Add and express the sum as a mixed fraction:

Solution:

(i) We have $\frac{-12}{2}+\frac{43}{10}$.

L.C.M. of the denominators 5 and 10 is $10 .$

Now, we will express $\frac{-12}{5}$ in the form in which it takes the denominator 10 .

$\frac{-12 \times 2}{5 \times 2}=\frac{-24}{10}$

$\therefore \frac{-12}{5}+\frac{43}{10}=\frac{-24}{10}+\frac{43}{10}$

$=\frac{-24+43}{10}$

$=\frac{19}{10}$

$=1 \frac{9}{10}$

(ii) We have $\frac{24}{7}+\frac{-11}{4}$

L.C.M. of the denominators 7 and 4 is 28 .

Now, we will express $\frac{24}{7}$ and $\frac{-11}{4}$ in the form in which they take the denominator 28 .

$\frac{24 \times 4}{7 \times 4}=\frac{96}{28}$

$\frac{-11 \times 7}{4 \times 7}=\frac{-77}{28}$

$\therefore \frac{24}{7}+\frac{-11}{4}=\frac{96}{28}+\frac{-77}{28}$

$=\frac{96-77}{28}$

$=\frac{19}{28}$

(iii) We have $\frac{-31}{6}+\frac{-27}{8}$.

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

Now, we will express $\frac{-31}{6}$ and $\frac{-27}{8}$ in the form in which they take the denominator 24 .

$\frac{-31 \times 4}{6 \times 4}=\frac{-124}{24}$

$\frac{-27 \times 3}{8 \times 3}=\frac{-81}{24}$

$\therefore \frac{-31}{6}+\frac{-27}{8}=\frac{-124}{24}+\frac{-81}{24}$

$=\frac{-124-81}{24}$

$=\frac{-205}{24}$

$=-8 \frac{13}{24}$

(iv) We have $\frac{101}{6}+\frac{7}{8}$

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

Now, we will express $\frac{101}{6}$ and $\frac{7}{8}$ in the form in which they take the denominator 24 .

$\frac{101 \times 4}{6 \times 4}=\frac{404}{24}$

$\frac{7 \times 3}{8 \times 3}=\frac{21}{24}$

$\therefore \frac{101}{6}+\frac{7}{8}=\frac{404}{24}+\frac{21}{24}$

$=\frac{404+21}{24}$

$=\frac{425}{24}$

$=17 \frac{17}{24}$