Let $`n\text{’}$ be the total number of tests taken by a $ \text{CAT}$ aspirant, and his average score be $`x\text{’}.$

His average score increase by $1,$ if the first $10$ tests are not considered, and his average score for the first $10$ tests is $20.$

So$, (n-10) (x+1) + 10 \times 20 = n \times x $

$ \Rightarrow nx + n – 10x – 10 + 200 = nx $

$ \Rightarrow -10x + n + 190 = 0 $

$ \Rightarrow 10x - n - 190 = 0 \quad \longrightarrow (1) $

His average score decreased by $1$ if the last tests are not considered, and his average score for the last $10$ tests is $30.$

So, $ (n-10)(x-1) + 10 \times 30 = n \times x $

$ \Rightarrow nx – n – 10x + 10 + 300 = nx $

$ \Rightarrow -10x – n + 310 = 0 $

$ \Rightarrow 10x + n – 310 = 0 \quad \longrightarrow (2) $

Adding the equation $(1)$ and $(2),$ we get

$\begin{array} {cc}10x - n - 190 = 0 \\ 10x + n – 310 = 0 \\\hline 20x – 500 = 0 \end{array}$

$ \Rightarrow 20x = 500 $

$ \Rightarrow x = \frac{500}{20} $

$ \Rightarrow \boxed{x = 25}$

From, the equation $(2),$ we get

$ 10x + n – 310 = 0 $

$ \Rightarrow 10(25) + n – 310 = 0 $

$ \Rightarrow 250 + n – 310 = 0 $

$ \Rightarrow \boxed {n = 60} $

$\therefore$ The number of tests taken by $\text{CAT}$ aspirant is $60.$

Correct Answer $: 60 $