AMC 10 Step by Step

2007 AMC 10B Problem 5

P1-10~2 minPrint

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?

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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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.