[Solution from source] In alkaline medium MnO₄⁻ → MnO₂ (gain 3e⁻) and I⁻ → IO₃⁻ (loss 6e⁻): 2MnO₄⁻ + I⁻ + H₂O → 2MnO₂ + IO₃⁻ + 2OH⁻. So one mole KI reduces two moles KMnO₄. 'One fifth' wrongly assumes MnO₄⁻ goes to Mn²⁺ (5 electrons) with I⁻ giving only 1 electron. Answer (4).