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2007 AMC 10B Problem 5

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2007 AMC 10B Problem 5AMC 10 Step by Step
Name:                             Date: September 13, 2026
Logic Puzzles · difficulty 1 of 5 · about 2 min · position P1-10

In a certain land, all Arogs are Brafs, all Crups are Brafs, all Dramps are Arogs, and all Crups are Dramps. Which of the following statements is implied by these facts?

  1. A)

    All Dramps are Brafs and are Crups.\text{All Dramps are Brafs and are Crups.}

  2. B)

    All Brafs are Crups and are Dramps.\text{All Brafs are Crups and are Dramps.}

  3. C)

    All Arogs are Crups and are Dramps.\text{All Arogs are Crups and are Dramps.}

  4. D)

    All Crups are Arogs and are Brafs.\text{All Crups are Arogs and are Brafs.}

  5. E)

    All Arogs are Dramps and some Arogs may not be Crups.\text{All Arogs are Dramps and some Arogs may not be Crups.}

Answer

D

Key insight

The facts chain into nested sets Crups inside Dramps inside Arogs inside Brafs, so anything about Crups being Arogs or Brafs is forced.

Solution

Translate each "all XX are YY" into a subset relation X⊆YX \subseteq Y:
Arogs⊆Brafs,Crups⊆Brafs,Dramps⊆Arogs,Crups⊆Dramps.\text{Arogs} \subseteq \text{Brafs},\quad \text{Crups} \subseteq \text{Brafs},\quad \text{Dramps} \subseteq \text{Arogs},\quad \text{Crups} \subseteq \text{Dramps}.
Chaining them gives a single nested picture:
Crups⊆Dramps⊆Arogs⊆Brafs.\text{Crups} \subseteq \text{Dramps} \subseteq \text{Arogs} \subseteq \text{Brafs}.

Now test the choices. The innermost set, Crups, is contained in everything, so "all Crups are Arogs and are Brafs" is certainly true.

The others each claim a larger set sits inside a smaller one: (A) Dramps inside Crups, (B) Brafs inside Crups, (C) Arogs inside Crups, (E) Arogs inside Dramps. None of these follow; the facts allow, for instance, a Braf that is nothing else.

The answer is (D) All Crups are Arogs and are Brafs.\boxed{\textbf{(D)}\ \text{All Crups are Arogs and are Brafs.}}

Why this works

"All XX are YY" is a one-directional subset statement, and subsets chain by transitivity. Drawing the four sets as nested circles makes the valid conclusions visible: only statements moving from an inner set outward are implied.

The trap

Reversing an inclusion, such as concluding all Arogs are Dramps from 'all Dramps are Arogs'.

Common mistakes

  • Reversing an inclusion, such as concluding all Arogs are Dramps from 'all Dramps are Arogs'.
  • Picking (E) because "some Arogs may not be Crups" sounds safe, without noticing its first half, "all Arogs are Dramps," is not implied.

Techniques

Set up the equation/formula and compute; no special trick needed