Rational Numbers

A rational number is a number that can be expressed in the form $\frac{p}{q}$, where p and q are integers and q ≠ 0. In $\frac{p}{q}$, we have numerator = p and denominator = q.

Examples of Positive Rational Numbers:

A rational number is positive if its numerator and denominator are both of the same sign (both positive or both negative).

(1) $\frac{3}{4}$

(2) $\frac{11}{20}$

(3) 5 (which is $\frac{5}{1}$)

(4) $\frac{-7}{-9}$

Examples of Negative Rational Numbers:

A rational number is negative if its numerator and denominator are of opposite signs.

(1) $-\frac{1}{2}$

(2) $\frac{-5}{8}$

(3) $\frac{4}{-7}$

(4)-10 (which is $\frac{-10}{1}$)

Example of a rational number which is neither positive nor negative is 0.

(i) $\frac{5}{15}$

⇒ It is in $\frac{p}{q}$ form where both 5 and 15 are integers and q ≠ 0.

Hence, it is a rational number.

(ii) $\frac{-6}{23}$

⇒ Both -6 and 23 are integers and the denominator is not zero.

Hence, it is a rational number.

(iii) 17

⇒ An integer that can be written as $\frac{17}{1}$.

Hence, it is a rational number.

(iv) -25

⇒ A negative integer that can be written as $\frac{-25}{1}$.

Hence, it is a rational number.

(v) 0

⇒ Zero is a rational number because it can be written as $\frac{0}{1}$.

Hence, it is a rational number.

(vi) $\frac{8}{0}$

⇒ Division by zero is undefined; the denominator q cannot be 0.

Hence, it is not a rational number.

(vii) $\frac{0}{0}$

⇒ The denominator is zero, which is not allowed in the definition of a rational number.

Hence, it is not a rational number.

(viii) $\frac{0}{8}$

⇒ The numerator can be 0 as long as the denominator is a non-zero integer.

Hence, it is a rational number.

(ix) $\frac{-23}{-37}$

⇒ Both are integers and the denominator is not zero (this simplifies to $\frac{23}{37}$).

Hence, it is a rational number.

(x) $\frac{-1}{7}$

⇒ Both are integers and the denominator is not zero.

Hence, it is a rational number.

(i) We have:

$\frac{12}{17}$

Numerator = 12, Denominator = 17

(ii) We have:

$\frac{6}{-23}$

Numerator = 6, Denominator = -23

(iii) We have:

$\frac{-21}{5}$

Numerator = -21, Denominator = 5

(iv) We have:

7

It can be written as $\frac{7}{1}$

Numerator = 7, Denominator = 1

(v) We have:

-8

It can be written as $\frac{-8}{1}$

Numerator = -8, Denominator = 1

(i) $\frac{-7}{8}$

⇒ The numerator is negative and the denominator is positive (opposite signs), making it a negative rational number.

Hence, it is not a positive rational number.

(ii) $\frac{-13}{17}$

⇒ The signs are opposite, so it is a negative rational number.

Hence, it is not a positive rational number.

(iii) $\frac{-8}{-11}$

⇒ Both the numerator and denominator are negative, the signs cancel out ($\frac{-8}{-11} = \frac{8}{11}$), making the value positive.

Hence, it is a positive rational number.

(iv) $\frac{0}{8}$

⇒ The value is 0, and zero is neither positive nor negative.

Hence, it is not a positive rational number.

(v) $\frac{0}{-7}$

⇒ The value is 0, and zero is neither positive nor negative.

Hence, it is not a positive rational number.

(i) $\frac{-16}{5}$

⇒ The numerator is negative and the denominator is positive (opposite signs).

Hence, it is a negative rational number.

(ii) $\frac{-10}{-11}$

⇒ Both are negative, so the signs cancel out ($\frac{10}{11}$), making it positive.

Hence, it is not a negative rational number.

(iii) -21

⇒ It can be written as $\frac{-21}{1}$. The signs are opposite.

Hence, it is a negative rational number.

(iv) $\frac{0}{-3}$

⇒ The value is 0, which is neutral (neither positive nor negative).

Hence, it is not a negative rational number.

(v) 17

⇒ It is a positive integer (can be written as $\frac{17}{1}$).

Hence, it is not a negative rational number.

(i) $\frac{3}{10}$

We have:

$\frac{3}{10} = \frac{3 \times 2}{10 \times 2} = \frac{3 \times 3}{10 \times 3} = \frac{3 \times 4}{10 \times 4} = \frac{3 \times 5}{10 \times 5}$ [Multiply the numerator and denominator by 2, 3, 4, and 5]

$\Rightarrow \frac{3}{10} = \frac{6}{20} = \frac{9}{30} = \frac{12}{40} = \frac{15}{50}$

∴ Four rational numbers equivalent to $\frac{3}{10}$ are:

$\frac{6}{20}, \frac{9}{30}, \frac{12}{40} , \frac{15}{50}$

(ii) $\frac{-5}{9}$

We have:

$\frac{-5}{9} = \frac{(-5) \times 2}{9 \times 2} = \frac{(-5) \times 3}{9 \times 3} = \frac{(-5) \times 4}{9 \times 4} = \frac{(-5) \times 5}{9 \times 5}$ [Multiply the numerator and denominator by 2, 3, 4, and 5]

$\Rightarrow \frac{-5}{9} = \frac{-10}{18} = \frac{-15}{27} = \frac{-20}{36} = \frac{-25}{45}$

∴ Four rational numbers equivalent to $\frac{-5}{9}$ are:

$\frac{-10}{18}, \frac{-15}{27}, \frac{-20}{36} , \frac{-25}{45}$

(iii) $\frac{6}{-13}$

We have:

$\frac{6}{-13} = \frac{6 \times 2}{(-13) \times 2} = \frac{6 \times 3}{(-13) \times 3} = \frac{6 \times 4}{(-13) \times 4} = \frac{6 \times 5}{(-13) \times 5}$ [Multiply the numerator and denominator by 2, 3, 4, and 5]

$\Rightarrow \frac{6}{-13} = \frac{12}{-26} = \frac{18}{-39} = \frac{24}{-52} = \frac{30}{-65}$

∴ Four rational numbers equivalent to $\frac{6}{-13}$ are:

$\frac{12}{-26}, \frac{18}{-39}, \frac{24}{-52} , \frac{30}{-65}$

(iv) 9

Since $9 = \frac{9}{1}$, we have:

$\frac{9}{1} = \frac{9 \times 2}{1 \times 2} = \frac{9 \times 3}{1 \times 3} = \frac{9 \times 4}{1 \times 4} = \frac{9 \times 5}{1 \times 5}$ [Multiply the numerator and denominator by 2, 3, 4, and 5]

$\Rightarrow 9 = \frac{18}{2} = \frac{27}{3} = \frac{36}{4} = \frac{45}{5}$

∴ Four rational numbers equivalent to 9 are:

$\frac{18}{2}, \frac{27}{3}, \frac{36}{4} , \frac{45}{5}$

(v) -1

Since $-1 = \frac{-1}{1}$, we have

$\frac{-1}{1} = \frac{(-1) \times 2}{1 \times 2} = \frac{(-1) \times 3}{1 \times 3} = \frac{(-1) \times 4}{1 \times 4} = \frac{(-1) \times 5}{1 \times 5}$ [Multiply the numerator and denominator by 2, 3, 4, and 5]

$\Rightarrow -1 = \frac{-2}{2} = \frac{-3}{3} = \frac{-4}{4} = \frac{-5}{5}$

∴ Four rational numbers equivalent to -1 are:

$\frac{-2}{2}, \frac{-3}{3}, \frac{-4}{4}, \frac{-5}{5}$

(i) We have:

$\frac{16}{-21} = \frac{16 \times (-1)}{-21 \times (-1)} = \frac{-16}{21}$

∴ The rational number with a positive denominator is $\frac{-16}{21}$.

(ii) We have:

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

∴ The rational number with a positive denominator is $\frac{-1}{5}$.

(iii) We have:

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

∴ The rational number with a positive denominator is $\frac{7}{12}$.

(iv) We have:

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

∴ The rational number with a positive denominator is $\frac{-5}{1}$.

(v) We have:

$\frac{-6}{-1} = \frac{(-6) \times (-1)}{-1 \times (-1)} = \frac{6}{1}$

∴ The rational number with a positive denominator is $\frac{6}{1}$.

(i) Numerator 24

$\frac{4}{9} = \frac{24}{x}$

4 × x = 24 × 9 $\quad$ [cross multiplication]

x = $\frac{{\overset{6}{\cancel{24}}} × 9}{\underset{1}{\cancel{4}}}$

x = $\frac{6 × 9}{1}$

x = 54

∴ The rational number with numerator 24 is $\frac{24}{54}$.

(ii) Numerator -20

$\frac{4}{9} = \frac{-20}{x}$

4 × x = -20 × 9 $\quad$ [cross multiplication]

x = $\frac{{\overset{-5}{\cancel{-20}}} × 9}{\underset{1}{\cancel{4}}}$

x = $\frac{-5 × 9}{1}$

x = -45

∴ The rational number with numerator -20 is $\frac{-20}{-45}$.

(i) Denominator 48

$\frac{3}{8} = \frac{x}{48}$

8 × x = 3 × 48 $\quad$ [cross multiplication]

x = $\frac{3 × {\overset{6}{\cancel{48}}}}{{\underset{1}{\cancel{8}}}}$

x = $\frac{3 × 6}{1}$

x = 18

∴ The rational number with denominator 48 is $\frac{18}{48}$.

(ii) Denominator -32

$\frac{3}{8} = \frac{x}{-32}$

8 × x = 3 × -32 $\quad$ [cross multiplication]

x = $\frac{3 × {\overset{-4}{\cancel{-32}}}}{{\underset{1}{\cancel{8}}}}$

x = $\frac{3 × -4}{1}$

x = -12

∴ The rational number with denominator -32 is $\frac{-12}{-32}$.

(i) Numerator -36

$\frac{-6}{11} = \frac{-36}{x}$

-6 × x = -36 × 11 $\quad$ [cross multiplication]

x = $\frac{{\overset{6}{\cancel{-36}}} × 11}{\underset{1}{\cancel{-6}}}$

x = $\frac{6 × 11}{1}$

x = 66

∴ The rational number with numerator -36 is $\frac{-36}{66}$.

(ii) Numerator 42

$\frac{-6}{11} = \frac{42}{x}$

-6 × x = 42 × 11 $\quad$ [cross multiplication]

x = $\frac{{\overset{7}{\cancel{42}}} × 11}{\underset{-1}{\cancel{-6}}}$

x = $\frac{7 × 11}{-1}$

x = -77

∴ The rational number with numerator 42 is $\frac{42}{-77}$.

(i) Denominator 42

$\frac{2}{-7} = \frac{x}{42}$

-7 × x = 2 × 42 $\quad$ [cross multiplication]

x = $\frac{2 × {\overset{6}{\cancel{42}}}}{{\underset{-1}{\cancel{-7}}}}$

x = $\frac{2 × 6}{-1}$

x = -12

∴ The rational number with denominator 42 is $\frac{-12}{42}$.

(ii) Denominator -28

$\frac{2}{-7} = \frac{x}{-28}$

-7 × x = 2 × -28 $\quad$ [cross multiplication]

x = $\frac{2 × {\overset{4}{\cancel{-28}}}}{{\underset{1}{\cancel{-7}}}}$

x = $\frac{2 × 4}{1}$

x = 8

∴ The rational number with denominator -28 is $\frac{8}{-28}$.

(i) Numerator -4

$\frac{-48}{36} = \frac{-4}{x}$

-48 × x = -4 × 36 $\quad$ [cross multiplication]

x = $\frac{-4 × 36}{-48}$

x = $\frac{{\overset{1}{\cancel{-4}}} × 36}{\underset{12}{\cancel{-48}}}$

x = $\frac{1 × {\overset{3}{\cancel{36}}}}{\underset{1}{\cancel{12}}}$

x = $\frac{1 × 3}{1}$

x = 3

∴ The rational number with numerator -4 is $\frac{-4}{3}$.

(ii) Numerator 8

$\frac{-48}{36} = \frac{8}{x}$

-48 × x = 8 × 36 $\quad$ [cross multiplication]

x = $\frac{8 × 36}{-48}$

x = $\frac{{\overset{1}{\cancel{8}}} × 36}{\underset{-6}{\cancel{-48}}}$

x = $\frac{1 × {\overset{6}{\cancel{36}}}}{{\underset{-1}{\cancel{-6}}}}$

x = $\frac{1 × 6}{-1}$

x = -6

∴ The rational number with numerator 8 is $\frac{8}{-6}$.

(i) Numerator -6

$\frac{78}{-117} = \frac{-6}{x}$

78 × x = -6 × (-117) $\quad$ [cross multiplication]

x = $\frac{-6 × (-117)}{78}$

x = $\frac{{\overset{-1}{\cancel{-6}}} × -117}{\underset{13}{\cancel{78}}}$

x = $\frac{-1 × {\overset{-9}{\cancel{-117}}}}{{\underset{1}{\cancel{13}}}}$

x = $\frac{-1 × -9}{1}$

x = 9

∴ The rational number with numerator -6 is $\frac{-6}{9}$.

(ii) Numerator 2

$\frac{78}{-117} = \frac{2}{x}$

78 × x = 2 × (-117) $\quad$ [cross multiplication]

x = $\frac{2 × (-117)}{78}$

x = $\frac{{\overset{1}{\cancel{2}}} × {-117}}{\underset{39}{\cancel{78}}}$

x = $\frac{1 × {\overset{-3}{\cancel{-117}}}}{{\underset{1}{\cancel{39}}}}$

x = $\frac{1 × -3}{1}$

x = -3

∴ The rational number with numerator 2 is $\frac{2}{-3}$.

(i) $\frac{56}{32}$

The HCF of 56 and 32 is 8.

Divide the numerator and denominator by 8:

$\frac{56}{32} = \frac{56 \div 8}{32 \div 8} = \frac{7}{4}$

∴ The standard form is $\frac{7}{4}$.

(ii) $\frac{16}{-40}$

First, multiply the numerator and denominator by -1 to make the denominator positive:

$\frac{16 \times (-1)}{-40 \times (-1)} = \frac{-16}{40}$

The HCF of 16 and 40 is 8.

Divide the numerator and denominator by 8:

$\frac{-16 \div 8}{40 \div 8} = \frac{-2}{5}$

∴ The standard form is $\frac{-2}{5}$.

(iii) $\frac{-36}{54}$

The HCF of 36 and 54 is 18.

Divide the numerator and denominator by 18:

$\frac{-36 \div 18}{54 \div 18} = \frac{-2}{3}$

∴ The standard form is $\frac{-2}{3}$.

(iv) $\frac{-22}{-77}$

First, multiply the numerator and denominator by -1 to make the denominator positive:

$\frac{-22 \times (-1)}{-77 \times (-1)} = \frac{22}{77}$

The HCF of 22 and 77 is 11.

$\frac{22 \div 11}{77 \div 11} = \frac{2}{7}$

∴ The standard form is $\frac{2}{7}$.

(v) $\frac{78}{-65}$

First, multiply the numerator and denominator by -1 to make the denominator positive:

$\frac{78 \times (-1)}{-65 \times (-1)}$ = $\frac{-78}{65}$

The HCF of 78 and 65 is 13.

$\frac{-78 \div 13}{65 \div 13} = \frac{-6}{5}$

∴ The standard form is $\frac{-6}{5}$.

(vi) $\frac{-95}{114}$

The HCF of 95 and 114 is 19.

$\frac{-95 \div 19}{114 \div 19} = \frac{-5}{6}$

∴ The standard form is $\frac{-5}{6}$.

(vii) $\frac{-69}{115}$

The HCF of 69 and 115 is 23.

$\frac{-69 \div 23}{115 \div 23} = \frac{-3}{5}$

∴ The standard form is $\frac{-3}{5}$.

(viii) $\frac{155}{-217}$

First, multiply the numerator and denominator by -1 to make the denominator positive:

$\frac{155 \times (-1)}{-217 \times (-1)}$ = $\frac{-155}{217}$

The HCF of 155 and 217 is 31.

$\frac{-155 \div 31}{217 \div 31} = \frac{-5}{7}$

∴ The standard form is $\frac{-5}{7}$.

(i) $\frac{-2}{3} = \frac{14}{x}$

We have,

$$\begin{aligned}\frac{-2}{3} = \frac{14}{x} \\[1em] \Rightarrow (-2) \times x = 3 \times 14 \quad \text{[cross multiplication]} \\[1em] \Rightarrow x = \frac{3 \times 14}{-2} \\[1em] \Rightarrow x = 3 \times (-7) = -21\end{aligned}$$

Hence, x = -21.

(ii) $\frac{8}{-3} = \frac{x}{6}$

We have,

$$\begin{aligned}\frac{8}{-3} = \frac{x}{6} \\[1em] \Rightarrow (-3) \times x = 8 \times 6 \quad \text{[cross multiplication]} \\[1em] \Rightarrow x = \frac{8 \times 6}{-3} \\[1em] \Rightarrow x = 8 \times (-2) = -16\end{aligned}$$

Hence, x = -16.

(iii) $\frac{5}{9} = \frac{x}{-27}$

We have,

$$\begin{aligned}\frac{5}{9} = \frac{x}{-27} \\[1em] \Rightarrow 9 \times x = 5 \times (-27) \quad \text{[cross multiplication]} \\[1em] \Rightarrow x = \frac{5 \times (-27)}{9} \\[1em] \Rightarrow x = 5 \times (-3) = -15\end{aligned}$$

Hence, x = -15.

(iv) $\frac{11}{6} = \frac{-55}{x}$

We have,

$$\begin{aligned}\frac{11}{6} = \frac{-55}{x} \\[1em] \Rightarrow 11 \times x = 6 \times (-55) \quad \text{[cross multiplication]} \\[1em] \Rightarrow x = \frac{6 \times (-55)}{11} \\[1em] \Rightarrow x = 6 \times (-5) = -30\end{aligned}$$

Hence, x = -30.

(v) $\frac{15}{x} = -3$

Since $-3 = \frac{-3}{1}$, we have

$$\begin{aligned}\frac{15}{x} = \frac{-3}{1} \\[1em] \Rightarrow (-3) \times x = 15 \times 1 \quad \text{[cross multiplication]} \\[1em] \Rightarrow x = \frac{15}{-3} = -5\end{aligned}$$

Hence, x = -5.

(vi) $\frac{-36}{x} = 2$

Since $2 = \frac{2}{1}$, we have

$$\begin{aligned}\frac{-36}{x} = \frac{2}{1} \\[1em] \Rightarrow 2 \times x = (-36) \times 1 \quad \text{[cross multiplication]} \\[1em] \Rightarrow x = \frac{-36}{2} = -18 \\[1em]\end{aligned}$$

Hence, x = -18.

(i) False
Reason — While a rational number is defined as the quotient of two integers $\frac{p}{q}$, the denominator q cannot be zero. If the divisor (the second integer) is zero, the quotient is undefined and is not a rational number.

(ii) False
Reason — Fractions typically refer to a part of a whole and are usually represented as a ratio of two natural numbers. Rational numbers include negative values (like $-\frac{2}{3}$) and integers (like -5), which are not traditionally considered fractions in their standard form.

(iii) False
Reason — Rational numbers include negative values. Therefore, any negative rational number (such as -1 or $-\frac{1}{2}$) is smaller than zero. Because the set of rational numbers extends infinitely in the negative direction, there is no "smallest" rational number.

(iv) True
Reason — A fraction is represented as $\frac{a}{b}$ where a and b are whole numbers and b ≠ 0. Since whole numbers are also integers, every fraction fits the definition of a rational number.

(i) We have:

$\frac{3}{-7}$ and $\frac{1}{7}$

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

The other number = $\frac{1}{7}$.

Since -3 < 1, therefore $\frac{-3}{7} \lt \frac{1}{7}$

Hence, $\frac{1}{7}$ is greater.

(ii) We have:

$\frac{11}{-18}$ and $\frac{-5}{18}$

One number = $\frac{11}{-18} = \frac{-11 \times (-1)}{18 \times (-1)} = \frac{-11}{18}$

The other number = $\frac{-5}{18}$.

Since -11 < -5, therefore $\frac{-11}{18} \lt \frac{-5}{18}$.

Hence, $\frac{-5}{18}$ is greater.

(iii) We have:

$\frac{7}{10}$ and $\frac{-9}{10}$

Since 7 > -9, therefore $\frac{7}{10} \gt \frac{-9}{10}$.

Hence, $\frac{7}{10}$ is greater.

(iv) We have:

0 and $\frac{-3}{4}$

Since $\frac{-3}{4}$ is a negative rational number, we have $\frac{-3}{4} \lt 0$.

Hence, 0 is greater.

(v) We have:

$\frac{1}{12}$ and 0

Since $\frac{1}{12}$ is a positive rational number, we have $\frac{1}{12} \gt 0$.

Hence, $\frac{1}{12}$ is greater.

(vi) We have:

$\frac{18}{-19}$ and 0

$\frac{18}{-19} = \frac{-18 \times (-1)}{19 \times (-1)} = \frac{-18}{19}$

Since, $\frac{-18}{19}$ is a negative rational number, we have $\frac{-18}{19} \lt 0$.

Hence, 0 is greater.

(vii) We have:

$\frac{7}{8}$ and $\frac{11}{16}$

L.C.M. of denominators 8 and 16 is 16.

$\frac{7}{8} = \frac{7 \times 2}{8 \times 2} = \frac{14}{16}$.

Now, we have:

$\frac{14}{16}$ and $\frac{11}{16}$

Clearly 14 > 11, and so $\frac{14}{16} \gt \frac{11}{16}$ i.e., $\frac{7}{8} \gt \frac{11}{16}$

Hence, $\frac{7}{8}$ is greater.

(viii) We have:

$\frac{11}{-12}$ and $\frac{-10}{11}$

$\frac{11}{-12} = \frac{11 \times (-1)}{-12 \times (-1)} = \frac{-11}{12}$

L.C.M. of denominators 12 and 11 is 132.

Now, expressing each fraction with denominator 132:

$$\begin{aligned}\frac{-11}{12} = \frac{-11 \times 11}{12 \times 11} = \frac{-121}{132} \\[1em] \frac{-10}{11} = \frac{-10 \times 12}{11 \times 12} = \frac{-120}{132}\end{aligned}$$.

Now, we have:

$\frac{-121}{132}$ and $\frac{-120}{132}$

Since -121 < -120, and so $\frac{-121}{132} \lt \frac{-120}{132}$ i.e., $\frac{11}{-12} \lt \frac{-10}{11}$

Hence, $\frac{-10}{11}$ is greater.

(ix) We have:

$\frac{-13}{5}$ and $\frac{-4}{1}$

L.C.M. of denominators 5 and 1 is 5.

= $\frac{-4 \times 5}{1 \times 5} = \frac{-20}{5}$.

Now, we have:

$\frac{-13}{5}$ and $\frac{-20}{5}$

Since -13 > -20, and so $\frac{-13}{5} \gt \frac{-20}{5}$ i.e., $\frac{-13}{5} \gt \frac{-4}{1}$

Hence, $\frac{-13}{5}$ is greater.

(x) We have:

$\frac{17}{-6}$ and $\frac{-13}{4}$

$\frac{17}{-6} = \frac{17 \times (-1)}{-6 \times (-1)} = \frac{-17}{6}$

L.C.M. of denominators 6 and 4 is 12.

$$\begin{aligned}\frac{-17}{6} = \frac{-17 \times 2}{6 \times 2} = \frac{-34}{12} \\[1em] \frac{-13}{4} = \frac{-13 \times 3}{4 \times 3} = \frac{-39}{12}\end{aligned}$$.

Now, we have:

$\frac{-34}{12}$ and $\frac{-39}{12}$

Since -34 > -39, and so $\frac{-34}{12} \gt \frac{-39}{12}$ i.e., $\frac{17}{-6} \gt \frac{-13}{4}$

Hence, $\frac{17}{-6}$ is greater.

(xi) We have:

$\frac{7}{-9}$ and $\frac{-5}{8}$

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

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

Now, expressing each fraction with denominator 72:

$$\begin{aligned}\frac{-7}{9} = \frac{-7 \times 8}{9 \times 8} = \frac{-56}{72} \\[1em] \frac{-5}{8} = \frac{-5 \times 9}{8 \times 9} = \frac{-45}{72}\end{aligned}$$

Now, we have:

$\frac{-56}{72}$ and $\frac{-45}{72}$

Since -56 < -45, and so $\frac{-56}{72} \lt \frac{-45}{72}$ i.e., $\frac{7}{-9} \lt \frac{-5}{8}$

Hence, $\frac{-5}{8}$ is greater.

(xii) We have:

$\frac{-3}{-8}$ and $\frac{5}{9}$

$\frac{-3 \times -1}{-8 \times -1} = \frac{3}{8}$.

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

$$\begin{aligned}\frac{3}{8} = \frac{3 \times 9}{8 \times 9} = \frac{27}{72} \\[1em] \frac{5}{9} = \frac{5 \times 8}{9 \times 8} = \frac{40}{72}\end{aligned}$$.

Now, we have:

$\frac{27}{72}$ and $\frac{40}{72}$

Since 27 < 40, and so $\frac{27}{72} \lt \frac{40}{72}$ i.e.,$\frac{-3}{-8} \lt \frac{5}{9}$

Hence, $\frac{5}{9}$ is greater.

(i) $\frac{-17}{4}$ < $\frac{-15}{4}$

(ii) 0 < $\frac{-1}{-2}$

(iii) $\frac{4}{-3}$ < $\frac{-8}{7}$

(iv) $\frac{-5}{12}$ > $\frac{7}{-16}$

(v) $\frac{-7}{8}$ > $\frac{-8}{9}$

(vi) $\frac{1}{-10}$ < $\frac{-4}{-5}$

Explanation

(i) Since the denominators are the same, we compare the numerators where -17 is less than -15.

(ii) $\frac{-1}{-2}$ simplifies to the positive rational number $\frac{1}{2}$, and zero is always less than any positive number.

(iii) L.C.M. of 3 and 7 is 21. After converting to a common denominator of 21, we compare $\frac{-28}{21}$ and $\frac{-24}{21}$, where -28 < -24.

(iv) L.C.M. of 12 and 16 is 48. After converting to a common denominator of 48, the fractions are $\frac{-20}{48}$ and $\frac{-21}{48}$, and since -20 > -21, the first is greater.

(v) L.C.M. of 8 and 9 is 72. After converting to a common denominator of 72, the fractions are $\frac{-63}{72}$ and $\frac{-64}{72}$, and -63 > -64 so $\frac{-63}{72}$ > $\frac{-64}{72}$.

(vi) $\frac{1}{-10}$ is a negative rational number while $\frac{-4}{-5}$ is positive, and any negative number is less than a positive one.

(i) We have:

$\frac{3}{4}, \frac{5}{8}, \frac{11}{16}, \frac{21}{32}$

First we find the L.C.M.

L.C.M. of denominators 4, 8, 16, and 32 is 32.

Now, expressing each fraction with denominator 32:

$$\begin{aligned}\frac{3}{4} = \frac{3 \times 8}{4 \times 8} = \frac{24}{32} \\[1em] \frac{5}{8} = \frac{5 \times 4}{8 \times 4} = \frac{20}{32} \\[1em] \frac{11}{16} = \frac{11 \times 2}{16 \times 2} = \frac{22}{32} \\[1em] \frac{21}{32} = \frac{21 \times 1}{32 \times 1} = \frac{21}{32} \\[1em]\end{aligned}$$

Clearly, $\frac{20}{32} \lt \frac{21}{32} \lt \frac{22}{32} \lt \frac{24}{32}$.

Therefore $\frac{5}{8} \lt \frac{21}{32} \lt \frac{11}{16} \lt \frac{3}{4}$

Hence, the ascending order is: $\frac{5}{8}, \frac{21}{32}, \frac{11}{16}, \frac{3}{4}$.

(ii) We have:

$\frac{-2}{5}, \frac{7}{-10}, \frac{-8}{15}, \frac{17}{-30}$

First, express each rational number with a positive denominator: $\frac{-2}{5}, \frac{-7}{10}, \frac{-8}{15}, \frac{-17}{30}$.

The L.C.M. of denominators 5, 10, 15, and 30 is 30.

Now, expressing each fraction with denominator 30:

$$\begin{aligned}\frac{-2}{5} = \frac{-2 \times 6}{5 \times 6} = \frac{-12}{30} \\[1em] \frac{-7}{10} = \frac{-7 \times 3}{10 \times 3} = \frac{-21}{30} \\[1em] \frac{-8}{15} = \frac{-8 \times 2}{15 \times 2} = \frac{-16}{30} \\[1em] \frac{-17}{30} = \frac{-17 \times 1}{30 \times 1} = \frac{-17}{30}\end{aligned}$$

Clearly, $\frac{-21}{30} \lt \frac{-17}{30} \lt \frac{-16}{30} \lt \frac{-12}{30}$.

Therefore $\frac{-7}{10} \lt \frac{-17}{30} \lt \frac{-8}{15} \lt \frac{-2}{5}$

Hence, the ascending order is: $\frac{7}{-10}, \frac{17}{-30}, \frac{-8}{15}, \frac{-2}{5}$.

(iii) We have:

$\frac{5}{-12}, \frac{-2}{3}, \frac{-7}{9}, \frac{11}{-18}$

Expressing with positive denominators: $\frac{-5}{12}, \frac{-2}{3}, \frac{-7}{9}, \frac{-11}{18}$.

The L.C.M. denominators of 12, 3, 9, and 18 is 36.

Now, expressing each fraction with denominator 36:

$$\begin{aligned}\frac{-5}{12} = \frac{-5 \times 3}{12 \times 3} = \frac{-15}{36} \\[1em] \frac{-2}{3} = \frac{-2 \times 12}{3 \times 12} = \frac{-24}{36} \\[1em] \frac{-7}{9} = \frac{-7 \times 4}{9 \times 4} = \frac{-28}{36} \\[1em] \frac{-11}{18} = \frac{-11 \times 2}{18 \times 2} = \frac{-22}{36}\end{aligned}$$

Clearly, $\frac{-28}{36} \lt \frac{-24}{36} \lt \frac{-22}{36} \lt \frac{-15}{36}$.

Therefore $\frac{-7}{9} \lt \frac{-2}{3} \lt \frac{-11}{18} \lt \frac{-5}{12}$.

Hence, the ascending order is: $\frac{-7}{9}, \frac{-2}{3}, \frac{11}{-18}, \frac{5}{-12}$.

(iv) We have:

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

Expressing with positive denominators: $\frac{-4}{7}, \frac{-13}{28}, \frac{9}{14}, \frac{23}{42}$.

The L.C.M. of denominators 7, 28, 14, and 42 is 84.

Now, expressing each fraction with denominator 84:

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

Clearly, $\frac{-48}{84} \lt \frac{-39}{84} \lt \frac{46}{84} \lt \frac{54}{84}$.

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

Hence, the ascending order is: $\frac{-4}{7}, \frac{13}{-28}, \frac{23}{42}, \frac{9}{14}$.

(i) We have:

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

The L.C.M. of denominators 12, 18, 6, and 9 is 36.

Now, expressing each fraction with denominator 36:

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

Clearly, $\frac{33}{36} \gt \frac{30}{36} \gt \frac{28}{36} \gt \frac{26}{36}$. Therefore $\frac{11}{12} \gt \frac{5}{6} \gt \frac{7}{9} \gt \frac{13}{18}$.

Hence, the descending order is: $\frac{11}{12}, \frac{5}{6}, \frac{7}{9}, \frac{13}{18}$.

(ii) We have:

$\frac{-11}{20}, \frac{3}{-10}, \frac{17}{-30}, \frac{-7}{15}$

First, express each with a positive denominator: $\frac{-11}{20}, \frac{-3}{10}, \frac{-17}{30}, \frac{-7}{15}$.

The L.C.M. of denominators 20, 10, 30, and 15 is 60.

Now, expressing each fraction with denominator 60:

$$\begin{aligned}\frac{-11}{20} = \frac{-11 \times 3}{20 \times 3} = \frac{-33}{60} \\[1em] \frac{-3}{10} = \frac{-3 \times 6}{10 \times 6} = \frac{-18}{60} \\[1em] \frac{-17}{30} = \frac{-17 \times 2}{30 \times 2} = \frac{-34}{60} \\[1em] \frac{-7}{15} = \frac{-7 \times 4}{15 \times 4} = \frac{-28}{60}\end{aligned}$$

Clearly, $\frac{-18}{60} \gt \frac{-28}{60} \gt \frac{-33}{60} \gt \frac{-34}{60}$. Therefore $\frac{-3}{10} \gt \frac{-7}{15} \gt \frac{-11}{20} \gt \frac{-17}{30}$.

Hence, the descending order is: $\frac{3}{-10}, \frac{-7}{15}, \frac{-11}{20}, \frac{17}{-30}$.

(iii) We have:

$\frac{9}{-24}, -1, \frac{2}{-3}, \frac{-7}{-6}$

Expressing with positive denominators: $\frac{-9}{24}, \frac{-1}{1}, \frac{-2}{3}, \frac{7}{6}$.

The L.C.M. of denominators 24, 1, 3, and 6 is 24.

Now, expressing each fraction with denominator 24:

$$\begin{aligned}\frac{-9}{24} = \frac{-9 \times 1}{24 \times 1} = \frac{-9}{24} \\[1em] -1 = \frac{-1 \times 24}{1 \times 24} = \frac{-24}{24} \\[1em] \frac{-2}{3} = \frac{-2 \times 8}{3 \times 8} = \frac{-16}{24} \\[1em] \frac{7}{6} = \frac{7 \times 4}{6 \times 4} = \frac{28}{24}\end{aligned}$$

Clearly, $\frac{28}{24} \gt \frac{-9}{24} \gt \frac{-16}{24} \gt \frac{-24}{24}$. Therefore $\frac{7}{6} \gt \frac{-9}{24} \gt \frac{-2}{3} \gt -1$.

Hence, the descending order is: $\frac{-7}{-6}, \frac{9}{-24}, \frac{2}{-3}, -1$.

(iv) We have:

$\frac{7}{-10}, \frac{11}{15}, \frac{-17}{-30}, \frac{-2}{5}$

Expressing with positive denominators: $\frac{-7}{10}, \frac{11}{15}, \frac{17}{30}, \frac{-2}{5}$.

The L.C.M. of denominators 10, 15, 30, and 5 is 30.

Now, expressing each fraction with denominator 30:

$$\begin{aligned}\frac{-7}{10} = \frac{-7 \times 3}{10 \times 3} = \frac{-21}{30} \\[1em] \frac{11}{15} = \frac{11 \times 2}{15 \times 2} = \frac{22}{30} \\[1em] \frac{17}{30} = \frac{17 \times 1}{30 \times 1} = \frac{17}{30} \\[1em] \frac{-2}{5} = \frac{-2 \times 6}{5 \times 6} = \frac{-12}{30}\end{aligned}$$

Clearly, $\frac{22}{30} \gt \frac{17}{30} \gt \frac{-12}{30} \gt \frac{-21}{30}$. Therefore $\frac{11}{15} \gt \frac{17}{30} \gt \frac{-2}{5} \gt \frac{-7}{10}$.

Hence, the descending order is: $\frac{11}{15}, \frac{-17}{-30}, \frac{-2}{5}, \frac{7}{-10}$.

(i) $\frac{5}{11}$ and $\frac{4}{11}$

We have:

$\frac{5}{11} + \frac{4}{11} = \frac{5 + 4}{11} = \frac{9}{11}$

Hence, the answer is $\frac{9}{11}$

(ii) $\frac{-3}{8}$ and $\frac{5}{8}$

We have:

$\frac{-3}{8} + \frac{5}{8} = \frac{-3 + 5}{8} = \frac{2}{8}$

Hence, the answer is $\frac{2}{8}$

(iii) $\frac{-6}{13}$ and $\frac{8}{13}$

We have:

$\frac{-6}{13} + \frac{8}{13} = \frac{-6 + 8}{13} = \frac{2}{13}$

Hence, the answer is $\frac{2}{13}$

(iv) $\frac{-8}{15}$ and $\frac{-7}{15}$

We have:

$\frac{-8}{15} + \frac{-7}{15} = \frac{(-8) + (-7)}{15} = \frac{-15}{15}$

Hence, the answer is $\frac{-15}{15}$

(v) $\frac{-13}{20}$ and $\frac{17}{20}$

We have:

$\frac{-13}{20} + \frac{17}{20} = \frac{-13 + 17}{20} = \frac{4}{20}$

Hence, the answer is $\frac{4}{20}$

(vi) $\frac{-3}{8}$ and $\frac{5}{-8}$

First, express $\frac{5}{-8}$ with a positive denominator:

$\frac{5}{-8} = \frac{5 \times (-1)}{-8 \times (-1)} = \frac{-5}{8}$

Now, add the numbers:

$\frac{-3}{8} + \frac{-5}{8} = \frac{(-3) + (-5)}{8} = \frac{-8}{8}$

Hence, the answer is $\frac{-8}{8}$

(i) $\frac{-2}{3}$ and $\frac{3}{4}$

We have:

$\frac{-2}{3}$ + $\frac{3}{4}$

Let us find L.C.M. of denominators 3 and 4

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

L.C.M. = 2 x 2 x 3 = 12

Now, expressing each fraction with denominator 12:

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

Hence, the answer is $\frac{1}{12}$

(ii) $\frac{-4}{9}$ and $\frac{5}{6}$

We have:

$\frac{-4}{9}$ + $\frac{5}{6}$

Let us find L.C.M. of denominators 9 and 6

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

L.C.M. = 3 x 3 x 2 = 18

Now, expressing each fraction with denominator 18:

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

Hence, the answer is $\frac{7}{18}$

(iii) $\frac{-5}{18}$ and $\frac{11}{27}$

We have:

$\frac{-5}{18}$ + $\frac{11}{27}$

Let us find LCM of denominators 18 and 27

$$\begin{array}{r|l} 3 & 18, 27 \\ \hline 3 & 6, 9 \\ \hline 3 & 2, 3 \\ \hline 2 & 2, 1 \\ \hline & 1, 1 \end{array}$$

L.C.M. = 3 x 3 x 3 x 2 = 54

Now, expressing each fraction with denominator 54:

$$\begin{aligned}\frac{-5}{18} = \frac{-5 \times 3}{18 \times 3} = \frac{-15}{54} \\[1em] \frac{11}{27} = \frac{11 \times 2}{27 \times 2} = \frac{22}{54} \\[1em] \therefore \frac{-5}{18} + \frac{11}{27} = \frac{-15}{54} + \frac{22}{54} \\[1em] = \frac{(-15) + 22}{54} = \frac{7}{54}\end{aligned}$$

Hence, the answer is $\frac{7}{54}$

(iv) $\frac{-7}{12}$ and $\frac{-5}{24}$

We have:

$\frac{-7}{12}$ + $\frac{-5}{24}$

Let us find LCM of denominators 12 and 24

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

L.C.M. = 2 x 2 x 3 x 2 = 24

Now, expressing each fraction with denominator 24:

$$\begin{aligned}\frac{-7}{12} = \frac{-7 \times 2}{12 \times 2} = \frac{-14}{24} \\[1em] \frac{-5}{24} = \frac{-5 \times 1}{24 \times 1} = \frac{-5}{24} \\[1em] \therefore \frac{-7}{12} + \frac{-5}{24} = \frac{-14}{24} + \frac{-5}{24} \\[1em] = \frac{(-14) + (-5)}{24} = \frac{-19}{24}\end{aligned}$$

Hence, the answer is $\frac{-19}{24}$

(v) $\frac{-1}{18}$ and $\frac{-7}{27}$

We have:

$\frac{-1}{18}$ + $\frac{-7}{27}$

LCM of denominators 18 and 27

$$\begin{array}{r|l} 3 & 18, 27 \\ \hline 3 & 6, 9 \\ \hline 3 & 2, 3 \\ \hline 2 & 2, 1 \\ \hline & 1, 1 \end{array}$$

L.C.M. = 3 x 3 x 3 x 2 = 54

Now, expressing each fraction with denominator 54

$$\begin{aligned}\frac{-1}{18} = \frac{-1 \times 3}{18 \times 3} = \frac{-3}{54} \\[1em] \frac{-7}{27} = \frac{-7 \times 2}{27 \times 2} = \frac{-14}{54} \\[1em] \therefore \frac{-1}{18} + \frac{-7}{27} = \frac{-3}{54} + \frac{-14}{54} \\[1em] = \frac{(-3) + (-14)}{54} = \frac{-17}{54}\end{aligned}$$.

Hence, the answer is $\frac{-17}{54}$

(vi) $\frac{21}{-4}$ and $\frac{-11}{8}$

We have:

$\frac{21}{-4}$ + $\frac{-11}{8}$

First, multiply the numerator and denominator of $\frac{21}{-4}$ by -1 to make denominator positive:

$\frac{21 \times (-1)}{-4 \times (-1)} = \frac{-21}{4}$.

LCM of denominators 4 and 8

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

L.C.M. = 2 x 2 x 2 = 8

Now, expressing each fraction with denominator 8:

$$\begin{aligned}\frac{-21}{4} = \frac{-21 \times 2}{4 \times 2} = \frac{-42}{8} \\[1em] \frac{-11}{8} = \frac{-11 \times 1}{8 \times 1} = \frac{-11}{8} \\[1em] \therefore \frac{-21}{4} + \frac{-11}{8} = \frac{-42}{8} + \frac{-11}{8} \\[1em] = \frac{(-42) + (-11)}{8} = \frac{-53}{8}\end{aligned}$$

Hence, the answer is $\frac{-53}{8}$

(i) $\frac{2}{-3} + \frac{-4}{9}$

First, express $\frac{2}{-3}$ with a positive denominator: $\frac{2 \times (-1)}{-3 \times (-1)} = \frac{-2}{3}$.

Let us find LCM of denominators 3 and 9

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

L.C.M. = 3 x 3 = 9

Now,

$$\begin{aligned}\frac{-2}{3} = \frac{-2 \times 3}{3 \times 3} = \frac{-6}{9} \\[1em] \therefore \frac{-6}{9} + \frac{-4}{9} = \frac{(-6) + (-4)}{9} \\[1em] = \frac{-10}{9}\end{aligned}$$

Hence, the answer is $\frac{-10}{9}$

(ii) $\frac{-1}{2} + \frac{-3}{4}$

Let us find LCM of denominators 2 and 4.

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

L.C.M. = 2 x 2 = 4

Now,

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

Hence, the answer is $\frac{-5}{4}$

(iii) $\frac{7}{-9} + \frac{-5}{6}$

First, express $\frac{7}{-9}$ with a positive denominator: $\frac{7 \times -1}{-9 \times -1} = \frac{-7}{9}$.

Let us find LCM of denominators 9 and 6.

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

L.C.M. = 3 x 3 x 2 = 18

Now, expressing each fraction with denominator 18:

$$\begin{aligned}\frac{-7}{9} = \frac{-7 \times 2}{9 \times 2} = \frac{-14}{18} \\[1em] \frac{-5}{6} = \frac{-5 \times 3}{6 \times 3} = \frac{-15}{18} \\[1em] \therefore \frac{-14}{18} + \frac{-15}{18} \\[1em] = \frac{(-14) + (-15)}{18} \\[1em] = \frac{-29}{18}\end{aligned}$$

Hence, the answer is $\frac{-29}{18}$

(iv) $2 + \frac{-3}{4}$

Express 2 as $\frac{2}{1}$.

LCM of denominators 1 and 4 is 4.

Now,

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

Hence, the answer is $\frac{5}{4}$

(v) $3 + \frac{-5}{6}$

Express 3 as $\frac{3}{1}$.

LCM of denominators 1 and 6 is 6.

Now,

$$\begin{aligned}\frac{3}{1} = \frac{3 \times 6}{1 \times 6} = \frac{18}{6} \\[1em] \therefore \frac{18}{6} + \frac{-5}{6} \\[1em] = \frac{18 + (-5)}{6} \\[1em] = \frac{13}{6}\end{aligned}$$

Hence, the answer is $\frac{13}{6}$

(vi) $-4 + \frac{2}{3}$

Express -4 as $\frac{-4}{1}$.

LCM of denominators 1 and 3 is 3.

Now,

$$\begin{aligned}\frac{-4}{1} = \frac{-4 \times 3}{1 \times 3} = \frac{-12}{3} \\[1em] \therefore \frac{-12}{3} + \frac{2}{3} \\[1em] = \frac{(-12) + 2}{3} \\[1em] = \frac{-10}{3}\end{aligned}$$

Hence, the answer is $\frac{-10}{3}$

(i) $\frac{-3}{8} + \frac{5}{8} + \frac{7}{8}$

Since the denominators are already the same and positive, we simply add the numerators.

$$\begin{aligned}\frac{-3 + 5 + 7}{8} \\[1em] = \frac{2 + 7}{8} \\[1em] = \frac{9}{8}\end{aligned}$$

Hence, the answer is $\frac{9}{8}$

(ii) $\frac{11}{3} + \frac{-5}{3} + \frac{-2}{3}$

Since the denominators are already the same and positive, we simply add the numerators.

$$\begin{aligned}\frac{11 + (-5) + (-2)}{3} \\[1em] = \frac{6 + (-2)}{3} \\[1em] = \frac{4}{3}\end{aligned}$$

Hence, the answer is $\frac{4}{3}$

(iii) $-1 + \frac{2}{-3} + \frac{5}{6}$

Express numbers as positive denominators: $\frac{-1}{1} + \frac{-2}{3} + \frac{5}{6}$.

LCM of denominators = LCM (1, 3, 6):

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

LCM = 3 x 2 = 6.

Now, expressing each fraction with denominator 6:

$$\begin{aligned}\frac{-1 \times 6}{1 \times 6} = \frac{-6}{6} \\[1em] \frac{-2 \times 2}{3 \times 2} = \frac{-4}{6} \\[1em] \frac{5 \times 1}{6 \times 1} = \frac{5}{6} \\[1em] \Rightarrow \frac{-6}{6} + \frac{-4}{6} + \frac{5}{6} \\[1em] \Rightarrow \frac{-6 + (-4) + 5}{6} \\[1em] \Rightarrow \frac{-10 + 5}{6} \\[1em] \Rightarrow \frac{-5}{6}\end{aligned}$$

Hence, the answer is $\frac{-5}{6}$

(iv) $\frac{7}{26} + \frac{-11}{13} + 2$

LCM of denominators = LCM (26, 13, 1):

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

LCM = 13 x 2 = 26.

Now, expressing each fraction with denominator 26:

$$\begin{aligned}\frac{7 \times 1}{26 \times 1} = \frac{7}{26} \\[1em] \frac{-11 \times 2}{13 \times 2} = \frac{-22}{26} \\[1em] \frac{2 \times 26}{1 \times 26} = \frac{52}{26} \\[1em] \Rightarrow \frac{7}{26} + \frac{-22}{26} + \frac{52}{26} \\[1em] \Rightarrow \frac{7 + (-22) + 52}{26} \\[1em] \Rightarrow \frac{-15 + 52}{26} \\[1em] \Rightarrow \frac{37}{26}\end{aligned}$$

Hence, the answer is $\frac{37}{26}$

(v) $3 + \frac{-7}{8} + \frac{-3}{4}$

LCM of denominators = LCM (1, 8, 4):

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

LCM = 2 x 2 x 2 = 8.

Now, expressing each fraction with denominator 8:

$$\begin{aligned}\frac{3 \times 8}{1 \times 8} = \frac{24}{8} \\[1em] \frac{-7 \times 1}{8 \times 1} = \frac{-7}{8} \\[1em] \frac{-3 \times 2}{4 \times 2} = \frac{-6}{8} \\[1em] \Rightarrow \frac{24}{8} + \frac{-7}{8} + \frac{-6}{8} \\[1em] \Rightarrow \frac{24 + (-7) + (-6)}{8} \\[1em] \Rightarrow \frac{17 + (-6)}{8} \\[1em] \Rightarrow \frac{11}{8}\end{aligned}$$

Hence, the answer is $\frac{11}{8}$

(vi) $\frac{-13}{8} + \frac{7}{16} + \frac{-3}{4}$

LCM of denominators = LCM (8, 16, 4):

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

LCM = 2 x 2 x 2 x 2 = 16.

Now, expressing each fraction with denominator 16:

$$\begin{aligned}\frac{-13 \times 2}{8 \times 2} = \frac{-26}{16} \\[1em] \frac{7 \times 1}{16 \times 1} = \frac{7}{16} \\[1em] \frac{-3 \times 4}{4 \times 4} = \frac{-12}{16} \\[1em] \Rightarrow \frac{-26}{16} + \frac{7}{16} + \frac{-12}{16} \\[1em] \Rightarrow \frac{-26 + 7 + (-12)}{16} \\[1em] \Rightarrow \frac{-19 + (-12)}{16} \\[1em] \Rightarrow \frac{-31}{16}\end{aligned}$$.

Hence, the answer is $\frac{-31}{16}$

(i) 9

Since, 9 + (-9) = 0

∴ Additive inverse of 9 is -9.

(ii) -11

Since, -11 + 11 = 0

∴ Additive inverse of -11 is 11.

(iii) $\frac{-8}{13}$

For a rational number $\frac{a}{b}$, the additive inverse is $\frac{-a}{b}$.

$\frac{-8}{13}$ + $\frac{8}{13}$ = 0

∴ Additive inverse of $\frac{-8}{13}$ is $\frac{8}{13}$.

(iv) $\frac{5}{-6}$

First, express the number with a positive denominator:

$\frac{5}{-6} = \frac{5 \times (-1)}{-6 \times (-1)} = \frac{-5}{6}$.

Now, find the additive inverse:

$\frac{-5}{6}$ + $\frac{5}{6}$ = 0

∴ Additive inverse of $\frac{5}{-6} = \frac{5}{6}$.

(v) 0

Zero is its own additive inverse because 0 + 0 = 0.

∴ Additive inverse of 0 is 0.

(i) $\frac{3}{5}$ from $\frac{1}{2}$

We have:

$$\begin{aligned}\phantom{=} \Big(\frac{1}{2} - \frac{3}{5}\Big) \\[1em] = \frac{1}{2} + \Big(\text{additive inverse of } \frac{3}{5}\Big) \\[1em] = \frac{1}{2} + \frac{-3}{5}\end{aligned}$$

The L.C.M. of 2 and 5 is 10.

Now, expressing each fraction with denominator 10:

$$\begin{aligned}= \frac{1 \times 5}{2 \times 5} + \frac{-3 \times 2}{5 \times 2} \\[1em] = \frac{5}{10} + \frac{-6}{10} \\[1em] = \frac{5 + (-6)}{10} \\[1em] = \frac{-1}{10}\end{aligned}$$

Hence, the answer is $\frac{-1}{10}$

(ii) $\frac{-4}{7}$ from $\frac{2}{3}$

We have:

$$\begin{aligned}\phantom{=} \Big(\frac{2}{3} - \frac{-4}{7}\Big) \\[1em] = \frac{2}{3} + \Big(\text{additive inverse of } \frac{-4}{7}\Big) \\[1em] = \frac{2}{3} + \frac{4}{7}\end{aligned}$$

The L.C.M. of 3 and 7 is 21.

Now, expressing each fraction with denominator 21:

$$\begin{aligned}= \frac{2 \times 7}{3 \times 7} + \frac{4 \times 3}{7 \times 3} \\[1em] = \frac{14}{21} + \frac{12}{21} \\[1em] = \frac{14 + 12}{21} \\[1em] = \frac{26}{21}\end{aligned}$$

Hence, the answer is $\frac{26}{21}$

(iii) $\frac{-5}{6}$ from $\frac{-3}{4}$

We have:

$$\begin{aligned}\phantom{=} \Big(\frac{-3}{4} - \frac{-5}{6}\Big) \\[1em] = \frac{-3}{4} + \Big(\text{additive inverse of } \frac{-5}{6}\Big) \\[1em] = \frac{-3}{4} + \frac{5}{6}\end{aligned}$$

The L.C.M. of 4 and 6 is 12.

Now, expressing each fraction with denominator 12:

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

Hence, the answer is $\frac{1}{12}$

(iv) $\frac{-7}{9}$ from 0

We have:

$$\begin{aligned}\phantom{=} \Big(0 - \frac{-7}{9}\Big) \\[1em] = 0 + \Big(\text{additive inverse of } \frac{-7}{9}\Big) \\[1em] = 0 + \frac{7}{9} \\[1em] = \frac{7}{9}\end{aligned}$$

Hence, the answer is $\frac{7}{9}$

(v) 4 from $\frac{-6}{11}$

we have:

$$\begin{aligned}\phantom{=} \Big(\frac{-6}{11} - 4\Big) \\[1em] = \frac{-6}{11} + \Big(\text{additive inverse of } 4\Big) \\[1em] = \frac{-6}{11} + \frac{-4}{1}\end{aligned}$$

The L.C.M. of 11 and 1 is 11.

Now, expressing each fraction with denominator 11:

$$\begin{aligned}= \frac{-6}{11} + \frac{-4 \times 11}{1 \times 11} \\[1em] = \frac{-6}{11} + \frac{-44}{11} \\[1em] = \frac{-6 + (-44)}{11} \\[1em] = \frac{-50}{11}\end{aligned}$$

Hence, the answer is $\frac{-50}{11}$

(vi) $\frac{3}{8}$ from $\frac{-5}{6}$

we have:

$$\begin{aligned}\phantom{=} \Big(\frac{-5}{6} - \frac{3}{8}\Big) \\[1em] = \frac{-5}{6} + \Big(\text{additive inverse of } \frac{3}{8}\Big) \\[1em] = \frac{-5}{6} + \frac{-3}{8}\end{aligned}$$

The L.C.M. of 6 and 8 is 24.

Now, expressing each fraction with denominator 24:

$$\begin{aligned}= \frac{-5 \times 4}{6 \times 4} + \frac{-3 \times 3}{8 \times 3} \\[1em] = \frac{-20}{24} + \frac{-9}{24} \\[1em] = \frac{-20 + (-9)}{24} \\[1em] = \frac{-29}{24}\end{aligned}$$

Hence, the answer is $\frac{-29}{24}$

(i) $\frac{5}{6} - \frac{7}{8}$

We have:

$$\begin{aligned}\phantom{=} \Big(\frac{5}{6} - \frac{7}{8}\Big) \\[1em] = \frac{5}{6} + \Big(\text{additive inverse of } \frac{7}{8}\Big) \\[1em] = \frac{5}{6} + \frac{-7}{8}\end{aligned}$$

L.C.M. of 6 and 8 is 24.

Now, expressing each fraction with denominator 24:

$$\begin{aligned}= \frac{5 \times 4}{6 \times 4} + \frac{-7 \times 3}{8 \times 3} \\[1em] = \frac{20}{24} + \frac{-21}{24} \\[1em] = \frac{20 + (-21)}{24} \\[1em] = \frac{-1}{24}\end{aligned}$$

Hence, the answer is $\frac{-1}{24}$

(ii) $\frac{5}{12} - \frac{17}{18}$

We have:

$$\begin{aligned}\phantom{=} \Big(\frac{5}{12} - \frac{17}{18}\Big) \\[1em] = \frac{5}{12} + \Big(\text{additive inverse of } \frac{17}{18}\Big) \\[1em] = \frac{5}{12} + \frac{-17}{18}\end{aligned}$$

L.C.M. of 12 and 18 is 36.

Now, expressing each fraction with denominator 36:

$$\begin{aligned}= \frac{5 \times 3}{12 \times 3} + \frac{-17 \times 2}{18 \times 2} \\[1em] = \frac{15}{36} + \frac{-34}{36} \\[1em] = \frac{15 + (-34)}{36} \\[1em] = \frac{-19}{36}\end{aligned}$$

Hence, the answer is $\frac{-19}{36}$

(iii) $\frac{11}{15} - \frac{13}{20}$

we have:

$$\begin{aligned}\phantom{=} \Big(\frac{11}{15} - \frac{13}{20}\Big) \\[1em] = \frac{11}{15} + \Big(\text{additive inverse of } \frac{13}{20}\Big) \\[1em] = \frac{11}{15} + \frac{-13}{20}\end{aligned}$$

L.C.M. of 15 and 20 is 60.

Now, expressing each fraction with denominator 60:

$$\begin{aligned}= \frac{11 \times 4}{15 \times 4} + \frac{-13 \times 3}{20 \times 3} \\[1em] = \frac{44}{60} + \frac{-39}{60}\\[1em] = \frac{44 + (-39)}{60} \\[1em] = \frac{5}{60}\end{aligned}$$

Hence, the answer is $\frac{5}{60}$

(iv) $\frac{-5}{9} - \frac{-2}{3}$

We have:

$$\begin{aligned}\phantom{=} \Big(\frac{-5}{9} - \frac{-2}{3}\Big) \\[1em] = \frac{-5}{9} + \Big(\text{additive inverse of } \frac{-2}{3}\Big) \\[1em] = \frac{-5}{9} + \frac{2}{3}\end{aligned}$$

L.C.M. of 9 and 3 is 9.

Now, expressing each fraction with denominator 9:

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

Hence, the answer is $\frac{1}{9}$

(v) $\frac{6}{11} - \frac{-3}{4}$

We have:

$$\begin{aligned}\phantom{=} \Big(\frac{6}{11} - \frac{-3}{4}\Big) \\[1em] = \frac{6}{11} + \Big(\text{additive inverse of } \frac{-3}{4}\Big) \\[1em] = \frac{6}{11} + \frac{3}{4}\end{aligned}$$

L.C.M. of 11 and 4 is 44.

Now, expressing each fraction with denominator 44:

$$\begin{aligned}= \frac{6 \times 4}{11 \times 4} + \frac{3 \times 11}{4 \times 11} \\[1em] = \frac{24}{44} + \frac{33}{44} \\[1em] = \frac{24 + 33}{44} \\[1em] = \frac{57}{44}\end{aligned}$$

Hence, the answer is $\frac{57}{44}$

(vi) $\frac{-2}{3} - \frac{3}{4}$

We have:

$$\begin{aligned}\phantom{=} \Big(\frac{-2}{3} - \frac{3}{4}\Big) \\[1em] = \frac{-2}{3} + \Big(\text{additive inverse of } \frac{3}{4}\Big) \\[1em] = \frac{-2}{3} + \frac{-3}{4}\end{aligned}$$

L.C.M. of 3 and 4 is 12.

Now, expressing each fraction with denominator 12:

$$\begin{aligned}= \frac{-2 \times 4}{3 \times 4} + \frac{-3 \times 3}{4 \times 3} \\[1em] = \frac{-8}{12} + \frac{-9}{12} \\[1em] = \frac{-8 + (-9)}{12} \\[1em] = \frac{-17}{12}\end{aligned}$$

Hence, the answer is $\frac{-17}{12}$

Given:

Let p and q be two rational numbers.

One rational number = p = $\frac{7}{16}$

Other rational number = q = ?

Sum of two rational numbers = (p + q) = $\frac{-5}{8}$

q = $\frac{-5}{8}$ - p

Substituting the values in above, we get:

$$\begin{aligned}q = \frac{-5}{8} - \frac{7}{16} \\[1em] = \frac{-5}{8} + \Big(\text{additive inverse of } \frac{7}{16}\Big) \\[1em] q = \frac{-5}{8} + \frac{-7}{16}\end{aligned}$$

L.C.M. of 8 and 16 is 16.

Now, expressing each fraction with denominator 16:

$$\begin{aligned}\frac{-5 \times 2}{8 \times 2} + \frac{-7 \times 1}{16 \times 1} \\[1em] = \frac{-10}{16} + \frac{-7}{16} \\[1em] = \frac{-10 + (-7)}{16} \\[1em] = \frac{-17}{16}\end{aligned}$$

The other rational number q is $\frac{-17}{16}$.

Let p and q be two rational numbers.

One rational number = p = $\frac{-3}{5}$

Other rational number = q = ?

Sum of two rational numbers = (p + q) = -4

q = -4 - p

Substituting the values in above, we get:

$$\begin{aligned}q = -4 - \Big(\frac{-3}{5}\Big) \\[1em] = -4 + \Big(\text{additive inverse of } \frac{-3}{5}\Big) \\[1em] q = \frac{-4}{1} + \frac{3}{5}\end{aligned}$$

L.C.M. of 1 and 5 is 5.

Now, expressing each fraction with denominator 5:

$$\begin{aligned}\frac{-4 \times 5}{1 \times 5} + \frac{3 \times 1}{5 \times 1} \\[1em] = \frac{-20}{5} + \frac{3}{5} \\[1em] = \frac{-20 + 3}{5} \\[1em] = \frac{-17}{5}\end{aligned}$$

The other rational number q is $\frac{-17}{5}$.

Let p and q be two rational numbers.

One rational number = p = -3

Other rational number = q = ?

Sum of two rational numbers = (p + q) = $\frac{-5}{4}$

q = $\frac{-5}{4}$ - p

Substituting the values in above, we get:

$$\begin{aligned}q = \frac{-5}{4} - (-3) \\[1em] = \frac{-5}{4} + \Big(\text{additive inverse of } -3\Big) \\[1em] q = \frac{-5}{4} + \frac{3}{1}\end{aligned}$$

L.C.M. of 4 and 1 is 4.

Now, expressing each fraction with denominator 4:

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

The other rational number q is $\frac{7}{4}$.

Let the required number be x. Then,

$$\begin{aligned}\frac{-5}{6} + x = \frac{-2}{3} \\[1em] \Rightarrow x = \frac{-2}{3} - \left(\frac{-5}{6}\right) \\[1em] = \frac{-2}{3} + \left(\text{additive inverse of } \frac{-5}{6}\right) \\[1em] = \frac{-2}{3} + \frac{5}{6}\end{aligned}$$

L.C.M. of denominators 3 and 6 is 6.

Now, expressing each fraction with denominator 6:

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

The required number is $\frac{1}{6}$.

Let the required number be x. Then,

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

L.C.M. of denominators 1 and 5 is 5.

Now, expressing each fraction with denominator 5:

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

The required number is $\frac{-7}{5}$.

Let the required number be x. Then,

$$\begin{aligned}\frac{-3}{4} - x = \frac{-5}{6} \\[1em] \Rightarrow \frac{-3}{4} = \frac{-5}{6} + x \\[1em] \Rightarrow x = \frac{-3}{4} - \left(\frac{-5}{6}\right) \\[1em] = \frac{-3}{4} + \left(\text{additive inverse of } \frac{-5}{6}\right) \\[1em] x = \frac{-3}{4} + \frac{5}{6}\end{aligned}$$

L.C.M. of denominators 4 and 6 is 12.

Now, expressing each fraction with denominator 12:

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

The required number is $\frac{1}{12}$.

Let the required number be x. Then,

$$\begin{aligned}\frac{-2}{3} - x = 1 \\[1em] \Rightarrow \frac{-2}{3} = 1 + x \\[1em] \Rightarrow x = \frac{-2}{3} - 1 \\[1em] = \frac{-2}{3} + (\text{additive inverse of } 1) \\[1em] = \frac{-2}{3} + \frac{-1}{1}\end{aligned}$$

L.C.M. of denominators 3 and 1 is 3.

Now, expressing each fraction with denominator 3:

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

The required number is $\frac{-5}{3}$.

(i) $\frac{2}{3}$ by $\frac{4}{5}$

We have:

$$\begin{aligned}\phantom{=} \frac{2}{3} \times \frac{4}{5} \\[1em] = \frac{2 \times 4}{3 \times 5} \\[1em] = \frac{8}{15}\end{aligned}$$

Hence the answer is $\frac{8}{15}$

(ii) $\frac{7}{6}$ by $\frac{9}{2}$

We have:

$$\begin{aligned}\phantom{=} \frac{7}{6} \times \frac{9}{2} \\[1em] = \frac{7 \times 9}{6 \times 2} \\[1em] = \frac{63}{12} \\[1em] = \frac{21}{4} \quad \text{[Dividing both by 3]}\end{aligned}$$

Hence the answer is $\frac{21}{4}$

(iii) $\frac{5}{6}$ by 30

We have:

$$\begin{aligned}\phantom{=} \frac{5}{6} \times 30 \\[1em] = \frac{5}{6} \times \frac{30}{1} \\[1em] = \frac{5 \times 30}{6 \times 1} \\[1em] = \frac{150}{6} \\[1em] = 25\end{aligned}$$

Hence the answer is 25

(iv) $\frac{-3}{4}$ by $\frac{8}{7}$

We have:

$$\begin{aligned}\phantom{=} \frac{-3}{4} \times \frac{8}{7} \\[1em] = \frac{-3 \times 8}{4 \times 7} \\[1em] = \frac{-24}{28} \\[1em] = \frac{-6}{7} \quad \text{[Dividing both by 4]}\end{aligned}$$

Hence the answer is $\frac{-6}{7}$

(v) $\frac{-16}{9}$ by $\frac{12}{-5}$

We have:

$$\begin{aligned}\phantom{=} \frac{-16}{9} \times \frac{12 \times (-1)}{-5 \times (-1)} \\[1em] \text{[Making denominator positive]} \\[1em] = \frac{-16}{9} \times \frac{-12}{5} \\[1em] = \frac{(-16) \times (-12)}{9 \times 5} \\[1em] = \frac{192}{45} \\[1em] = \frac{64}{15} \quad \text{[Dividing both by 3]}\end{aligned}$$

Hence the answer is $\frac{64}{15}$

(vi) $\frac{35}{-8}$ by $\frac{12}{-5}$

We have:

$$\begin{aligned}\phantom{=} \frac{35 \times (-1)}{-8 \times (-1)} \times \frac{12 \times (-1)}{-5 \times (-1)} \\[1em] \text{[Making denominator positive]} \\[1em] = \frac{-35}{8} \times \frac{-12}{5} \\[1em] = \frac{35 \times 12}{(8) \times (5)} \\[1em] = \frac{420}{40} \\[1em] = \frac{42}{4} \\[1em] = \frac{21}{2} \quad \text{[Dividing both by 2]}\end{aligned}$$

Hence the answer is $\frac{21}{2}$

(vii) $\frac{-3}{10}$ by $\frac{-40}{9}$

We have:

$$\begin{aligned}\phantom{=} \frac{-3}{10} \times \frac{-40}{9} \\[1em] = \frac{(-3) \times (-40)}{10 \times 9} \\[1em] = \frac{120}{90} \\[1em] = \frac{12}{9} \\[1em] = \frac{4}{3} \quad \text{[Dividing both by 3]}\end{aligned}$$

Hence the answer is $\frac{4}{3}$

(viii) $\frac{-32}{5}$ by $\frac{15}{-16}$

We have:

$$\begin{aligned}\phantom{=} \frac{-32}{5} \times \frac{15 \times (-1)}{-16 \times (-1)} \\[1em] = \frac{(-32) \times -15}{5 \times (16)} \\[1em] = \frac{480}{80} \\[1em] = 6 \hspace{2cm}\text{[Dividing both by 80]}\end{aligned}$$

Hence the answer is 6

(ix) $\frac{-8}{15}$ by $\frac{-25}{32}$

We have:

$$\begin{aligned}\phantom{=} \frac{-8}{15} \times \frac{-25}{32} \\[1em] = \frac{(-8) \times (-25)}{15 \times 32} \\[1em] = \frac{200}{480} \\[1em] = \frac{20}{48} \\[1em] = \frac{5}{12} \quad \text{[Dividing both by 4]}\end{aligned}$$

Hence the answer is $\frac{5}{12}$

(i) $\frac{7}{15} \times \frac{5}{6}$

We have:

$$\begin{aligned}\phantom{=} \frac{7}{15} \times \frac{5}{6} \\[1em] = \frac{7}{3} \times \frac{1}{6} \\[1em] = \frac{7 \times 1}{3 \times 6} \\[1em] = \frac{7}{18}\end{aligned}$$

Hence the answer is $\frac{7}{18}$

(ii) $\frac{-5}{24} \times \frac{6}{25}$

We have:

$$\begin{aligned}\phantom{=} \frac{-5}{24} \times \frac{6}{25} \\[1em] = \frac{-1}{24} \times \frac{6}{5} \\[1em] = \frac{-1}{4} \times \frac{1}{5} \\[1em] = \frac{-1 \times 1}{4 \times 5} \\[1em] = \frac{-1}{20}\end{aligned}$$

Hence the answer is $\frac{-1}{20}$

(iii) $\frac{7}{-18} \times \frac{-9}{14}$

First, express $\frac{7}{-18}$ with a positive denominator: $\frac{7 \times (-1)}{18 \times (-1)} = \frac{-7}{18}$.

We have:

$$\begin{aligned}\phantom{=} \frac{-7}{18} \times \frac{-9}{14} \\[1em] = \frac{-1}{18} \times \frac{-9}{2} \\[1em] = \frac{(-1) \times (-9)}{18 \times 2} \\[1em] = \frac{(1) \times (9)}{18 \times 2} \\[1em] = \frac{9}{36} \\[1em] = \frac{1}{4}\end{aligned}$$

Hence the answer is $\frac{1}{4}$

(iv) $\frac{-9}{5} \times \frac{-10}{3}$

We have:

$$\begin{aligned}\phantom{=} \frac{-9}{5} \times \frac{-10}{3} \\[1em] = \frac{-3}{5} \times \frac{-10}{1} \\[1em] = \frac{-3}{1} \times \frac{-2}{1} \\[1em] = \frac{(-3) \times (-2)}{1 \times 1} \\[1em] = \frac{3 \times 2}{1 \times 1} \\[1em] = \frac{6}{1} = 6\end{aligned}$$

Hence the answer is 6

(v) $-28 \times \frac{-8}{7}$

Express -28 as $\frac{-28}{1}$

We have:

$$\begin{aligned}\phantom{=} \frac{-28}{1} \times \frac{-8}{7} \\[1em] = \frac{-4}{1} \times \frac{-8}{1} \\[1em] = \frac{(-4) \times (-8)}{1 \times 1} \\[1em] = \frac{(4) \times (8)}{1 \times 1} \\[1em] = \frac{32}{1} = 32\end{aligned}$$

Hence the answer is 32

(vi) $\frac{8}{-21} \times \frac{-14}{3}$

First, express $\frac{8}{-21}$ with a positive denominator: $\frac{8 \times (-1)}{-21 \times (-1)} = \frac{-8}{21}$.

We have:

$$\begin{aligned}\phantom{=} \frac{-8}{21} \times \frac{-14}{3} \\[1em] = \frac{-8}{3} \times \frac{-2}{3} \\[1em] = \frac{(-8) \times (-2)}{3 \times 3} \\[1em] = \frac{8 \times 2}{3 \times 3} \\[1em] = \frac{16}{9}\end{aligned}$$

Hence the answer is $\frac{16}{9}$

(i) $\frac{5}{12} \times (-36)$

Express -36 as $\frac{-36}{1}$

We have:

$$\begin{aligned}\phantom{=} \frac{5}{12} \times \frac{-36}{1} \\[1em] = \frac{5}{1} \times \frac{-3}{1} \\[1em] = \frac{5 \times (-3)}{1 \times 1} \\[1em] = \frac{-15}{1} = -15\end{aligned}$$

Hence the answer is -15

(ii) $\frac{-17}{18} \times 12$

Express 12 as $\frac{12}{1}$

We have:

$$\begin{aligned}\phantom{=} \frac{-17}{18} \times \frac{12}{1} \\[1em] = \frac{-17}{3} \times \frac{2}{1} \\[1em] = \frac{-17 \times 2}{3 \times 1} \\[1em] = \frac{-34}{3}\end{aligned}$$

Hence the answer is $\frac{-34}{3}$

(iii) $\frac{-5}{6} \times \frac{6}{5}$

We have:

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

Hence the answer is -1

(iv) $-14 \times \frac{9}{28}$

Express -14 as $\frac{-14}{1}$

We have:

$$\begin{aligned}\phantom{=} \frac{-14}{1} \times \frac{9}{28} \\[1em] = \frac{-1}{1} \times \frac{9}{2} \\[1em] = \frac{-1 \times 9}{1 \times 2} \\[1em] = \frac{-9}{2}\end{aligned}$$

Hence the answer is $\frac{-9}{2}$

(v) $-4\frac{4}{5} \times \Big(-7\frac{1}{2}\Big)$

We have:

$$\begin{aligned}\phantom{=} \frac{-24}{5} \times \frac{-15}{2} \\[1em] \text{[Converting mixed to improper fraction]} \\[1em] = \frac{-12}{5} \times \frac{-15}{1} \\[1em] = \frac{-12}{1} \times \frac{-3}{1} \\[1em] = \frac{-12 \times (-3)}{1 \times 1} \\[1em] = \frac{12 \times 3}{1 \times 1} \\[1em] = \frac{36}{1} = 36\end{aligned}$$

Hence the answer is 36

(vi) $\frac{-8}{15} \times \frac{-25}{32}$

We have:

$$\begin{aligned}\phantom{=} \frac{-8}{15} \times \frac{-25}{32} \\[1em] = \frac{-1}{15} \times \frac{-25}{4} \\[1em] = \frac{-1}{3} \times \frac{-5}{4} \\[1em] = \frac{-1 \times (-5)}{3 \times 4} \\[1em] = \frac{1 \times 5}{3 \times 4} \\[1em] = \frac{5}{12}\end{aligned}$$

Hence the answer is $\frac{5}{12}$

(i) $\Big(\frac{2}{5} \times \frac{5}{8}\Big) + \Big(\frac{-3}{7} \times \frac{14}{-15}\Big)$

First, express $\frac{14}{-15}$ with a positive denominator: $\frac{14 \times (-1)}{-15 \times (-1)} = \frac{-14}{15}$.

We have:

$$\begin{aligned}\phantom{=} \Big(\frac{2}{5} \times \frac{5}{8}\Big) + \Big(\frac{-3}{7} \times \frac{-14}{15}\Big) \\[1em] = \Big(\frac{1}{1} \times \frac{1}{4}\Big) + \Big(\frac{-1}{1} \times \frac{-2}{5}\Big) \\[1em] = \Big(\frac{1}{1} \times \frac{1}{4}\Big) + \Big(\frac{-1}{1} \times \frac{-2}{5}\Big) \\[1em] = \frac{1}{4} + \frac{2}{5}\end{aligned}$$

L.C.M. of 4 and 5 is 20.

Now, expressing each fraction with denominator 20:

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

Hence the answer is $\frac{13}{20}$

(ii) $\Big(\frac{-14}{3} \times \frac{-12}{7}\Big) + \Big(\frac{-6}{25} \times \frac{15}{8}\Big)$

We have:

$$\begin{aligned}\phantom{=} \Big(\frac{-14}{3} \times \frac{-12}{7}\Big) + \Big(\frac{-6}{25} \times \frac{15}{8}\Big) \\[1em] = \Big(\frac{-2}{1} \times \frac{-4}{1}\Big) + \Big(\frac{-3}{5} \times \frac{3}{4}\Big) \\[1em] = \frac{8}{1} + \frac{-9}{20}\end{aligned}$$

L.C.M. of 1 and 20 is 20.

Now, expressing each fraction with denominator 20:

$$\begin{aligned}= \frac{8 \times 20}{1 \times 20} + \frac{-9}{20} \\[1em] = \frac{160}{20} + \frac{-9}{20} \\[1em] = \frac{160 - 9}{20} \\[1em] = \frac{151}{20}\end{aligned}$$

Hence the answer is $\frac{151}{20}$

(iii) $\Big(\frac{6}{25} \times \frac{-15}{8}\Big) - \Big(\frac{13}{100} \times \frac{-25}{26}\Big)$

We have:

$$\begin{aligned}\phantom{=} \Big(\frac{6}{25} \times \frac{-15}{8}\Big) - \Big(\frac{13}{100} \times \frac{-25}{26}\Big) \\[1em] = \Big(\frac{3}{5} \times \frac{-3}{4}\Big) - \Big(\frac{1}{4} \times \frac{-1}{2}\Big) \\[1em] = \Big(\frac{3 \times (-3)}{5 \times 4}\Big) - \Big(\frac{1 \times (-1)}{4 \times 2}\Big) \\[1em] = \Big(\frac{-9}{20}\Big) - \Big(\frac{-1}{8}\Big)\end{aligned}$$

L.C.M. of 20 and 8 is 40.

Now, expressing each fraction with denominator 40:

$$\begin{aligned}= \frac{-9 \times 2}{20 \times 2} - \frac{-1 \times 5}{8 \times 5} \\[1em] = \frac{-18}{40} - \Big(\frac{-5}{40}\Big) \\[1em] = \frac{-18}{40} + \frac{5}{40} \\[1em] = \frac{-18 + 5}{40} \\[1em] = \frac{-13}{40}\end{aligned}$$

Hence the answer is $\frac{-13}{40}$.

(iv) $\Big(\frac{-14}{5} \times \frac{-10}{7}\Big) - \Big(\frac{-8}{9} \times \frac{3}{16}\Big)$

We have:

$$\begin{aligned}\phantom{=} \Big(\frac{-14}{5} \times \frac{-10}{7}\Big) - \Big(\frac{-8}{9} \times \frac{3}{16}\Big) \\[1em] = \Big(\frac{-2}{1} \times \frac{-2}{1}\Big) - \Big(\frac{-1}{3} \times \frac{1}{2}\Big) \\[1em] = \Big(\frac{-2 \times (-2)}{1 \times 1}\Big) - \Big(\frac{-1 \times 1}{3 \times 2}\Big) \\[1em] = \frac{4}{1} - \Big(\frac{-1}{6}\Big) \\[1em] = \frac{4}{1} + \frac{1}{6}\end{aligned}$$

L.C.M. of 1 and 6 is 6.

Now, expressing each fraction with denominator 6:

$$\begin{aligned}= \frac{4 \times 6}{1 \times 6} + \frac{1}{6} \\[1em] \frac{24}{6} + \frac{1}{6} \\[1em] = \frac{24 + 1}{6} \\[1em] = \frac{25}{6}\end{aligned}$$

Hence the answer is $\frac{25}{6}$.

Given:

Quantity of rice = $3\frac{1}{3}$ kg = $\frac{10}{3}$ kg

Cost per kg = ₹ $40\frac{1}{2}$ = ₹ $\frac{81}{2}$

Cost of $3\frac{1}{3}$ kg of rice = ?

Cost of $3\frac{1}{3}$ kg of rice = (Quantity of rice) x (Cost per kg)

Substituting the values in above, we get:

Cost of $3\frac{1}{3}$ kg of rice = $\frac{10}{3}$ kg x ₹ $\frac{81}{2}$

$$\begin{aligned}= ₹ \frac{10}{3} \times \frac{81}{2} \\[1em] = ₹ \frac{5}{1} \times \frac{27}{1} \\[1em] = ₹ \frac{5 \times 27}{1 \times 1} \\[1em] = ₹ \frac{135}{1} \\[1em] = ₹ 135\end{aligned}$$

Hence, the cost of $3\frac{1}{3}$ kg of rice is ₹ 135.

Given:

Time taken = $2\frac{2}{5}$ hours = $\frac{12}{5}$ hours

Speed of the car = $46\frac{2}{3}$ km per hour = $\frac{140}{3}$ km per hour

Total distance = ?

We know the formula,

Distance = Speed x Time

Substituting the values in above, we get:

Distance = $\frac{140}{3}$ km per hour x $\frac{12}{5}$ hours

$$\begin{aligned}= \frac{140}{3} \times \frac{12}{5} \text{ km} \\[1em] = \frac{28}{1} \times \frac{4}{1} \text{ km} \\[1em] = \frac{28 \times 4}{1 \times 1} \text{ km} \\[1em] = \frac{112}{1} \text{ km} \\[1em] = 112 \text{ km}\end{aligned}$$

Hence, the distance covered by the car is 112 km.

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

Since, $\frac{5}{6}$ x $\frac{6}{5}$ = 1

∴ The multiplicative inverse of $\frac{5}{6}$ is $\frac{6}{5}$.

(ii) $\frac{-3}{7}$

Since, $\frac{-3}{7}$ x $\frac{-7}{3}$ = 1

∴ The multiplicative inverse of $\frac{-3}{7}$ is $\frac{-7}{3}$.

(iii) -8

Express -8 as $\frac{-8}{1}$

Since, $\frac{-8}{1}$ x $\frac{-1}{8}$ = 1

∴ The multiplicative inverse of $\frac{-8}{1}$ is $\frac{-1}{8}$.

(iv) $\frac{-11}{3}$

Since, $\frac{-11}{3}$ x $\frac{-3}{11}$ = 1

∴ The multiplicative inverse of $\frac{-11}{3}$ is $\frac{-3}{11}$.

(v) $\frac{-1}{8}$

Since, $\frac{-1}{8}$ x $\frac{-8}{1}$ = 1

∴ The multiplicative inverse of $\frac{-1}{8}$ is -8.

(i) $\frac{6}{25}$

Since, $\frac{6}{25}$ x $\frac{25}{6}$ = 1

∴ The multiplicative inverse of $\frac{6}{25}$ is $\frac{25}{6}$.

(ii) $-2\frac{3}{11}$

First, convert the mixed fraction to an improper fraction: $\frac{-(2 \times 11 + 3)}{11} = \frac{-25}{11}$.

Since, $\frac{-25}{11}$ x $\frac{-11}{25}$ = 1

∴ The multiplicative inverse of $\frac{-25}{11}$ is $\frac{-11}{25}$.

(iii) $\frac{-8}{9}$

Since, $\frac{-8}{9}$ x $\frac{-9}{8}$ = 1

∴ The multiplicative inverse of $\frac{-8}{9}$ is $\frac{-9}{8}$.

(iv) $\frac{-23}{16}$

Since, $\frac{-23}{16}$ x $\frac{-16}{23}$ = 1

∴ The multiplicative inverse of $\frac{-23}{16}$ is $\frac{-16}{23}$.

(v) 12

Express 12 as $\frac{12}{1}$

Since, $\frac{12}{1}$ x $\frac{1}{12}$ = 1

∴ The multiplicative inverse of $\frac{12}{1}$ is $\frac{1}{12}$.

(vi) $\frac{1}{10}$

Since, $\frac{1}{10}$ x $\frac{10}{1}$ = 1

∴ The multiplicative inverse of $\frac{1}{10}$ is $\frac{10}{1}$, which is 10.

(vii) -6

Express -6 as $\frac{-6}{1}$.

Since, $\frac{-6}{1}$ x $\frac{-1}{6}$ = 1

∴ The multiplicative inverse of $\frac{-6}{1}$ is $\frac{-1}{6}$.

(viii) -1

Express -1 as $\frac{-1}{1}$.

Since, $\frac{-1}{1}$ x $\frac{-1}{1}$ = 1

∴ The multiplicative inverse of $\frac{-1}{1}$ is $\frac{-1}{1}$, which is still -1.

(ix) $\frac{-1}{5}$

Since, $\frac{-1}{5}$ x $\frac{-5}{1}$ = 1

∴ The multiplicative inverse of $\frac{-1}{5}$ is $\frac{-5}{1}$, which is -5.

(x) $\frac{-7}{-9}$

First, simplify the fraction: $\frac{-7}{-9} = \frac{7}{9}$.

Since, $\frac{7}{9}$ x $\frac{9}{7}$ = 1

∴ The multiplicative inverse of $\frac{7}{9}$ is $\frac{9}{7}$.

(i) $\frac{7}{12} ÷ \frac{-4}{3}$

We have:

$$\begin{aligned}\frac{7}{12} \div \frac{-4}{3} \\[1em] = \frac{7}{12} \times \frac{3}{-4} \quad \left[\text{Reciprocal of } \frac{-4}{3} \text{ is } \frac{3}{-4}\right] \\[1em] =\frac{7}{12} \times \frac{3 \times (-1)}{-4 \times (-1)} \\[1em] = \frac{7}{12} \times \frac{-3}{4} \\[1em] = \frac{7 \times (-3)}{12 \times 4} \\[1em] = \frac{7 \times (-1)}{4 \times 4} \\[1em] = \frac{-7}{16}\end{aligned}$$

Hence, the answer is $\frac{-7}{16}$

(ii) $\frac{-12}{25} ÷ \frac{-5}{6}$

We have:

$$\begin{aligned}\frac{-12}{25} \div \frac{-5}{6} \\[1em] = \frac{-12}{25} \times \frac{6}{-5} \quad \left[\text{Reciprocal of } \frac{-5}{6} \text{ is } \frac{6}{-5}\right] \\[1em] = \frac{-12}{25} \times \frac{6 \times (-1)}{-5 \times (-1)} \\[1em] = \frac{-12}{25} \times \frac{-6}{5} \\[1em] = \frac{(-12) \times (-6)}{25 \times 5} \\[1em] = \frac{72}{125}\end{aligned}$$

Hence, the answer is $\frac{72}{125}$

(iii) $\frac{-27}{32} ÷ \frac{-9}{16}$

We have:

$$\begin{aligned}\frac{-27}{32} \div \frac{-9}{16} \\[1em] = \frac{-27}{32} \times \frac{16}{-9} \quad \left[\text{Reciprocal of } \frac{-9}{16} \text{ is } \frac{16}{-9}\right] \\[1em] = \frac{-27}{32} \times \frac{16 \times (-1)}{-9 \times (-1)} \\[1em] = \frac{-27}{32} \times \frac{-16}{9} \\[1em] = \frac{-3}{32} \times \frac{-16}{1} \\[1em] = \frac{-3}{2} \times \frac{-1}{1} \\[1em] = \frac{(-3) \times (-1)}{2 \times 1} \\[1em] = \frac{3}{2}\end{aligned}$$

Hence, the answer is $\frac{3}{2}$

(iv) $-2\frac{4}{7} ÷ \frac{6}{35}$

We have:

$$\begin{aligned}\phantom{=} -2\frac{4}{7} ÷ \frac{6}{35} \\[1em] = -\frac{18}{7} ÷ \frac{6}{35} \\[1em] = \frac{-18}{7} \times \frac{35}{6} \quad \left[\text{Reciprocal of } \frac{6}{35} \text{ is } \frac{35}{6}\right] \\[1em] = \frac{-3}{7} \times \frac{35}{1} \\[1em] = \frac{-3}{1} \times \frac{5}{1} \\[1em] = \frac{-3 \times 5}{1 \times 1} \\[1em] = \frac{-15}{1} \\[1em] = -15\end{aligned}$$

Hence, the answer is -15

(v) $26 ÷ \frac{-1}{13}$

Express 26 as $\frac{26}{1}$

We have:

$$\begin{aligned}\frac{26}{1} \div \frac{-1}{13} \\[1em] = \frac{26}{1} \times \frac{13}{-1} \quad \left[\text{Reciprocal of } \frac{-1}{13} \text{ is } \frac{13}{-1}\right] \\[1em] \text{Make denominator positive} \\[1em] = \frac{26}{1} \times \frac{13 \times (-1)}{-1 \times (-1)} \\[1em] = \frac{26}{1} \times \frac{-13}{1} \\[1em] = \frac{26 \times (-13)}{1 \times 1} \\[1em] = \frac{-338}{1} \\[1em] = -338\end{aligned}$$

Hence, the answer is -338

(vi) $\frac{1}{25} ÷ -5$

Express -5 as $\frac{-5}{1}$

$$\begin{aligned}\frac{1}{25} \div \frac{-5}{1} \\[1em] = \frac{1}{25} \times \frac{1}{-5} \quad \left[\text{Reciprocal of } \frac{-5}{1} \text{ is } \frac{1}{-5}\right] \\[1em] = \frac{1}{25} \times \frac{1 \times (-1)}{-5 \times (-1)} \\[1em] = \frac{1}{25} \times \frac{-1}{5} \\[1em] = \frac{1 \times (-1)}{25 \times (5)} \\[1em] = \frac{-1}{125} \\[1em]\end{aligned}$$

Hence, the answer is $\frac{-1}{125}$

Let p and q be two rational numbers.

One rational number = p = $\frac{-8}{25}$

Other rational number = ?

Product of two rational numbers = p x q = $\frac{2}{5}$

q = $\frac{2}{5}$ ÷ p

Substituting the values in above, we get:

$$\begin{aligned}\text{q} = \frac{2}{5} \div \frac{-8}{25} \\[1em] = \frac{2}{5} \times \frac{25}{-8} \quad \left[\text{Reciprocal of } \frac{-8}{25} \text{ is } \frac{25}{-8}\right] \\[1em] = \frac{\overset{1}{\cancel{2}}}{\underset{1}{\cancel{5}}} \times \frac{\overset{-5}{\cancel{-25}}}{\underset{4}{\cancel{8}}} \quad \left[\because \frac{25}{-8} = \frac{25 \times (-1)}{-8 \times (-1)} = \frac{-25}{8}\right] \\[1em] = \frac{1}{1} \times \frac{-5}{4} \\[1em] = \frac{1 \times -5}{1 \times 4} \\[1em] = \frac{-5}{4}\end{aligned}$$

The other rational number q is $\frac{-5}{4}$.

Let p and q be two rational numbers.

One rational number = p = $\frac{16}{39}$

Other rational number = q = ?

Product of two rational numbers = p x q = $\frac{-2}{3}$

q = $\frac{-2}{3}$ ÷ p

Substituting the values in above, we get:

$$\begin{aligned}\text{q} = \frac{-2}{3} \div \frac{16}{39} \\[1em] = \frac{-2}{3} \times \frac{39}{16} \quad \left[\text{Reciprocal of } \frac{16}{39} \text{ is } \frac{39}{16}\right] \\[1em] = \frac{\overset{-1}{\cancel{-2}}}{\underset{1}{\cancel{3}}} \times \frac{\overset{13}{\cancel{39}}}{\underset{8}{\cancel{16}}} \\[1em] = \frac{-1}{1} \times \frac{13}{8} \\[1em] = \frac{-1 \times 13}{1 \times 8} \\[1em] = \frac{-13}{8}\end{aligned}$$

The other rational number q is $\frac{-13}{8}$.

Let the required number be x. Then,

$$\begin{aligned}\frac{-9}{35} \times x = \frac{3}{5} \\[1em] \Rightarrow x = \frac{3}{5} \div \frac{-9}{35} \\[1em] = \frac{3}{5} \times \frac{35}{-9} \quad \left[\text{Reciprocal of } \frac{-9}{35} \text{ is } \frac{35}{-9}\right] \\[1em] = \frac{3}{5} \times \frac{-35}{9} \quad \left[\because \frac{35}{-9} = \frac{35 \times (-1)}{-9 \times (-1)} = \frac{-35}{9}\right] \\[1em] = \frac{\overset{1}{\cancel{3}}}{\underset{1}{\cancel{5}}} \times \frac{\overset{-7}{\cancel{-35}}}{\underset{3}{\cancel{9}}} \\[1em] = \frac{1}{1} \times \frac{-7}{3} \quad \text{[Dividing 3 and 9 by 3, 35 and 5 by 5]} \\[1em] = \frac{1 \times (-7)}{1 \times 3} \\[1em] = \frac{-7}{3}\end{aligned}$$

Hence, the required number is $\frac{-7}{3}$.

Let the required number be x. Then,

$$\begin{aligned}\frac{25}{8} \times x = \frac{-20}{7} \\[1em] \Rightarrow x = \frac{-20}{7} \div \frac{25}{8} \\[1em] = \frac{-20}{7} \times \frac{8}{25} \quad \left[\text{Reciprocal of } \frac{25}{8} \text{ is } \frac{8}{25}\right] \\[1em] = \frac{-4}{7} \times \frac{8}{5} \quad \text{[Dividing 20 and 25 by 5]} \\[1em] = \frac{-4 \times 8}{7 \times 5} \\[1em] = \frac{-32}{35}\end{aligned}$$

The required number is $\frac{-32}{35}$.

Given:

Total cost = ₹ $59\frac{1}{2}$ = ₹ $\frac{119}{2}$

Total pencils = 17

Cost of 1 pencil = ?

Cost of 1 pencil = Total cost ÷ Total pencils

Substituting the values in above, we get:

Cost of 1 pencil = ₹ $\frac{119}{2} ÷ 17$

$$\begin{aligned}= ₹ \frac{119}{2} \div \frac{17}{1} \\[1em] = ₹ \frac{119}{2} \times \frac{1}{17} \quad \left[\text{Reciprocal of } \frac{17}{1} \text{ is } \frac{1}{17}\right] \\[1em] = ₹ \frac{7}{2} \times \frac{1}{1} \quad \text{[Dividing 119 by 17]} \\[1em] = ₹ \frac{7 \times 1}{2 \times 1} \\[1em] = ₹ \frac{7}{2} = ₹ 3\frac{1}{2}\end{aligned}$$  

The cost of each pencil is ₹ $3\frac{1}{2}$.

Given:

Total cost = ₹ 335

Total length = 20 metres

Cost of 1 metre of ribbon = ?

Cost of 1 metre of ribbon = Total cost ÷ Total length

Substituting the values in above, we get:

Cost of 1 metre of ribbon = ₹ 335 ÷ 20 metres

$$\begin{aligned}= ₹ \frac{335}{20} \\[1em] = ₹ \frac{67}{4} \quad \text{[Dividing 335 and 20 by 5]} \\[1em] = ₹ 16\frac{3}{4} \quad \text{[Converting to mixed fraction]}\end{aligned}$$

The cost of each metre is ₹ $16\frac{3}{4}$.

Given:

Total length of rope = 66 m

Length of each piece = $2\frac{3}{4}$ m = $\frac{11}{4}$ m

Number of pieces = ?

Number of pieces = (Total length ÷ length of each piece)

Substituting the values in above, we get:

Number of pieces = 66 m ÷ $\frac{11}{4}$ m

$$\begin{aligned}= \frac{66}{1} \times \frac{4}{11} \quad \left[\text{Reciprocal of }\frac{11}{4} \text{ is } \frac{4}{11}\right] \\[1em] = \frac{6}{1} \times \frac{4}{1} \quad \text{[Dividing 66 by 11]} \\[1em] = \frac{6 \times 4}{1 \times 1} \\[1em] = \frac{24}{1} = 24\end{aligned}$$

Hence, 24 pieces can be cut from the rope.

(i) 25 ÷ $\Big(\frac{-5}{6}\Big)$ = -30

(ii) 6 ÷ (-8) = $\Big(\frac{-3}{4}\Big)$

(iii) $\Big(\frac{-15}{14}\Big)$ ÷ $\Big(\frac{\bold{-3}}{\bold{7}}\Big)$ = $\frac{5}{2}$

(iv) (-16) ÷ $\Big(\frac{\bold{-8}}{\bold{3}}\Big)$ = 6

Explanation

(i) To find the dividend, multiply the quotient by the divisor: $-30 \times \frac{-5}{6} = 25$.

(ii) The dividend is found by multiplying the quotient and divisor: $\frac{-3}{4} \times -8 = 6$.

(iii) To find the divisor, divide the dividend by the quotient: $\frac{-15}{14} \div \frac{5}{2} = \frac{-3}{7}$.

(iv) The missing divisor is calculated by dividing the dividend by the result: $-16 \div 6 = \frac{-8}{3}$.

$\frac{2}{3}$

Draw a number line and mark 0, on the right side of 0 mark 1. Divide the distance between 0 and 1 into 3 equal parts. The 2nd mark from 0 is point P, which represents $\frac{2}{3}$.

$-\frac{5}{7}$

Draw a number line and mark 0, on the left side of 0 mark -1. Divide the distance between 0 and −1 into 7 equal parts. The 5th mark to the left of 0 is P, which represents $-\frac{5}{7}$.

$\frac{1}{6}$

Draw a number line and mark 0, on the right side of 0 mark 1. Divide the distance between 0 and 1 into 6 equal parts. The 1st mark from 0 is P, which represents $\frac{1}{6}$.

$-\frac{3}{8}$

Draw a number line and mark 0, on the left side of 0 mark -1. Divide the distance between 0 and −1 into 8 equal parts. The 3rd mark to the left of 0 P, which represents $-\frac{3}{8}$.

$\frac{22}{7}$

First convert to a mixed fraction to see which integers they fall between.

$\frac{22}{7} = 3\frac{1}{7}$

This lies between 3 and 4. Draw a number line and mark 0, on the right side of 0 mark the points 1, 2, 3, 4. Divide the distance between 3 and 4 into 7 equal parts.The 1st mark after 3 is point P, which represents $\frac{22}{7}$.

$\frac{23}{-5}$

First convert to a mixed fraction to see which integers they fall between.

$\frac{23}{-5} = -4\frac{3}{5}$

This lies between -4 and -5. Draw a number line and mark 0, on the left side of 0, mark the points -1, -2, -3, -4, -5. Divide the space between -4 and -5 into 5 equal parts. The 3rd mark to the left of -4 is P, which represents $-\frac{23}{5}$.

$-\frac{3}{4}$

Draw a number line and mark 0, on the left side of 0 mark -1. Divide the distance between 0 and −1 into 4 equal parts. The 3rd mark to the left of 0 is P, which represents $-\frac{3}{4}$.

$\frac{-12}{5}$

First convert to a mixed fraction to see which integers they fall between.

$\frac{-12}{5} = -2\frac{2}{5}$

This lies between -2 and -3. Draw a number line and mark 0, on the left side of 0, mark the points -1, -2, -3. Divide the space between -2 and -3 into 5 equal parts. The 2nd mark to the left of -2 is P, which represents $-\frac{12}{5}$.

$\frac{13}{6}$

First convert to a mixed fraction to see which integers they fall between.

$\frac{13}{6} = 2\frac{1}{6}$

This lies between 2 and 3. Draw a number line and mark 0, on the right side of 0, mark the points 1, 2, 3. Divide the space between 2 and 3 into 6 equal parts.The 1st mark after 2 is P, which represents $\frac{13}{6}$.

(i) $\frac{11}{16}$

The given number is $\frac{11}{16}$.

Its denominator is $16 = 2^4$.

Thus, the denominator of $\frac{11}{16}$ has no prime factor other than 2.

∴ $\frac{11}{16}$ is expressible as a terminating decimal.

(ii) $\frac{17}{20}$

The given number is $\frac{17}{20}$.

Its denominator is $20 = 2^2 \times 5^1$.

Thus, the denominator of $\frac{17}{20}$ has no prime factor other than 2 and 5.

∴ $\frac{17}{20}$ is expressible as a terminating decimal.

(iii) $\frac{44}{125}$

The given number is $\frac{44}{125}$.

Its denominator is $125 = 5^3$.

Thus, the denominator of $\frac{44}{125}$ has no prime factor other than 5.

∴ $\frac{44}{125}$ is expressible as a terminating decimal.

(iv) $\frac{9}{80}$

The given number is $\frac{9}{80}$.

Its denominator is $80 = 2^4 \times 5^1$.

Thus, the denominator of $\frac{9}{80}$ has no prime factor other than 2 and 5.

∴ $\frac{9}{80}$ is expressible as a terminating decimal.

(v) $\frac{123}{200}$

The given number is $\frac{123}{200}$.

Its denominator is $200 = 2^3 \times 5^2$.

Thus, the denominator of $\frac{123}{200}$ has no prime factor other than 2 and 5.

∴ $\frac{123}{200}$ is expressible as a terminating decimal.

(vi) $\frac{129}{320}$

The given number is $\frac{129}{320}$.

Its denominator is $320 = 2^6 \times 5^1$.

Thus, the denominator of $\frac{129}{320}$ has no prime factor other than 2 and 5.

∴ $\frac{129}{320}$ is expressible as a terminating decimal.

(vii) $\frac{431}{500}$

The given number is $\frac{431}{500}$.

Its denominator is $500 = 2^2 \times 5^3$.

Thus, the denominator of $\frac{431}{500}$ has no prime factor other than 2 and 5.

∴ $\frac{431}{500}$ is expressible as a terminating decimal.

(viii) $\frac{807}{1250}$

The given number is $\frac{807}{1250}$.

Its denominator is $1250 = 2^1 \times 5^4$.

Thus, the denominator of $\frac{807}{1250}$ has no prime factor other than 2 and 5.

∴ $\frac{807}{1250}$ is expressible as a terminating decimal.

(i) $\frac{11}{8}$

By actual division, we have:

$$\begin{array}{r} 1.375 \\ 8 \overline{) 11.000} \\ \underline{-8\phantom{.000}} \\ 30\phantom{00} \\ \underline{-24\phantom{00}} \\ 60\phantom{0} \\ \underline{-56\phantom{0}} \\ 40 \\ \underline{-40} \\ 0 \end{array}$$

∴ $\frac{11}{8}$ = 1.375

(ii) $\frac{23}{16}$

By actual division, we have:

$$\begin{array}{r} 1.4375 \\ 16 \overline{) 23.0000} \\ \underline{-16\phantom{.0000}} \\ 70\phantom{000} \\ \underline{-64\phantom{000}} \\ 60\phantom{00} \\ \underline{-48\phantom{00}} \\ 120\phantom{0} \\ \underline{-112\phantom{0}} \\ 80 \\ \underline{-80} \\ 0 \end{array}$$

∴ $\frac{23}{16}$ = 1.4375

(iii) $\frac{76}{125}$

By actual division, we have:

$$\begin{array}{r} 0.608 \\ 125 \overline{) 76.000} \\ \underline{-0\phantom{.000}} \\ 760\phantom{00} \\ \underline{-750\phantom{00}} \\ 100\phantom{0} \\ \underline{-0\phantom{0}} \\ 1000 \\ \underline{-1000} \\ 0 \end{array}$$

∴ $\frac{76}{125}$ = 0.608

(iv) $\frac{103}{40}$

By actual division, we have:

$$\begin{array}{r} 2.575 \\ 40 \overline{) 103.000} \\ \underline{-80\phantom{.000}} \\ 230\phantom{00} \\ \underline{-200\phantom{00}} \\ 300\phantom{0} \\ \underline{-280\phantom{0}} \\ 200 \\ \underline{-200} \\ 0 \end{array}$$

∴ $\frac{103}{40}$ = 2.575

(v) $\frac{17}{80}$

By actual division, we have:

$$\begin{array}{r} 0.2125 \\ 80 \overline{) 17.0000} \\ \underline{-0\phantom{.0000}} \\ 170\phantom{000} \\ \underline{-160\phantom{000}} \\ 100\phantom{00} \\ \underline{-80\phantom{00}} \\ 200\phantom{0} \\ \underline{-160\phantom{0}} \\ 400 \\ \underline{-400} \\ 0 \end{array}$$

∴ $\frac{17}{80}$ = 0.2125

(vi) $\frac{2}{25}$

By actual division, we have:

$$\begin{array}{r} 0.08 \\ 25 \overline{) 2.00} \\ \underline{-0\phantom{.00}} \\ 20\phantom{0} \\ \underline{-0\phantom{.0}} \\ 200 \\ \underline{-200} \\ 0 \end{array}$$

∴ $\frac{2}{25}$ = 0.08

(vii) $\frac{1}{125}$

By actual division, we have:

$$\begin{array}{r} 0.008 \\ 125 \overline{) 1.000} \\ \underline{-0\phantom{.000}} \\ 10\phantom{00} \\ \underline{-0\phantom{.00}} \\ 100\phantom{0} \\ \underline{-0\phantom{0}} \\ 1000 \\ \underline{-1000} \\ 0 \end{array}$$

∴ $\frac{1}{125}$ = 0.008

(viii) $\frac{309}{1250}$

By actual division, we have:

$$\begin{array}{r} 0.2472 \\ 1250 \overline{) 309.0000} \\ \underline{-0\phantom{.0000}} \\ 3090\phantom{000} \\ \underline{-2500\phantom{000}} \\ 5900\phantom{00} \\ \underline{-5000\phantom{00}} \\ 9000\phantom{0} \\ \underline{-8750\phantom{0}} \\ 2500 \\ \underline{-2500} \\ 0 \end{array}$$

∴ $\frac{309}{1250}$ = 0.2472

(i) $\frac{23}{24}$

The given rational number is $\frac{23}{24}$.

Its denominator is $24 = 2^3 \times 3$.

Thus, the denominator of $\frac{23}{24}$ has at least one prime factor, namely 3, other than 2 and 5.

∴ $\frac{23}{24}$ is expressible as a repeating decimal.

(ii) $\frac{79}{30}$

The given rational number is $\frac{79}{30}$.

Its denominator is $30 = 2 \times 3 \times 5$.

Thus, the denominator of $\frac{79}{30}$ has a prime factor, namely 3, other than 2 and 5.

∴ $\frac{79}{30}$ is expressible as a repeating decimal.

(iii) $\frac{100}{9}$

The given rational number is $\frac{100}{9}$.

Its denominator is $9 = 3^2$.

Thus, the denominator of $\frac{100}{9}$ has at least one prime factor, namely 3, which is different from 2 and 5.

∴ $\frac{100}{9}$ is expressible as a repeating decimal.

(iv) $\frac{205}{27}$

The given rational number is $\frac{205}{27}$.

Its denominator is $27 = 3^3$.

Thus, the denominator of $\frac{205}{27}$ has at least one prime factor, namely 3, which is different from 2 and 5.

∴ $\frac{205}{27}$ is expressible as a repeating decimal.

(v) $\frac{461}{60}$

The given rational number is $\frac{461}{60}$.

Its denominator is $60 = 2^2 \times 3 \times 5$.

Thus, the denominator has at least one prime factor, namely 3, other than 2 and 5.

∴ $\frac{461}{60}$ is expressible as a repeating decimal.

(vi) $\frac{1003}{112}$

The given rational number is $\frac{1003}{112}$.

Its denominator is $112 = 2^4 \times 7$.

Thus, the denominator has at least one prime factor, namely 7, other than 2 and 5.

∴ $\frac{1003}{112}$ is expressible as a repeating decimal.

(vii) $\frac{127}{225}$

The given rational number is $\frac{127}{225}$.

Its denominator is $225 = 3^2 \times 5^2$.

Thus, the denominator has at least one prime factor, namely 3, which is different from 2 and 5.

∴ $\frac{127}{225}$ is expressible as a repeating decimal.

(viii) $\frac{219}{440}$

The given rational number is $\frac{219}{440}$.

Its denominator is $440 = 2^3 \times 5 \times 11$.

Thus, the denominator has a prime factor, namely 11, other than 2 and 5.

∴ $\frac{219}{440}$ is expressible as a repeating decimal.

(i) $\frac{103}{9}$

By actual division, we have:

$$\begin{array}{r} 11.444... \\ 9 \overline{) 103.000} \\ \underline{-9\phantom{.000}} \\ 13\phantom{.000} \\ \underline{-9\phantom{.000}} \\ 4.0\phantom{00} \\ \underline{-3.6\phantom{00}} \\ 40\phantom{0} \\ \underline{-36\phantom{0}} \\ 40 \\ \underline{-36} \\ 4... \end{array}$$

∴ $\frac{103}{9} = 11.444... = 11.\overline{4}$

(ii) $\frac{7}{12}$

By actual division, we have:

$$\begin{array}{r} 0.5833... \\ 12 \overline{) 7.0000} \\ \underline{-0\phantom{.0000}} \\ 70\phantom{000} \\ \underline{-60\phantom{000}} \\ 100\phantom{00} \\ \underline{-96\phantom{00}} \\ 40\phantom{0} \\ \underline{-36\phantom{0}} \\ 40 \\ \underline{-36} \\ 4... \end{array}$$

∴ $\frac{7}{12} = 0.5833... = 0.58\overline{3}$

(iii) $\frac{101}{15}$

By actual division, we have:

$$\begin{array}{r} 6.733... \\ 15 \overline{) 101.000} \\ \underline{-90\phantom{.000}} \\ 110\phantom{00} \\ \underline{-105\phantom{00}} \\ 50\phantom{0} \\ \underline{-45\phantom{0}} \\ 50 \\ \underline{-45} \\ 5... \end{array}$$

∴ $\frac{101}{15} = 6.733... = 6.7\overline{3}$

(iv) $\frac{303}{11}$

By actual division, we have:

$$\begin{array}{r} 27.5454... \\ 11 \overline{) 303.0000} \\ \underline{-22\phantom{.0000}} \\ 83\phantom{.0000} \\ \underline{-77\phantom{.0000}} \\ 60\phantom{000} \\ \underline{-55\phantom{000}} \\ 50\phantom{00} \\ \underline{-44\phantom{00}} \\ 60\phantom{0} \\ \underline{-55\phantom{0}} \\ 50 \\ \underline{-44} \\ 6... \end{array}$$

∴ $\frac{303}{11} = 27.5454... = 27.\overline{54}$

(v) $\frac{212}{143}$

By actual division, we have:

$$\begin{array}{r} 1.482517... \\ 143 \overline{) 212.000000} \\ \underline{-143\phantom{.000000}} \\ 690\phantom{00000} \\ \underline{-572\phantom{00000}} \\ 1180\phantom{0000} \\ \underline{-1144\phantom{0000}} \\ 360\phantom{000} \\ \underline{-286\phantom{000}} \\ 740\phantom{00} \\ \underline{-715\phantom{00}} \\ 250\phantom{0} \\ \underline{-143\phantom{0}} \\ 1070 \\ \underline{-1001} \\ 69... \end{array}$$

∴ $\frac{212}{143} = 1.482517482517... = 1.\overline{482517}$

(vi) $\frac{16}{7}$

By actual division, we have:

$$\begin{array}{r} 2.285714... \\ 7 \overline{) 16.000000} \\ \underline{-14\phantom{.000000}} \\ 20\phantom{00000} \\ \underline{-14\phantom{00000}} \\ 60\phantom{0000} \\ \underline{-56\phantom{0000}} \\ 40\phantom{000} \\ \underline{-35\phantom{000}} \\ 50\phantom{00} \\ \underline{-49\phantom{00}} \\ 10\phantom{0} \\ \underline{-7\phantom{0}} \\ 30 \\ \underline{-28} \\ 2... \end{array}$$

∴ $\frac{16}{7} = 2.285714... = 2.\overline{285714}$

(vii) $\frac{227}{30}$

By actual division, we have:

$$\begin{array}{r} 7.566... \\ 30 \overline{) 227.000} \\ \underline{-210\phantom{.000}} \\ 170\phantom{00} \\ \underline{-150\phantom{00}} \\ 200\phantom{0} \\ \underline{-180\phantom{0}} \\ 200 \\ \underline{-180} \\ 20... \end{array}$$

∴ $\frac{227}{30} = 7.566... = 7.5\overline{6}$

(viii) $\frac{2000}{33}$

By actual division, we have:

$$\begin{array}{r} 60.6060... \\ 33 \overline{) 2000.0000} \\ \underline{-198\phantom{0.0000}} \\ 20\phantom{.0000} \\ \underline{-0\phantom{.0000}} \\ 200\phantom{000} \\ \underline{-198\phantom{000}} \\ 20\phantom{00} \\ \underline{-0\phantom{00}} \\ 200\phantom{0} \\ \underline{-198\phantom{0}} \\ 20 \\ \underline{-0} \\ 20... \end{array}$$

∴ $\frac{2000}{33} = 60.6060... = 60.\overline{60}$

(i) $\frac{2}{3}$ = $0.\overline{6}$

(ii) $\frac{11}{30}$ = $0.3\overline{6}$

(iii) $\frac{13}{11}$ = $1.\overline{18}$

(iv) $\frac{23}{55}$ = $0.4\overline{18}$

Explanation

(i) $\frac{2}{3}$

By actual division:

$$\begin{array}{r} 0.666... \\ 3 \overline{) 2.000} \\ \underline{-0\phantom{.000}} \\ 2.0\phantom{00} \\ \underline{-1.8\phantom{00}} \\ 20\phantom{0} \\ \underline{-18\phantom{0}} \\ 20 \\ \underline{-18} \\ 2... \end{array}$$

∴ $\frac{2}{3} = 0.666... = 0.\overline{6}$

(ii) $\frac{11}{30}$

By actual division:

$$\begin{array}{r} 0.366... \\ 30 \overline{) 11.000} \\ \underline{-0\phantom{.000}} \\ 110\phantom{00} \\ \underline{-90\phantom{00}} \\ 200\phantom{0} \\ \underline{-180\phantom{0}} \\ 200 \\ \underline{-180} \\ 20... \end{array}$$

∴ $\frac{11}{30} = 0.366... = 0.3\overline{6}$

(iii) $\frac{13}{11}$

By actual division:

$$\begin{array}{r} 1.1818... \\ 11 \overline{) 13.0000} \\ \underline{-11\phantom{.0000}} \\ 20\phantom{000} \\ \underline{-11\phantom{000}} \\ 90\phantom{00} \\ \underline{-88\phantom{00}} \\ 20\phantom{0} \\ \underline{-11\phantom{0}} \\ 90 \\ \underline{-88} \\ 2... \end{array}$$

∴ $\frac{13}{11} = 1.1818... = 1.\overline{18}$

(iv) $\frac{23}{55}$

By actual division:

$$\begin{array}{r} 0.41818... \\ 55 \overline{) 23.00000} \\ \underline{-0\phantom{.00000}} \\ 230\phantom{0000} \\ \underline{-220\phantom{0000}} \\ 100\phantom{000} \\ \underline{-55\phantom{000}} \\ 450\phantom{00} \\ \underline{-440\phantom{00}} \\ 100\phantom{0} \\ \underline{-55\phantom{0}} \\ 450 \\ \underline{-440} \\ 10... \end{array}$$

∴ $\frac{23}{55} = 0.41818... = 0.4\overline{18}$

The additive inverse of a number a is -a, such that their sum is 0. Therefore the additive inverse of $\frac{5}{9}$ is $\frac{-5}{9}$.

Hence, option 2 is the correct option.

Move the negative sign to the numerator: $\frac{-32}{40}$.

Divide both by their HCF (8): $\frac{-32 \div 8}{40 \div 8} = \frac{-4}{5}$.

Hence, option 4 is the correct option.

Let the number be x.

$$\begin{aligned}x + \Big(\frac{-3}{16}\Big) = \frac{5}{8} \\[1em] x = \frac{5}{8} + \frac{3}{16} \\[1em]\end{aligned}$$

L.C.M of 8 and 16 is 16.

Now, expressing each fraction with denominator 16:
$$\begin{aligned}x = \frac{5 \times 2}{8 \times 2} + \frac{3 \times 1}{16 \times 1} \\[1em] x = \frac{10}{16} + \frac{3}{16} \\[1em] x = \frac{13}{16}\end{aligned}$$

Hence, option 4 is the correct option.

The multiplicative inverse (reciprocal) of $\frac{a}{b}$ is $\frac{b}{a}$. Therefore the reciprocal of $\frac{-3}{7}$ is $\frac{7}{-3}$, which is $\frac{-7}{3}$.

Hence, option 1 is the correct option.

Multiplicative inverse of $-\frac{1}{3}$ is -3.

Sum = $-\frac{1}{3} + (-3) = -\frac{1}{3} - \frac{3}{1}$

L.C.M. of 3 and 1 is 3.

Now, expressing each fraction with denominator 3:

$$\begin{aligned}= -\frac{1 \times 1}{3 \times 1} - \frac{3 \times 3}{1 \times 3} \\[1em] = -\frac{1}{3} - \frac{9}{3} \\[1em] = \frac{-1 - 9}{3} \\[1em] = \frac{-10}{3} \\[1em] = -3\frac{1}{3}\end{aligned}$$

Hence, option 4 is the correct option.

Additive inverse of $-\frac{1}{3}$ is $\frac{1}{3}$.

Product =

$$\begin{aligned}\Big(-\frac{1}{3}\Big) \times \Big(\frac{1}{3}\Big) \\[1em] = -\frac{1 \times 1}{3 \times 3} \\[1em] = -\frac{1}{9}\end{aligned}$$

Hence, option 3 is the correct option.

Reduce each option to standard form:

$$\begin{aligned}\frac{-14}{21}=\frac{-14 \div 7}{21 \div 7}=\frac{-2}{3} \\[1em] \frac{-8}{14}=\frac{-8 \div 2}{14 \div 2}=\frac{-4}{7} \\[1em] \frac{-14}{49}=\frac{-14 \div 7}{49 \div 7}=\frac{-2}{7} \\[1em] \frac{-6}{28}=\frac{-6 \div 2}{28 \div 2}=\frac{-3}{14} \\[1em]\end{aligned}$$

Only option 3 reduces to $\frac{-2}{7}$

Hence, option 3 is the correct option.

Given:

Total length of cloth = 30 m

Required length for one suit = $3\frac{3}{4}$ m = $\frac{15}{4}$ m

Total number of suits = ?

Total number of suits = Total length of cloth ÷ Required length for one suit

Substituting the values in above, we get:

Total number of suits = 30 m ÷ $\frac{15}{4}$ m

$$\begin{aligned}= 30 \times \frac{4}{15} \quad \left[\text{Reciprocal of } \frac{15}{4} \text{ is } \frac{4}{15}\right] \\[1em] = \frac{30 \times 4}{15} \\[1em] = \frac{2 \times 4}{1} \quad \text{[Dividing 30 and 15 by 15]} \\[1em] = \frac{8}{1} = 8\end{aligned}$$

∴ 8 suits can be prepared from 30 m of cloth.

Hence, option 3 is the correct option.

(i) The multiplicative inverse of a rational number is also called its reciprocal.

(ii) Every negative rational number is less or smaller than 0.

(iii) A rational number $\frac{p}{q}$ is said to be in standard form, if q is positive and p and q have no common divisor other than 1.

(iv) $\frac{-5}{-9}$ is a positive rational number.

(v) The additive inverse of a rational number $\frac{a}{b}$ is $\Big(-\frac{a}{b}\Big)$.

(i) True
Reason — Zero (0) is a rational number that is neither positive nor negative.

(ii) False
Reason — While most rational numbers have a multiplicative inverse, zero (0) does not, because division by zero is undefined.

(iii) False
Reason — In standard form, the denominator must be positive, but it can be smaller than the numerator (for example, $\frac{5}{2}$ is in standard form).

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

(v) The product of a rational number and its multiplicative inverse is always 1.

(vi) True
Reason — Equivalent rational numbers like $\frac{2}{4}$ and $\frac{3}{6}$ both reduce to the same standard form, which is $\frac{1}{2}$.

(vii) True
Reason — According to closure property of multiplication for rational numbers, the product of any two rational numbers is also a rational number.

(viii) False
Reason — A rational number divided by zero does not give a rational number, as division by zero is undefined.

(ix) True
Reason — Every rational number corresponds to a unique point on the number line.

(x) True
Reason — On a number line, values decrease as you move to the left; therefore, all numbers smaller than $\frac{p}{q}$ lie to its left.

Assertion (A) is false but Reason (R) is true.

Explanation

The assertion is false because rational numbers with different numerators can still be equal. For example, $\frac{1}{2}$ and $\frac{2}{4}$ have different numerators but represent the same value.

The reason is true because it is the correct definition of the standard form of a rational number.

Hence, option 4 is the correct option.

Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

Explanation

The assertion is true because we can always find a smaller rational number by moving further to the left on the number line.

The reason is also true because numbers to the left of 0 are negative, but this does not explain why the smallest rational number does not exist.

Hence, option 2 is the correct option.

Checking option 1: $\frac{-3}{7}$ and $\frac{-3}{5}$

LCM of 7 and 5 = 35

$$\begin{aligned}\frac{-3}{7} = \frac{-3 \times 5}{7 \times 5} = \frac{-15}{35} \\[1em] \frac{-3}{5} = \frac{-3 \times 7}{5 \times 7} = \frac{-21}{35}\end{aligned}$$

Since −15 > −21, we have $\frac{-15}{35} \gt \frac{-21}{35}$, i.e., $\frac{-3}{7} \gt \frac{-3}{5}$.

∴ Option 1 is correct.

Checking option 2: $-9\frac{2}{3}$ and $-9\frac{4}{5}$

Compare the fractional parts $\frac{2}{3}$ and $\frac{4}{5}$.

LCM of 3 and 5 = 15

$\frac{2}{3} = \frac{10}{15}$ and $\frac{4}{5} = \frac{12}{15}$

Since $\frac{10}{15} \lt \frac{12}{15}$, we have $\frac{2}{3} \lt \frac{4}{5}$.

So, $-9\frac{2}{3} \gt -9\frac{4}{5}$

[Greater fractional part makes the negative mixed number smaller]

∴ Option 2 is correct.

Checking option 3: $-11\frac{1}{4}$ and $-11\frac{3}{5}$

Compare the fractional parts $\frac{1}{4}$ and $\frac{3}{5}$.

LCM of 4 and 5 = 20

$\frac{1}{4} = \frac{5}{20}$ and $\frac{3}{5} = \frac{12}{20}$

Since $\frac{5}{20} \lt \frac{12}{20}$, we have $\frac{1}{4} \lt \frac{3}{5}$.

So, $-11\frac{1}{4} \gt -11\frac{3}{5}$.

But the given statement says $-11\frac{1}{4} \lt -11\frac{3}{5}$, which is false.

∴ Option 3 is incorrect.

Checking option 4: $-13\frac{1}{3}$ and $-13\frac{1}{5}$

Compare the fractional parts $\frac{1}{3}$ and $\frac{1}{5}$.

Since $\frac{1}{3} \gt \frac{1}{5}$

[Same numerator, smaller denominator gives a greater fraction]

So, $-13\frac{1}{3} \lt -13\frac{1}{5}$.

∴ Option 4 is correct.

Hence, the incorrect statement is in option 3.

Hence, option 3 is the correct option.

Given:

$\frac{x}{y} = \frac{9}{8}$

So, we can take x = 9 and y = 8.

Step 1: Find the value of $\frac{y-x}{y+x}$

$y - x = 8 - 9 = -1$

$y + x = 8 + 9 = 17$

$\frac{y-x}{y+x} = \frac{-1}{17}$

Step 2: Add $\frac{6}{7}$ and $\frac{-1}{17}$

$\frac{6}{7} + \left(\frac{-1}{17}\right)$

LCM of 7 and 17 = 119

$$\begin{aligned}= \frac{102}{119} + \frac{-7}{119} \\[1em] = \frac{102 + (-7)}{119} \\[1em] = \frac{95}{119}\end{aligned}$$

∴ The required value is $\frac{95}{119}$.

Hence, option 2 is the correct option.

To compare the given rational numbers, express $\frac{1}{2}$ and $\frac{3}{4}$ with denominator 16.

$\frac{1}{2} = \frac{1 \times 8}{2 \times 8} = \frac{8}{16}$

$\frac{3}{4} = \frac{3 \times 4}{4 \times 4} = \frac{12}{16}$

So, we need rational numbers lying between $\frac{8}{16}$ and $\frac{12}{16}$, i.e., numerators between 8 and 12.

Checking option 1: $\frac{9}{16}$

Since 8 < 9 < 12, $\frac{9}{16}$ lies between $\frac{1}{2}$ and $\frac{3}{4}$.

Checking option 2: $\frac{13}{16}$

Since 13 > 12, $\frac{13}{16} \gt \frac{12}{16} = \frac{3}{4}$.

So, $\frac{13}{16}$ does not lie between $\frac{1}{2}$ and $\frac{3}{4}$.

Checking option 3: $\frac{10}{16}$

Since 8 < 10 < 12, $\frac{10}{16}$ lies between $\frac{1}{2}$ and $\frac{3}{4}$.

Checking option 4: $\frac{11}{16}$

Since 8 < 11 < 12, $\frac{11}{16}$ lies between $\frac{1}{2}$ and $\frac{3}{4}$.

∴ $\frac{13}{16}$ is not a rational number between $\frac{1}{2}$ and $\frac{3}{4}$.

Hence, option 2 is the correct option.

Step 1: Convert the mixed fractions into improper fractions

$$\begin{aligned}-18\frac{1}{3} = \frac{-(18 \times 3 + 1)}{3} = \frac{-55}{3} \\[1em] 2\frac{8}{11} = \frac{2 \times 11 + 8}{11} = \frac{30}{11} \\[1em] 4\frac{5}{7} = \frac{4 \times 7 + 5}{7} = \frac{33}{7} \\[1em] 2\frac{1}{3} = \frac{2 \times 3 + 1}{3} = \frac{7}{3} \\[1em]\end{aligned}$$

Step 2: Evaluate the numerator

$$\begin{aligned}-18\frac{1}{3} \times 2\frac{8}{11} = \left(\frac{-55}{3} \times \frac{30}{11}\right) = \left(\frac{(-55) \times 30}{3 \times 11}\right) = \frac{-1650}{33} = -50 \\[1em] 4\frac{5}{7} \times 2\frac{1}{3} = \left(\frac{33}{7} \times \frac{7}{3}\right) = \left(\frac{33 \times 7}{7 \times 3}\right) = \frac{231}{21} = 11\end{aligned}$$

Numerator = $-50 - 11 = -61$

Step 3: Evaluate the denominator

First term: $\left|\frac{3}{5} + \left(\frac{-9}{10}\right)\right|$

LCM of 5 and 10 = 10

$$\begin{aligned}\frac{3}{5} = \frac{3 \times 2}{5 \times 2} = \frac{6}{10} \\[1em] \frac{3}{5} + \frac{-9}{10} = \frac{6}{10} + \frac{-9}{10} = \frac{6 + (-9)}{10} = \frac{-3}{10} \\[1em] \left|\frac{-3}{10}\right| = \frac{3}{10} \\[1em] \text{[Modulus of a number is its non-negative value]}\end{aligned}$$

Second term: $\left|-\left(\frac{-3}{5}\right)\right|$

$$\begin{aligned}-\left(\frac{-3}{5}\right) = \frac{3}{5} \\[1em] \left|\frac{3}{5}\right| = \frac{3}{5} = \frac{6}{10}\end{aligned}$$

Denominator = $\frac{3}{10} + \frac{6}{10} = \frac{9}{10}$

Step 4: Divide the numerator by the denominator

$\frac{-61}{\frac{9}{10}} = -61 \times \frac{10}{9} = \frac{-610}{9}$

[Dividing by a fraction is the same as multiplying by its reciprocal]

Step 5: Convert into a mixed fraction

610 ÷ 9 = 67 with remainder 7

So, $\frac{-610}{9} = -67\frac{7}{9}$

∴ The required value is $-67\frac{7}{9}$.

Hence, option 3 is the correct option.

Given:

Fraction from India = $\frac{3}{20}$

Fraction from Britain = $\frac{1}{4}$

Fraction from Germany = $\frac{3}{10}$

Total members = 1200

Step 1: Find the fraction of non-American members

Fraction of non-Americans = $\frac{3}{20} + \frac{1}{4} + \frac{3}{10}$

LCM of 20, 4 and 10 = 20

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

Fraction of non-Americans:

$$\begin{aligned}\phantom{=}\frac{3}{20} + \frac{5}{20} + \frac{6}{20} \\[1em] = \frac{3 + 5 + 6}{20} \\[1em] = \frac{14}{20} \\[1em] = \frac{7}{10}\end{aligned}$$

Step 2: Find the fraction of Americans

Fraction of Americans = $1 - \frac{7}{10} = \frac{10}{10} - \frac{7}{10} = \frac{3}{10}$

Step 3: Find the number of Americans

Number of Americans:

$$\begin{aligned}\phantom{=}\frac{3}{10} \times 1200 \\[1em] = \frac{3 \times 1200}{10} \\[1em] = \frac{3600}{10} \\[1em] = 360\end{aligned}$$

∴ There are 360 Americans in the delegation.

Hence, option 3 is the correct option.

Let the number be x.

Correct answer = $\frac{5}{17} \times x = \frac{5x}{17}$

Wrong answer = $\frac{17}{5} \times x = \frac{17x}{5}$

According to the question, the equation can be written as:

$\frac{17x}{5} - \frac{5x}{17} = \frac{264}{119}$ $\hspace{2cm}$[Wrong answer − Correct answer = Excess]

LCM of 5 and 17 = 85

$$\begin{aligned}\frac{17x}{5} = \frac{17x \times 17}{5 \times 17} = \frac{289x}{85} \\[1em] \frac{5x}{17} = \frac{5x \times 5}{17 \times 5} = \frac{25x}{85} \\[1em] \frac{289x}{85} - \frac{25x}{85} = \frac{264}{119} \\[1em] \frac{289x - 25x}{85} = \frac{264}{119} \\[1em] \frac{264x}{85} = \frac{264}{119} \\[1em] x = \frac{264}{119} \times \frac{85}{264} \hspace{2cm}\text{[Multiplying both sides by} \frac{85}{264}] \\[1em] x = \frac{85}{119} \\[1em] \frac{85 \div 17}{119 \div 17} = \frac{5}{7} \hspace{2cm}\text{[HCF of 85 and 119 is 17]}\end{aligned}$$

∴ The required number is $\frac{5}{7}$.

Hence, option 2 is the correct option.

Checking (i):

Any rational number with a negative numerator can be rewritten with a positive numerator by multiplying its numerator and denominator by −1.

For example, $\frac{-3}{5} = \frac{(-3) \times (-1)}{5 \times (-1)} = \frac{3}{-5}$.

So, every rational number can be expressed with a positive numerator.

∴ (i) is True.

Checking (ii):

The given numbers are $\frac{-11}{5}, \frac{1}{6}, \frac{3}{10}, \frac{6}{10}$.

$\frac{-11}{5}$ is negative, while $\frac{1}{6}, \frac{3}{10}$ and $\frac{6}{10}$ are positive.

So $\frac{-11}{5}$ is the smallest. Now compare the three positive numbers.

LCM of 6, 10 and 10 = 30

$$\begin{aligned}\frac{1}{6} = \frac{1 \times 5}{6 \times 5} = \frac{5}{30} \\[1em] \frac{3}{10} = \frac{3 \times 3}{10 \times 3} = \frac{9}{30} \\[1em] \frac{6}{10} = \frac{6 \times 3}{10 \times 3} = \frac{18}{30} \\[1em]\end{aligned}$$

Since $\frac{5}{30} \lt \frac{9}{30} \lt \frac{18}{30}$, we get $\frac{1}{6} \lt \frac{3}{10} \lt \frac{6}{10}$.

So the ascending order is $\frac{-11}{5} \lt \frac{1}{6} \lt \frac{3}{10} \lt \frac{6}{10}$, which matches the given order.

∴ (ii) is True.

Checking (iii):

$$\begin{aligned}\frac{8}{5} \times \frac{-3}{2} = \frac{8 \times (-3)}{5 \times 2} = \frac{-24}{10} = \frac{-12}{5} \\[1em] \frac{3}{10} \times \frac{11}{8} = \frac{3 \times 11}{10 \times 8} = \frac{33}{80}\end{aligned}$$

Now, add $\frac{-12}{5}$ and $\frac{33}{80}$.

LCM of 5 and 80 = 80

$$\begin{aligned}\frac{-12}{5} = \frac{-12 \times 16}{5 \times 16} = \frac{-192}{80} \\[1em] \frac{-192}{80} + \frac{33}{80} = \frac{-192 + 33}{80} = \frac{-159}{80}\end{aligned}$$

Converting into a mixed fraction:

159 ÷ 80 = 1 with remainder 79

So, $\frac{-159}{80} = -1\frac{79}{80}$.

∴ (iii) is True.

Checking (iv):

$\frac{15}{10}$ and $\frac{12}{8}$ are equivalent if 15 x 8 = 10 x 12. $\hspace{0.5cm}$[$\frac{a}{b} = \frac{c}{d} \Leftrightarrow a \times d = b \times c$]

15 x 8 = 120 and 10 x 12 = 120

Since both products are equal, $\frac{15}{10}$ and $\frac{12}{8}$ are equivalent rational numbers.

∴ (iv) is True.

So, the answers are T, T, T, T.

Hence, option 3 is the correct option.

Step 1: Simplify $\frac{9}{5} - \frac{5}{2}$

LCM of 5 and 2 = 10

$$\begin{aligned}\frac{9}{5} = \frac{9 \times 2}{5 \times 2} = \frac{18}{10} \\[1em] \frac{5}{2} = \frac{5 \times 5}{2 \times 5} = \frac{25}{10} \\[1em] \frac{9}{5} - \frac{5}{2} = \frac{18}{10} - \frac{25}{10} = \frac{18 - 25}{10} = \frac{-7}{10}\end{aligned}$$

Step 2: Set up the equation

Let the required number be x.

According to the question, the equation can be written as:

$$\begin{aligned}\frac{-7}{10} \div x = \frac{7}{10} \\[1em] \frac{-7}{10} \times \frac{1}{x} = \frac{7}{10} \\[1em] \frac{1}{x} = \frac{7}{10} \times \frac{10}{-7} \hspace{2cm}\text{[Multiplying both sides by} \frac{10}{-7}] \\[1em] \frac{1}{x} = \frac{7 \times 10}{10 \times (-7)} \\[1em] \frac{1}{x} = \frac{70}{-70} \\[1em] \frac{1}{x} = \frac{1}{-1} \\[1em] x = -1\end{aligned}$$

∴ The required number is −1.

Hence, option 2 is the correct option.

Two rational numbers $\frac{a}{b}$ and $\frac{c}{d}$ are equivalent if and only if $a \times d = b \times c$.

Given:

$\frac{2}{9}$ and $\frac{1}{b}$ are equivalent.

According to the question, the equation can be written as:

2 x b = 9 x 1

2b = 9

$b = \frac{9}{2}$

Verification:

$\frac{1}{b} = \frac{1}{\frac{9}{2}} = \frac{2}{9}$

LHS = RHS

∴ The required value of b is $\frac{9}{2}$.

Hence, option 4 is the correct option.

Given:

x is a positive rational number, so x > 0.

y is a negative rational number, so y < 0.

Combining the two, we get y < 0 < x, i.e., x > y.

Checking (i): x < y

A positive number is always greater than a negative number, not less than. So, x < y is false.

Checking (ii): x = y

A positive number can never be equal to a negative number. So, x = y is false.

Checking (iii): x > y

A positive number is always greater than a negative number. So, x > y is true.

∴ Only (iii) is correct.

Hence, option 3 is the correct option.

The rational number $-\frac{1}{2}$ is a negative number.

So, it must lie to the left of 0 on the number line.

[Every negative rational number is less than 0]

Also, since $-\frac{1}{2}$ is the midpoint between -1 and 0, the point P must lie exactly halfway between -1 and 0.

Checking option 1:

The number line shows the segment from -1 to 0, divided into equal parts. The point P is marked exactly at the midpoint between -1 and 0.

So, P represents $-\frac{1}{2}$.

Checking option 2:

The number line shows the segment from -1 to 0, but the point P is marked closer to 0, not at the midpoint.

So, P does not represent $-\frac{1}{2}$.

Checking option 3:

The number line shows the segment from 0 to 1, so the point P lies to the right of 0.

Since $-\frac{1}{2}$ is negative, P cannot represent $-\frac{1}{2}$.

Checking option 4:

The number line shows the segment from 0 to 1, so the point P lies to the right of 0.

Since $-\frac{1}{2}$ is negative, P cannot represent $-\frac{1}{2}$.

∴ Only option 1 represents $-\frac{1}{2}$ correctly.

Hence, option 1 is the correct option.