If $M$ and $N$ are two natural numbers, then $\frac{LCM(M,N)}{HCF(M,N)}$ is

The following are the steps involved in converting $0.\overline{23}$ into $\frac{p}{q}$ form where $q\ne 0$. Arrange them in sequential order from the first to the last.

$\left(A\right)\therefore 100x=23.232323......$ $\left(2\right)$

$\left(B\right)$ As periodicity is $2$, multiply the Eq. $\left(1\right)$ with $100$.

$\left(C\right)\therefore 99x=23\Rightarrow x=\frac{23}{99}$

$\left(D\right)$ Let $x=0.\overline{23}=0.232323......$$\left(1\right)$

$\left(E\right)$ Subtract equation $\left(1\right)$ from equation $\left(2\right)$

The following are the steps involved in finding the positive value of $x$ from the equation $x^2=12.96$. Arrange them in sequential order from the first to the last.

$\left(A\right) x^2=\frac{(36{)}^2}{{10}^2}$

$\left(B\right) \therefore x=3.6$

$\left(C\right) x^2=12.96=\frac{1296}{100}$

$\left(D\right) x=\frac{36}{10}$

The following are the steps involved in finding the positive value of $x$ from the equation $x^2=\frac{8.1}{36.1}$. Arrange them in sequential order from the first to the last.

$\left(A\right) x=\frac{9}{19}$

$\left(B\right) x^2=\frac{9^2}{(19{)}^2}$

$\left(C\right) x^2=\frac{8.1}{36.1}=\frac{81}{361}$