Question

$\frac{5}{6}+\frac{2}{3}-\frac{4}{9}=?$
(a) $1\frac{1}{3}$
(b) $1\frac{1}{6}$
(c) $1\frac{1}{9}$
(d) $1\frac{1}{18}$

Answer 1

(d) $1\frac{1}{18}$

Explanation:
$$\begin{gathered} 3\underline{\overline{)3,6,9}} \\ \underline{2\underline{\overline{)1,2}},3} \\ 3\overline{)\underline{1,1,3}} \\ \overline{)\underline{1,1,1}} \end{gathered}$$

$$\begin{gathered} \frac{5}{6}+\frac{2}{3}-\frac{4}{9}(\mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}3,6\mathrm{and}9=\left(2\times 3\times 3\right)=18) \\ =\frac{\left(15+12-8\right)}{18}\left\{\left[18÷6=3,3\times 5=15\right],\left[18÷3=6,6\times 2=12\right]\mathrm{and}\left[18÷9=2,2\times 4=8\right]\right\} \\ =\frac{\left(27-8\right)}{18}=\frac{19}{18}=1\frac{1}{18} \end{gathered}$$