# Molar ratio of two oxides of the same metal

A certain "metal oxide" is made of the oxides $A$ and $B$ of the same metal. In the oxide $A$, the metal is divalent, and in B, the metal is tetravalent. By reduction with hydrogen of a quantity $a$ of the "metal oxide", $B$ is transformed into metal, and $A$ remains unchanged, and $a$ drops by $0.15 g$. By treating the same mass $a$ of the "metal oxide" with dilute sulfuric acid, only $A$ reacts and becomes an insoluble sulfate and the mass $a$ increases by $0.75 g$.
What is the molar ratio between $A$ and $B$?

So far, I have come up with this solution:

1. Let $Me$ be the metal.
2. The reduction reaction of $B$ goes like this: $$\ce{MeO2 + 2H2 -> Me + 2H2O}$$ The mass drop of $0.15 g$ must be given by the loss of oxygen that goes into water.
3. When reacted with dilute sulfuric acid (for $A$): $$\ce{MeO + H2SO4 -> MeSO4 + H2O}$$ The mass increase must be given by the addition of $0.75 g$ $\ce{SO4^2-}$.

Am I right up to this point? What should I do next? How do I use this information?