$50 \ cm^{3}$ of $0.2 \ N \ HCl$ is titrated against $0.1 \ N \ NaOH$ solution. The titration was discontinued after adding $50 \ cm^{3}$ of $NaOH$. The remaining titration is completed by adding $0.5 \ N \ KOH$. The volume of $KOH$ required for completing the titration is (in $cm^{3}$)
- A$12$
- B$10$
- C$21.0$
- D$16.2$
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