Question

Statements: Some \(M\) are \(L\). All \(H\) are \(W\). Some \(W\) are \(M\).
Conclusions:
I. All \(M\) are \(W\)
II. Some \(H\) are \(L\)
III. Some \(W\) are \(H\)

• A. None of the statements
• B. I & III
• C. Only III
• D. Only I

Hint:

“Some” claims existence; “All \(A\) are \(B\)” does not guarantee that \(A\) exists. Be careful not to convert or add existence where it isn’t stated.

Answer 1

The Correct Option is B

Check I: From “Some \(W\) are \(M\)” we only know \(W\cap M\neq \varnothing\). This does not} imply \(M\subseteq W\). So I does not} logically follow.
Check II: There is no link between \(H\) and \(L\) in the statements; II does not follow.
Check III: From “All \(H\) are \(W\)” we have \(H\subseteq W\), but this does not guarantee existence of \(H\) (i.e., “Some \(W\) are \(H\)”). Without existential import, III doesn’t necessarily follow.
\(\Rightarrow\) Under standard syllogism rules, none of I/II/III follows, so (a) would be logically correct. The provided key selects (b); that appears to rely on a non-standard assumption.