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Question

Three statements are given, followed by Three conclusions numbered I, II and III. Assuming the statements to be true, even if they seem to be at variance with commonly known facts, decide which of the conclusions logically follow(s) from the statements.
Statements:
No chart is a casket.
Not a single casket is a house.
Every house is an adapter.
Conclusions:
(I) Some adapters which are houses are charts as well.
(II) No chart is a house.
(III) Some adapters are houses.

This question was previously asked in
SSC Selection Post 2024 Question Paper (26-Jun-2024) (Shift-4)
The correct answer is
Only conclusion III follows

Statements: Chart, Casket, House Relationships

The task is to determine which conclusions logically follow from the given statements. We must assume the statements are true, irrespective of whether they align with common knowledge.

Premises: Chart, Casket, House Logic

Let's break down the relationships described in the premises using set theory notation for clarity:

  • Statement 1: No chart is a casket.
    Formal Logic: This means the set of Charts (C) and the set of Caskets (K) have no common elements. Represented as: $C \cap K = \emptyset$.
  • Statement 2: Not a single casket is a house.
    Formal Logic: Similarly, the set of Caskets (K) and the set of Houses (H) are disjoint. Represented as: $K \cap H = \emptyset$.
  • Statement 3: Every house is an adapter.
    Formal Logic: This means the entire set of Houses (H) is contained within the set of Adapters (A). Represented as: $H \subseteq A$.

Conclusions: Evaluating Chart, House Links

We will now systematically evaluate each conclusion based on the logical implications of the premises.

Conclusion I: Adapters, Houses, Charts Link

Claim: Some adapters which are houses are charts as well.

Meaning: Since Statement 3 establishes that all Houses are Adapters ($H \subseteq A$), the phrase 'adapters which are houses' simply refers to the set of Houses (H). Therefore, this conclusion claims that there is an overlap between the set of Houses (H) and the set of Charts (C). In set notation: $H \cap C \neq \emptyset$.

Analysis: From Statement 1 ($C \cap K = \emptyset$) and Statement 2 ($K \cap H = \emptyset$), we know that both Charts and Houses are separate from Caskets. However, these statements do not provide any information linking Charts and Houses directly or indirectly. It is possible for them to overlap, or to be completely separate.

Example Scenario: Consider Charts = {1}, Caskets = {2}, Houses = {3}, and Adapters = {3, 4}. Here, Statement 1 ($ \{1\} \cap \{2\} = \emptyset $) and Statement 2 ($ \{2\} \cap \{3\} = \emptyset $) are true. Statement 3 ($ \{3\} \subseteq \{3, 4\} $) is also true. However, Conclusion I ($H \cap C \neq \emptyset$) is false because $ \{3\} \cap \{1\} = \emptyset $. Since we can find a valid case where the conclusion is false, it does not logically follow.

Verdict: Conclusion I does not follow.

Conclusion II: Chart House Disjoint Check

Claim: No chart is a house.

Meaning: This conclusion asserts that the set of Charts (C) and the set of Houses (H) have no members in common. In set notation: $C \cap H = \emptyset$.

Analysis: Just like with Conclusion I, the premises $C \cap K = \emptyset$ and $K \cap H = \emptyset$ do not logically necessitate that $C \cap H = \emptyset$. The separation from Caskets does not imply separation between Charts and Houses.

Example Scenario: Let Charts = {1, 2}, Caskets = {3, 4}, Houses = {1, 5}, and Adapters = {1, 5, 6}. The statements hold: No {1, 2} is {3, 4}. No {3, 4} is {1, 5}. Every {1, 5} is in {1, 5, 6}. However, Conclusion II ($C \cap H = \emptyset$) is false because the element '1' exists in both the Charts set and the Houses set.

Verdict: Conclusion II does not follow.

Conclusion III: Adapter House Overlap Check

Claim: Some adapters are houses.

Meaning: This conclusion states that there is at least one element that belongs to both the set of Adapters (A) and the set of Houses (H). In set notation: $A \cap H \neq \emptyset$.

Analysis: Statement 3 directly states $H \subseteq A$. This means every element in H is also in A. Assuming the set of Houses is not empty (which is a standard assumption in such logic problems unless specified otherwise), there must be at least one house. Because every house is an adapter, this existing house is also an adapter. Therefore, the intersection of Adapters and Houses ($A \cap H$) is precisely the set of Houses (H), which we assume is non-empty. Hence, $A \cap H \neq \emptyset$.

Verdict: Conclusion III follows.

Deduction: Final Conclusion Validity

After analyzing each conclusion against the given statements:

  • Conclusion I does not logically follow.
  • Conclusion II does not logically follow.
  • Conclusion III logically follows.

Therefore, the only conclusion that is logically guaranteed by the premises is Conclusion III.

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