The ratio of the equivalent mass of KMnO4 in strongly acidic medium (reduced to Mn^2+) to its equivalent mass in neutral medium (reduced to MnO2) is:

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The ratio of the equivalent mass of $KMnO X_{4}$ in strongly acidic medium (reduced to $Mn X^{2 +}$ ) to its equivalent mass in neutral medium (reduced to $MnO X_{2}$ ) is:

The n-factors are 5 (acidic) and 3 (neutral). Equivalent mass is

M / n

, so the ratio is

(M /5) : (M /3) = 3 : 5