These questions refer to the Venn diagram problem we discussed previously.

These questions refer to the Venn diagram problem we discussed previously.

As established in the previous analysis, the numbers provided in the problem statement are contradictory, making it impossible to construct a consistent Venn diagram. Both standard interpretations of the data lead to negative numbers for certain regions, which is not possible for the number of students.

Specifically:

1. If we interpret $n(A \cap B)$ as the total intersection (including $n(A \cap B \cap C)$), then $n(BDT \cap SS) = 4$ and $n(BDT \cap S) = 3$ are both less than $n(S \cap SS \cap BDT) = 6$. This means the number of students learning BDT and SS only would be $4 - 6 = -2$, and BDT and Science only would be $3 - 6 = -3$. These are impossible.
2. If we interpret $n(A \cap B)$ as meaning only those two subjects (e.g., $n(BDT \cap SS \text{ only}) = 4$), and $n(S \cap SS \cap BDT) = 6$, then calculating the number of students learning Social Studies only leads to: $n(SS \text{ only}) = n(SS) - [n(S \cap SS \text{ only}) + n(BDT \cap SS \text{ only}) + n(S \cap SS \cap BDT)]$ $n(SS \text{ only}) = 15 - (6 + 4 + 6) = 15 - 16 = -1$. This is also impossible.

Since the given data is contradictory, a mathematically sound solution for the number of students learning B.D.T Only or Science Only cannot be determined. The problem, as stated, is unsolvable.