Question

Express 0.6+0.7bar+0.47bar in the form of p/q where p and q are integers and q not = 0?

Answer 1

$$\begin{gathered} \mathrm{Let}x=0.\overline{6}.............\left(1\right) \\ \mathrm{Multiplying}\left(1\right)\mathrm{by}10,\mathrm{we}\mathrm{get}, \\ 10x=10\times 0.\overline{6} \\ \Rightarrow 10x=6.\overline{6}............\left(2\right) \\ \mathrm{Now}\mathrm{subtracting}\left(1\right)\mathrm{from}\left(2\right),\mathrm{we}\mathrm{get}, \\ 10x-x=6.\overline{6}-0.\overline{6} \\ \Rightarrow 9x=6 \\ \Rightarrow x=\frac{6}{9}=\frac{2}{3} \\ \mathrm{Let}y=0.\overline{7}............\left(3\right) \\ \mathrm{Multiplying}\left(3\right)\mathrm{by}10 \\ 10\times y=10\times 0.\overline{7} \\ \Rightarrow 10y=7.\overline{7}.............\left(4\right) \\ \mathrm{Subtracting}\left(3\right)\mathrm{from}\left(4\right),\mathrm{we}\mathrm{get}, \\ \Rightarrow 10y-y=7.\overline{7}-0.\overline{7} \\ \Rightarrow 9y=7 \\ \Rightarrow y=\frac{7}{9} \\ \mathrm{Let}z=0.\overline{47}............\left(5\right) \\ \mathrm{Multiplying}\left(5\right)\mathrm{by}100 \\ 100\times z=100\times 0.\overline{47} \\ \Rightarrow 100z=47.\overline{47}............\left(6\right) \\ \mathrm{Subtracting}\left(5\right)\mathrm{from}\left(6\right),\mathrm{we}\mathrm{get} \\ 100z-z=47.\overline{47}-0.\overline{47} \\ \Rightarrow 99z=47 \\ \Rightarrow z=\frac{47}{99} \\ \mathrm{Therefore},x+y+z=0.\overline{6}+0.\overline{7}+0.\overline{47}=\frac{2}{3}+\frac{7}{9}+\frac{47}{99} \\ =\frac{66+77+47}{99} \\ =\frac{190}{99} \end{gathered}$$