Question

A. No barbarian is gentleman.
B. Some gentlemen are barbarians.
C. Some gentlemen are rude.
D. No gentlemen are rude.
E. Some barbarians are not rude.
F. All barbarians are rude.

• A. ABE
• B. BCE
• C. ADF
• D. BDE

Hint:

When solving syllogism sets, watch out for contradictory universals (e.g., “All X are Y” vs. “No X are Y”) — they can help eliminate options quickly.

Answer 1

The Correct Option is C

We evaluate the logical consistency of option (c): ADF A: No barbarian is gentleman.
D: No gentlemen are rude.
F: All barbarians are rude.
Step 1: Consider A and F together. - A says no barbarian is a gentleman. - F says all barbarians are rude. This implies a logical classification where: \[ \text{Barbarian} \subset \text{Rude} \quad \text{and} \quad \text{Barbarian} \cap \text{Gentleman} = \emptyset \] Step 2: Now consider D: No gentlemen are rude. - If all rude people are not gentlemen (D), and barbarians are all rude (F), then: \[ \text{Barbarian} \cap \text{Gentleman} = \emptyset \] Which is exactly what A asserts. Conclusion: Statements A, D, and F are logically aligned and reinforce each other. They form a consistent logical group. Check other options briefly: - (a) ABE: E contradicts F, and B contradicts A. - (b) BCE: Contradictory — B and C don’t connect to E clearly. - (d) BDE: B contradicts A; inconsistent with D.