Question

The value of $0.\overline{2}+0.\overline{3}+0.\overline{4}+0.\overline{9}+0.\overline{39}$ is

• A. $0.\overline{57}$
• B. $1\frac{20}{33}$
• C. $2\frac{1}{3}$
• D. $2\frac{13}{33}$

Answer 1

Answer: (D)

$2\frac{13}{33}$

Let $x=0.\overline{2}\dots \dots \left(i\right)$

$x=\mathrm{0.22222.............}\dots \dots \left(ii\right)$

On multiplying eq(ii) by 10 both sides, we get

$10x=\mathrm{2.222222.......}\dots \dots \left(iii\right)$

On subtracting $eq.\left(ii\right)$ from eq.$\left(iii\right),$ we get

$10x-x=\mathrm{2.2222....}-\mathrm{0.2222....}$

$9x=2$

$x=\frac{2}{9}$

$\therefore 0.\overline{2}=\frac{2}{9}$

Similarly, $0.\overline{3}=\frac{3}{9},0.\overline{4}=\frac{4}{9},0.\overline{9}=\frac{9}{9},$ $0.\overline{39}=\frac{39}{99}$

Now,

$0.\overline{2}+0.\overline{3}+0.\overline{4}+0.\overline{9}+0.\overline{39}$

$=\frac{2}{9}+\frac{3}{9}+\frac{4}{9}+\frac{9}{9}+\frac{39}{99}$

$=\frac{22+33+44+99+39}{99}$

$=\frac{237}{99}$

$=\frac{79}{33}$

$=2\frac{13}{33}$