A 100ml solution of 0.1N HCl was titrated with 0.2N NaOH solution. The titration was discontinued after adding 30ml of NaOH solution. The remaining titration was completed by adding 0.25N KOH solution. The volume of KOH required for completing the titration is

Initial volume of HCl taken = 100ml
I:  HCl
  V_a = ?
  N_a = 0.1N
  NaOH
  V_b = 30ml
  N_b = 0.2N
  V_aN_a = V_bN_b
  V_a(0.1) = 30(0.2)
  V_a = 60 ml
  Volume of 0.1N HCl neutralized by 30 ml of 0.2N NaOH = 60ml
  Volume of 0.1N HCl left over = (100-60)ml = 40ml
II:  HCl
  V_a = 40ml
  N_a = 0.1N
  KOH
  V_b = ?
  N_b = 0.25N
  V_aN_a = V_bN_b
  40(0.1) = V_b(0.2)
  V_b = 16 ml
  Volume of 0.25 KOH required to neutralize 40ml of 0.1N HCl = 16 ml