Question 16.

Case 1 : $x \ge \frac{40}{3}$

we get $3x - 20 + 3x - 40 = 20$

$6x = 80$

$x = \frac{80}{6} = \frac{40}{3} = 13.33$

Case 2 : $\frac{20}{3} \le x \lt \frac{40}{3}$

we get $3x - 20 + 40 - 3x = 20$

we get $20 = 20$

So we get $x \in \left[ \frac{20}{3}, \frac{40}{3} \right]$

Case 3 : $x \lt \frac{20}{3}$

we get $20 - 3x + 40 - 3x = 20$

$40 = 6x$

$x = \frac{20}{3}$

but this is not possible

So we get from case 1, 2, and 3:

$\frac{20}{3} \le x \le \frac{40}{3}$

Now looking at options,

we can say only option C satisfies for all $x$.

Hence 7 < x < 12.