Rational Numbers

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

$\frac{5}{8}$ → numerator and denominator both positive → positive

$\frac{-3}{11}$ → numerator negative, denominator positive → negative

$\frac{0}{5}$ → equal to 0 → neither positive nor negative

$7 = \frac{7}{1}$ → both positive → positive

-4 → negative

$\frac{-3}{-13}$ → both negative → positive

$\frac{-17}{-6}$ → both negative → positive

$\frac{9}{-20}$ → numerator positive, denominator negative → negative

Hence, the positive rational numbers are $\frac{5}{8}, 7, \frac{-3}{-13}$ and $\frac{-17}{-6}$.

A rational number is negative if one of its numerator and denominator is a positive integer and the other is a negative integer.

$\frac{-5}{7}$ → numerator negative, denominator positive → negative

$\frac{4}{-3}$ → numerator positive, denominator negative → negative

$\frac{-3}{-11}$ → both negative → positive

-6 → negative

9 → positive

0 → neither positive nor negative

$\frac{-28}{5}$ → numerator negative, denominator positive → negative

$\frac{31}{7}$ → both positive → positive

Hence, the negative rational numbers are $\frac{-5}{7}, \frac{4}{-3}, -6$ and $\frac{-28}{5}$.

(i) Multiplying the numerator and the denominator of $\frac{3}{-7}$ by 2, 3, 4 and 5, we get:

$$\begin{aligned}\frac{3 \times 2}{-7 \times 2} = \frac{6}{-14} \\[1em] \frac{3 \times 3}{-7 \times 3} = \frac{9}{-21} \\[1em] \frac{3 \times 4}{-7 \times 4} = \frac{12}{-28} \\[1em] \frac{3 \times 5}{-7 \times 5} = \frac{15}{-35}\end{aligned}$$

Hence, four rational numbers equivalent to $\frac{3}{-7}$ are $\frac{6}{-14}, \frac{9}{-21}, \frac{12}{-28}$ and $\frac{15}{-35}$.

(ii) We have $\frac{-5}{-9} = \frac{-5 \times (-1)}{-9 \times (-1)} = \frac{5}{9}$

Multiplying the numerator and the denominator of $\frac{5}{9}$ by 2, 3 and 4, we get:

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

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

To get a positive denominator, multiply the numerator and the denominator by -1.

(i) Solving,

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

∴ $\frac{4}{-9} = \frac{-4}{9}$

(ii) Solving,

$\frac{17}{-33} = \frac{17 \times (-1)}{-33 \times (-1)} = \frac{-17}{33}$

∴ $\frac{17}{-33} = \frac{-17}{33}$

(iii) Solving,

$\frac{-15}{-38} = \frac{-15 \times (-1)}{-38 \times (-1)} = \frac{15}{38}$

∴ $\frac{-15}{-38} = \frac{15}{38}$

(i) To get 20 from 5, we multiply 5 by 4.

$\frac{5}{-9} = \frac{5 \times 4}{-9 \times 4} = \frac{20}{-36}$

∴ $\frac{5}{-9} = \frac{20}{-36}$

(ii) To get -35 from 5, we multiply 5 by -7.

$\frac{5}{-9} = \frac{5 \times (-7)}{-9 \times (-7)} = \frac{-35}{63}$

∴ $\frac{5}{-9} = \frac{-35}{63}$

(iii) To get -54 from -9, we multiply -9 by 6.

$\frac{5}{-9} = \frac{5 \times 6}{-9 \times 6} = \frac{30}{-54}$

∴ $\frac{5}{-9} = \frac{30}{-54}$

(iv) To get 72 from -9, we multiply -9 by -8.

$\frac{5}{-9} = \frac{5 \times (-8)}{-9 \times (-8)} = \frac{-40}{72}$

∴ $\frac{5}{-9} = \frac{-40}{72}$

First reduce $\frac{-80}{112}$ to standard form. HCF of 80 and 112 is 16.

$\frac{-80}{112} = \frac{-80 \div 16}{112 \div 16} = \frac{-5}{7}$

(i) The numerator is already -5.

∴ $\frac{-80}{112} = \frac{-5}{7}$

(ii) To get -14 from 7, we multiply 7 by -2.

$\frac{-5}{7} = \frac{-5 \times (-2)}{7 \times (-2)} = \frac{10}{-14}$

∴ $\frac{-80}{112} = \frac{10}{-14}$

Two rational numbers $\frac{p}{q}$ and $\frac{r}{s}$ are equal if and only if $p \times s = q \times r$.

(i) For $\frac{-7}{21}$ and $\frac{3}{9}$:

-7 × 9 = -63 and 21 × 3 = 63

As -63 ≠ 63, the pair does not represent the same rational number.

(ii) For $\frac{-16}{20}$ and $\frac{20}{-25}$:

-16 × (-25) = 400 and 20 × 20 = 400

As 400 = 400, the pair represents the same rational number.

(iii) For $\frac{-3}{5}$ and $\frac{-12}{20}$:

-3 × 20 = -60 and 5 × (-12) = -60

As -60 = -60, the pair represents the same rational number.

(iv) For $\frac{8}{-5}$ and $\frac{-24}{15}$:

8 × 15 = 120 and -5 × (-24) = 120

As 120 = 120, the pair represents the same rational number.

Hence, the pairs (ii), (iii) and (iv) represent the same rational number.

(i) Solving,

$\frac{5}{4} = \frac{...}{16}$ : To get 16 from 4, multiply by 4. So, 5 × 4 = 20.

$\frac{5}{4} = \frac{25}{...}$ : To get 25 from 5, multiply by 5. So, 4 × 5 = 20.

$\frac{5}{4} = \frac{-15}{...}$ : To get -15 from 5, multiply by -3. So, 4 × (-3) = -12.

$\frac{5}{4} = \frac{20}{16} = \frac{25}{20} = \frac{-15}{-12}$

Hence, the blanks are 20, 20 and -12.

(ii) Solving,

$\frac{-3}{7} = \frac{...}{14}$ : To get 14 from 7, multiply by 2. So, -3 × 2 = -6.

$\frac{-3}{7} = \frac{9}{...}$ : To get 9 from -3, multiply by -3. So, 7 × (-3) = -21.

$\frac{-3}{7} = \frac{-6}{...}$ : To get -6 from -3, multiply by 2. So, 7 × 2 = 14.

$\frac{-3}{7} = \frac{-6}{14} = \frac{9}{-21} = \frac{-6}{14}$

Hence, the blanks are -6, -21 and 14.

(i) The denominator is positive. HCF of 45 and 30 is 15.

$\frac{-45}{30} = \frac{-45 \div 15}{30 \div 15} = \frac{-3}{2}$

∴ Standard form of $\frac{-45}{30}$ is $\frac{-3}{2}$.

(ii) Making the denominator positive:

$\frac{16}{-36} = \frac{16 \times (-1)}{-36 \times (-1)} = \frac{-16}{36}$

HCF of 16 and 36 is 4.

$\frac{-16}{36} = \frac{-16 \div 4}{36 \div 4} = \frac{-4}{9}$

∴ Standard form of $\frac{16}{-36}$ is $\frac{-4}{9}$.

(iii) Making the denominator positive:

$\frac{-3}{-15} = \frac{-3 \times (-1)}{-15 \times (-1)} = \frac{3}{15}$

HCF of 3 and 15 is 3.

$\frac{3}{15} = \frac{3 \div 3}{15 \div 3} = \frac{1}{5}$

∴ Standard form of $\frac{-3}{-15}$ is $\frac{1}{5}$.

(iv) Making the denominator positive:

$\frac{68}{-119} = \frac{68 \times (-1)}{-119 \times (-1)} = \frac{-68}{119}$

HCF of 68 and 119 is 17.

$\frac{-68}{119} = \frac{-68 \div 17}{119 \div 17} = \frac{-4}{7}$

∴ Standard form of $\frac{68}{-119}$ is $\frac{-4}{7}$.

(i) To represent $\frac{3}{8}$, divide the gap between 0 and 1 into 8 equal parts and take the 3rd point to the right of 0.

(ii) To represent $\frac{-3}{4}$, divide the gap between 0 and -1 into 4 equal parts and take the 3rd point to the left of 0.

(iii) To represent $\frac{-7}{8}$, divide the gap between 0 and -1 into 8 equal parts and take the 7th point to the left of 0.

(iv) Since $\frac{-17}{8} = -2\frac{1}{8}$, it lies between -2 and -3. Divide the gap between -2 and -3 into 8 equal parts and take the 1st point to the left of -2.

From the number line, A and B are at 2 and 3, and the gap AB is divided into 3 equal parts by P and Q, so AP = PQ = QB = $\frac{1}{3}$.

$$\begin{aligned}P = 2 + \frac{1}{3} = \frac{7}{3} \\[1em] Q = 2 + \frac{2}{3} = \frac{8}{3}\end{aligned}$$

Also, T and U are at -1 and -2, and the gap TU is divided into 3 equal parts by R and S, so TR = RS = SU = $\frac{1}{3}$.

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

Hence, P, Q, R and S represent $\frac{7}{3}, \frac{8}{3}, \frac{-4}{3}$ and $\frac{-5}{3}$ respectively.

(i) $\frac{-13}{-5} = \frac{13}{5}$, which is a positive rational number, so it lies to the right of zero.

Hence, the statement is True.

(ii) $\frac{-5}{-7} = \frac{5}{7}$ is positive (right of zero) and $\frac{7}{-9} = \frac{-7}{9}$ is negative (left of zero). So, they lie on opposite sides of zero.

Hence, the statement is True.

(iii) $\frac{-17}{6}$ is negative (left of zero) and $\frac{8}{-15} = \frac{-8}{15}$ is also negative (left of zero). So, they lie on the same side of zero.

Hence, the statement is False.

(i) $\frac{-4}{7}$ is a negative rational number and every negative rational number is less than zero.

Hence, the greater rational number is 0.

(ii) $\frac{5}{-9} = \frac{-5}{9}$ is negative and $\frac{3}{7}$ is positive. Every positive rational number is greater than every negative rational number.

Hence, the greater rational number is $\frac{3}{7}$.

(iii) $\frac{-9}{-5} = \frac{9}{5}$ is a positive rational number and every positive rational number is greater than zero.

Hence, the greater rational number is $\frac{-9}{-5}$.

(iv) $\frac{7}{-5} = \frac{-7}{5}$ is negative and $\frac{-21}{-23} = \frac{21}{23}$ is positive. Every positive rational number is greater than every negative rational number.

Hence, the greater rational number is $\frac{-21}{-23}$.

(i) LCM of 5 and 7 is 35.

$\frac{-4 \times 7}{5 \times 7} = \frac{-28}{35}, \quad \frac{-5 \times 5}{7 \times 5} = \frac{-25}{35}$

As -28 < -25, so $\frac{-4}{5}$ < $\frac{-5}{7}$.

∴ $\frac{-4}{5}$ < $\frac{-5}{7}$

(ii) LCM of 5 and 4 is 20.

$\frac{-8 \times 4}{5 \times 4} = \frac{-32}{20}, \quad \frac{-7 \times 5}{4 \times 5} = \frac{-35}{20}$

As -32 > -35, so $\frac{-8}{5}$ > $\frac{-7}{4}$.

∴ $\frac{-8}{5}$ > $\frac{-7}{4}$

(iii) $\frac{42}{-48} = \frac{-42}{48} = \frac{-7}{8}$

∴ $\frac{-7}{8}$ = $\frac{42}{-48}$

(iv) $\frac{1}{-3} = \frac{-1}{3}$. LCM of 3 and 4 is 12.

$\frac{-1 \times 4}{3 \times 4} = \frac{-4}{12}, \quad \frac{-1 \times 3}{4 \times 3} = \frac{-3}{12}$

As -4 < -3, so $\frac{1}{-3}$ < $\frac{-1}{4}$.

∴ $\frac{1}{-3}$ < $\frac{-1}{4}$

(v) LCM of 8 and 7 is 56.

$\frac{-3 \times 7}{8 \times 7} = \frac{-21}{56}, \quad \frac{-2 \times 8}{7 \times 8} = \frac{-16}{56}$

As -21 < -16, so $-\frac{3}{8}$ < $-\frac{2}{7}$.

∴ $-\frac{3}{8}$ < $-\frac{2}{7}$

(vi) LCM of 3 and 2 is 6.

$\frac{-4 \times 2}{3 \times 2} = \frac{-8}{6}, \quad \frac{-3 \times 3}{2 \times 3} = \frac{-9}{6}$

As -8 > -9, so $\frac{-4}{3}$ > $-\frac{3}{2}$.

∴ $\frac{-4}{3}$ > $-\frac{3}{2}$

(i) LCM of 7, 2 and 4 is 28.

$$\begin{aligned}\frac{-3 \times 4}{7 \times 4} = \frac{-12}{28} \\[1em] \frac{-3 \times 14}{2 \times 14} = \frac{-42}{28} \\[1em] \frac{-3 \times 7}{4 \times 7} = \frac{-21}{28}\end{aligned}$$

As -42 < -21 < -12,

$\frac{-42}{28} \lt \frac{-21}{28} \lt \frac{-12}{28}$

Hence, the ascending order is $\frac{-3}{2}, \frac{-3}{4}, \frac{-3}{7}$.

(ii) Writing each with a positive denominator:

$\frac{5}{-12} = \frac{-5}{12}, \quad \frac{9}{-24} = \frac{-9}{24} = \frac{-3}{8}$

So the numbers are $\frac{-3}{4}, \frac{-5}{12}, \frac{-3}{8}, \frac{-7}{16}$. LCM of 4, 12, 8 and 16 is 48.

$$\begin{aligned}\frac{-3 \times 12}{4 \times 12} = \frac{-36}{48} \\[1em] \frac{-5 \times 4}{12 \times 4} = \frac{-20}{48} \\[1em] \frac{-3 \times 6}{8 \times 6} = \frac{-18}{48} \\[1em] \frac{-7 \times 3}{16 \times 3} = \frac{-21}{48}\end{aligned}$$

As -36 < -21 < -20 < -18,

$\frac{-36}{48} \lt \frac{-21}{48} \lt \frac{-20}{48} \lt \frac{-18}{48}$

Hence, the ascending order is $\frac{-3}{4}, \frac{-7}{16}, \frac{5}{-12}, \frac{9}{-24}$.

(i) Writing each with a positive denominator, $\frac{17}{-30} = \frac{-17}{30}$.

So the numbers are $\frac{-3}{10}, \frac{-11}{20}, \frac{-7}{15}, \frac{-17}{30}$. LCM of 10, 20, 15 and 30 is 60.

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

As -18 > -28 > -33 > -34,

$\frac{-18}{60} \gt \frac{-28}{60} \gt \frac{-33}{60} \gt \frac{-34}{60}$

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

(ii) Writing each with a positive denominator, $\frac{2}{-5} = \frac{-2}{5}$ and $\frac{19}{-30} = \frac{-19}{30}$.

So the numbers are $\frac{-7}{10}, \frac{-11}{15}, \frac{-2}{5}, \frac{-19}{30}$. LCM of 10, 15, 5 and 30 is 30.

$$\begin{aligned}\frac{-7 \times 3}{10 \times 3} = \frac{-21}{30} \\[1em] \frac{-11 \times 2}{15 \times 2} = \frac{-22}{30} \\[1em] \frac{-2 \times 6}{5 \times 6} = \frac{-12}{30} \\[1em] \frac{-19 \times 1}{30 \times 1} = \frac{-19}{30}\end{aligned}$$

As -12 > -19 > -21 > -22,

$\frac{-12}{30} \gt \frac{-19}{30} \gt \frac{-21}{30} \gt \frac{-22}{30}$

Hence, the descending order is $\frac{2}{-5}, \frac{19}{-30}, \frac{-7}{10}, \frac{-11}{15}$.

(i) Solving,

$3\frac{5}{8} = 3 + \frac{5}{8} = 3 + \frac{5 \times 125}{8 \times 125} = 3 + \frac{625}{1000} = 3 + 0.625 = 3.625$

∴ $3\frac{5}{8} = 3.625$

(ii) Solving,

$\frac{-21}{125} = \frac{-21 \times 8}{125 \times 8} = \frac{-168}{1000} = -0.168$

∴ $\frac{-21}{125} = -0.168$

(iii) Solving,

$-\frac{3}{32} = -\frac{3 \times 3125}{32 \times 3125} = -\frac{9375}{100000} = -0.09375$

∴ $-\frac{3}{32} = -0.09375$

(iv) Solving,

$-7\frac{11}{80} = -\left(7 + \frac{11}{80}\right) = -\left(7 + \frac{11 \times 125}{80 \times 125}\right) = -\left(7 + \frac{1375}{10000}\right) = -(7 + 0.1375) = -7.1375$

∴ $-7\frac{11}{80} = -7.1375$

(i) Solving,

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

∴ $\frac{3}{11} + \frac{-5}{11} = \frac{-2}{11}$

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

$\frac{4}{9} + \frac{-5}{9} = \frac{4 + (-5)}{9} = \frac{-1}{9}$

∴ $\frac{4}{9} + \frac{5}{-9} = \frac{-1}{9}$

(iii) $\frac{5}{-7} = \frac{-5}{7}$ and $\frac{-2}{-7} = \frac{2}{7}$

$\frac{-5}{7} + \frac{2}{7} = \frac{-5 + 2}{7} = \frac{-3}{7}$

∴ $\frac{5}{-7} + \frac{-2}{-7} = \frac{-3}{7}$

(iv) LCM of 5 and 4 is 20.

$$\begin{aligned}\frac{-2 \times 4}{5 \times 4} + \frac{3 \times 5}{4 \times 5} \\[1em] = \frac{-8}{20} + \frac{15}{20} \\[1em] = \frac{-8 + 15}{20} \\[1em] = \frac{7}{20}\end{aligned}$$

∴ $\frac{-2}{5} + \frac{3}{4} = \frac{7}{20}$

(i) $\frac{27}{-4} = \frac{-27}{4}$. LCM of 4 and 8 is 8.

$$\begin{aligned}\Rightarrow \frac{-27 \times 2}{4 \times 2} + \frac{-15 \times 1}{8 \times 1} \\[1em] = \frac{-54}{8} + \frac{-15}{8} \\[1em] = \frac{-54 + (-15)}{8} \\[1em] = \frac{-69}{8}\end{aligned}$$

∴ $\frac{27}{-4} + \frac{-15}{8} = \frac{-69}{8}$

(ii) LCM of 18 and 8 is 72.

$$\begin{aligned}\Rightarrow \frac{-1 \times 4}{18 \times 4} + \frac{-3 \times 9}{8 \times 9} \\[1em] = \frac{-4}{72} + \frac{-27}{72} \\[1em] = \frac{-4 + (-27)}{72} \\[1em] = \frac{-31}{72}\end{aligned}$$

∴ $\frac{-1}{18} + \frac{-3}{8} = \frac{-31}{72}$

(iii) $-3\frac{1}{6} = \frac{-19}{6}$ and $2\frac{3}{8} = \frac{19}{8}$. LCM of 6 and 8 is 24.

$$\begin{aligned}\Rightarrow \frac{-19 \times 4}{6 \times 4} + \frac{19 \times 3}{8 \times 3} \\[1em] = \frac{-76}{24} + \frac{57}{24} \\[1em] = \frac{-76 + 57}{24} \\[1em] = \frac{-19}{24}\end{aligned}$$

∴ $-3\frac{1}{6} + 2\frac{3}{8} = \frac{-19}{24}$

(iv) $-2\frac{4}{5} = \frac{-14}{5}$ and $4\frac{3}{10} = \frac{43}{10}$.

LCM of 5 and 10 is 10.

$$\begin{aligned}\Rightarrow \frac{-14 \times 2}{5 \times 2} + \frac{43 \times 1}{10 \times 1} \\[1em] = \frac{-28}{10} + \frac{43}{10} \\[1em] = \frac{-28 + 43}{10} \\[1em] = \frac{15}{10} = \frac{3}{2} = 1\frac{1}{2}\end{aligned}$$

∴ $-2\frac{4}{5} + 4\frac{3}{10} = 1\frac{1}{2}$

(i) Solving,

$\frac{4}{13} - \frac{-6}{13} = \frac{4}{13} + \frac{6}{13} = \frac{4 + 6}{13} = \frac{10}{13}$

∴ The result is $\frac{10}{13}$.

(ii) LCM of 3 and 2 is 6.

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

∴ The result is $\frac{-1}{6}$.

(iii) LCM of 3 and 9 is 9.

$$\begin{aligned}\Rightarrow \frac{-2}{3} - \frac{5}{9} = \frac{-2 \times 3}{3 \times 3} - \frac{5 \times 1}{9 \times 1} \\[1em] = \frac{-6}{9} - \frac{5}{9} \\[1em] = \frac{-6 - 5}{9} \\[1em] = \frac{-11}{9}\end{aligned}$$

∴ The result is $\frac{-11}{9}$.

(i)

$\frac{5}{63} - \left(\frac{-6}{21}\right) = \frac{5}{63} + \frac{6}{21}$

LCM of 63 and 21 is 63.

$$\begin{aligned}\Rightarrow \frac{5 \times 1}{63 \times 1} + \frac{6 \times 3}{21 \times 3} \\[1em] = \frac{5}{63} + \frac{18}{63} \\[1em] = \frac{5 + 18}{63} \\[1em] = \frac{23}{63}\end{aligned}$$

∴ $\frac{5}{63} - \left(\frac{-6}{21}\right) = \frac{23}{63}$

(ii)

$\frac{-6}{13} - \left(\frac{-7}{15}\right) = \frac{-6}{13} + \frac{7}{15}$

LCM of 13 and 15 is 195.

$$\begin{aligned}\Rightarrow \frac{-6 \times 15}{13 \times 15} + \frac{7 \times 13}{15 \times 13} \\[1em] = \frac{-90}{195} + \frac{91}{195} \\[1em] = \frac{-90 + 91}{195} \\[1em] = \frac{1}{195}\end{aligned}$$

∴ $\frac{-6}{13} - \left(\frac{-7}{15}\right) = \frac{1}{195}$

(iii) $3\frac{1}{8} = \frac{25}{8}$ and $-1\frac{5}{6} = \frac{-11}{6}$.

$\frac{25}{8} - \left(\frac{-11}{6}\right) = \frac{25}{8} + \frac{11}{6}$

LCM of 8 and 6 is 24.

$$\begin{aligned}\Rightarrow \frac{25 \times 3}{8 \times 3} + \frac{11 \times 4}{6 \times 4} \\[1em] = \frac{75}{24} + \frac{44}{24} \\[1em] = \frac{75 + 44}{24} \\[1em] = \frac{119}{24} = 4\frac{23}{24}\end{aligned}$$

∴ $3\frac{1}{8} - \left(-1\frac{5}{6}\right) = 4\frac{23}{24}$

Let the other number be $x$.

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

LCM of 5 and 7 is 35.

$$\begin{aligned}\Rightarrow x = \frac{2 \times 7}{5 \times 7} + \frac{4 \times 5}{7 \times 5} \\[1em] \Rightarrow x = \frac{14}{35} + \frac{20}{35} \\[1em] \Rightarrow x = \frac{14 + 20}{35} \\[1em] \Rightarrow x = \frac{34}{35}\end{aligned}$$

Hence, the other number is $\frac{34}{35}$.

Let the required number be $x$.

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

LCM of 8 and 12 is 24.

$$\begin{aligned}\Rightarrow x = \frac{-7 \times 3}{8 \times 3} + \frac{5 \times 2}{12 \times 2} \\[1em] \Rightarrow x = \frac{-21}{24} + \frac{10}{24} \\[1em] \Rightarrow x = \frac{-21 + 10}{24} \\[1em] \Rightarrow x = \frac{-11}{24}\end{aligned}$$

Hence, the required number is $\frac{-11}{24}$.

Let the required number be $x$.

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

LCM of 3 and 6 is 6.

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

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

(i) Solving,

$\frac{2}{3} \times \frac{-7}{8} = \frac{2 \times (-7)}{3 \times 8} = \frac{-14}{24} = \frac{-7}{12}$

∴ $\frac{2}{3} \times \frac{-7}{8} = \frac{-7}{12}$

(ii) Solving,

$\frac{-6}{7} \times \frac{5}{7} = \frac{-6 \times 5}{7 \times 7} = \frac{-30}{49}$

∴ $\frac{-6}{7} \times \frac{5}{7} = \frac{-30}{49}$

(iii) Solving,

$\frac{-2}{9} \times (-5) = \frac{-2}{9} \times \frac{-5}{1} = \frac{-2 \times (-5)}{9 \times 1} = \frac{10}{9}$

∴ $\frac{-2}{9} \times (-5) = \frac{10}{9}$

(iv) Solving,

$\frac{-5}{11} \times \frac{11}{-5} = \frac{-5 \times 11}{11 \times (-5)} = \frac{-55}{-55} = 1$

∴ $\frac{-5}{11} \times \left(\frac{11}{-5}\right) = 1$

(v) Solving,

$\frac{8}{35} \times \frac{21}{-32} = \frac{8 \times 21}{35 \times (-32)} = \frac{168}{-1120} = \frac{-3}{20}$

∴ $\frac{8}{35} \times \frac{21}{-32} = \frac{-3}{20}$

(vi) $-1\frac{29}{35} = \frac{-64}{35}$

$\frac{-105}{128} \times \frac{-64}{35} = \frac{-105 \times (-64)}{128 \times 35} = \frac{6720}{4480} = \frac{3}{2} = 1\frac{1}{2}$

∴ $\frac{-105}{128} \times \left(-1\frac{29}{35}\right) = 1\frac{1}{2}$

(i) Solving,

$(-6) \div \frac{2}{5} = -6 \times \frac{5}{2} = \frac{-6 \times 5}{2} = \frac{-30}{2} = -15$

∴ $(-6) \div \frac{2}{5} = -15$

(ii) Solving,

$\frac{-1}{10} \div \frac{-8}{5} = \frac{-1}{10} \times \frac{5}{-8} = \frac{-1 \times 5}{10 \times (-8)} = \frac{-5}{-80} = \frac{1}{16}$

∴ $\frac{-1}{10} \div \frac{-8}{5} = \frac{1}{16}$

(iii) Solving,

$\frac{-65}{14} \div \frac{13}{-7} = \frac{-65}{14} \times \frac{-7}{13} = \frac{-65 \times (-7)}{14 \times 13} = \frac{455}{182} = \frac{5}{2} = 2\frac{1}{2}$

∴ $\frac{-65}{14} \div \frac{13}{-7} = 2\frac{1}{2}$

(iv) $3\frac{3}{5} = \frac{18}{5}$

$(-6) \div \frac{18}{5} = -6 \times \frac{5}{18} = \frac{-6 \times 5}{18} = \frac{-30}{18} = \frac{-5}{3} = -1\frac{2}{3}$

∴ $(-6) \div 3\frac{3}{5} = -1\frac{2}{3}$

(v) Solving,

$\frac{-48}{49} \div \frac{72}{-35} = \frac{-48}{49} \times \frac{-35}{72} = \frac{-48 \times (-35)}{49 \times 72} = \frac{1680}{3528} = \frac{10}{21}$

∴ $\frac{-48}{49} \div \frac{72}{-35} = \frac{10}{21}$

(vi) $3\frac{1}{7} = \frac{22}{7}$

$\frac{22}{7} \div \frac{-33}{34} = \frac{22}{7} \times \frac{34}{-33} = \frac{22 \times 34}{7 \times (-33)} = \frac{748}{-231} = \frac{-68}{21} = -3\frac{5}{21}$

∴ $3\frac{1}{7} \div \left(\frac{-33}{34}\right) = -3\frac{5}{21}$

Let the other number be $x$.

$$\begin{aligned}x \times \frac{-2}{5} = \frac{18}{35} \\[1em] \Rightarrow x = \frac{18}{35} \div \frac{-2}{5} \\[1em] \Rightarrow x = \frac{18}{35} \times \frac{5}{-2} \\[1em] \Rightarrow x = \frac{18 \times 5}{35 \times (-2)} \\[1em] \Rightarrow x = \frac{90}{-70} \\[1em] \Rightarrow x = \frac{-9}{7} = -1\frac{2}{7}\end{aligned}$$

Hence, the other number is $-1\frac{2}{7}$.

(i) First solve the bracket:

$\frac{13}{21} \div \frac{39}{42} = \frac{13}{21} \times \frac{42}{39} = \frac{13 \times 42}{21 \times 39} = \frac{546}{819} = \frac{2}{3}$

Now,

$\frac{2}{3} \times \frac{-3}{5} = \frac{2 \times (-3)}{3 \times 5} = \frac{-6}{15} = \frac{-2}{5}$

∴ $\left(\frac{13}{21} \div \frac{39}{42}\right) \times \left(\frac{-3}{5}\right) = \frac{-2}{5}$

(ii) First solve the bracket:

$\frac{7}{11} \times \frac{5}{12} = \frac{7 \times 5}{11 \times 12} = \frac{35}{132}$

Also, $-5\frac{5}{21} = \frac{-110}{21}$. Now,

$\frac{-110}{21} \div \frac{35}{132} = \frac{-110}{21} \times \frac{132}{35} = \frac{-110 \times 132}{21 \times 35} = \frac{-14520}{735} = \frac{-968}{49} = -19\frac{37}{49}$

∴ $\left(-5\frac{5}{21}\right) \div \left(\frac{7}{11} \times \frac{5}{12}\right) = -19\frac{37}{49}$

(i) First find the value:

$\frac{3}{13} \div \frac{-4}{65} = \frac{3}{13} \times \frac{65}{-4} = \frac{3 \times 65}{13 \times (-4)} = \frac{195}{-52} = \frac{-15}{4}$

Reciprocal of $\frac{-15}{4}$ is $\frac{-4}{15}$.

Hence, the reciprocal is $\frac{-4}{15}$.

(ii) First find the value:

$$\begin{aligned}\left(-5 \times \frac{12}{15}\right) = \frac{-60}{15} = -4 \\[1em] \left(-3 \times \frac{2}{9}\right) = \frac{-6}{9} = \frac{-2}{3}\end{aligned}$$

So,

$-4 - \left(\frac{-2}{3}\right) = -4 + \frac{2}{3} = \frac{-12}{3} + \frac{2}{3} = \frac{-12 + 2}{3} = \frac{-10}{3}$

Reciprocal of $\frac{-10}{3}$ is $\frac{-3}{10}$.

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

(i) Two rational numbers are called equivalent if they have the same value.

(ii) $\frac{-5}{-11} = \frac{5}{11}$, which is positive, so it lies to the right of zero on the number line.

(iii) Making the denominator positive, $\frac{14}{-12} = \frac{-14}{12}$. HCF of 14 and 12 is 2, so $\frac{-14}{12} = \frac{-7}{6}$.

The standard form is $\frac{-7}{6}$.

(iv) $-3\frac{1}{5} = \frac{-16}{5}$. The multiplicative inverse is $\frac{-5}{16}$.

(v) If p and q are positive integers, then $\frac{p}{q}$ is a positive rational number and $\frac{p}{-q}$ is a negative rational number.

(i) False. Negative rational numbers are smaller than zero, so zero is not the smallest rational number.

(ii) True. Every integer can be written in the form $\frac{p}{q}$ with $q = 1$.

(iii) False. For example, $\frac{2}{3}$ is a rational number but not an integer.

(iv) True. Every fraction is a rational number.

(v) False. For example, $\frac{-2}{3}$ is a rational number but not a fraction.

(vi) True. The reciprocal of -1 is -1.

(vii) True. The difference of two rational numbers is always a rational number.

(viii) False. Division by zero is not defined, so the quotient of two integers is not always a rational number.

(ix) True. The value of a rational number remains the same if both numerator and denominator are multiplied (or divided) by the same non-zero integer.

Making the denominator positive, $\frac{110}{-132} = \frac{-110}{132}$. HCF of 110 and 132 is 22.

$\frac{-110}{132} = \frac{-110 \div 22}{132 \div 22} = \frac{-5}{6}$

Hence, Option 3 is the correct option.

Reducing each to standard form:

$\frac{21}{-56} = \frac{-21}{56} = \frac{-3}{8}$

$\frac{-15}{40} = \frac{-3}{8}$

$\frac{-6}{16} = \frac{-3}{8}$

$\frac{18}{48} = \frac{3}{8}$

The first three equal $\frac{-3}{8}$, but $\frac{18}{48} = \frac{3}{8}$ is different.

Hence, Option 4 is the correct option.

The multiplicative inverse of $\frac{-4}{9}$ is $\frac{9}{-4} = \frac{-9}{4}$.

Hence, Option 2 is the correct option.

$-2\frac{3}{7} = \frac{-17}{7}$. Its reciprocal is $\frac{7}{-17} = -\frac{7}{17}$.

Hence, Option 3 is the correct option.

The product of any non-zero rational number and its multiplicative inverse is always 1.

Hence, Option 1 is the correct option.

The additive inverse of $\frac{-2}{3}$ is $\frac{2}{3}$.

$\frac{-2}{3} \times \frac{2}{3} = \frac{-2 \times 2}{3 \times 3} = \frac{-4}{9}$

Hence, Option 4 is the correct option.

The reciprocal of $\frac{-1}{3}$ is -3.

$\frac{-1}{3} + (-3) = \frac{-1}{3} + \frac{-9}{3} = \frac{-1 + (-9)}{3} = \frac{-10}{3}$

Hence, Option 3 is the correct option.

$\frac{-3}{5} - \left(\frac{-2}{15}\right) = \frac{-3}{5} + \frac{2}{15}$

LCM of 5 and 15 is 15.

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

Hence, Option 3 is the correct option.

$-5\frac{1}{3} = \frac{-16}{3}$ and $-1\frac{7}{8} = \frac{-15}{8}$.

$\frac{-16}{3} \times \frac{-15}{8} = \frac{-16 \times (-15)}{3 \times 8} = \frac{240}{24} = 10$

Hence, Option 1 is the correct option.

$-2\frac{1}{3} = \frac{-7}{3}$ and $2\frac{11}{12} = \frac{35}{12}$.

$\frac{-7}{3} \div \frac{35}{12} = \frac{-7}{3} \times \frac{12}{35} = \frac{-7 \times 12}{3 \times 35} = \frac{-84}{105} = -\frac{4}{5}$

Hence, Option 1 is the correct option.

In the standard form of a rational number, the denominator is always a positive integer.

Hence, Option 4 is the correct option.

Let the other number be $x$.

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

Hence, Option 2 is the correct option.

Statement I: $\frac{-2}{-3} = \frac{2}{3} = \frac{4}{6}$, which is correct. So Statement I is true.

Statement II: A rational number is in standard form only if its denominator is positive. Since $\frac{-2}{-3}$ has a negative denominator, it is not in standard form. So Statement II is false.

Hence, Option 1 is the correct option.

Statement I: $\frac{304}{12589}$ is a positive rational number, while $\frac{4782}{-844921}$ is a negative rational number. Every positive rational number is greater than every negative rational number, so $\frac{304}{12589} \gt \frac{4782}{-844921}$. So Statement I is true.

Statement II: Both numbers are of the form $\frac{p}{q}$ where p and q are integers and $q \neq 0$, so both are rational numbers. So Statement II is true.

Hence, Option 3 is the correct option.

Statement I: Since there is no rational number which when multiplied by 0 gives 1, the multiplicative inverse (reciprocal) of 0 is not defined. So Statement I is false.

Statement II: 0 + 0 = 0, so the additive inverse of 0 is 0. So Statement II is true.

Hence, Option 2 is the correct option.

Statement I: Every positive rational number is greater than zero. So Statement I is true.

Statement II: Negative rational numbers are smaller than 0, so 0 is not the smallest rational number. So Statement II is false.

Hence, Option 1 is the correct option.

First write $\frac{5}{-11}$ with a positive denominator: $\frac{5}{-11} = \frac{-5}{11}$.

Multiplying the numerator and the denominator by 2, 3, 4 and 5:

$$\begin{aligned}\frac{-5 \times 2}{11 \times 2} = \frac{-10}{22} \\[1em] \frac{-5 \times 3}{11 \times 3} = \frac{-15}{33} \\[1em] \frac{-5 \times 4}{11 \times 4} = \frac{-20}{44} \\[1em] \frac{-5 \times 5}{11 \times 5} = \frac{-25}{55}\end{aligned}$$

Hence, five rational numbers equivalent to $\frac{5}{-11}$ are $\frac{-5}{11}, \frac{-10}{22}, \frac{-15}{33}, \frac{-20}{44}$ and $\frac{-25}{55}$.

First reduce $\frac{9}{-15}$ to standard form. $\frac{9}{-15} = \frac{-9}{15} = \frac{-3}{5}$ (dividing by HCF 3).

(i) The denominator is already 5.

$\frac{9}{-15} = \frac{-3}{5}$

∴ $\frac{9}{-15} = \frac{-3}{5}$

(ii) To get -12 from -3, multiply by 4.

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

∴ $\frac{9}{-15} = \frac{-12}{20}$

(iii) To get 30 from 5, multiply by 6.

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

∴ $\frac{9}{-15} = \frac{-18}{30}$

(i) Making the denominator positive:

$\frac{78}{-91} = \frac{-78}{91}$

HCF of 78 and 91 is 13.

$\frac{-78}{91} = \frac{-78 \div 13}{91 \div 13} = \frac{-6}{7}$

∴ Standard form of $\frac{78}{-91}$ is $\frac{-6}{7}$.

(ii) The denominator is positive. HCF of 216 and 162 is 54.

$\frac{-216}{162} = \frac{-216 \div 54}{162 \div 54} = \frac{-4}{3}$

∴ Standard form of $\frac{-216}{162}$ is $\frac{-4}{3}$.

(iii) Making the denominator positive:

$\frac{-195}{-520} = \frac{195}{520}$

HCF of 195 and 520 is 65.

$\frac{195}{520} = \frac{195 \div 65}{520 \div 65} = \frac{3}{8}$

∴ Standard form of $\frac{-195}{-520}$ is $\frac{3}{8}$.

(i) $\frac{60}{-195} = \frac{-60}{195} = \frac{-4}{13}$ (dividing by HCF 15). So $\frac{-4}{13} = \frac{60}{-195}$. Equivalent.

(ii) $\frac{7}{-15} = \frac{-7}{15}$ and $\frac{-35}{-75} = \frac{35}{75} = \frac{7}{15}$. Since $\frac{-7}{15} \neq \frac{7}{15}$, they are not equivalent.

(iii) $\frac{16}{-20} = \frac{-16}{20} = \frac{-4}{5}$ and $\frac{-56}{70} = \frac{-4}{5}$ (dividing by HCF 14). So they are equivalent.

Hence, the pairs (i) and (iii) are pairs of equivalent rational numbers.

(i) Writing each with a positive denominator: $\frac{23}{-24} = \frac{-23}{24}$ and $\frac{-11}{-13} = \frac{11}{13}$.

So the numbers are $\frac{-5}{6}, \frac{-17}{18}, \frac{-23}{24}, \frac{11}{13}$. The first three are negative and the last is positive.

LCM of 6, 18 and 24 is 72.

$$\begin{aligned}\frac{-5 \times 12}{6 \times 12} = \frac{-60}{72} \\[1em] \frac{-17 \times 4}{18 \times 4} = \frac{-68}{72} \\[1em] \frac{-23 \times 3}{24 \times 3} = \frac{-69}{72}\end{aligned}$$

As -69 < -68 < -60, the negatives in ascending order are $\frac{-23}{24}, \frac{-17}{18}, \frac{-5}{6}$, and the positive $\frac{11}{13}$ is the greatest.

Hence, the ascending order is $\frac{23}{-24}, \frac{-17}{18}, \frac{-5}{6}, \frac{-11}{-13}$.

(ii) Writing each with a positive denominator: $\frac{15}{-4} = \frac{-15}{4}$.

So the numbers are $\frac{-25}{6}, \frac{-15}{4}, \frac{-17}{8}, \frac{-53}{12}$. LCM of 6, 4, 8 and 12 is 24.

$$\begin{aligned}\frac{-25 \times 4}{6 \times 4} = \frac{-100}{24} \\[1em] \frac{-15 \times 6}{4 \times 6} = \frac{-90}{24} \\[1em] \frac{-17 \times 3}{8 \times 3} = \frac{-51}{24} \\[1em] \frac{-53 \times 2}{12 \times 2} = \frac{-106}{24}\end{aligned}$$

As -106 < -100 < -90 < -51,

$\frac{-106}{24} \lt \frac{-100}{24} \lt \frac{-90}{24} \lt \frac{-51}{24}$

Hence, the ascending order is $\frac{-53}{12}, \frac{-25}{6}, \frac{15}{-4}, \frac{-17}{8}$.

Writing each with a positive denominator: $\frac{5}{-8} = \frac{-5}{8}$ and $\frac{2}{-3} = \frac{-2}{3}$.

So the numbers are $\frac{-7}{10}, \frac{-5}{8}, \frac{-2}{3}, \frac{-1}{4}, \frac{-3}{5}$. LCM of 10, 8, 3, 4 and 5 is 120.

$$\begin{aligned}\frac{-7 \times 12}{10 \times 12} = \frac{-84}{120} \\[1em] \frac{-5 \times 15}{8 \times 15} = \frac{-75}{120} \\[1em] \frac{-2 \times 40}{3 \times 40} = \frac{-80}{120} \\[1em] \frac{-1 \times 30}{4 \times 30} = \frac{-30}{120} \\[1em] \frac{-3 \times 24}{5 \times 24} = \frac{-72}{120}\end{aligned}$$

As -30 > -72 > -75 > -80 > -84,

$\frac{-30}{120} \gt \frac{-72}{120} \gt \frac{-75}{120} \gt \frac{-80}{120} \gt \frac{-84}{120}$

Hence, the descending order is $\frac{-1}{4}, \frac{-3}{5}, \frac{5}{-8}, \frac{2}{-3}, \frac{-7}{10}$.

(i) $\frac{5}{-7} = \frac{-5}{7}$. LCM of 3 and 7 is 21.

$$\begin{aligned}\frac{-2 \times 7}{3 \times 7} + \frac{-5 \times 3}{7 \times 3} \\[1em] = \frac{-14}{21} + \frac{-15}{21} \\[1em] = \frac{-14 + (-15)}{21} \\[1em] = \frac{-29}{21}\end{aligned}$$

∴ $\frac{-2}{3} + \left(\frac{5}{-7}\right) = \frac{-29}{21}$

(ii) $-1\frac{1}{12} = \frac{-13}{12}$. LCM of 12 and 9 is 36.

$$\begin{aligned}\frac{-13 \times 3}{12 \times 3} + \frac{-5 \times 4}{9 \times 4} \\[1em] = \frac{-39}{36} + \frac{-20}{36} \\[1em] = \frac{-39 + (-20)}{36} \\[1em] = \frac{-59}{36} = -1\frac{23}{36}\end{aligned}$$

∴ $-1\frac{1}{12} + \frac{-5}{9} = -1\frac{23}{36}$

(iii) $2\frac{2}{5} = \frac{12}{5}$ and $-4\frac{3}{10} = \frac{-43}{10}$. LCM of 5 and 10 is 10.

$$\begin{aligned}\frac{12 \times 2}{5 \times 2} + \frac{-43 \times 1}{10 \times 1} \\[1em] = \frac{24}{10} + \frac{-43}{10} \\[1em] = \frac{24 + (-43)}{10} \\[1em] = \frac{-19}{10} = -1\frac{9}{10}\end{aligned}$$

∴ $2\frac{2}{5} + \left(-4\frac{3}{10}\right) = -1\frac{9}{10}$

(i)

$\frac{-5}{36} - \frac{-11}{24} = \frac{-5}{36} + \frac{11}{24}$

LCM of 36 and 24 is 72.

$$\begin{aligned}= \frac{-5 \times 2}{36 \times 2} + \frac{11 \times 3}{24 \times 3} \\[1em] = \frac{-10}{72} + \frac{33}{72} \\[1em] = \frac{-10 + 33}{72} \\[1em] = \frac{23}{72}\end{aligned}$$

∴ The result is $\frac{23}{72}$.

(ii) $-1\frac{2}{5} = \frac{-7}{5}$.

$\frac{-7}{5} - \frac{-8}{15} = \frac{-7}{5} + \frac{8}{15}$

LCM of 5 and 15 is 15.

$$\begin{aligned}= \frac{-7 \times 3}{5 \times 3} + \frac{8 \times 1}{15 \times 1} \\[1em] = \frac{-21}{15} + \frac{8}{15} \\[1em] = \frac{-21 + 8}{15} \\[1em] = \frac{-13}{15}\end{aligned}$$

∴ The result is $\frac{-13}{15}$.

(iii) $-3\frac{5}{12} = \frac{-41}{12}$ and $-2\frac{2}{9} = \frac{-20}{9}$.

$\frac{-41}{12} - \frac{-20}{9} = \frac{-41}{12} + \frac{20}{9}$

LCM of 12 and 9 is 36.

$$\begin{aligned}= \frac{-41 \times 3}{12 \times 3} + \frac{20 \times 4}{9 \times 4} \\[1em] = \frac{-123}{36} + \frac{80}{36} \\[1em] = \frac{-123 + 80}{36} \\[1em] = \frac{-43}{36} = -1\frac{7}{36}\end{aligned}$$

∴ The result is $-1\frac{7}{36}$.

Let the required number be $x$.

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

LCM of 4 and 8 is 8.

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

Hence, $\frac{-1}{8}$ should be subtracted.

(i)

$\frac{-3}{11} \times (-7) = \frac{-3}{11} \times \frac{-7}{1} = \frac{-3 \times (-7)}{11 \times 1} = \frac{21}{11} = 1\frac{10}{11}$

∴ $\frac{-3}{11} \times (-7) = \frac{21}{11}$

(ii)

$\frac{-5}{12} \times \frac{16}{25} = \frac{-5 \times 16}{12 \times 25} = \frac{-80}{300} = \frac{-4}{15}$

∴ $\frac{-5}{12} \times \frac{16}{25} = \frac{-4}{15}$

(iii) $-2\frac{5}{8} = \frac{-21}{8}$ and $1\frac{3}{7} = \frac{10}{7}$.

$\frac{-21}{8} \times \frac{10}{7} = \frac{-21 \times 10}{8 \times 7} = \frac{-210}{56} = \frac{-15}{4} = -3\frac{3}{4}$

∴ $\left(-2\frac{5}{8}\right) \times \left(1\frac{3}{7}\right) = -3\frac{3}{4}$

(i)

$\frac{-8}{13} \div \frac{3}{-26} = \frac{-8}{13} \times \frac{-26}{3} = \frac{-8 \times (-26)}{13 \times 3} = \frac{208}{39} = \frac{16}{3} = 5\frac{1}{3}$

∴ $\frac{-8}{13} \div \frac{3}{-26} = \frac{16}{3}$

(ii) $3\frac{1}{7} = \frac{22}{7}$.

$\frac{22}{7} \div \frac{11}{-12} = \frac{22}{7} \times \frac{-12}{11} = \frac{22 \times (-12)}{7 \times 11} = \frac{-264}{77} = \frac{-24}{7} = -3\frac{3}{7}$

∴ $3\frac{1}{7} \div \frac{11}{-12} = -3\frac{3}{7}$

(iii) First solve the bracket:

$\frac{-3}{7} \times \frac{-2}{3} = \frac{-3 \times (-2)}{7 \times 3} = \frac{6}{21} = \frac{2}{7}$

Now,

$\frac{2}{7} \div \frac{16}{-21} = \frac{2}{7} \times \frac{-21}{16} = \frac{2 \times (-21)}{7 \times 16} = \frac{-42}{112} = \frac{-3}{8}$

∴ $\left(\frac{-3}{7} \times \frac{-2}{3}\right) \div \frac{16}{-21} = \frac{-3}{8}$

Length of rope cut off first $= 4\frac{1}{3}$ m $= \frac{13}{3}$ m.

Remaining length after first cut:

$15 - \frac{13}{3} = \frac{45}{3} - \frac{13}{3} = \frac{45 - 13}{3} = \frac{32}{3} \text{ m}$

Now $\frac{3}{5}$ of the remaining is cut off, so the part left is $\left(1 - \frac{3}{5}\right) = \frac{2}{5}$ of $\frac{32}{3}$.

$\frac{2}{5} \times \frac{32}{3} = \frac{2 \times 32}{5 \times 3} = \frac{64}{15} = 4\frac{4}{15} \text{ m}$

Hence, the length of the remaining part of the rope is $4\frac{4}{15}$ m.

Perimeter of the square = 40 m.

$\frac{2}{5} \text{ of perimeter of square} = \frac{2}{5} \times 40 = 16 \text{ m}$

Perimeter of rectangle = 16 - 2 = 14 m.

Let the length be $l$. Then breadth $= \frac{1}{3}l$.

$$\begin{aligned}2(l + b) = 14 \\[1em] \Rightarrow l + \frac{1}{3}l = 7 \\[1em] \Rightarrow \frac{3l + l}{3} = 7 \\[1em] \Rightarrow \frac{4l}{3} = 7 \\[1em] \Rightarrow l = \frac{7 \times 3}{4} = \frac{21}{4} \text{ m}\end{aligned}$$

Breadth:

$b = \frac{1}{3} \times \frac{21}{4} = \frac{21}{12} = \frac{7}{4} \text{ m}$

Hence, the length of the rectangle is $\frac{21}{4}$ m and the breadth is $\frac{7}{4}$ m.