Implication check – What can be concluded from “All beggars are poor”? Assume standard category meanings for “beggar”, “poor”, and “rich”.

Introduction / Context:We test typical aptitude-exam reading of socioeconomic labels: “All beggars are poor” is taken with the everyday background that “rich” and “poor” are mutually exclusive. Within that convention, the question asks for a safe implication.

Given Data / Assumptions:

• Premise: Beggar ⊆ Poor.
• Background convention in such questions: Poor ∩ Rich = ∅ (disjoint economic categories).

Concept / Approach:From Beggar ⊆ Poor and Poor ∩ Rich = ∅, it follows that Beggar ∩ Rich = ∅. Hence any specific beggar cannot be rich. Be careful not to invert the subset (which would claim all poor are beggars).

Step-by-Step Solution:Take any individual x. If x is a Beggar, then x is Poor (by premise).If x is Poor, then x is not Rich (by background disjointness).Therefore, if x is a Beggar, x is not Rich.

Verification / Alternative check:Venn diagram: Beggar region sits entirely inside Poor; Rich is disjoint from Poor. Hence Beggar and Rich do not overlap.

Why Other Options Are Wrong:

• “All poor are beggars” reverses the subset.
• “If A is rich then not a beggar” requires Rich ⇒ not Poor; while commonly assumed, the logically safest choice focused on the given premise is the beggar-focused conditional (option D).
• “If A is not rich then not a beggar” is invalid; non-rich includes poor and middle-income, and could include beggars.

Common Pitfalls:Converse and inverse errors when handling subset statements.

Final Answer:If A is a beggar, then A is not rich.