Solve $\frac{7x+3}{4}+\frac{9x-5}{8}=\frac{16x-3}{16}$

The following steps are involved in solving the above problem. Arrange them in sequential order.

$$\begin{gathered} A) x=\frac{-5}{30}=\frac{-1}{6} \\ B) \frac{7x+3}{4}+\frac{9x-5}{8}=\frac{16x-3}{16} \\ \Rightarrow \frac{14x+6+9x-5}{8}=\frac{16x-3}{16} \\ C) \frac{23x+1}{8}=\frac{16x-3}{16} \\ \Rightarrow 23x+1=\frac{16x-3}{2} \\ D) 46x+2=16x-3 \\ \Rightarrow 30x=5 \end{gathered}$$  

One-third of a certain, number exceeds ${\frac{1}{9}}^{th}$ of the number by $20$. Find the number. The following steps are involved in solving the above problem. Arrange them in sequential order.

$A)$ Let the number be $x$.

$$\begin{gathered} B) \frac{2x}{9}=20 \\ \Rightarrow \frac{x}{9}=10 \\ \Rightarrow x=90 \end{gathered}$$

$C)$ Given $\frac{x}{3}-\frac{x}{9}=20$

$D) \frac{3x-x}{9}=20$

Select the correct answer from the given options:

Select the correct answer from the given options:

  If $\frac{9x-5}{7}+\frac{6-3x}{2}=2$, then $x=$

Select the correct answer from the given options:

  If $3y+1\frac{1}{2}+6\left(4-5y\right)=12,$, then $y=$

Select the correct answer from the given options:

  If $\frac{z+5}{7}+\frac{4\left(z-11\right)}{9}+3=0$, then $z=$

A number is doubled and half of the number is added to it. If $10$ is subtracted from the result, then we get a number which is one less than the original number. Find the original number.

 If a number is multiplied by $5$ and $5$ is added to it, then the result is equal to $50$. Find the number.

 If seven times a number is added to one-fifth of itself, then five-sixth of the sum is equal to $30$. Find the number.