8}{-20} \gt \frac{-13}{30}$.

(i) $\frac{5}{8} \text{ and } \frac{3}{10}$ are both positive, so they are on the same side of zero.

(ii) $-\frac{5}{8}$ is negative and $\frac{3}{10}$ is positive, so they are on the opposite sides of zero.

(iii) $-\frac{5}{8} \text{ and } -\frac{3}{10}$ are both negative, so they are on the same side of zero.

(iv) $\frac{5}{8}$ is positive and $-\frac{3}{10}$ is negative, so they are on the opposite sides of zero.

(i) By Division Method,

$$\begin{array}{l|r} 2 & 3, 4 \\ \hline 2 & 3, 2 \\ \hline 3 & 3, 1 \\ \hline & 1, 1 \end{array}$$

LCM of 3 and 4 = 2 × 2 × 3 = 12

$-\frac{2}{3} = \frac{-2 \times 4}{3 \times 4} = \frac{-8}{12} \text{ and } \frac{3}{4} = \frac{3 \times 3}{4 \times 3} = \frac{9}{12}$

Now, $\frac{-8}{12} \lt \frac{-7}{12} \lt \frac{-6}{12} \lt \frac{-5}{12} \lt \frac{9}{12}$

Hence, $\frac{-7}{12}, \frac{-6}{12} \text{ and } \frac{-5}{12}$ are three rational numbers between $-\frac{2}{3} \text{ and } \frac{3}{4}$.

(ii) By Division Method,

$$\begin{array}{l|r} 3 & 7, 9 \\ \hline 3 & 7, 3 \\ \hline 7 & 7, 1 \\ \hline & 1, 1 \end{array}$$

LCM of 7 and 9 = 3 × 3 × 7 = 63

$$\begin{aligned}\frac{5}{7} = \frac{5 \times 9}{7 \times 9} = \frac{45}{63} \\[1em] \frac{7}{9} = \frac{7 \times 7}{9 \times 7} = \frac{49}{63}\end{aligned}$$

Now, $\frac{45}{63} \lt \frac{46}{63} \lt \frac{47}{63} \lt \frac{48}{63} \lt \frac{49}{63}$

Hence, $\frac{46}{63}, \frac{47}{63} \text{ and } \frac{48}{63}$ are three rational numbers between $\frac{5}{7} \text{ and } \frac{7}{9}$.

(iii) By Division Method,

$$\begin{array}{l|r} 2 & 8, 6 \\ \hline 2 & 4, 3 \\ \hline 2 & 2, 3 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}$$

LCM of 8 and 6 is 2 × 2 × 2 × 3 = 24

$$\begin{aligned}-\frac{5}{8} = \frac{-5 \times 3}{8 \times 3} = \frac{-15}{24} \\[1em] -\frac{1}{6} = \frac{-1 \times 4}{6 \times 4} = \frac{-4}{24}\end{aligned}$$

Now, $\frac{-15}{24} \lt \frac{-14}{24} \lt \frac{-13}{24} \lt \frac{-12}{24} \lt \frac{-4}{24}$

Hence, $\frac{-14}{24}, \frac{-13}{24} \text{ and } \frac{-12}{24}$ are three rational numbers between $-\frac{5}{8} \text{ and } -\frac{1}{6}$.

(i) $\frac{5}{-6} = -\frac{5}{6}$

By Division Method,

$$\begin{array}{l|r} 2 & 8, 6 \\ \hline 2 & 4, 3 \\ \hline 2 & 2, 3 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}$$

LCM of 8 and 6 = 2 × 2 × 2 × 3 = 24

$$\begin{aligned}-\frac{5}{6} = \frac{-5 \times 4}{6 \times 4} = \frac{-20}{24} \\[1em] -\frac{3}{8} = \frac{-3 \times 3}{8 \times 3} = \frac{-9}{24}\end{aligned}$$

Now, $\frac{-20}{24} \lt \frac{-19}{24} \lt \frac{-18}{24} \lt \frac{-17}{24} \lt \frac{-16}{24} \lt \frac{-9}{24}$

Hence, $\frac{-19}{24}, \frac{-18}{24}, \frac{-17}{24} \text{ and } \frac{-16}{24}$ are four rational numbers between $-\frac{3}{8} \text{ and } -\frac{5}{6}$.

(ii) By Division Method,

$$\begin{array}{l|r} 3 & 5, 3 \\ \hline 5 & 5, 1 \\ \hline & 1, 1 \end{array}$$

LCM of 5 and 3 = 3 × 5 = 15

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

Now, $\frac{-12}{15} \lt \frac{-11}{15} \lt \frac{-10}{15} \lt \frac{-9}{15} \lt \frac{-8}{15} \lt \frac{10}{15}$

Hence, $\frac{-11}{15}, \frac{-10}{15}, \frac{-9}{15} \text{ and } \frac{-8}{15}$ are four rational numbers between $-\frac{4}{5} \text{ and } \frac{2}{3}$.

(iii) The integers between -3 and 6 are -2, -1, 0, 1, 2, 3, 4 and 5, each of which is a rational number.

Hence, -2, -1, 0 and 1 are four rational numbers between -3 and 6.

(iv) The integers between 0 and 6 are 1, 2, 3, 4 and 5, each of which is a rational number.

Hence, 1, 2, 3 and 4 are four rational numbers between 0 and 6.

Solving,

$$\begin{aligned}\Rightarrow \frac{7}{5} + \frac{2}{5} \\[1em] = \frac{7 + 2}{5} \\[1em] = \frac{9}{5} \\[1em] = 1\frac{4}{5}\end{aligned}$$

Hence, $\frac{7}{5} + \frac{2}{5} = 1\frac{4}{5}$

Solving,

$$\begin{aligned}\Rightarrow \frac{-4}{9} + \frac{2}{9} \\[1em] = \frac{-4 + 2}{9} \\[1em] = \frac{-2}{9}\end{aligned}$$

Hence, $\frac{-4}{9} + \frac{2}{9} = \frac{-2}{9}$

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

Solving,

$$\begin{aligned}\Rightarrow \frac{-5}{12} + \frac{1}{12} \\[1em] = \frac{-5 + 1}{12} \\[1em] = \frac{-4}{12} \\[1em] = \frac{-1}{3}\end{aligned}$$

Hence, $\frac{5}{-12} + \frac{1}{12} = \frac{-1}{3}$

$\frac{4}{-15} = \frac{-4}{15} \text{ and } \frac{-7}{-15} = \frac{7}{15}$

Solving,

$$\begin{aligned}\Rightarrow \frac{-4}{15} + \frac{7}{15} \\[1em] = \frac{-4 + 7}{15} \\[1em] = \frac{3}{15} \\[1em] = \frac{1}{5}\end{aligned}$$

Hence, $\frac{4}{-15} + \frac{-7}{-15} = \frac{1}{5}$

$\frac{9}{-25} = \frac{-9}{25}$

Solving,

$$\begin{aligned}\Rightarrow \frac{-7}{25} + \frac{-9}{25} \\[1em] = \frac{-7 + (-9)}{25} \\[1em] = \frac{-16}{25}\end{aligned}$$

Hence, $\frac{-7}{25} + \frac{9}{-25} = \frac{-16}{25}$

$\frac{7}{-26} = \frac{-7}{26}$

Solving,

$$\begin{aligned}\Rightarrow \frac{-7}{26} + \frac{-7}{26} \\[1em] = \frac{-7 + (-7)}{26} \\[1em] = \frac{-14}{26} \\[1em] = \frac{-7}{13}\end{aligned}$$

Hence, $\frac{-7}{26} + \frac{7}{-26} = \frac{-7}{13}$

By Division Method,

$$\begin{array}{l|r} 5 & 5, 7 \\ \hline 7 & 1, 7 \\ \hline & 1, 1 \end{array}$$

LCM of 5 and 7 = 5 × 7 = 35

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

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

By Division Method,

$$\begin{array}{l|r} 2 & 6, 9 \\ \hline 3 & 3, 9 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}$$

LCM of 6 and 9 = 2 × 3 × 3 = 18

$$\begin{aligned}\Rightarrow \frac{-5 \times 3}{6 \times 3} + \frac{4 \times 2}{9 \times 2} \\[1em] = \frac{-15}{18} + \frac{8}{18} \\[1em] = \frac{-15 + 8}{18} \\[1em] = \frac{-7}{18}\end{aligned}$$

Hence, $\frac{-5}{6} + \frac{4}{9} = \frac{-7}{18}$

Solving,

$\frac{-3}{1} + \frac{2}{3}$

LCM of 1 and 3 = 3

$$\begin{aligned}\Rightarrow \frac{-3 \times 3}{1 \times 3} + \frac{2 \times 1}{3 \times 1} \\[1em] = \frac{-9}{3} + \frac{2}{3} \\[1em] = \frac{-9 + 2}{3} \\[1em] = \frac{-7}{3} \\[1em] = -2\frac{1}{3}\end{aligned}$$

Hence, $-3 + \frac{2}{3} = -2\frac{1}{3}$

By Division Method,

$$\begin{array}{l|r} 2 & 9, 18 \\ \hline 3 & 9, 9 \\ \hline 3 & 3, 3 \\ \hline & 1, 1 \end{array}$$

LCM of 9 and 18 = 2 × 3 × 3 = 18

$$\begin{aligned}\Rightarrow \frac{-5 \times 2}{9 \times 2} + \frac{7 \times 1}{18 \times 1} \\[1em] = \frac{-10}{18} + \frac{7}{18} \\[1em] = \frac{-10 + 7}{18} \\[1em] = \frac{-3}{18} \\[1em] = \frac{-1}{6}\end{aligned}$$

Hence, $\frac{-5}{9} + \frac{7}{18} = \frac{-1}{6}$

By Division Method,

$$\begin{array}{l|r} 2 & 24, 48 \\ \hline 2 & 12, 24 \\ \hline 2 & 6, 12 \\ \hline 2 & 3, 6 \\ \hline 3 & 3, 3 \\ \hline & 1, 1 \end{array}$$

LCM of 24 and 48 = 2 × 2 × 2 × 2 × 3 = 48

$$\begin{aligned}\Rightarrow \frac{-7 \times 2}{24 \times 2} + \frac{-5 \times 1}{48 \times 1} \\[1em] = \frac{-14}{48} + \frac{-5}{48} \\[1em] = \frac{-14 + (-5)}{48} \\[1em] = \frac{-19}{48}\end{aligned}$$

Hence, $\frac{-7}{24} + \frac{-5}{48} = \frac{-19}{48}$

$\frac{1}{-18} = \frac{-1}{18} \text{ and } \frac{5}{-27} = \frac{-5}{27}$

By Division Method,

$$\begin{array}{l|r} 2 & 18, 27 \\ \hline 3 & 9, 27 \\ \hline 3 & 3, 9 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}$$

LCM of 18 and 27 = 2 × 3 × 3 × 3 = 54

$$\begin{aligned}\Rightarrow \frac{-1 \times 3}{18 \times 3} + \frac{-5 \times 2}{27 \times 2} \\[1em] = \frac{-3}{54} + \frac{-10}{54} \\[1em] = \frac{-3 + (-10)}{54} \\[1em] = \frac{-13}{54}\end{aligned}$$

Hence, $\frac{1}{-18} + \frac{5}{-27} = \frac{-13}{54}$

$\frac{1}{-75} = \frac{-1}{75}$

By Division Method,

$$\begin{array}{l|r} 3 & 25, 75 \\ \hline 5 & 25, 25 \\ \hline 5 & 5, 5 \\ \hline & 1, 1 \end{array}$$

LCM of 25 and 75 = 3 × 5 × 5 = 75

$$\begin{aligned}\Rightarrow \frac{-9 \times 3}{25 \times 3} + \frac{-1 \times 1}{75 \times 1} \\[1em] = \frac{-27}{75} + \frac{-1}{75} \\[1em] = \frac{-27 + (-1)}{75} \\[1em] = \frac{-28}{75}\end{aligned}$$

Hence, $\frac{-9}{25} + \frac{1}{-75} = \frac{-28}{75}$

$\frac{13}{-16} = \frac{-13}{16}$

By Division Method,

$$\begin{array}{l|r} 2 & 16, 24 \\ \hline 2 & 8, 12 \\ \hline 2 & 4, 6 \\ \hline 2 & 2, 3 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}$$

LCM of 16 and 24 = 2 × 2 × 2 × 2 × 3 = 48

$$\begin{aligned}\Rightarrow \frac{-13 \times 3}{16 \times 3} + \frac{-11 \times 2}{24 \times 2} \\[1em] = \frac{-39}{48} + \frac{-22}{48} \\[1em] = \frac{-39 + (-22)}{48} \\[1em] = \frac{-61}{48} \\[1em] = -1\frac{13}{48}\end{aligned}$$

Hence, $\frac{13}{-16} + \frac{-11}{24} = -1\frac{13}{48}$

$\frac{-9}{-16} = \frac{9}{16}$

By Division Method,

$$\begin{array}{l|r} 2 & 16, 8 \\ \hline 2 & 8, 4 \\ \hline 2 & 4, 2 \\ \hline 2 & 2, 1 \\ \hline & 1, 1 \end{array}$$

LCM of 16 and 8 = 2 × 2 × 2 × 2 = 16

$$\begin{aligned}\Rightarrow \frac{9 \times 1}{16 \times 1} + \frac{-11 \times 2}{8 \times 2} \\[1em] = \frac{9}{16} + \frac{-22}{16} \\[1em] = \frac{9 + (-22)}{16} \\[1em] = \frac{-13}{16}\end{aligned}$$

Hence, $\frac{-9}{-16} + \frac{-11}{8} = \frac{-13}{16}$

Solving,

$$\begin{aligned}\Rightarrow \frac{-2}{5} + \frac{3}{5} + \frac{-1}{5} \\[1em] = \frac{-2 + 3 + (-1)}{5} \\[1em] = \frac{0}{5} \\[1em] = 0\end{aligned}$$

Hence, $\frac{-2}{5} + \frac{3}{5} + \frac{-1}{5} = 0$

Solving,

$$\begin{aligned}\Rightarrow \frac{-8}{9} + \frac{4}{9} + \frac{-2}{9} \\[1em] = \frac{-8 + 4 + (-2)}{9} \\[1em] = \frac{-6}{9} \\[1em] = \frac{-2}{3}\end{aligned}$$

Hence, $\frac{-8}{9} + \frac{4}{9} + \frac{-2}{9} = \frac{-2}{3}$

Solving,

$\frac{5}{-24} + \frac{-1}{8} + \frac{3}{16}$

By division method:

$$\begin{array}{l|r} 2 & 24, 8, 16 \\ \hline 2 & 12, 4, 8 \\ \hline 2 & 6, 2, 4 \\ \hline 2 & 3, 1, 2 \\ \hline 3 & 3, 1, 1 \\ \hline & 1, 1, 1 \end{array}$$

LCM of 24, 8 and 16 = 2 × 2 × 2 × 2 × 3 = 48

$$\begin{aligned}\frac{-5 \times 2}{24 \times 2} + \frac{-1 \times 6}{8 \times 6} + \frac{3 \times 3}{16 \times 3} \\[1em] = \frac{-10}{48} + \frac{-6}{48} + \frac{9}{48} \\[1em] = \frac{-10 + (-6) + 9}{48} \\[1em] = \frac{-7}{48}\end{aligned}$$

Hence, $\frac{5}{-24} + \frac{-1}{8} + \frac{3}{16} = \frac{-7}{48}$

$\frac{4}{-15} = \frac{-4}{15} \text{ and } \frac{-4}{-30} = \frac{4}{30}$

LCM of 6, 15 and 30 = 30

$$\begin{aligned}\Rightarrow \frac{-7 \times 5}{6 \times 5} + \frac{-4 \times 2}{15 \times 2} + \frac{4 \times 1}{30 \times 1} \\[1em] = \frac{-35}{30} + \frac{-8}{30} + \frac{4}{30} \\[1em] = \frac{-35 + (-8) + 4}{30} \\[1em] = \frac{-39}{30} \\[1em] = \frac{-13}{10} \\[1em] = -1\frac{3}{10}\end{aligned}$$

Hence, $\frac{-7}{6} + \frac{4}{-15} + \frac{-4}{-30} = -1\frac{3}{10}$

$\frac{-2}{1} + \frac{2}{5} + \frac{-2}{15}$

LCM of 1, 5 and 15 = 15

$$\begin{aligned}\Rightarrow \frac{-2 \times 15}{1 \times 15} + \frac{2 \times 3}{5 \times 3} + \frac{-2 \times 1}{15 \times 1} \\[1em] = \frac{-30}{15} + \frac{6}{15} + \frac{-2}{15} \\[1em] = \frac{-30 + 6 + (-2)}{15} \\[1em] = \frac{-26}{15} \\[1em] = -1\frac{11}{15}\end{aligned}$$

Hence, $-2 + \frac{2}{5} + \frac{-2}{15} = -1\frac{11}{15}$

Solving,

By division method,

$$\begin{array}{l|r} 2 & 12, 16, 8 \\ \hline 2 & 6, 8, 4 \\ \hline 2 & 3, 4, 2 \\ \hline 2 & 3, 2, 1 \\ \hline 3 & 3, 1, 1 \\ \hline & 1, 1, 1 \end{array}$$

LCM of 12, 16 and 8 = 2 × 2 × 2 × 2 × 3 = 48

$$\begin{aligned}\Rightarrow \frac{-11 \times 4}{12 \times 4} + \frac{5 \times 3}{16 \times 3} + \frac{-3 \times 6}{8 \times 6} \\[1em] = \frac{-44}{48} + \frac{15}{48} + \frac{-18}{48} \\[1em] = \frac{-44 + 15 + (-18)}{48} \\[1em] = \frac{-47}{48}\end{aligned}$$

Hence, $\frac{-11}{12} + \frac{5}{16} + \frac{-3}{8} = \frac{-47}{48}$

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

By Division Method,

$$\begin{array}{l|r} 2 & 18, 9, 3 \\ \hline 3 & 9, 9, 3 \\ \hline 3 & 3, 3, 1 \\ \hline & 1, 1, 1 \end{array}$$

LCM of 18, 9 and 3 = 2 × 3 × 3 = 18

$$\begin{aligned}\Rightarrow \frac{-11 \times 1}{18 \times 1} + \frac{-3 \times 2}{9 \times 2} + \frac{-2 \times 6}{3 \times 6} \\[1em] = \frac{-11}{18} + \frac{-6}{18} + \frac{-12}{18} \\[1em] = \frac{-11 + (-6) + (-12)}{18} \\[1em] = \frac{-29}{18} \\[1em] = -1\frac{11}{18}\end{aligned}$$

Hence, $-\frac{11}{18} + \frac{-3}{9} + \frac{2}{-3} = -1\frac{11}{18}$

By Division Method,

$$\begin{array}{l|r} 2 & 4, 3, 6 \\ \hline 2 & 2, 3, 3 \\ \hline 3 & 1, 3, 3 \\ \hline & 1, 1, 1 \end{array}$$

LCM of 4, 3 and 6 = 2 × 2 × 3 = 12

$$\begin{aligned}\Rightarrow \frac{-9 \times 3}{4 \times 3} + \frac{13 \times 4}{3 \times 4} + \frac{25 \times 2}{6 \times 2} \\[1em] = \frac{-27}{12} + \frac{52}{12} + \frac{50}{12} \\[1em] = \frac{-27 + 52 + 50}{12} \\[1em] = \frac{75}{12} \\[1em] = \frac{25}{4} \\[1em] = 6\frac{1}{4}\end{aligned}$$

Hence, $\frac{-9}{4} + \frac{13}{3} + \frac{25}{6} = 6\frac{1}{4}$

$\frac{5}{-8} = \frac{-5}{8}\text{ and }\frac{-5}{-12} = \frac{5}{12}$

By Division Method,

$$\begin{array}{l|r} 2 & 1, 8, 12 \\ \hline 2 & 1, 4, 6 \\ \hline 2 & 1, 2, 3 \\ \hline 3 & 1, 1, 3 \\ \hline & 1, 1, 1 \end{array}$$

LCM of 1, 8 and 12 = 2 × 2 × 2 × 3 = 24

$$\begin{aligned}\Rightarrow \frac{-5 \times 24}{1 \times 24} + \frac{-5 \times 3}{8 \times 3} + \frac{5 \times 2}{12 \times 2} \\[1em] = \frac{-120}{24} + \frac{-15}{24} + \frac{10}{24} \\[1em] = \frac{-120 + (-15) + 10}{24} \\[1em] = \frac{-125}{24} \\[1em] = -5\frac{5}{24}\end{aligned}$$

Hence, $-5 + \frac{5}{-8} + \frac{-5}{-12} = -5\frac{5}{24}$

By Division Method,

$$\begin{array}{l|r} 2 & 3, 2, 1 \\ \hline 3 & 3, 1, 1 \\ \hline & 1, 1, 1 \end{array}$$

LCM of 3, 2 and 1 = 2 × 3 = 6

$$\begin{aligned}\Rightarrow \frac{-2 \times 2}{3 \times 2} + \frac{5 \times 3}{2 \times 3} + \frac{2 \times 6}{1 \times 6} \\[1em] = \frac{-4}{6} + \frac{15}{6} + \frac{12}{6} \\[1em] = \frac{-4 + 15 + 12}{6} \\[1em] = \frac{23}{6} \\[1em] = 3\frac{5}{6}\end{aligned}$$

Hence, $-\frac{2}{3} + \frac{5}{2} + 2 = 3\frac{5}{6}$

By Division Method,

$$\begin{array}{l|r} 2 & 1, 4, 8 \\ \hline 2 & 1, 2, 4 \\ \hline 2 & 1, 1, 2 \\ \hline & 1, 1, 1 \end{array}$$

LCM of 1, 4 and 8 = 2 × 2 × 2 = 8

$$\begin{aligned}\Rightarrow \frac{5 \times 8}{1 \times 8} + \frac{-3 \times 2}{4 \times 2} + \frac{-5 \times 1}{8 \times 1} \\[1em] = \frac{40}{8} + \frac{-6}{8} + \frac{-5}{8} \\[1em] = \frac{40 + (-6) + (-5)}{8} \\[1em] = \frac{29}{8} \\[1em] = 3\frac{5}{8}\end{aligned}$$

Hence, $5 + \frac{-3}{4} + \frac{-5}{8} = 3\frac{5}{8}$

Solving,

$$\begin{aligned}\Rightarrow \frac{5}{9} - \frac{2}{9} \\[1em] = \frac{5 - 2}{9} \\[1em] = \frac{3}{9} \\[1em] = \frac{1}{3}\end{aligned}$$

Hence, subtracting $\frac{2}{9}$ from $\frac{5}{9}$ gives $\frac{1}{3}$.

$\frac{-3}{-11} = \frac{3}{11}$

Solving,

$$\begin{aligned}\Rightarrow \frac{3}{11} - \frac{-6}{11} \\[1em] = \frac{3 - (-6)}{11} \\[1em] = \frac{3 + 6}{11} \\[1em] = \frac{9}{11}\end{aligned}$$

Hence, subtracting $\frac{-6}{11}$ from $\frac{-3}{-11}$ gives $\frac{9}{11}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{-8}{15} - \frac{-2}{15} \\[1em] = \frac{-8 - (-2)}{15} \\[1em] = \frac{-8 + 2}{15} \\[1em] = \frac{-6}{15} \\[1em] = \frac{-2}{5}\end{aligned}$$

Hence, subtracting $\frac{-2}{15}$ from $\frac{-8}{15}$ gives $\frac{-2}{5}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{-5}{18} - \frac{11}{18} \\[1em] = \frac{-5 - 11}{18} \\[1em] = \frac{-16}{18} \\[1em] = \frac{-8}{9}\end{aligned}$$

Hence, subtracting $\frac{11}{18}$ from $\frac{-5}{18}$ gives $\frac{-8}{9}$.

Solving,

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

LCM of 1 and 11 = 11

$$\begin{aligned}= \frac{-2 \times 11}{1 \times 11} + \frac{4 \times 1}{11 \times 1} \\[1em] = \frac{-22}{11} + \frac{4}{11} \\[1em] = \frac{-22 + 4}{11} \\[1em] = \frac{-18}{11} \\[1em] = -1\frac{7}{11}\end{aligned}$$

Hence, subtracting $\frac{-4}{11}$ from $-2$ gives $-1\frac{7}{11}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{1}{5} - \left(-\frac{3}{10}\right) \\[1em] = \frac{1}{5} + \frac{3}{10}\end{aligned}$$

By Division Method,

$$\begin{array}{l|r} 2 & 5, 10 \\ \hline 5 & 5, 5 \\ \hline & 1, 1 \end{array}$$

LCM of 5 and 10 = 2 × 5 = 10

$$\begin{aligned}= \frac{1 \times 2}{5 \times 2} + \frac{3 \times 1}{10 \times 1} \\[1em] = \frac{2}{10} + \frac{3}{10} \\[1em] = \frac{2 + 3}{10} \\[1em] = \frac{5}{10} \\[1em] = \frac{1}{2}\end{aligned}$$

Hence, the difference = $\frac{1}{2}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{-8}{5} - \frac{-6}{25} \\[1em] = \frac{-8}{5} + \frac{6}{25}\end{aligned}$$

By Division Method,

$$\begin{array}{l|r} 5 & 5, 25 \\ \hline 5 & 1, 5 \\ \hline & 1, 1 \end{array}$$

LCM of 5 and 25 = 5 × 5 = 25

$$\begin{aligned}= \frac{-8 \times 5}{5 \times 5} + \frac{6 \times 1}{25 \times 1} \\[1em] = \frac{-40}{25} + \frac{6}{25} \\[1em] = \frac{-40 + 6}{25} \\[1em] = \frac{-34}{25} \\[1em] = -1\frac{9}{25}\end{aligned}$$

Hence, the difference = $-1\frac{9}{25}$.

Solving,

$$\begin{aligned}\Rightarrow -2 - \frac{-7}{4} \\[1em] = \frac{-2}{1} + \frac{7}{4}\end{aligned}$$

By Division Method,

$$\begin{array}{l|r} 2 & 1, 4 \\ \hline 2 & 1, 2 \\ \hline & 1, 1 \end{array}$$

LCM of 1 and 4 = 2 × 2 = 4

$$\begin{aligned}= \frac{-2 \times 4}{1 \times 4} + \frac{7 \times 1}{4 \times 1} \\[1em] = \frac{-8}{4} + \frac{7}{4} \\[1em] = \frac{-8 + 7}{4} \\[1em] = \frac{-1}{4}\end{aligned}$$

Hence, the difference = $\frac{-1}{4}$.

Solving,

$$\begin{aligned}\Rightarrow 1 - \frac{-16}{21} \\[1em] = \frac{1}{1} + \frac{16}{21}\end{aligned}$$

By Division Method,

$$\begin{array}{l|r} 3 & 1, 21 \\ \hline 7 & 1, 7 \\ \hline & 1, 1 \end{array}$$

LCM of 1 and 21 = 3 × 7 = 21

$$\begin{aligned}= \frac{1 \times 21}{1 \times 21} + \frac{16 \times 1}{21 \times 1} \\[1em] = \frac{21}{21} + \frac{16}{21} \\[1em] = \frac{21 + 16}{21} \\[1em] = \frac{37}{21} \\[1em] = 1\frac{16}{21}\end{aligned}$$

Hence, the difference = $1\frac{16}{21}$.

Solving,

$$\begin{aligned}\Rightarrow 0 - \frac{-8}{15} \\[1em] = 0 + \frac{8}{15} \\[1em] = \frac{8}{15}\end{aligned}$$

Hence, the difference = $\frac{8}{15}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{-3}{8} - 0 \\[1em] = \frac{-3}{8}\end{aligned}$$

Hence, the difference = $\frac{-3}{8}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{-3}{10} - (-2) \\[1em] = \frac{-3}{10} + \frac{2}{1}\end{aligned}$$

By Division Method,

$$\begin{array}{l|r} 2 & 10, 1 \\ \hline 5 & 5, 1 \\ \hline & 1, 1 \end{array}$$

LCM of 10 and 1 = 2 × 5 = 10

$$\begin{aligned}= \frac{-3 \times 1}{10 \times 1} + \frac{2 \times 10}{1 \times 10} \\[1em] = \frac{-3}{10} + \frac{20}{10} \\[1em] = \frac{-3 + 20}{10} \\[1em] = \frac{17}{10} \\[1em] = 1\frac{7}{10}\end{aligned}$$

Hence, the difference = $1\frac{7}{10}$.

By Division Method,

$$\begin{array}{l|r} 2 & 16, 8 \\ \hline 2 & 8, 4 \\ \hline 2 & 4, 2 \\ \hline 2 & 2, 1 \\ \hline & 1, 1 \end{array}$$

LCM of 16 and 8 = 2 × 2 × 2 × 2 = 16

Solving,

$$\begin{aligned}\Rightarrow \frac{-5}{16} - \frac{5}{8} \\[1em] = \frac{-5 \times 1}{16 \times 1} - \frac{5 \times 2}{8 \times 2} \\[1em] = \frac{-5}{16} - \frac{10}{16} \\[1em] = \frac{-5 - 10}{16} \\[1em] = \frac{-15}{16}\end{aligned}$$

Hence, the difference = $\frac{-15}{16}$.

Solving,

$$\begin{aligned}\Rightarrow -\frac{3}{13} - 4 \\[1em] = \frac{-3}{13} - \frac{4}{1}\end{aligned}$$

LCM of 13 and 1 = 13

$$\begin{aligned}= \frac{-3 \times 1}{13 \times 1} - \frac{4 \times 13}{1 \times 13} \\[1em] = \frac{-3}{13} - \frac{52}{13} \\[1em] = \frac{-3 - 52}{13} \\[1em] = \frac{-55}{13} \\[1em] = -4\frac{3}{13}\end{aligned}$$

Hence, the difference = $-4\frac{3}{13}$.

Let the other number be x.

According to the question,

$\text{x} + \frac{3}{8} = \frac{11}{24}$

$\text{x} = \frac{11}{24} - \frac{3}{8}$

By Division Method,

$$\begin{array}{l|r} 2 & 24, 8 \\ \hline 2 & 12, 4 \\ \hline 2 & 6, 2 \\ \hline 3 & 3, 1 \\ \hline & 1, 1 \end{array}$$

LCM of 24 and 8 = 2 × 2 × 2 × 3 = 24

$$\begin{aligned}\text{x} = \frac{11 \times 1}{24 \times 1} - \frac{3 \times 3}{8 \times 3} \\[1em] = \frac{11}{24} - \frac{9}{24} \\[1em] = \frac{11 - 9}{24} \\[1em] = \frac{2}{24} \\[1em] = \frac{1}{12}\end{aligned}$$

Hence, the other rational number is $\frac{1}{12}$.

Let the other number be x.

According to the question,

$\text{x} + \frac{13}{24} = \frac{-7}{12}$

$\text{x} = \frac{-7}{12} - \frac{13}{24}$

By Division Method,

$$\begin{array}{l|r} 2 & 12, 24 \\ \hline 2 & 6, 12 \\ \hline 2 & 3, 6 \\ \hline 3 & 3, 3 \\ \hline & 1, 1 \end{array}$$

LCM of 12 and 24 = 2 × 2 × 2 × 3 = 24

$$\begin{aligned}\text{x} = \frac{-7 \times 2}{12 \times 2} - \frac{13 \times 1}{24 \times 1} \\[1em] = \frac{-14}{24} - \frac{13}{24} \\[1em] = \frac{-14 - 13}{24} \\[1em] = \frac{-27}{24} \\[1em] = \frac{-9}{8} \\[1em] = -1\frac{1}{8}\end{aligned}$$

Hence, the other rational number is $-1\frac{1}{8}$.

Let the other number be x.

According to the question,

$$\begin{aligned}\text{x} + \frac{-13}{12} = -4 \\[1em] \text{x} = -4 - \left(-\frac{13}{12}\right) \\[1em] = \frac{-4}{1} + \frac{13}{12}\end{aligned}$$

By Division Method,

$$\begin{array}{l|r} 2 & 1, 12 \\ \hline 2 & 1, 6 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}$$

LCM of 1 and 12 = 2 × 2 × 3 = 12

$$\begin{aligned}\text{x} = \frac{-4 \times 12}{1 \times 12} + \frac{13 \times 1}{12 \times 1} \\[1em] = \frac{-48}{12} + \frac{13}{12} \\[1em] = \frac{-48 + 13}{12} \\[1em] = \frac{-35}{12} \\[1em] = -2\frac{11}{12}\end{aligned}$$

Hence, the other rational number is $-2\frac{11}{12}$.

Let x be added to $-\frac{3}{32}$.

$$\begin{aligned}\Rightarrow -\frac{3}{32} + x = \frac{53}{96} \\[1em] \Rightarrow x = \frac{53}{96} - \left(-\frac{3}{32}\right) \\[1em] \Rightarrow x = \frac{53}{96} + \frac{3}{32}\end{aligned}$$

By Division Method,

$$\begin{array}{l|r} 2 & 96, 32 \\ \hline 2 & 48, 16 \\ \hline 2 & 24, 8 \\ \hline 2 & 12, 4 \\ \hline 2 & 6, 2 \\ \hline 3 & 3, 1 \\ \hline & 1, 1 \end{array}$$

LCM of 96 and 32 = 2 × 2 × 2 × 2 × 2 × 3 = 96

$$\begin{aligned}\Rightarrow x = \frac{53 \times 1}{96 \times 1} + \frac{3 \times 3}{32 \times 3} \\[1em] \Rightarrow x = \frac{53}{96} + \frac{9}{96} \\[1em] \Rightarrow x = \frac{53 + 9}{96} \\[1em] \Rightarrow x = \frac{62}{96} \\[1em] \Rightarrow x = \frac{31}{48}\end{aligned}$$

Hence, $\frac{31}{48}$ should be added to $-\frac{3}{32}$ to get $\frac{53}{96}$.

$2\frac{9}{20} = \frac{49}{20}$

Let x be added to $\frac{-3}{20}$ to get $2\frac{9}{20}$.

$$\begin{aligned}\Rightarrow \frac{-3}{20} + x = \frac{49}{20} \\[1em] \Rightarrow x = \frac{49}{20} - \frac{-3}{20} \\[1em] \Rightarrow x = \frac{49 - (-3)}{20} \\[1em] \Rightarrow x = \frac{49 + 3}{20} \\[1em] \Rightarrow x = \frac{52}{20} \\[1em] \Rightarrow x = \frac{13}{5} \\[1em] \Rightarrow x = 2\frac{3}{5}\end{aligned}$$

Hence, $2\frac{3}{5}$ should be added to $\frac{-3}{20}$ to get $2\frac{9}{20}$.

Let x be subtracted from $\frac{-4}{5}$.

$$\begin{aligned}\Rightarrow \frac{-4}{5} - x = 1 \\[1em] \Rightarrow x = \frac{-4}{5} - 1 \\[1em] \Rightarrow x = \frac{-4}{5} - \frac{1}{1}\end{aligned}$$

LCM of 5 and 1 = 5

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

Hence, $-1\frac{4}{5}$ should be subtracted from $\frac{-4}{5}$ to get 1.

Let the other number be x.

According to the question,

$$\begin{aligned}\Rightarrow \text{x} + (-2) = \frac{-6}{5} \\[1em] \Rightarrow \text{x} = -\frac{6}{5} - (-2) \\[1em] = \frac{-6}{5} + \frac{2}{1}\end{aligned}$$

LCM of 5 and 1 = 5

$$\begin{aligned}\Rightarrow \text{x} = \frac{-6 \times 1}{5 \times 1} + \frac{2 \times 5}{1 \times 5} \\[1em] = \frac{-6}{5} + \frac{10}{5} \\[1em] = \frac{-6 + 10}{5} \\[1em] = \frac{4}{5}\end{aligned}$$

Hence, the other number is $\frac{4}{5}$.

Let x be added to $\frac{-7}{12}$ to get $\frac{3}{8}$.

$$\begin{aligned}\Rightarrow \frac{-7}{12} + x = \frac{3}{8} \\[1em] \Rightarrow x = \frac{3}{8} - \frac{-7}{12} \\[1em] \Rightarrow x = \frac{3}{8} + \frac{7}{12}\end{aligned}$$

By Division Method,

$$\begin{array}{l|r} 2 & 8, 12 \\ \hline 2 & 4, 6 \\ \hline 2 & 2, 3 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}$$

LCM of 8 and 12 = 2 × 2 × 2 × 3 = 24

$$\begin{aligned}\Rightarrow x = \frac{3 \times 3}{8 \times 3} + \frac{7 \times 2}{12 \times 2} \\[1em] \Rightarrow x = \frac{9}{24} + \frac{14}{24} \\[1em] \Rightarrow x = \frac{9 + 14}{24} \\[1em] \Rightarrow x = \frac{23}{24}\end{aligned}$$

Hence, $\frac{23}{24}$ should be added to $\frac{-7}{12}$ to get $\frac{3}{8}$.

Let x be subtracted from $\frac{5}{9}$ to get $\frac{9}{5}$.

$$\begin{aligned}\frac{5}{9} - x = \frac{9}{5} \\[1em] \Rightarrow x = \frac{5}{9} - \frac{9}{5}\end{aligned}$$

By Division Method,

$$\begin{array}{l|r} 3 & 9, 5 \\ \hline 3 & 3, 5 \\ \hline 5 & 1, 5 \\ \hline & 1, 1 \end{array}$$

LCM of 9 and 5 = 3 × 3 × 5 = 45

$$\begin{aligned}\Rightarrow x = \frac{5 \times 5}{9 \times 5} - \frac{9 \times 9}{5 \times 9} \\[1em] \Rightarrow x = \frac{25}{45} - \frac{81}{45} \\[1em] \Rightarrow x = \frac{25 - 81}{45} \\[1em] \Rightarrow x = \frac{-56}{45} \\[1em] \Rightarrow x = -1\frac{11}{45}\end{aligned}$$

Hence, $-1\frac{11}{45}$ should be subtracted from $\frac{5}{9}$ to get $\frac{9}{5}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{5}{4} \times \frac{3}{7} \\[1em] = \frac{5 \times 3}{4 \times 7} \\[1em] = \frac{15}{28}\end{aligned}$$

Hence, $\frac{5}{4} \times \frac{3}{7} = \frac{15}{28}$

Solving,

$$\begin{aligned}\Rightarrow \frac{2}{3} \times -\frac{6}{7} \\[1em] = \frac{2 \times (-6)}{3 \times 7} \\[1em] = \frac{-12}{21} \\[1em] = \frac{-4}{7}\end{aligned}$$

Hence, $\frac{2}{3} \times -\frac{6}{7} = \frac{-4}{7}$

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{-12}{5}\right) \times \left(\frac{10}{-3}\right) \\[1em] = \frac{-12 \times 10}{5 \times (-3)} \\[1em] = \frac{-120}{-15} \\[1em] = 8\end{aligned}$$

Hence, $\left(\frac{-12}{5}\right) \times \left(\frac{10}{-3}\right) = 8$

Solving,

$$\begin{aligned}\Rightarrow -\frac{45}{39} \times \frac{-13}{15} \\[1em] = \frac{-45 \times (-13)}{39 \times 15} \\[1em] = \frac{585}{585} \\[1em] = 1\end{aligned}$$

Hence, $-\frac{45}{39} \times \frac{-13}{15} = 1$

Solving,

$$\begin{aligned}\Rightarrow 3\frac{1}{8} \times \left(-2\frac{2}{5}\right) \\[1em] = \frac{25}{8} \times \left(-\frac{12}{5}\right) \\[1em] = \frac{25 \times (-12)}{8 \times 5} \\[1em] = \frac{-300}{40} \\[1em] = \frac{-15}{2} \\[1em] = -7\frac{1}{2}\end{aligned}$$

Hence, $3\frac{1}{8} \times \left(-2\frac{2}{5}\right) = -7\frac{1}{2}$

Solving,

$$\begin{aligned}\Rightarrow 2\frac{14}{25} \times \left(\frac{-5}{16}\right) \\[1em] = \frac{64}{25} \times \left(\frac{-5}{16}\right) \\[1em] = \frac{64 \times (-5)}{25 \times 16} \\[1em] = \frac{-320}{400} \\[1em] = \frac{-4}{5}\end{aligned}$$

Hence, $2\frac{14}{25} \times \left(\frac{-5}{16}\right) = \frac{-4}{5}$

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{-8}{9}\right) \times \left(\frac{-3}{16}\right) \\[1em] = \frac{-8 \times (-3)}{9 \times 16} \\[1em] = \frac{24}{144} \\[1em] = \frac{1}{6}\end{aligned}$$

Hence, $\left(\frac{-8}{9}\right) \times \left(\frac{-3}{16}\right) = \frac{1}{6}$

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{5}{-27}\right) \times \left(\frac{-9}{20}\right) \\[1em] = \left(\frac{-5}{27}\right) \times \left(\frac{-9}{20}\right) \\[1em] = \frac{-5 \times (-9)}{27 \times 20} \\[1em] = \frac{45}{540} \\[1em] = \frac{1}{12}\end{aligned}$$

Hence, $\left(\frac{5}{-27}\right) \times \left(\frac{-9}{20}\right) = \frac{1}{12}$

Solving,

$$\begin{aligned}\Rightarrow \frac{3}{25} \times \frac{4}{5} \\[1em] = \frac{3 \times 4}{25 \times 5} \\[1em] = \frac{12}{125}\end{aligned}$$

Hence, $\frac{3}{25} \times \frac{4}{5} = \frac{12}{125}$

Solving,

$$\begin{aligned}\Rightarrow 1\frac{1}{8} \times 10\frac{2}{3} \\[1em] = \frac{9}{8} \times \frac{32}{3} \\[1em] = \frac{9 \times 32}{8 \times 3} \\[1em] = \frac{288}{24} \\[1em] = 12\end{aligned}$$

Hence, $1\frac{1}{8} \times 10\frac{2}{3} = 12$

Solving,

$$\begin{aligned}\Rightarrow 6\frac{2}{3} \times \frac{-3}{8} \\[1em] = \frac{20}{3} \times \frac{-3}{8} \\[1em] = \frac{20 \times (-3)}{3 \times 8} \\[1em] = \frac{-60}{24} \\[1em] = \frac{-5}{2} \\[1em] = -2\frac{1}{2}\end{aligned}$$

Hence, $6\frac{2}{3} \times \frac{-3}{8} = -2\frac{1}{2}$

Solving,

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

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

Solving,

$$\begin{aligned}\Rightarrow 1\frac{1}{6} \times 18 \\[1em] = \frac{7}{6} \times \frac{18}{1} \\[1em] = \frac{7 \times 18}{6 \times 1} \\[1em] = \frac{126}{6} \\[1em] = 21\end{aligned}$$

Hence, $1\frac{1}{6} \times 18 = 21$

Solving,

$$\begin{aligned}\Rightarrow 2\frac{1}{14} \times (-7) \\[1em] = \frac{29}{14} \times \frac{-7}{1} \\[1em] = \frac{29 \times (-7)}{14 \times 1} \\[1em] = \frac{-203}{14} \\[1em] = \frac{-29}{2} \\[1em] = -14\frac{1}{2}\end{aligned}$$

Hence, $2\frac{1}{14} \times (-7) = -14\frac{1}{2}$

Solving,

$$\begin{aligned}\Rightarrow 5\frac{1}{8} \times (-16) \\[1em] = \frac{41}{8} \times \frac{-16}{1} \\[1em] = \frac{41 \times (-16)}{8 \times 1} \\[1em] = \frac{-656}{8} \\[1em] = -82\end{aligned}$$

Hence, $5\frac{1}{8} \times (-16) = -82$

Solving,

$$\begin{aligned}\Rightarrow 35 \times \frac{-18}{25} \\[1em] = \frac{35}{1} \times \frac{-18}{25} \\[1em] = \frac{35 \times (-18)}{1 \times 25} \\[1em] = \frac{-630}{25} \\[1em] = \frac{-126}{5} \\[1em] = -25\frac{1}{5}\end{aligned}$$

Hence, $35 \times \frac{-18}{25} = -25\frac{1}{5}$

Solving,

$$\begin{aligned}\Rightarrow 6\frac{2}{3} \times -\frac{3}{8} \\[1em] = \frac{20}{3} \times \frac{-3}{8} \\[1em] = \frac{20 \times (-3)}{3 \times 8} \\[1em] = \frac{-60}{24} \\[1em] = \frac{-5}{2} \\[1em] = -2\frac{1}{2}\end{aligned}$$

Hence, $6\frac{2}{3} \times -\frac{3}{8} = -2\frac{1}{2}$

Solving,

$$\begin{aligned}\Rightarrow 3\frac{3}{5} \times (-10) \\[1em] = \frac{18}{5} \times \frac{-10}{1} \\[1em] = \frac{18 \times (-10)}{5 \times 1} \\[1em] = \frac{-180}{5} \\[1em] = -36\end{aligned}$$

Hence, $3\frac{3}{5} \times (-10) = -36$

Solving,

$$\begin{aligned}\Rightarrow \frac{27}{28} \times (-14) \\[1em] = \frac{27}{28} \times \frac{-14}{1} \\[1em] = \frac{27 \times (-14)}{28 \times 1} \\[1em] = \frac{-378}{28} \\[1em] = \frac{-27}{2} \\[1em] = -13\frac{1}{2}\end{aligned}$$

Hence, $\frac{27}{28} \times (-14) = -13\frac{1}{2}$

Solving,

$$\begin{aligned}\Rightarrow -24 \times \frac{5}{16} \\[1em] = \frac{-24}{1} \times \frac{5}{16} \\[1em] = \frac{-24 \times 5}{1 \times 16} \\[1em] = \frac{-120}{16} \\[1em] = \frac{-15}{2} \\[1em] = -7\frac{1}{2}\end{aligned}$$

Hence, $-24 \times \frac{5}{16} = -7\frac{1}{2}$

Solving,

$$\begin{aligned}\Rightarrow \left(-6 \times \frac{5}{18}\right) - \left(-4\frac{2}{9}\right) \\[1em] = \left(\frac{-6 \times 5}{1 \times 18}\right) - \left(\frac{-38}{9}\right) \\[1em] = \frac{-30}{18} + \frac{38}{9} \\[1em] = \frac{-5}{3} + \frac{38}{9}\end{aligned}$$

LCM of 3 and 9 = 9

$$\begin{aligned}= \frac{-5 \times 3}{3 \times 3} + \frac{38 \times 1}{9 \times 1} \\[1em] = \frac{-15}{9} + \frac{38}{9} \\[1em] = \frac{-15 + 38}{9} \\[1em] = \frac{23}{9} \\[1em] = 2\frac{5}{9}\end{aligned}$$

Hence, $\left(-6 \times \frac{5}{18}\right) - \left(-4\frac{2}{9}\right) = 2\frac{5}{9}$

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{7}{8} \times \frac{8}{7}\right) + \left(\frac{-5}{9} \times \frac{6}{-25}\right) \\[1em] = \frac{7 \times 8}{8 \times 7} + \frac{-5 \times 6}{9 \times (-25)} \\[1em] = \frac{56}{56} + \frac{-30}{-225} \\[1em] = 1 + \frac{2}{15}\end{aligned}$$

LCM of 1 and 15 = 15

$$\begin{aligned}= \frac{1 \times 15}{1 \times 15} + \frac{2 \times 1}{15 \times 1} \\[1em] = \frac{15}{15} + \frac{2}{15} \\[1em] = \frac{15 + 2}{15} \\[1em] = \frac{17}{15} \\[1em] = 1\frac{2}{15}\end{aligned}$$

Hence, $\left(\frac{7}{8} \times \frac{8}{7}\right) + \left(\frac{-5}{9}\right) \times \left(\frac{6}{-25}\right) = 1\frac{2}{15}$

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{11}{-9} \times \frac{21}{44}\right) + \left(\frac{-5}{9} \times \frac{63}{-100}\right) \\[1em] = \frac{11 \times 21}{-9 \times 44} + \frac{-5 \times 63}{9 \times (-100)} \\[1em] = \frac{231}{-396} + \frac{-315}{-900} \\[1em] = \frac{-7}{12} + \frac{7}{20}\end{aligned}$$

LCM of 12 and 20 = 60

$$\begin{aligned}= \frac{-7 \times 5}{12 \times 5} + \frac{7 \times 3}{20 \times 3} \\[1em] = \frac{-35}{60} + \frac{21}{60} \\[1em] = \frac{-35 + 21}{60} \\[1em] = \frac{-14}{60} \\[1em] = \frac{-7}{30}\end{aligned}$$

Hence, $\left(\frac{11}{-9} \times \frac{21}{44}\right) + \left(\frac{-5}{9}\right) \times \left(\frac{63}{-100}\right) = \frac{-7}{30}$

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{-5}{9} \times \frac{6}{-25}\right) + \left(\frac{24}{21} \times \frac{7}{8}\right) \\[1em] = \frac{-5 \times 6}{9 \times (-25)} + \frac{24 \times 7}{21 \times 8} \\[1em] = \frac{-30}{-225} + \frac{168}{168} \\[1em] = \frac{2}{15} + 1\end{aligned}$$

LCM of 15 and 1 = 15

$$\begin{aligned}= \frac{2 \times 1}{15 \times 1} + \frac{1 \times 15}{1 \times 15} \\[1em] = \frac{2}{15} + \frac{15}{15} \\[1em] = \frac{2 + 15}{15} \\[1em] = \frac{17}{15} \\[1em] = 1\frac{2}{15}\end{aligned}$$

Hence, $\left(\frac{-5}{9} \times \frac{6}{-25}\right) + \left(\frac{24}{21} \times \frac{7}{8}\right) = 1\frac{2}{15}$

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{-35}{39} \times \frac{-13}{7}\right) - \left(\frac{7}{90} \times \frac{-18}{14}\right) \\[1em] = \frac{-35 \times (-13)}{39 \times 7} - \frac{7 \times (-18)}{90 \times 14} \\[1em] = \frac{455}{273} - \frac{-126}{1260} \\[1em] = \frac{5}{3} - \left(\frac{-1}{10}\right) \\[1em] = \frac{5}{3} + \frac{1}{10}\end{aligned}$$

LCM of 3 and 10 = 30

$$\begin{aligned}= \frac{5 \times 10}{3 \times 10} + \frac{1 \times 3}{10 \times 3} \\[1em] = \frac{50}{30} + \frac{3}{30} \\[1em] = \frac{50 + 3}{30} \\[1em] = \frac{53}{30} \\[1em] = 1\frac{23}{30}\end{aligned}$$

Hence, $\left(\frac{-35}{39} \times \frac{-13}{7}\right) - \left(\frac{7}{90} \times \frac{-18}{14}\right) = 1\frac{23}{30}$

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{-4}{5} \times \frac{3}{2}\right) + \left(\frac{9}{-5} \times \frac{10}{3}\right) - \left(\frac{-3}{2} \times \frac{-1}{4}\right) \\[1em] = \frac{-4 \times 3}{5 \times 2} + \frac{9 \times 10}{-5 \times 3} - \frac{-3 \times (-1)}{2 \times 4} \\[1em] = \frac{-12}{10} + \frac{90}{-15} - \frac{3}{8} \\[1em] = \frac{-6}{5} + (-6) - \frac{3}{8}\end{aligned}$$

LCM of 5, 1 and 8 = 40

$$\begin{aligned}= \frac{-6 \times 8}{5 \times 8} + \frac{-6 \times 40}{1 \times 40} - \frac{3 \times 5}{8 \times 5} \\[1em] = \frac{-48}{40} + \frac{-240}{40} - \frac{15}{40} \\[1em] = \frac{-48 + (-240) - 15}{40} \\[1em] = \frac{-303}{40} \\[1em] = -7\frac{23}{40}\end{aligned}$$

Hence, $\left(\frac{-4}{5} \times \frac{3}{2}\right) + \left(\frac{9}{-5} \times \frac{10}{3}\right) - \left(\frac{-3}{2} \times \frac{-1}{4}\right) = -7\frac{23}{40}$

Cost of 1 m cloth = ₹ $325\frac{1}{2}$ = ₹ $\frac{651}{2}$

Cost of $3\frac{1}{2}$ m cloth = $\frac{651}{2} \times 3\frac{1}{2}$

$$\begin{aligned}= \frac{651}{2} \times \frac{7}{2} \\[1em] = \frac{651 \times 7}{2 \times 2} \\[1em] = \frac{4557}{4} \\[1em] = 1139\frac{1}{4}\end{aligned}$$

Hence, the cost of $3\frac{1}{2}$ m cloth is ₹ $1139\frac{1}{4}$.

Speed = $65\frac{1}{2}$ km per hour = $\frac{131}{2}$ km per hour

Time = $1\frac{1}{3}$ hours = $\frac{4}{3}$ hours

Distance covered = Speed × Time

$$\begin{aligned}= \frac{131}{2} \times \frac{4}{3} \\[1em] = \frac{131 \times 4}{2 \times 3} \\[1em] = \frac{524}{6} \\[1em] = \frac{262}{3} \\[1em] = 87\frac{1}{3}\end{aligned}$$

Hence, the bus will cover $87\frac{1}{3}$ km in $1\frac{1}{3}$ hours.

Solving,

$$\begin{aligned}\Rightarrow \frac{15}{28} ÷ \frac{3}{4} \\[1em] = \frac{15}{28} \times \frac{4}{3} \\[1em] = \frac{15 \times 4}{28 \times 3} \\[1em] = \frac{60}{84} \\[1em] = \frac{5}{7}\end{aligned}$$

Hence, $\frac{15}{28} ÷ \frac{3}{4} = \frac{5}{7}$

Solving,

$$\begin{aligned}\Rightarrow \frac{-20}{9} ÷ \frac{-5}{9} \\[1em] = \frac{-20}{9} \times \frac{9}{-5} \\[1em] = \frac{-20 \times 9}{9 \times (-5)} \\[1em] = \frac{-180}{-45} \\[1em] = 4\end{aligned}$$

Hence, $\frac{-20}{9} ÷ \frac{-5}{9} = 4$

Solving,

$$\begin{aligned}\Rightarrow \frac{16}{-5} ÷ \frac{-8}{7} \\[1em] = \frac{-16}{5} \times \frac{7}{-8} \\[1em] = \frac{-16 \times 7}{5 \times (-8)} \\[1em] = \frac{-112}{-40} \\[1em] = \frac{14}{5} \\[1em] = 2\frac{4}{5}\end{aligned}$$

Hence, $\frac{16}{-5} ÷ \frac{-8}{7} = 2\frac{4}{5}$

Solving,

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

Hence, $-7 ÷ \frac{-14}{5} = 2\frac{1}{2}$

Solving,

$$\begin{aligned}\Rightarrow -14 ÷ \frac{7}{-2} \\[1em] = \frac{-14}{1} \times \frac{-2}{7} \\[1em] = \frac{-14 \times (-2)}{1 \times 7} \\[1em] = \frac{28}{7} \\[1em] = 4\end{aligned}$$

Hence, $-14 ÷ \frac{7}{-2} = 4$

Solving,

$$\begin{aligned}\Rightarrow \frac{-22}{9} ÷ \frac{11}{18} \\[1em] = \frac{-22}{9} \times \frac{18}{11} \\[1em] = \frac{-22 \times 18}{9 \times 11} \\[1em] = \frac{-396}{99} \\[1em] = -4\end{aligned}$$

Hence, $\frac{-22}{9} ÷ \frac{11}{18} = -4$

Solving,

$$\begin{aligned}\Rightarrow 35 ÷ \frac{-7}{9} \\[1em] = \frac{35}{1} \times \frac{9}{-7} \\[1em] = \frac{35 \times 9}{1 \times (-7)} \\[1em] = \frac{315}{-7} \\[1em] = -45\end{aligned}$$

Hence, $35 ÷ \frac{-7}{9} = -45$

Solving,

$$\begin{aligned}\Rightarrow \frac{21}{44} ÷ -\frac{11}{9} \\[1em] = \frac{21}{44} \times \frac{9}{-11} \\[1em] = \frac{21 \times 9}{44 \times (-11)} \\[1em] = \frac{189}{-484} \\[1em] = \frac{-189}{484}\end{aligned}$$

Hence, $\frac{21}{44} ÷ -\frac{11}{9} = \frac{-189}{484}$

Solving,

$$\begin{aligned}\Rightarrow 3\frac{5}{12} + 1\frac{2}{3} \\[1em] = \frac{41}{12} + \frac{5}{3}\end{aligned}$$

LCM of 12 and 3 is 2 × 2 × 3 = 12

$$\begin{aligned}= \frac{41 \times 1}{12 \times 1} + \frac{5 \times 4}{3 \times 4} \\[1em] = \frac{41}{12} + \frac{20}{12} \\[1em] = \frac{41 + 20}{12} \\[1em] = \frac{61}{12} \\[1em] = 5\frac{1}{12}\end{aligned}$$

Hence, $3\frac{5}{12} + 1\frac{2}{3} = 5\frac{1}{12}$

Solving,

$$\begin{aligned}\Rightarrow 3\frac{5}{12} - 1\frac{2}{3} \\[1em] = \frac{41}{12} - \frac{5}{3}\end{aligned}$$

LCM of 12 and 3 = 12

$$\begin{aligned}= \frac{41 \times 1}{12 \times 1} - \frac{5 \times 4}{3 \times 4} \\[1em] = \frac{41}{12} - \frac{20}{12} \\[1em] = \frac{41 - 20}{12} \\[1em] = \frac{21}{12} \\[1em] = \frac{7}{4} \\[1em] = 1\frac{3}{4}\end{aligned}$$

Hence, $3\frac{5}{12} - 1\frac{2}{3} = 1\frac{3}{4}$

Solving,

From parts (i) and (ii):

$3\frac{5}{12} + 1\frac{2}{3} = \frac{61}{12} \text{ and } 3\frac{5}{12} - 1\frac{2}{3} = \frac{21}{12}$

Solving,

$$\begin{aligned}\Rightarrow \left(3\frac{5}{12} + 1\frac{2}{3}\right) \div \left(3\frac{5}{12} - 1\frac{2}{3}\right) \\[1em] = \frac{61}{12} ÷ \frac{21}{12} \\[1em] = \frac{61}{12} \times \frac{12}{21} \\[1em] = \frac{61 \times 12}{12 \times 21} \\[1em] = \frac{61}{21} \\[1em] = 2\frac{19}{21}\end{aligned}$$

Hence, $\left(3\frac{5}{12} + 1\frac{2}{3}\right) \div \left(3\frac{5}{12} - 1\frac{2}{3}\right) = 2\frac{19}{21}$

Product of two numbers = 14

Let the other number = x

One number = $\frac{-8}{7}$

According to question,

$$\begin{aligned}\Rightarrow x \times \frac{-8}{7} = 14\\[1em] \Rightarrow x = 14 ÷ \frac{-8}{7}\\[1em] = \frac{14}{1} \times \frac{7}{-8} \\[1em] = \frac{14 \times 7}{1 \times (-8)} \\[1em] = \frac{98}{-8} \\[1em] = \frac{-49}{4} \\[1em] = -12\frac{1}{4}\end{aligned}$$

Hence, the other number is $-12\frac{1}{4}$.

Cost of 11 pens = ₹ $24\frac{3}{4} = ₹ \frac{99}{4}$

Cost of one pen = $\frac{99}{4} ÷ 11$

$$\begin{aligned}= \frac{99}{4} \times \frac{1}{11} \\[1em] = \frac{99 \times 1}{4 \times 11} \\[1em] = \frac{99}{44} \\[1em] = \frac{9}{4} \\[1em] = 2\frac{1}{4}\end{aligned}$$

Hence, the cost of one pen is ₹ $2\frac{1}{4}$.

Cost of 6 articles = ₹ $2\frac{6}{17} = ₹ \frac{40}{17}$

Cost of each article = $\frac{40}{17} ÷ 6$

$$\begin{aligned}= \frac{40}{17} \times \frac{1}{6} \\[1em] = \frac{40 \times 1}{17 \times 6} \\[1em] = \frac{40}{102} \\[1em] = \frac{20}{51}\end{aligned}$$

Hence, the cost of each article is ₹ $\frac{20}{51}$.

Let the required number be x.

$$\begin{aligned}\Rightarrow \frac{-3}{8} \times x = \frac{-9}{16} \\[1em] \Rightarrow x = \frac{-9}{16} ÷ \frac{-3}{8} \\[1em] \Rightarrow x = \frac{-9}{16} \times \frac{8}{-3} \\[1em] \Rightarrow x = \frac{-9 \times 8}{16 \times (-3)} \\[1em] \Rightarrow x = \frac{-72}{-48} \\[1em] \Rightarrow x = \frac{3}{2} \\[1em] \Rightarrow x = 1\frac{1}{2}\end{aligned}$$

Hence, $\frac{-3}{8}$ should be multiplied by $1\frac{1}{2} \text{ to get } \frac{-9}{16}$.

Let the required number be x.

$$\begin{aligned}\Rightarrow \frac{-5}{7} ÷ x = \frac{-15}{28} \\[1em] \Rightarrow \frac{-5}{7} \times \frac{1}{x} = \frac{-15}{28} \\[1em] \Rightarrow \frac{1}{x} = \frac{-15}{28} ÷ \frac{-5}{7} \\[1em] \Rightarrow \frac{1}{x} = \frac{-15}{28} \times \frac{7}{-5} \\[1em] \Rightarrow \frac{1}{x} = \frac{-15 \times 7}{28 \times (-5)} \\[1em] \Rightarrow \frac{1}{x} = \frac{-105}{-140} \\[1em] \Rightarrow \frac{1}{x} = \frac{3}{4} \\[1em] \Rightarrow x = \frac{4}{3} \\[1em] \Rightarrow x = 1\frac{1}{3}\end{aligned}$$

Hence, $\frac{-5}{7}$ should be divided by $1\frac{1}{3}$ to get $\frac{-15}{28}$.

By Division Method,

$$\begin{array}{l|r} 3 & 15, 5 \\ \hline 5 & 5, 5 \\ \hline & 1, 1 \end{array}$$

LCM of 15 and 5 = 3 × 5 = 15

$$\begin{aligned}\Rightarrow \frac{32}{15} + \frac{8}{5} = \frac{32 \times 1}{15 \times 1} + \frac{8 \times 3}{5 \times 3} = \frac{32}{15} + \frac{24}{15} = \frac{56}{15} \\[1em] \Rightarrow \frac{32}{15} - \frac{8}{5} = \frac{32 \times 1}{15 \times 1} - \frac{8 \times 3}{5 \times 3} = \frac{32}{15} - \frac{24}{15} = \frac{8}{15}\end{aligned}$$

Solving,

$$\begin{aligned}\Rightarrow\left(\frac{32}{15} + \frac{8}{5}\right) \div \left(\frac{32}{15} - \frac{8}{5}\right) \\[1em] = \frac{56}{15} ÷ \frac{8}{15} \\[1em] = \frac{56}{15} \times \frac{15}{8} \\[1em] = \frac{56 \times 15}{15 \times 8} \\[1em] = \frac{56}{8} \\[1em] = 7\end{aligned}$$

Hence, $\left(\frac{32}{15} + \frac{8}{5}\right) \div \left(\frac{32}{15} - \frac{8}{5}\right) = 7$

Length of rope = $21\frac{5}{7} \text{ m} = \frac{152}{7}$ m

Length of each piece = $\frac{152}{7} ÷ 7$

$$\begin{aligned}= \frac{152}{7} \times \frac{1}{7} \\[1em] = \frac{152 \times 1}{7 \times 7} \\[1em] = \frac{152}{49} \\[1em] = 3\frac{5}{49}\end{aligned}$$

Hence, the length of each piece is $3\frac{5}{49}$ m.

By Division Method,

$$\begin{array}{l|r} 2 & 3, 4 \\ \hline 2 & 3, 2 \\ \hline 3 & 3, 1 \\ \hline & 1, 1 \end{array}$$

LCM of 3 and 4 = 2 × 2 × 3 = 12

Solving,

$$\begin{aligned}\Rightarrow \frac{-2 \times 4}{3 \times 4} + \frac{3 \times 3}{4 \times 3} \\[1em] = \frac{-8}{12} + \frac{9}{12} \\[1em] = \frac{-8 + 9}{12} \\[1em] = \frac{1}{12}\end{aligned}$$

Hence, $\frac{-2}{3} + \frac{3}{4} = \frac{1}{12}$

By Division Method,

$$\begin{array}{l|r} 2 & 27, 18 \\ \hline 3 & 27, 9 \\ \hline 3 & 9, 3 \\ \hline 3 & 3, 1 \\ \hline & 1, 1 \end{array}$$

LCM of 27 and 18 is 2 × 3 × 3 × 3 = 54

Solving,

$$\begin{aligned}\Rightarrow \frac{7}{-27} + \frac{11}{18} \\[1em] = \frac{-7}{27} + \frac{11}{18} \\[1em] = \frac{-7 \times 2}{27 \times 2} + \frac{11 \times 3}{18 \times 3} \\[1em] = \frac{-14}{54} + \frac{33}{54} \\[1em] = \frac{-14 + 33}{54} \\[1em] = \frac{19}{54}\end{aligned}$$

Hence, $\frac{7}{-27} + \frac{11}{18} = \frac{19}{54}$

By Division Method,

$$\begin{array}{l|r} 2 & 8, 12 \\ \hline 2 & 4, 6 \\ \hline 2 & 2, 3 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}$$

LCM of 8 and 12 is 2 × 2 × 2 × 3 = 24

Solving,

$$\begin{aligned}\Rightarrow \frac{-3 \times 3}{8 \times 3} + \frac{-5 \times 2}{12 \times 2} \\[1em] = \frac{-9}{24} + \frac{-10}{24} \\[1em] = \frac{-9 + (-10)}{24} \\[1em] = \frac{-19}{24}\end{aligned}$$

Hence, $\frac{-3}{8} + \frac{-5}{12} = \frac{-19}{24}$

By Division Method,

$$\begin{array}{l|r} 2 & 16, 12 \\ \hline 2 & 8, 6 \\ \hline 2 & 4, 3 \\ \hline 2 & 2, 3 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}$$

LCM of 16 and 12 is 2 × 2 × 2 × 2 × 3 = 48

Solving,

$$\begin{aligned}\Rightarrow \frac{9}{-16} + \frac{-5}{-12} \\[1em] = \frac{-9}{16} + \frac{5}{12} \\[1em] = \frac{-9 \times 3}{16 \times 3} + \frac{5 \times 4}{12 \times 4} \\[1em] = \frac{-27}{48} + \frac{20}{48} \\[1em] = \frac{-27 + 20}{48} \\[1em] = \frac{-7}{48}\end{aligned}$$

Hence, $\frac{9}{-16} + \frac{-5}{-12} = \frac{-7}{48}$

Solving,

By Division Method,

$$\begin{array}{l|r} 2 & 9, 12, 18 \\ \hline 2 & 9, 6, 9 \\ \hline 3 & 9, 3, 9 \\ \hline 3 & 3, 1, 3 \\ \hline & 1, 1, 1 \end{array}$$

LCM of 9, 12 and 18 is 2 × 2 × 3 × 3 = 36

$$\begin{aligned}\Rightarrow \frac{-5 \times 4}{9 \times 4} + \frac{-7 \times 3}{12 \times 3} + \frac{11 \times 2}{18 \times 2} \\[1em] = \frac{-20}{36} + \frac{-21}{36} + \frac{22}{36} \\[1em] = \frac{-20 + (-21) + 22}{36} \\[1em] = \frac{-19}{36}\end{aligned}$$

Hence, $\frac{-5}{9} + \frac{-7}{12} + \frac{11}{18} = \frac{-19}{36}$

By Division Method,

$$\begin{array}{l|r} 2 & 26, 39 \\ \hline 3 & 13, 39 \\ \hline 13 & 13, 13 \\ \hline & 1, 1 \end{array}$$

LCM of 26 and 39 is 2 × 3 × 13 = 78

Solving,

$$\begin{aligned}\Rightarrow \frac{7}{-26} + \frac{16}{39} \\[1em] = \frac{-7}{26} + \frac{16}{39} \\[1em] = \frac{-7 \times 3}{26 \times 3} + \frac{16 \times 2}{39 \times 2} \\[1em] = \frac{-21}{78} + \frac{32}{78} \\[1em] = \frac{-21 + 32}{78} \\[1em] = \frac{11}{78}\end{aligned}$$

Hence, $\frac{7}{-26} + \frac{16}{39} = \frac{11}{78}$

By Division Method,

$$\begin{array}{l|r} 3 & 3, 7 \\ \hline 7 & 1, 7 \\ \hline & 1, 1 \end{array}$$

LCM of 3 and 7 is 3 × 7 = 21

Solving,

$$\begin{aligned}\Rightarrow -\frac{2}{3} - \left(\frac{-5}{7}\right) \\[1em] = -\frac{2}{3} + \frac{5}{7} \\[1em] = \frac{-2 \times 7}{3 \times 7} + \frac{5 \times 3}{7 \times 3} \\[1em] = \frac{-14}{21} + \frac{15}{21} \\[1em] = \frac{-14 + 15}{21} \\[1em] = \frac{1}{21}\end{aligned}$$

Hence, $-\frac{2}{3} - \left(\frac{-5}{7}\right) = \frac{1}{21}$

By Division Method,

$$\begin{array}{l|r} 2 & 7, 8 \\ \hline 2 & 7, 4 \\ \hline 2 & 7, 2 \\ \hline 7 & 7, 1 \\ \hline & 1, 1 \end{array}$$

LCM of 7 and 8 is 2 × 2 × 2 × 7 = 56

Solving,

$$\begin{aligned}\Rightarrow -\frac{5}{7} - \left(-\frac{3}{8}\right) \\[1em] = -\frac{5}{7} + \frac{3}{8} \\[1em] = \frac{-5 \times 8}{7 \times 8} + \frac{3 \times 7}{8 \times 7} \\[1em] = \frac{-40}{56} + \frac{21}{56} \\[1em] = \frac{-40 + 21}{56} \\[1em] = \frac{-19}{56}\end{aligned}$$

Hence, $-\frac{5}{7} - \left(-\frac{3}{8}\right) = \frac{-19}{56}$

By Division Method,

$$\begin{array}{l|r} 2 & 26, 1, 13 \\ \hline 13 & 13, 1, 13 \\ \hline & 1, 1, 1 \end{array}$$

LCM of 26, 1 and 13 is 2 × 13 = 26

Solving,

$$\begin{aligned}\Rightarrow \frac{7}{26} + 2 + \frac{-11}{13} \\[1em] = \frac{7}{26} + \frac{2}{1} + \frac{-11}{13} \\[1em] = \frac{7 \times 1}{26 \times 1} + \frac{2 \times 26}{1 \times 26} + \frac{-11 \times 2}{13 \times 2} \\[1em] = \frac{7}{26} + \frac{52}{26} + \frac{-22}{26} \\[1em] = \frac{7 + 52 + (-22)}{26} \\[1em] = \frac{37}{26} \\[1em] = 1\frac{11}{26}\end{aligned}$$

Hence, $\frac{7}{26} + 2 + \frac{-11}{13} = 1\frac{11}{26}$

By Division Method,

$$\begin{array}{l|r} 2 & 1, 3, 6 \\ \hline 3 & 1, 3, 3 \\ \hline & 1, 1, 1 \end{array}$$

LCM of 1, 3 and 6 is 2 × 3 = 6

Solving,

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

Hence, $-1 + \frac{2}{-3} + \frac{5}{6} = \frac{-5}{6}$

Let x be the other number.

$$\begin{aligned}\Rightarrow \frac{3}{16} + x = \frac{-3}{8} \\[1em] \Rightarrow x = \frac{-3}{8} - \frac{3}{16}\end{aligned}$$

By Division Method,

$$\begin{array}{l|r} 2 & 8, 16 \\ \hline 2 & 4, 8 \\ \hline 2 & 2, 4 \\ \hline 2 & 1, 2 \\ \hline & 1, 1 \end{array}$$

LCM of 8 and 16 is 2 × 2 × 2 × 2 = 16

$$\begin{aligned}\Rightarrow x = \frac{-3 \times 2}{8 \times 2} - \frac{3 \times 1}{16 \times 1} \\[1em] \Rightarrow x = \frac{-6}{16} - \frac{3}{16} \\[1em] \Rightarrow x = \frac{-6 - 3}{16} \\[1em] \Rightarrow x = \frac{-9}{16}\end{aligned}$$

Hence, the other rational number is $\frac{-9}{16}$.

Let x be the other number.

According to question,

$$\begin{aligned}\Rightarrow \frac{-52}{25} + x = -5 \\[1em] \Rightarrow x = \frac{-5}{1} - \frac{-52}{25}\end{aligned}$$

By Division Method,

$$\begin{array}{l|r} 5 & 1, 25 \\ \hline 5 & 1, 5 \\ \hline & 1, 1 \end{array}$$

LCM of 1 and 25 is 5 × 5 = 25

$$\begin{aligned}\Rightarrow x = \frac{-5 \times 25}{1 \times 25} - \frac{-52 \times 1}{25 \times 1} \\[1em] \Rightarrow x = \frac{-125}{25} - \frac{-52}{25} \\[1em] \Rightarrow x = \frac{-125 - (-52)}{25} \\[1em] \Rightarrow x = \frac{-125 + 52}{25} \\[1em] \Rightarrow x = \frac{-73}{25} \\[1em] \Rightarrow x = -2\frac{23}{25}\end{aligned}$$

Hence, the other rational number is $-2\frac{23}{25}$.

Let x be added to $-\frac{3}{16}$.

$$\begin{aligned}\Rightarrow -\frac{3}{16} + x = \frac{11}{24} \\[1em] \Rightarrow x = \frac{11}{24} - \left(-\frac{3}{16}\right) \\[1em] \Rightarrow x = \frac{11}{24} + \frac{3}{16}\end{aligned}$$

By Division Method,

$$\begin{array}{l|r} 2 & 24, 16 \\ \hline 2 & 12, 8 \\ \hline 2 & 6, 4 \\ \hline 2 & 3, 2 \\ \hline 3 & 3, 1 \\ \hline & 1, 1 \end{array}$$

LCM of 24 and 16 is 2 × 2 × 2 × 2 × 3 = 48

$$\begin{aligned}\Rightarrow x = \frac{11 \times 2}{24 \times 2} + \frac{3 \times 3}{16 \times 3} \\[1em] \Rightarrow x = \frac{22}{48} + \frac{9}{48} \\[1em] \Rightarrow x = \frac{22 + 9}{48} \\[1em] \Rightarrow x = \frac{31}{48}\end{aligned}$$

Hence, $\frac{31}{48}$ should be added to $-\frac{3}{16}$ to get $\frac{11}{24}$.

Let x be added to $-\frac{3}{5}$.

$$\begin{aligned}\Rightarrow -\frac{3}{5} + x = 2 \\[1em] \Rightarrow x = \frac{2}{1} - \left(-\frac{3}{5}\right) \\[1em] \Rightarrow x = \frac{2}{1} + \frac{3}{5}\end{aligned}$$

LCM of 1 and 5 is 5.

$$\begin{aligned}\Rightarrow x = \frac{2 \times 5}{1 \times 5} + \frac{3 \times 1}{5 \times 1} \\[1em] \Rightarrow x = \frac{10}{5} + \frac{3}{5} \\[1em] \Rightarrow x = \frac{10 + 3}{5} \\[1em] \Rightarrow x = \frac{13}{5} \\[1em] \Rightarrow x = 2\frac{3}{5}\end{aligned}$$

Hence, $2\frac{3}{5}$ should be added to $-\frac{3}{5}$ to get 2.

Let $x$ be subtracted from $-\frac{5}{12}$.

$$\begin{aligned}\Rightarrow -\frac{5}{12} - x = \frac{5}{24} \\[1em] \Rightarrow x = -\frac{5}{12} - \frac{5}{24}\end{aligned}$$

By Division Method,

$$\begin{array}{l|r} 2 & 12, 24 \\ \hline 2 & 6, 12 \\ \hline 2 & 3, 6 \\ \hline 3 & 3, 3 \\ \hline & 1, 1 \end{array}$$

LCM of 12 and 24 is 2 × 2 × 2 × 3 = 24

$$\begin{aligned}\Rightarrow x = \frac{-5 \times 2}{12 \times 2} - \frac{5 \times 1}{24 \times 1} \\[1em] \Rightarrow x = \frac{-10}{24} - \frac{5}{24} \\[1em] \Rightarrow x = \frac{-10 - 5}{24} \\[1em] \Rightarrow x = \frac{-15}{24} \\[1em] \Rightarrow x = \frac{-5}{8}\end{aligned}$$

Hence, $\frac{-5}{8}$ should be subtracted from $-\frac{5}{12}$ to get $\frac{5}{24}$.

Let x be subtracted from $\frac{5}{8}$.

$$\begin{aligned}\Rightarrow \frac{5}{8} - x = \frac{8}{5} \\[1em] \Rightarrow x = \frac{5}{8} - \frac{8}{5}\end{aligned}$$

By Division Method,

$$\begin{array}{l|r} 2 & 8, 5 \\ \hline 2 & 4, 5 \\ \hline 2 & 2, 5 \\ \hline 5 & 1, 5 \\ \hline & 1, 1 \end{array}$$

LCM of 8 and 5 is 2 × 2 × 2 × 5 = 40

$$\begin{aligned}\Rightarrow x = \frac{5 \times 5}{8 \times 5} - \frac{8 \times 8}{5 \times 8} \\[1em] \Rightarrow x = \frac{25}{40} - \frac{64}{40} \\[1em] \Rightarrow x = \frac{25 - 64}{40} \\[1em] \Rightarrow x = \frac{-39}{40}\end{aligned}$$

Hence, $\frac{-39}{40}$ should be subtracted from $\frac{5}{8}$ to get $\frac{8}{5}$.

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{7}{8} \times \frac{24}{21}\right) + \left(\frac{-5}{9} \times \frac{6}{-25}\right) \\[1em] = \frac{7 \times 24}{8 \times 21} + \frac{-5 \times 6}{9 \times (-25)} \\[1em] = \frac{168}{168} + \frac{-30}{-225} \\[1em] = 1 + \frac{2}{15}\end{aligned}$$

By Division Method,

$$\begin{array}{l|r} 3 & 1, 15 \\ \hline 5 & 1, 5 \\ \hline & 1, 1 \end{array}$$

LCM of 1 and 15 is 3 × 5 = 15

$$\begin{aligned}= \frac{1 \times 15}{1 \times 15} + \frac{2 \times 1}{15 \times 1} \\[1em] = \frac{15}{15} + \frac{2}{15} \\[1em] = \frac{15 + 2}{15} \\[1em] = \frac{17}{15} \\[1em] = 1\frac{2}{15}\end{aligned}$$

Hence, $\left(\frac{7}{8} \times \frac{24}{21}\right) + \left(\frac{-5}{9} \times \frac{6}{-25}\right) = 1\frac{2}{15}$

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{8}{15} \times \frac{-25}{16}\right) + \left(\frac{-18}{35} \times \frac{5}{6}\right) \\[1em] = \frac{8 \times (-25)}{15 \times 16} + \frac{-18 \times 5}{35 \times 6} \\[1em] = \frac{-200}{240} + \frac{-90}{210} \\[1em] = \frac{-5}{6} + \frac{-3}{7}\end{aligned}$$

By Division Method,

$$\begin{array}{l|r} 2 & 6, 7 \\ \hline 3 & 3, 7 \\ \hline 7 & 1, 7 \\ \hline & 1, 1 \end{array}$$

LCM of 6 and 7 is 2 × 3 × 7 = 42

$$\begin{aligned}= \frac{-5 \times 7}{6 \times 7} + \frac{-3 \times 6}{7 \times 6} \\[1em] = \frac{-35}{42} + \frac{-18}{42} \\[1em] = \frac{-35 + (-18)}{42} \\[1em] = \frac{-53}{42} \\[1em] = -1\frac{11}{42}\end{aligned}$$

Hence, $\left(\frac{8}{15} \times \frac{-25}{16}\right) + \left(\frac{-18}{35} \times \frac{5}{6}\right) = -1\frac{11}{42}$

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{18}{33} \times \frac{-22}{27}\right) - \left(\frac{13}{25} \times \frac{-75}{26}\right) \\[1em] = \frac{18 \times (-22)}{33 \times 27} - \frac{13 \times (-75)}{25 \times 26} \\[1em] = \frac{-396}{891} - \frac{-975}{650} \\[1em] = \frac{-4}{9} - \left(\frac{-3}{2}\right) \\[1em] = \frac{-4}{9} + \frac{3}{2}\end{aligned}$$

By Division Method,

$$\begin{array}{l|r} 2 & 9, 2 \\ \hline 3 & 9, 1 \\ \hline 3 & 3, 1 \\ \hline & 1, 1 \end{array}$$

LCM of 9 and 2 is 2 × 3 × 3 = 18

$$\begin{aligned}= \frac{-4 \times 2}{9 \times 2} + \frac{3 \times 9}{2 \times 9} \\[1em] = \frac{-8}{18} + \frac{27}{18} \\[1em] = \frac{-8 + 27}{18} \\[1em] = \frac{19}{18} \\[1em] = 1\frac{1}{18}\end{aligned}$$

Hence, $\left(\frac{18}{33} \times \frac{-22}{27}\right) - \left(\frac{13}{25} \times \frac{-75}{26}\right) = 1\frac{1}{18}$

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{-13}{7} \times \frac{-35}{39}\right) - \left(\frac{-7}{45} \times \frac{9}{14}\right) \\[1em] = \frac{-13 \times (-35)}{7 \times 39} - \frac{-7 \times 9}{45 \times 14} \\[1em] = \frac{455}{273} - \frac{-63}{630} \\[1em] = \frac{5}{3} - \left(\frac{-1}{10}\right) \\[1em] = \frac{5}{3} + \frac{1}{10}\end{aligned}$$

By Division Method,

$$\begin{array}{l|r} 2 & 3, 10 \\ \hline 3 & 3, 5 \\ \hline 5 & 1, 5 \\ \hline & 1, 1 \end{array}$$

LCM of 3 and 10 is 2 × 3 × 5 = 30

$$\begin{aligned}= \frac{5 \times 10}{3 \times 10} + \frac{1 \times 3}{10 \times 3} \\[1em] = \frac{50}{30} + \frac{3}{30} \\[1em] = \frac{50 + 3}{30} \\[1em] = \frac{53}{30} \\[1em] = 1\frac{23}{30}\end{aligned}$$

Hence, $\left(\frac{-13}{7} \times \frac{-35}{39}\right) - \left(\frac{-7}{45} \times \frac{9}{14}\right) = 1\frac{23}{30}$

Product of two rational numbers = 24

One of them = $\frac{-36}{11}$

Let the other number = x

According to question,

x = $24 ÷ \frac{-36}{11}$

$$\begin{aligned}= \frac{24}{1} \times \frac{11}{-36} \\[1em] = \frac{24 \times 11}{1 \times (-36)} \\[1em] = \frac{264}{-36} \\[1em] = \frac{-22}{3} \\[1em] = -7\frac{1}{3}\end{aligned}$$

Hence, the other rational number is $-7\frac{1}{3}$.

Let the required rational number be $x$.

$$\begin{aligned}\Rightarrow \frac{20}{-9} \times x = \frac{-5}{9} \\[1em] \Rightarrow x = \frac{-5}{9} ÷ \frac{20}{-9} \\[1em] \Rightarrow x = \frac{-5}{9} ÷ \frac{-20}{9} \\[1em] \Rightarrow x = \frac{-5}{9} \times \frac{9}{-20} \\[1em] \Rightarrow x = \frac{-5 \times 9}{9 \times (-20)} \\[1em] \Rightarrow x = \frac{-45}{-180} \\[1em] \Rightarrow x = \frac{1}{4}\end{aligned}$$

Hence, $\frac{20}{-9}$ should be multiplied by $\frac{1}{4}$ to get $\frac{-5}{9}$.

(i) False.

The quotient of two integers is not always a rational number, because if the divisor (denominator) is 0, then the quotient is not defined and so it is not a rational number.

(ii) False.

$-\frac{6}{11}$ is a negative rational number and $\frac{4}{11}$ is a positive rational number. A negative rational number is always less than a positive rational number. So, $-\frac{6}{11} \lt \frac{4}{11}$.

(iii) False.

While adding rational numbers, we do not add the numerators and the denominators separately. Taking LCM of 32 and 23 as 736:

$$\begin{aligned}\Rightarrow -\frac{9}{32} + \frac{5}{23} \\[1em] = \frac{-9 \times 23}{32 \times 23} + \frac{5 \times 32}{23 \times 32} \\[1em] = \frac{-207}{736} + \frac{160}{736} \\[1em] = \frac{-47}{736}.\end{aligned}$$

Thus, sum is not equal to $\frac{-4}{55}$.

(iv) False.

$$\begin{aligned}1 - \frac{3}{15} = \frac{15}{15} - \frac{3}{15} = \frac{15 - 3}{15} \\[1em] = \frac{12}{15} = \frac{4}{5}\end{aligned}$$

which is not equal to $\frac{-2}{15}$.

(i) Solving,

$$\begin{aligned}\Rightarrow -\frac{5}{8} + x = \frac{7}{12} \\[1em] \Rightarrow x = \frac{7}{12} + \frac{5}{8}\end{aligned}$$

By Division Method,

$$\begin{array}{l|r} 2 & 12, 8 \\ \hline 2 & 6, 4 \\ \hline 2 & 3, 2 \\ \hline 3 & 3, 1 \\ \hline & 1, 1 \end{array}$$

LCM of 12 and 8 = 2 × 2 × 2 × 3 = 24

$$\begin{aligned}\Rightarrow x = \frac{7 \times 2}{12 \times 2} + \frac{5 \times 3}{8 \times 3} \\[1em] \Rightarrow x = \frac{14}{24} + \frac{15}{24} \\[1em] \Rightarrow x = \frac{14 + 15}{24} \\[1em] \Rightarrow x = \frac{29}{24} \\[1em] \Rightarrow x = 1\frac{5}{24}\end{aligned}$$

Hence, x = $1\frac{5}{24}$

(ii) Solving,

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

LCM of 1 and 5 is 5.

$$\begin{aligned}\Rightarrow x = \frac{-2 \times 5}{1 \times 5} - \frac{2 \times 1}{5 \times 1} \\[1em] \Rightarrow x = \frac{-10}{5} - \frac{2}{5} \\[1em] \Rightarrow x = \frac{-10 - 2}{5} \\[1em] \Rightarrow x = \frac{-12}{5} \\[1em] \Rightarrow x = -2\frac{2}{5}\end{aligned}$$

Hence, x = $-2\frac{2}{5}$

(iii) Solving,

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

LCM of 3 and 1 is 3.

$$\begin{aligned}\Rightarrow x = \frac{-2 \times 1}{3 \times 1} - \frac{2 \times 3}{1 \times 3} \\[1em] \Rightarrow x = \frac{-2}{3} - \frac{6}{3} \\[1em] \Rightarrow x = \frac{-2 - 6}{3} \\[1em] \Rightarrow x = \frac{-8}{3} \\[1em] \Rightarrow x = -2\frac{2}{3}\end{aligned}$$

Hence, x = $-2\frac{2}{3}$

(iv) Solving,

$$\begin{aligned}\Rightarrow 2\frac{1}{2}x = 33\frac{1}{3} \\[1em] \Rightarrow \frac{5}{2}x = \frac{100}{3} \\[1em] \Rightarrow x = \frac{100}{3} ÷ \frac{5}{2} \\[1em] \Rightarrow x = \frac{100}{3} \times \frac{2}{5} \\[1em] \Rightarrow x = \frac{100 \times 2}{3 \times 5} \\[1em] \Rightarrow x = \frac{200}{15} \\[1em] \Rightarrow x = \frac{40}{3} \\[1em] \Rightarrow x = 13\frac{1}{3}\end{aligned}$$

Hence, x = $13\frac{1}{3}$

(v) Solving,

$$\begin{aligned}\Rightarrow -\frac{9}{35}x = \frac{3}{5} \\[1em] \Rightarrow x = \frac{3}{5} ÷ \left(-\frac{9}{35}\right) \\[1em] \Rightarrow x = \frac{3}{5} \times \frac{35}{-9} \\[1em] \Rightarrow x = \frac{3 \times 35}{5 \times (-9)} \\[1em] \Rightarrow x = \frac{105}{-45} \\[1em] \Rightarrow x = \frac{-7}{3} \\[1em] \Rightarrow x = -2\frac{1}{3}\end{aligned}$$

Hence, x = $-2\frac{1}{3}$

Let the distance covered towards East be positive and towards West be negative.

Distance walked towards East = $\frac{8}{9}$ km

Distance walked towards West = $2\frac{1}{2}$ km = $\frac{5}{2}$ km

Final position from P = $\frac{8}{9} - \frac{5}{2}$

By Division Method,

$$\begin{array}{l|r} 2 & 9, 2 \\ \hline 3 & 9, 1 \\ \hline 3 & 3, 1 \\ \hline & 1, 1 \end{array}$$

LCM of 9 and 2 is 2 × 3 × 3 = 18

$$\begin{aligned}= \frac{8 \times 2}{9 \times 2} - \frac{5 \times 9}{2 \times 9} \\[1em] = \frac{16}{18} - \frac{45}{18} \\[1em] = \frac{16 - 45}{18} \\[1em] = \frac{-29}{18} \\[1em] = -1\frac{11}{18}\end{aligned}$$

The negative sign shows the final position is towards West.

Hence, Manish is $1\frac{11}{18}$ km towards West from the place P.

(i) False.

The statement $\frac{p}{q} = \frac{p \times m}{q \times m}$ holds only when m is a non-zero integer. If m = 0, then $\frac{p \times m}{q \times m} = \frac{0}{0}$, which is not defined. So, the statement is not true for every integer m.

(ii) True.

Since q is a positive integer, q ≠ 0, and p is an integer. So $\frac{p}{q}$ is of the form $\frac{p}{q}$ where p and q are integers and q ≠ 0. Hence, $\frac{p}{q}$ is a rational number.

Since $-\frac{3}{8} \text { and }\frac{x}{16}$ are equivalent rational numbers,

$$\begin{aligned}\Rightarrow -\frac{3}{8} = \frac{x}{16} \\[1em] \Rightarrow -3 \times 16 = 8 \times x \\[1em] \Rightarrow -48 = 8x \\[1em] \Rightarrow x = \frac{-48}{8} \\[1em] \Rightarrow x = -6\end{aligned}$$

Hence, x = -6

First, find $-\frac{17}{6} + \frac{-7}{8}$.

By Division Method,

$$\begin{array}{l|r} 2 & 6, 8 \\ \hline 2 & 3, 4 \\ \hline 2 & 3, 2 \\ \hline 3 & 3, 1 \\ \hline & 1, 1 \end{array}$$

LCM of 6 and 8 is 2 × 2 × 2 × 3 = 24

$$\begin{aligned}\Rightarrow -\frac{17}{6} + \frac{-7}{8}\\[1em] = \frac{-17 \times 4}{6 \times 4} + \frac{-7 \times 3}{8 \times 3} \\[1em] = \frac{-68}{24} + \frac{-21}{24}\\[1em] = \frac{-68 + (-21)}{24} \\[1em] = \frac{-89}{24}\end{aligned}$$

Let x be added to $\frac{-89}{24}$.

$$\begin{aligned}\Rightarrow \frac{-89}{24} + x = -2 \\[1em] \Rightarrow x = \frac{-2}{1} - \frac{-89}{24}\end{aligned}$$

By Division Method,

$$\begin{array}{l|r} 2 & 1, 24 \\ \hline 2 & 1, 12 \\ \hline 2 & 1, 6 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}$$

LCM of 1 and 24 is 2 × 2 × 2 × 3 = 24

$$\begin{aligned}\Rightarrow x = \frac{-2 \times 24}{1 \times 24} - \frac{-89}{24} \\[1em] \Rightarrow x = \frac{-48}{24} - \frac{-89}{24} \\[1em] \Rightarrow x = \frac{-48 - (-89)}{24} \\[1em] \Rightarrow x = \frac{-48 + 89}{24} \\[1em] \Rightarrow x = \frac{41}{24} \\[1em] \Rightarrow x = 1\frac{17}{24}\end{aligned}$$

Hence, $1\frac{17}{24}$ should be added to $\left(-\frac{17}{6} + \frac{-7}{8}\right)$ to get -2.

To get numerator -10 from -2, we multiply by 5.

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

Hence, Option 3 is the correct option.

By Division Method,

$$\begin{array}{l|r} 2 & 6, 24, 12 \\ \hline 2 & 3, 12, 6 \\ \hline 2 & 3, 6, 3 \\ \hline 3 & 3, 3, 3 \\ \hline & 1, 1, 1 \end{array}$$

LCM of 6, 24 and 12 is 2 × 2 × 2 × 3 = 24

Converting each rational number to denominator 24:

$$\begin{aligned}\Rightarrow \frac{-5}{6} = \frac{-5 \times 4}{6 \times 4} = \frac{-20}{24} \\[1em] \frac{-19}{24} = \frac{-19}{24} \\[1em] \frac{37}{-12} = \frac{-37}{12} = \frac{-37 \times 2}{12 \times 2} = \frac{-74}{24}\end{aligned}$$

Since $-74 \lt -20 \lt -19$, the largest rational number is $\frac{-19}{24}$.

Hence, Option 2 is the correct option.

$\frac{-88}{-136} = \frac{88}{136}$

HCF of 88 and 136 is 8.

$\frac{88}{136} = \frac{88 \div 8}{136 \div 8} = \frac{11}{17}$

Hence, Option 3 is the correct option.

By Division Method,

$$\begin{array}{l|r} 3 & 5, 3 \\ \hline 5 & 5, 1 \\ \hline & 1, 1 \end{array}$$

LCM of 5 and 3 is 3 × 5 = 15

$$\begin{aligned}\Rightarrow -\frac{3}{5} = \frac{-3 \times 3}{5 \times 3} = \frac{-9}{15} \\[1em] \Rightarrow -\frac{1}{3} = \frac{-1 \times 5}{3 \times 5} = \frac{-5}{15}\end{aligned}$$

Since -5 > -9, we have $\frac{-5}{15} \gt \frac{-9}{15}$.

Hence, $-\frac{1}{3} \gt -\frac{3}{5}$

Hence, Option 2 is the correct option.

From the number line, A lies between -2 and -1 at the point $-\frac{7}{4}$, and B lies between 2 and 3 at the point $\frac{7}{3}$.

Hence, $A = -\frac{7}{4} \text{ and } B = \frac{7}{3}$

Hence, Option 3 is the correct option.

A rational number between $-\frac{1}{3} \text { and } \frac{1}{3}$ is their mean.

$\frac{1}{2}\left(-\frac{1}{3} + \frac{1}{3}\right) = \frac{1}{2} \times 0 = 0$

Hence, 0 lies between $-\frac{1}{3} \text { and }\frac{1}{3}$.

Hence, Option 3 is the correct option.

Taking the direction towards north as positive and south as negative:

By Division Method,

$$\begin{array}{l|r} 2 & 3, 6 \\ \hline 3 & 3, 3 \\ \hline & 1, 1 \end{array}$$

LCM of 3 and 6 is 2 × 3 = 6

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

The negative sign shows the boy is $\frac{1}{6}$ km towards south of P.

Hence, Option 2 is the correct option.

Let x be added to $-\frac{3}{7}$.

$$\begin{aligned}\Rightarrow -\frac{3}{7} + x = -\frac{5}{7} \\[1em] \Rightarrow x = -\frac{5}{7} - \left(-\frac{3}{7}\right) \\[1em] \Rightarrow x = \frac{-5}{7} + \frac{3}{7} \\[1em] \Rightarrow x = \frac{-5 + 3}{7} \\[1em] \Rightarrow x = \frac{-2}{7}\end{aligned}$$

Hence, Option 3 is the correct option.

Let x be subtracted from $-\frac{2}{7}$.

$$\begin{aligned}\Rightarrow -\frac{2}{7} - x = -\frac{5}{7} \\[1em] \Rightarrow x = -\frac{2}{7} - \left(-\frac{5}{7}\right) \\[1em] \Rightarrow x = \frac{-2}{7} + \frac{5}{7} \\[1em] \Rightarrow x = \frac{-2 + 5}{7} \\[1em] \Rightarrow x = \frac{3}{7}\end{aligned}$$

Hence, Option 1 is the correct option.

Let x be the other number.

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

Hence, Option 1 is the correct option.

The cost of $2\frac{2}{7}$ m of cloth is obtained by multiplying the cost of 1 m by the length.

$\text{Cost} = 20\frac{5}{8} \times 2\frac{2}{7}$

Hence, Option 3 is the correct option.

Let x be the other rational number.

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

Hence, Option 2 is the correct option.

Distance = Speed × Time

$\text{Distance} = 5\frac{2}{5} \times 2\frac{2}{9} = \frac{27}{5} \times \frac{20}{9}$

The distance is obtained by multiplication, which is not given in options 1, 2 or 3.

Hence, Option 4 is the correct option.

$$\begin{aligned}\Rightarrow 2\frac{5}{8} \div 1\frac{1}{6} = \frac{21}{8} \div \frac{7}{6} \\[1em] = \frac{21}{8} \times \frac{6}{7} \\[1em] = \frac{21 \times 6}{8 \times 7} \\[1em] = \frac{126}{56} \\[1em] = \frac{9}{4} \\[1em] = 2\frac{1}{4}\end{aligned}$$

Hence, Option 3 is the correct option.

By Division Method,

$$\begin{array}{l|r} 3 & 3, 9, 3 \\ \hline 3 & 1, 3, 1 \\ \hline & 1, 1, 1 \end{array}$$

LCM of 3 and 9 = 3 × 3 = 9

Converting each rational number to denominator 9:

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

Comparing the numerators: -12 < -5 < -3

Hence, $\frac{-4}{3} \lt \frac{-5}{9} \lt -\frac{1}{3}$

Hence, Option 3 is the correct option.

Statement 1 is true, since 1.6666.... is a non-terminating recurring decimal which equals $\frac{5}{3}$, a rational number.

Statement 2 is true, as it correctly defines a non-terminating recurring decimal.

Hence, both the statements are true.

Hence, Option 1 is the correct option.

Let x be the other rational number.

$$\begin{aligned}\Rightarrow -\frac{3}{20} \times x = \frac{3}{8} \\[1em] \Rightarrow x = \frac{3}{8} \div \left(-\frac{3}{20}\right) \\[1em] \Rightarrow x = \frac{3}{8} \times \left(-\frac{20}{3}\right) \\[1em] \Rightarrow x = \frac{3 \times (-20)}{8 \times 3} \\[1em] \Rightarrow x = \frac{-60}{24} \\[1em] \Rightarrow x = -\frac{5}{2}\end{aligned}$$

So, Assertion (A) is true. Reason (R) correctly states the rule for multiplication of rational numbers, so it is also true.

Hence, both A and R are true.

Hence, Option 3 is the correct option.

By Division Method,

$$\begin{array}{l|r} 2 & 12, 3 \\ \hline 2 & 6, 3 \\ \hline 3 & 3, 3 \\ \hline & 1, 1 \end{array}$$

LCM of 12 and 3 is 2 × 2 × 3 = 12

Sum:

$\frac{65}{12} + \frac{8}{3} = \frac{65}{12} + \frac{8 \times 4}{3 \times 4} = \frac{65}{12} + \frac{32}{12} = \frac{97}{12}$

Difference:

$\frac{65}{12} - \frac{8}{3} = \frac{65}{12} - \frac{32}{12} = \frac{33}{12}$

Dividing the sum by the difference:

$\frac{97}{12} \div \frac{33}{12} = \frac{97}{12} \times \frac{12}{33} = \frac{97}{33}$

So, Assertion (A) is true. Reason (R) correctly states the rule for division of rational numbers, so it is also true.

Hence, both A and R are true.

Hence, Option 3 is the correct option.

A rational number is of the form $\frac{p}{q}$ where $q \neq 0$. Since $\frac{1}{0}$ has denominator 0, it is not a rational number. So, Assertion (A) is false.

Every integer is a rational number, but its converse "every rational number is an integer" is not true (for example, $\frac{1}{2}$ is rational but not an integer). So, Reason (R) is also false.

Hence, both A and R are false.

Hence, Option 4 is the correct option.

Reducing $\frac{34}{-51}$ to standard form:

HCF of 34 and 51 is 17.

$\frac{34}{-51} = \frac{34 \div 17}{-51 \div 17} = \frac{2}{-3} = \frac{-2}{3}$

So, $\frac{34}{-51} \text { and } \frac{-2}{3}$ are equivalent rational numbers and Assertion (A) is true. Reason (R) correctly states the property of equivalent rational numbers, so it is also true.

Hence, both A and R are true.

Hence, Option 3 is the correct option.