Question

Which of the following groups of puppies could be in pen 2?

• A. Gabino, Daffy, Custard, Earl
• B. Blakely, Gabino, Ashlen, Daffy
• C. Ashlen, Gabino, Earl, Custard
• D. Blakely, Custard, Earl, Fala
• E. Gabino, Ashlen, Fala, Earl
• A. Gabino is in pen 1 with Daffy
• B. Custard is in pen 2
• C. Blakely is in pen 2 and Fala is in pen 1
• D. Earl is in pen 1
• J. Gabino shares a pen with Blakely
• A. Fala shares a pen with Custard.
• B. Gabino shares a pen with Ashlen.
• C. Earl is in a higher-numbered pen than Blakely.
• D. Blakely shares pen 2 with Earl and Daffy.
• O. Custard is in a higher-numbered pen than Fala.

Answer 1

The Correct Option is D

Step 1: Assign shorthand Let us denote the puppies as A (Ashlen), B (Blakely), C (Custard), D (Daffy), E (Earl), F (Fala), G (Gabino).
Step 2: Apply given conditions
- If G is in Pen 1, then D is not in Pen 2 \(\Rightarrow\) G and D must be together in Pen 1.
- If D is not in Pen 2, then G must be in Pen 1 \(\Rightarrow\) confirms both G and D in Pen 1.
- If A is in Pen 2, then B cannot be in Pen 2 \(\Rightarrow\) A and B cannot be in the same pen.
- If B is in Pen 1, then A cannot be in Pen 1 \(\Rightarrow\) again, A and B must be in opposite pens.
Step 3: Construct possibilities
Pen 1 must contain G and D together. The third member of Pen 1 could be either A or B (depending on constraints). Checking all valid distributions, only in Case II (from the explanation table), Pen 2 is \(\{B, C, E, F\}\) \(\Rightarrow\) Blakely, Custard, Earl, Fala.
Step 4: Match with options
This matches exactly with Option D. \[ \boxed{\text{Blakely, Custard, Earl, Fala}} \]

Answer 2

N/A

Answer 3

N/A