CAT Exam 2021 Slot 2 Solved Paper

Case 1 :
$\mathrm{x}\ge \frac{40}{3}$
We get : $3\mathrm{x}-20+3\mathrm{x}-40=20$
$6\mathrm{x}=80$
$\mathrm{x}=\frac{80}{6}=\frac{40}{3}=13.33$
Case 2 :

We get : $3\mathrm{x}-20+40-3\mathrm{x}=20$
We get 20 = 20
So, we get :
$\mathrm{x}\mathrm{\epsilon }\left[\frac{20}{3},40,3\right]$
Case 3:

We get 20-3x + 40-3x = 20
$40=6\mathrm{x}$
$\mathrm{x}=\frac{20}{3}$ but this is not possible
so we get from case 1,2 and 3
Now looking at options
we can say only option C satisfies for all $\mathrm{x}$ .Hence .