The ratio of non-metal atoms to the number of atoms of two non-metallic oxides A and B are 0.33 and 0.25 respectively. Find the valencies of non-metals present in those non-metallic oxides.

Let's analyze the ratios given:

For oxide A, the ratio of non-metal atoms to the total number of atoms is 0.33, which can be expressed as 1/3. This means that for every 3 atoms in the oxide, 1 is a non-metal atom, and the remaining 2 are oxygen atoms (since it's an oxide). Therefore, the formula for oxide A can be represented as NM₂O₃ (where NM represents the non-metal). To balance the charges, the non-metal must have a valency of 6 (2 * 3 = 6, balancing the 6 negative charges from the three oxygen atoms with 2- charge each).

For oxide B, the ratio of non-metal atoms to the total number of atoms is 0.25, which is 1/4. This means that for every 4 atoms in the oxide, 1 is a non-metal atom, and the remaining 3 are oxygen atoms. The formula for oxide B is therefore NM₂O₃. In this case to balance the charges, the non-metal must have a valency of 4 (1 * 4 = 4, balancing the 6 negative charges from the three oxygen atoms with 2- charge each).

Therefore, the valencies of the non-metals in oxides A and B are 6 and 4 respectively. The correct answer is 6 and 4.