Which of the following diagrams indicates the best relation between dogs, pet…

2025

Which of the following diagrams indicates the best relation between dogs, pet animals and animals?

Show answer & explanation

In classification (Venn diagram) reasoning, the relationship between three classes is shown by how their circles are drawn relative to each other: two classes that share some but not all members are drawn as two circles that partially overlap; a class that is entirely part of another is drawn fully inside it; and classes with no members in common at all are drawn as separate, non-touching circles.

Here, dogs and pet animals overlap only partially: some dogs are kept as pets, but not every dog is a pet (there are also stray and wild dogs), and not every pet animal is a dog (cats, birds, fish, etc. are pets too). At the same time, both dogs and pet animals are themselves animals, so both of their circles must sit completely inside the larger 'animals' circle. The correct diagram is therefore one large circle for animals, fully enclosing two smaller circles for dogs and pet animals that partially overlap each other.

Checking each option against this:

  • The figure with the 'dogs'-'pet animals' overlapping pair sitting beside a separate, disconnected 'animals' circle is wrong: a disconnected animals circle would mean animals shares no members with dogs or pet animals, which is false since both are animals.

  • The figure with 'dogs' and 'pet animals' drawn as two fully separate circles inside 'animals' is wrong: keeping them fully apart would mean no dog is ever a pet animal, contradicting the premise that some dogs are.

  • The figure with 'dogs' drawn entirely inside 'pet animals' (or vice versa), next to a separate 'animals' circle, is wrong on two counts: it shows a full subset relation where only a partial overlap exists, and it fails to place either class inside animals at all.

  • Only the diagram with one large 'animals' circle enclosing two partially overlapping circles for dogs and pet animals satisfies both parts of the relation.

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