Question

Express $0.8+0.\overline{7}+0.4\overline{3}$ in the form $\frac{p}{q},$ where $p$ and $q$ are integers and $q\ne 0$.

• A. $\frac{182}{90}$
• B. $\frac{181}{90}$
• C. $\frac{180}{90}$
• D. $\frac{90}{182}$

Answer 1

The correct option is B

$\frac{181}{90}$

We have$0.8=\frac{8}{10}$
Let $x=0.\overline{7}=\mathrm{0.777...}$ ....(1)
$\therefore 10x=7.\overline{7}$ ....(2)
Subtracting (1) from (2), we get
$9x=7\Rightarrow x=\frac{7}{9}\text{or}0.\overline{7}=\frac{7}{9}$
Now, let $y=0.4\overline{3}=\mathrm{0.4333....}$
$\therefore 10y=4.\overline{3}$ .... (3)
And $100y=43.\overline{3}$ ....(4)
Subtracting (3) and (4), we get
$90y=39\Rightarrow y=\frac{39}{90}$
$\therefore 0.4\overline{3}=\frac{39}{90}$
Now, $0.8+0.\overline{7}+0.4\overline{3}=\frac{8}{10}+\frac{7}{9}+\frac{39}{90}$
$=\frac{72+70+39}{90}=\frac{181}{90}$
So, $\frac{181}{90}$ is of the form $\frac{p}{q}\text{and}q\ne 0$.