Question

The value of $\mathrm{1.999...}$ in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q\ne 0$, is:

• A. $\frac{1999}{1000}$
• B. $2$
• C. $\frac{1}{9}$
• D. $\frac{19}{10}$

Answer 1

Answer: (B)

Let $x=\mathrm{1.999...}$

Since, one digit is repeating, we multiply $x$ by $10$, we get

$10x=\mathrm{19.99...}$

So, $10x=18+\mathrm{1.999...}=18+x$
Therefore, $10x-x=18$, i.e., $9x=18$
i.e., $x=\frac{18}{9}=\frac{2}{1}=2$

Hence, option $D$ is correct answer.