Question

Add and express the sum as a mixed fraction:
(i) $\frac{-12}{5}\mathrm{and}\frac{43}{10}$

(ii) $\frac{24}{7}\mathrm{and}\frac{-11}{4}$

(iii) $\frac{-31}{6}\mathrm{and}\frac{-27}{8}$

(iv) $\frac{101}{6}\mathrm{and}\frac{7}{8}$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right) \\ \text{We have}\frac{-12}{5}+\frac{43}{10}. \\ \text{L.C.M. of the denominators 5 and 10 is 10.} \\ \text{Now, we will express}\frac{-12}{5}\text{in the form in which it takes the denominator 10.} \\ \frac{-12\times 2}{5\times 2}=\frac{-24}{10} \\ \therefore \frac{-12}{5}+\frac{43}{10}=\frac{-24}{10}+\frac{43}{10} \\ =\frac{-24+43}{10} \\ =\frac{19}{10} \\ =1\frac{9}{10} \\ \left(\mathrm{ii}\right) \\ \text{We have}\frac{24}{7}+\frac{-11}{4}. \\ \mathrm{L.C.M.\; of\; the}\text{denominators 7 and 4 is 28.} \\ \mathrm{Now,\; we\; will\; express}\frac{24}{7}\text{and}\frac{-11}{4}in\; the\; form\; in\; which\text{they take the denominator 28.} \\ \frac{24\times 4}{7\times 4}=\frac{96}{28} \\ \frac{-11\times 7}{4\times 7}=\frac{-77}{28} \\ \therefore \frac{24}{7}+\frac{-11}{4}=\frac{96}{28}+\frac{-77}{28} \\ =\frac{96-77}{28} \\ =\frac{19}{28} \\ \left(\mathrm{iii}\right) \\ \text{We have}\frac{-31}{6}+\frac{-27}{8}. \\ \mathrm{L.C.M.\; of\; the}\text{denominators 6 and 8 is 24.} \\ \mathrm{Now,\; we\; will\; express}\frac{-31}{6}\text{and}\frac{-27}{8}in\; the\; form\; in\; which\mathrm{they\; take\; the\; denominator}\text{24.} \\ \frac{-31\times 4}{6\times 4}=\frac{-124}{24} \\ \frac{-27\times 3}{8\times 3}=\frac{-81}{24} \\ \therefore \frac{-31}{6}+\frac{-27}{8}=\frac{-124}{24}+\frac{-81}{24} \\ =\frac{-124-81}{24} \\ =\frac{-205}{24} \\ =-8\frac{13}{24} \\ \left(\mathrm{iv}\right) \\ \text{We have}\frac{101}{6}+\frac{7}{8}. \\ \mathrm{L.C.M.\; of\; the}\text{denominators 6 and 8 is 24.} \\ \mathrm{Now,\; we\; will\; express}\frac{101}{6}\text{and}\frac{7}{8}in\; the\; form\; in\; which\mathrm{they\; take\; the\; denominator}\text{24.} \\ \frac{101\times 4}{6\times 4}=\frac{404}{24} \\ \frac{7\times 3}{8\times 3}=\frac{21}{24} \\ \therefore \frac{101}{6}+\frac{7}{8}=\frac{404}{24}+\frac{21}{24} \\ =\frac{404+21}{24} \\ =\frac{425}{24} \\ =17\frac{17}{24} \end{gathered}$$