Question

Statements: All the students are young. All the teens are young. Some men are teens.
Conclusions:
I. Some students are teens.
II. Some young are students.
III. Some young are men.

• A. Only (I) follows
• B. Only (II) and (III) follow
• C. Only (I) and (II) follow
• D. Only (I) and (III) follow

Hint:

In syllogisms, “All \(A\) are \(B\)” places \(A\) fully inside \(B\). If another set is also inside \(B\), the two need not intersect. Use subset chains (like Teens \(\subseteq\) Young) to deduce valid “some” conclusions.

Answer 1

The Correct Option is B

Step 1: Translate the premises into set relations.
- All students are young: \(S \subseteq Y\).
- All teens are young: \(T \subseteq Y\).
- Some men are teens: \(\exists\) men in \(T\). Since \(T \subseteq Y\), those men are also in \(Y\). Step 2: Test each conclusion.
(I) Some students are teens. This claims \(S \cap T \neq \varnothing\). From the premises we only know both \(S\) and \(T\) lie inside \(Y\); no overlap between \(S\) and \(T\) is guaranteed. \(\Rightarrow\) \textit{Does not follow}.
[4pt] (II) Some young are students. From \(S \subseteq Y\), if there exist students then those are young, so at least “some young” are students. Standard syllogism conversion from universal \(A\) (“All \(S\) are \(Y\)”) to particular \(I\) (“Some \(Y\) are \(S\)”) is accepted. \(\Rightarrow\) \textit{Follows}.
[4pt] (III) Some young are men. We have “Some men are teens” and all teens are young; hence those men are young. Therefore \(Y \cap M \neq \varnothing\). \(\Rightarrow\) \textit{Follows}. \[ \boxed{\text{Only (II) and (III) follow.}} \]