Question

Evaluate:

(i) $\frac{-3}{5}+\frac{7}{5}+\frac{-1}{5}$
(ii) $\frac{-12}{7}+\frac{3}{7}+\frac{-2}{7}$
(iii) $\frac{11}{-12}+\frac{3}{-8}+\frac{1}{4}$
(iv) $\frac{-16}{9}+\frac{-5}{12}+\frac{7}{18}$
(v) $-3+\frac{1}{8}+\frac{-2}{5}$
(vi) $\frac{-13}{8}+\frac{5}{16}+\frac{-1}{4}$

Answer 1

$\begin{array}{l}\left(i\right)\\ \frac{-3}{5}+\frac{7}{5}+\frac{-1}{5}\\ \text{L.C.M. of the given rational number is 5}.\\ \\ \frac{(-3)}{5}+\frac{7}{5}+\frac{(-1)}{5}\\ =\frac{-3+7-1}{5}\\ =\frac{-4+7}{5}\\ =\frac{3}{5}\end{array}$

(ii)
$\begin{array}{l}\frac{\begin{array}{l}\\ -12\end{array}}{7}+\frac{3}{7}+\frac{-2}{7}\\ =\frac{(-12)}{7}+\frac{3}{7}+\frac{(-2)}{7}\\ =\frac{-12+3-2}{7}\\ =\frac{-14+3}{7}\\ =\frac{-11}{7}\end{array}$


$\begin{array}{l}\left(\mathrm{iii}\right)\frac{11}{-12}+\frac{3}{-8}+\frac{1}{4}\\ \\ \text{We need a positive denominator.}\\ \\ \frac{11}{-12}\times \frac{-1}{-1}=\frac{-11}{12}\mathrm{and}\frac{3}{-8}\times \frac{-1}{-1}=\frac{-3}{8}\\ \text{L.C.M. of the denominators 12},\text{8 and}\text{4 is 24.}\\ \therefore \frac{-11\times 2}{12\times 2}=\frac{-22}{24}\\ \\ \frac{-3\times 3}{8\times 3}=\frac{-9}{24}\\ \\ \frac{1\times 6}{4\times 6}=\frac{6}{24}\\ \mathrm{Now},\frac{(-22)}{24}+\frac{(-9)}{24}+\frac{6}{24}\\ =\frac{-22-9+6}{24}\\ \text{=}\frac{-31+6}{24}\\ =\frac{-25}{24}\end{array}$
$\begin{array}{l}\left(\mathrm{iv}\right)\frac{-16}{9}+\frac{-5}{12}+\frac{7}{18}\\ \\ \text{L.C.M. of the denominators}\text{9},\text{12 and}\text{18 is 36.}\\ \frac{-16\times 4}{9\times 4}=\frac{-64}{36}\\ \frac{-5\times 3}{12\times 3}=\frac{-15}{36}\\ \frac{7\times 2}{18\times 2}=\frac{14}{36}\\ \mathrm{Now},\frac{(-64)}{36}+\frac{(-15)}{36}+\frac{14}{36}\\ =\frac{-64-15+14}{36}\\ \frac{-79+14}{36}\\ =\frac{-65}{36}\end{array}$

$\begin{array}{l}\left(v\right)-3+\frac{1}{8}=\frac{-2}{5}\\ \text{L.C.M. of the denominators}\text{1},\text{8 and 5 is 4}0.\\ \\ \frac{-3\times 40}{1\times 40}=\frac{-120}{40}\\ \\ \frac{1\times 5}{8\times 5}=\frac{5}{40}\\ \\ \frac{-2\times 8}{5\times 8}=\frac{-16}{40}\\ \mathrm{Now},\frac{(-120)}{40}+\frac{5}{40}+\frac{(-16)}{40}\\ =\frac{-120+5-16}{40}\\ =\frac{-136+5}{40}=\frac{-131}{40}\end{array}$

$\begin{array}{l}\left(vi\right)\frac{-13}{8}+\frac{5}{16}+\frac{-1}{4}\\ \text{L.C.M. of the denominator 8},\text{16 and 4 is 16.}\\ \frac{-13\times 2}{8\times 2}=\frac{-26}{16}\\ \\ \frac{5\times 1}{16\times 1}=\frac{5}{16}\\ \\ \frac{-1\times 4}{4\times 4}=\frac{-4}{16}\\ \mathrm{Now},\frac{(-26)}{16}+\frac{5}{16}+\frac{(-4)}{16}\\ =\frac{-26+5-4}{16}\\ \\ \mathrm{Now},\frac{-30+5}{16}=\frac{-25}{16}\end{array}$