Question

Express
$\left(a\right)0.\overline{6}\left(b\right)0.4\overline{7}\left(c\right)0.\overline{001}\left(d\right)0.2\overline{6}$
in the p/q form

Answer 1

we have to express $0.\overline{6}$ in $\frac{p}{q}$ from

Let $x=0.\overline{6}=\mathrm{0.6666....}\left(1\right)$

Multiplying equation $\left(1\right)$ with $10$

$10x=10\times \left(\mathrm{0.6666.....}\right)$

$10x=\mathrm{0.6666.....}\left(2\right)$

Subtract $\left(2\right)$ from$\left(1\right)$ i.e. $\left(2\right)-\left(1\right)$

$10X-X=\mathrm{6.666.....}-\mathrm{0.6666...}$

$9x=6\Rightarrow x=\frac{2}{3}$

Thus, $0.\overline{6}=2/3$

$\left(ii\right)0.\overline{47}=x=\mathrm{0.47777....}\left(1\right)$

Multiplying equation $\left(1\right)$ both $10$

$10x=10x\left(\mathrm{0.4777...}\right)$

$10x=\mathrm{4.777....}---\left(2\right)$

Subtract $\left(2\right)$ from $\left(1\right)$ i.e., $\left(2\right)-\left(1\right)$

$10x-x=\mathrm{4.777..}-\mathrm{0.4777...}$

$9x=4.3000$

$x=\frac{43}{90}=0.\overline{47}$

$\left(iii\right)x=0.\overline{001}=\mathrm{0.001001.....}\left(1\right)$

Multiplying equilibrium $\left(1\right)$ with $1000$

$1000x=1000x\left(\mathrm{0.001001.....}\right)$

$1000x=\mathrm{1.001001....}\left(2\right)$

Subtract $92)$ from $\left(1\right)$ i.e., $\left(2\right)-\left(1\right)$

$1000x-x=\mathrm{1.001001....}-\mathrm{0.001001...}$

$999x=1$

$x=\frac{1}{999}$

$\left(iv\right)x=0.\overline{26}=\mathrm{0.2666...}--\left(1\right)$

Multiplying equation $\left(1\right)$ with $10$

$10x=10x\left(\mathrm{0.2666...}\right)$

$10x=\mathrm{2.666...}--(20$

Subtract $\left(2\right)$ from $\left(1\right)$ i.e., $\left(2\right)-\left(1\right)$

$9x=2.4$

$x=\frac{2.4}{9}=\frac{8}{3\times 10}=\frac{4}{6}$