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2007 AMC 10B Problem 5

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2007 AMC 10B Problem 5AMC 10 Step by Step
Name:                             Date: October 28, 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 stack up like circles inside circles: Crups inside Dramps inside Arogs inside Brafs. So every Crup is also an Arog and a Braf.

Solution

"All Crups are Dramps" means the group of Crups sits entirely inside the group of Dramps. Draw each fact as a smaller circle inside a bigger one.

1. All Crups are Dramps: Crups circle inside the Dramps circle.
2. All Dramps are Arogs: Dramps circle inside the Arogs circle.
3. All Arogs are Brafs: Arogs circle inside the Brafs circle.
4. Put these together. The circles nest from smallest to largest:
Crups  →  Dramps  →  Arogs  →  Brafs.\text{Crups} \;\to\; \text{Dramps} \;\to\; \text{Arogs} \;\to\; \text{Brafs}.
(The fact "all Crups are Brafs" is already built in.)
5. A Crup is in the smallest circle, so it is inside every bigger circle. It is an Arog and a Braf. That is choice (D).
6. Each other choice tries to put a bigger circle inside a smaller one: (A) says every Dramp is a Crup, (B) every Braf is a Crup, (C) every Arog is a Crup, (E) every Arog is a Dramp. The facts never say any of these; there could be 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" only works in one direction: it puts XX inside YY, not the other way around. Once the circles are drawn, a true conclusion always moves from an inner circle to an outer one. Any statement that moves inward is not guaranteed.

The trap

Turning a fact around, for example reading 'all Dramps are Arogs' as if it also meant 'all Arogs are Dramps'.

Concise 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.}}

Review in the AoPS books

Techniques

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