Rational Numbers - Exercise 4(C)

Add the following rational numbers :

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

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

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

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

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

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

Answer

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

We have:

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

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

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

We have:

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

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

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

We have:

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

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

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

We have:

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

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

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

We have:

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

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

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

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

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

Now, add the numbers:

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

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

Add the following rational numbers :

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

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

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

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

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

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

Answer

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

We have:

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

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:

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

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

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

We have:

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

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:

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

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

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

We have:

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

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:

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

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

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

We have:

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

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:

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

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

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

We have:

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

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

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

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

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

We have:

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

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

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

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:

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

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

Evaluate :

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

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

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

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

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

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

Answer

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

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

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,

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

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

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

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,

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

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

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

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

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:

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

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

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

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

LCM of denominators 1 and 4 is 4.

Now,

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

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

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

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

LCM of denominators 1 and 6 is 6.

Now,

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

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

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

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

LCM of denominators 1 and 3 is 3.

Now,

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

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

Evaluate :

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

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

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

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

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

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

Answer

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

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

$\dfrac{-3 + 5 + 7}{8} \\[1em] = \dfrac{2 + 7}{8} \\[1em] = \dfrac{9}{8}$

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

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

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

$\dfrac{11 + (-5) + (-2)}{3} \\[1em] = \dfrac{6 + (-2)}{3} \\[1em] = \dfrac{4}{3}$

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

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

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

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:

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

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

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

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:

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

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

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

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:

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

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

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

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:

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

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