The value of $0.\overline{234}$ is

  • A
    $\frac{232}{990}$
  • B
    $\frac{232}{9990}$
  • C
    $\frac{232}{900}$
  • D
    $\frac{232}{9909}$

Explore More

Similar Questions

Let $p = 99$ and $q = 101$. Define $p_1 = \log_{10} \left(\frac{p+q}{2}\right)$ and $q_1 = \frac{1}{2}(\log_{10} p + \log_{10} q)$,and $p_2 = \log_{10} \left(\frac{p_1+q_1}{2}\right)$,$q_2 = \frac{1}{2}(\log_{10} p_1 + \log_{10} q_1)$. Then:

The number of solution pairs $(x, y)$ of the simultaneous equations $\log _{1 / 3}(x+y)+\log _3(x-y)=2$ and $2^{y^2}=512^{x+1}$ is

If for $x \in \left(0, \frac{\pi}{2}\right)$,$\log_{10} \sin x + \log_{10} \cos x = -1$ and $\log_{10}(\sin x + \cos x) = \frac{1}{2}(\log_{10} n - 1)$,$n > 0$,then the value of $n$ is equal to

Evaluate: $\frac{{[4 + \sqrt{15}]}^{3/2} + {[4 - \sqrt{15}]}^{3/2}}{{[6 + \sqrt{35}]}^{3/2} - {[6 - \sqrt{35}]}^{3/2}}$

If ${a^x} = {b^y} = {(ab)^{xy}}$,then $x + y = $

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo