Rational Numbers — Exercise 2(E)

Evaluate:

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

Answer

By Division Method,

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

LCM of 3 and 4 = 2 × 2 × 3 = 12

Solving,

$\Rightarrow \dfrac{-2 \times 4}{3 \times 4} + \dfrac{3 \times 3}{4 \times 3} \\[1em] = \dfrac{-8}{12} + \dfrac{9}{12} \\[1em] = \dfrac{-8 + 9}{12} \\[1em] = \dfrac{1}{12}$

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

Evaluate:

$\dfrac{7}{-27} + \dfrac{11}{18}$

Answer

By Division Method,

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

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

Solving,

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

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

Evaluate:

$\dfrac{-3}{8} + \dfrac{-5}{12}$

Answer

By Division Method,

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

LCM of 8 and 12 is 2 × 2 × 2 × 3 = 24

Solving,

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

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

Evaluate:

$\dfrac{9}{-16} + \dfrac{-5}{-12}$

Answer

By Division Method,

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

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

Solving,

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

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

Evaluate:

$\dfrac{-5}{9} + \dfrac{-7}{12} + \dfrac{11}{18}$

Answer

Solving,

By Division Method,

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

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

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

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

Evaluate:

$\dfrac{7}{-26} + \dfrac{16}{39}$

Answer

By Division Method,

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

LCM of 26 and 39 is 2 × 3 × 13 = 78

Solving,

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

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

Evaluate:

$-\dfrac{2}{3} - \left(\dfrac{-5}{7}\right)$

Answer

By Division Method,

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

LCM of 3 and 7 is 3 × 7 = 21

Solving,

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

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

Evaluate:

$-\dfrac{5}{7} - \left(-\dfrac{3}{8}\right)$

Answer

By Division Method,

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

LCM of 7 and 8 is 2 × 2 × 2 × 7 = 56

Solving,

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

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

Evaluate:

$\dfrac{7}{26} + 2 + \dfrac{-11}{13}$

Answer

By Division Method,

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

LCM of 26, 1 and 13 is 2 × 13 = 26

Solving,

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

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

Evaluate:

$-1 + \dfrac{2}{-3} + \dfrac{5}{6}$

Answer

By Division Method,

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

LCM of 1, 3 and 6 is 2 × 3 = 6

Solving,

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

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

The sum of two rational numbers is $\dfrac{-3}{8}$. If one of them is $\dfrac{3}{16}$, find the other.

Answer

Let x be the other number.

$\Rightarrow \dfrac{3}{16} + x = \dfrac{-3}{8} \\[1em] \Rightarrow x = \dfrac{-3}{8} - \dfrac{3}{16}$

By Division Method,

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

LCM of 8 and 16 is 2 × 2 × 2 × 2 = 16

$\Rightarrow x = \dfrac{-3 \times 2}{8 \times 2} - \dfrac{3 \times 1}{16 \times 1} \\[1em] \Rightarrow x = \dfrac{-6}{16} - \dfrac{3}{16} \\[1em] \Rightarrow x = \dfrac{-6 - 3}{16} \\[1em] \Rightarrow x = \dfrac{-9}{16}$

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

The sum of two rational numbers is -5. If one of them is $\dfrac{-52}{25}$, find the other.

Answer

Let x be the other number.

According to question,

$\Rightarrow \dfrac{-52}{25} + x = -5 \\[1em] \Rightarrow x = \dfrac{-5}{1} - \dfrac{-52}{25}$

By Division Method,

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

LCM of 1 and 25 is 5 × 5 = 25

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

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

What rational number should be added to $-\dfrac{3}{16}$ to get $\dfrac{11}{24}$?

Answer

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

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

By Division Method,

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

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

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

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

What rational number should be added to $-\dfrac{3}{5}$ to get 2?

Answer

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

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

LCM of 1 and 5 is 5.

$\Rightarrow x = \dfrac{2 \times 5}{1 \times 5} + \dfrac{3 \times 1}{5 \times 1} \\[1em] \Rightarrow x = \dfrac{10}{5} + \dfrac{3}{5} \\[1em] \Rightarrow x = \dfrac{10 + 3}{5} \\[1em] \Rightarrow x = \dfrac{13}{5} \\[1em] \Rightarrow x = 2\dfrac{3}{5}$

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

What rational number should be subtracted from $-\dfrac{5}{12}$ to get $\dfrac{5}{24}$?

Answer

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

$\Rightarrow -\dfrac{5}{12} - x = \dfrac{5}{24} \\[1em] \Rightarrow x = -\dfrac{5}{12} - \dfrac{5}{24}$

By Division Method,

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

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

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

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

What rational number should be subtracted from $\dfrac{5}{8}$ to get $\dfrac{8}{5}$?

Answer

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

$\Rightarrow \dfrac{5}{8} - x = \dfrac{8}{5} \\[1em] \Rightarrow x = \dfrac{5}{8} - \dfrac{8}{5}$

By Division Method,

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

LCM of 8 and 5 is 2 × 2 × 2 × 5 = 40

$\Rightarrow x = \dfrac{5 \times 5}{8 \times 5} - \dfrac{8 \times 8}{5 \times 8} \\[1em] \Rightarrow x = \dfrac{25}{40} - \dfrac{64}{40} \\[1em] \Rightarrow x = \dfrac{25 - 64}{40} \\[1em] \Rightarrow x = \dfrac{-39}{40}$

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

Evaluate:

$\left(\dfrac{7}{8} \times \dfrac{24}{21}\right) + \left(\dfrac{-5}{9} \times \dfrac{6}{-25}\right)$

Answer

Solving,

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

By Division Method,

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

LCM of 1 and 15 is 3 × 5 = 15

$= \dfrac{1 \times 15}{1 \times 15} + \dfrac{2 \times 1}{15 \times 1} \\[1em] = \dfrac{15}{15} + \dfrac{2}{15} \\[1em] = \dfrac{15 + 2}{15} \\[1em] = \dfrac{17}{15} \\[1em] = 1\dfrac{2}{15}$

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

Evaluate:

$\left(\dfrac{8}{15} \times \dfrac{-25}{16}\right) + \left(\dfrac{-18}{35} \times \dfrac{5}{6}\right)$

Answer

Solving,

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

By Division Method,

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

LCM of 6 and 7 is 2 × 3 × 7 = 42

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

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

Evaluate:

$\left(\dfrac{18}{33} \times \dfrac{-22}{27}\right) - \left(\dfrac{13}{25} \times \dfrac{-75}{26}\right)$

Answer

Solving,

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

By Division Method,

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

LCM of 9 and 2 is 2 × 3 × 3 = 18

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

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

Evaluate:

$\left(\dfrac{-13}{7} \times \dfrac{-35}{39}\right) - \left(\dfrac{-7}{45} \times \dfrac{9}{14}\right)$

Answer

Solving,

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

By Division Method,

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

LCM of 3 and 10 is 2 × 3 × 5 = 30

$= \dfrac{5 \times 10}{3 \times 10} + \dfrac{1 \times 3}{10 \times 3} \\[1em] = \dfrac{50}{30} + \dfrac{3}{30} \\[1em] = \dfrac{50 + 3}{30} \\[1em] = \dfrac{53}{30} \\[1em] = 1\dfrac{23}{30}$

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

The product of two rational numbers is 24. If one of them is $\dfrac{-36}{11}$, find the other.

Answer

Product of two rational numbers = 24

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

Let the other number = x

According to question,

x = $24 ÷ \dfrac{-36}{11}$

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

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

By what rational number should we multiply $\dfrac{20}{-9}$, so that the product is $\dfrac{-5}{9}$?

Answer

Let the required rational number be $x$.

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

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

State true or false:

(i) The quotient of two integers is always a rational number

(ii) $-\dfrac{6}{11}$ is greater than $\dfrac{4}{11}$

(iii) $-\dfrac{9}{32} + \dfrac{5}{23} = \dfrac{-9+5}{32+23} = \dfrac{-4}{55}$

(iv) $1 - \dfrac{3}{15} = \dfrac{-2}{15}$

Answer

(i) False.

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

(ii) False.

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

(iii) False.

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

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

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

(iv) False.

$1 - \dfrac{3}{15} = \dfrac{15}{15} - \dfrac{3}{15} = \dfrac{15 - 3}{15} \\[1em] = \dfrac{12}{15} = \dfrac{4}{5}$

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

Find x, if:

(i) $-\dfrac{5}{8} + x = \dfrac{7}{12}$

(ii) $\dfrac{2}{5} + x = -2$

(iii) $2 + x = -\dfrac{2}{3}$

(iv) $2\dfrac{1}{2}x = 33\dfrac{1}{3}$

(v) $-\dfrac{9}{35}x = \dfrac{3}{5}$

Answer

(i) Solving,

$\Rightarrow -\dfrac{5}{8} + x = \dfrac{7}{12} \\[1em] \Rightarrow x = \dfrac{7}{12} + \dfrac{5}{8}$

By Division Method,

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

LCM of 12 and 8 = 2 × 2 × 2 × 3 = 24

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

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

(ii) Solving,

$\Rightarrow \dfrac{2}{5} + x = -2 \\[1em] \Rightarrow x = \dfrac{-2}{1} - \dfrac{2}{5}$

LCM of 1 and 5 is 5.

$\Rightarrow x = \dfrac{-2 \times 5}{1 \times 5} - \dfrac{2 \times 1}{5 \times 1} \\[1em] \Rightarrow x = \dfrac{-10}{5} - \dfrac{2}{5} \\[1em] \Rightarrow x = \dfrac{-10 - 2}{5} \\[1em] \Rightarrow x = \dfrac{-12}{5} \\[1em] \Rightarrow x = -2\dfrac{2}{5}$

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

(iii) Solving,

$\Rightarrow 2 + x = -\dfrac{2}{3} \\[1em] \Rightarrow x = -\dfrac{2}{3} - \dfrac{2}{1}$

LCM of 3 and 1 is 3.

$\Rightarrow x = \dfrac{-2 \times 1}{3 \times 1} - \dfrac{2 \times 3}{1 \times 3} \\[1em] \Rightarrow x = \dfrac{-2}{3} - \dfrac{6}{3} \\[1em] \Rightarrow x = \dfrac{-2 - 6}{3} \\[1em] \Rightarrow x = \dfrac{-8}{3} \\[1em] \Rightarrow x = -2\dfrac{2}{3}$

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

(iv) Solving,

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

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

(v) Solving,

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

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

Manish walks $\dfrac{8}{9}$ km from a place P towards East. From there, he walks $2\dfrac{1}{2}$ km towards West. Find his final position from the place P.

Answer

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

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

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

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

By Division Method,

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

LCM of 9 and 2 is 2 × 3 × 3 = 18

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

The negative sign shows the final position is towards West.

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

State true or false:

(i) If $\dfrac{p}{q}$ is a rational number and m is an integer, then $\dfrac{p}{q} = \dfrac{p \times m}{q \times m}$.

(ii) If q is a positive integer and p and q are co-prime numbers, then $\dfrac{p}{q}$ is a rational number.

Answer

(i) False.

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

(ii) True.

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

Find x such that $-\dfrac{3}{8} \text { and }\dfrac{x}{16}$ are equivalent rational numbers.

Answer

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

$\Rightarrow -\dfrac{3}{8} = \dfrac{x}{16} \\[1em] \Rightarrow -3 \times 16 = 8 \times x \\[1em] \Rightarrow -48 = 8x \\[1em] \Rightarrow x = \dfrac{-48}{8} \\[1em] \Rightarrow x = -6$

Hence, x = -6

What should be added to $\left(-\dfrac{17}{6} + \dfrac{-7}{8}\right)$ to get -2?

Answer

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

By Division Method,

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

LCM of 6 and 8 is 2 × 2 × 2 × 3 = 24

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

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

$\Rightarrow \dfrac{-89}{24} + x = -2 \\[1em] \Rightarrow x = \dfrac{-2}{1} - \dfrac{-89}{24}$

By Division Method,

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

LCM of 1 and 24 is 2 × 2 × 2 × 3 = 24

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

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