If x ∈ W and $\dfrac{2x - 1}{3}$ ≥ 4, the solution set is:

1. {0, 1, 2, 3, 4, 5}

2. {0, 1, 2, 3, 4, 5, 6}

3. {7, 8, 9, 10, ……………}

4. {7, 8, 9, 10}

$\dfrac{2x - 1}{3}$ ≥ 4

On cross multiplying, we get

⇒ (2x - 1) ≥ 4 $\times$ 3

⇒ 2x - 1 ≥ 12

⇒ 2x ≥ 12 + 1

⇒ 2x ≥ 13

⇒ x ≥ $\dfrac{13}{2}$

⇒ x ≥ 6.5

∵ x is a whole number.

∵ Solution set = {7, 8, 9, 10……}

Hence, option 3 is the correct option.