The conductivity of 20 mL of 0.2 N NaOH solution is a S cm−1 . If 10 mL of 0.4 N HCl solution is added to this solution, the conductivity decreases by b S cm−1 . The resultant solution has equivalent conductivity in S cm2eq−1 is :
(1) a−b
(2) 4(a−b)×30
(3) 4(a−b)×3×104
(4) 3(a−b)×4