To find the odd one out among the given pairs, we first need to determine the meeting schedule for all 7 persons (A, B, C, D, E, F, G) based on the provided clues.

Determining the Meeting Schedule

Let's assign numbers to the days of the week, starting from Monday as Day 1:

• Monday: 1
• Tuesday: 2
• Wednesday: 3
• Thursday: 4
• Friday: 5
• Saturday: 6
• Sunday: 7

Now, let's apply the clues step-by-step:

1. B attends on Thursday: B is scheduled for Day 4.

   Schedule: _ _ _ B _ _ _

2. Only one person between B and G: This means G can be on Day 2 (Tuesday) or Day 5 (Friday).
   • Possibility 1: G is on Day 2. Schedule: G _ B _ _ _ _
   • Possibility 2: G is on Day 5. Schedule: _ _ _ B G _ _
3. A attends immediately after G:
   • If G is on Day 2 (Possibility 1), A is on Day 3. Schedule: G A B _ _ _ _ (G=2, A=3, B=4)
   • If G is on Day 5 (Possibility 2), A is on Day 6. Schedule: _ _ _ B G A _ (B=4, G=5, A=6)
4. The number of people before A is the same as the number of people after C:
   • Consider Possibility 1 (G=2, A=3, B=4): There are 2 people before A (Monday and G on Tuesday). So, there must be 2 people after C. This places C on Day 5. The schedule would be G A B _ C _ _. However, this conflicts with B being scheduled on Day 4. Therefore, Possibility 1 is invalid.
   • Consider Possibility 2 (B=4, G=5, A=6): There are 5 people before A (Monday, Tuesday, Wednesday, B on Thursday, G on Friday). So, there must be 5 people after C. With 7 total days, this places C on Day 1 (Monday). Schedule: C _ _ B G A _
5. Only one person attends between C and D: Since C is on Day 1, and there's one person between C and D, D must be on Day 3 (Wednesday).

   Schedule: C _ D B G A _

6. F attends the meeting after D: D is on Day 3. The available slots after Day 3 are Day 7 (Sunday), as Day 5 (G) and Day 6 (A) are already occupied. Therefore, F must be on Day 7 (Sunday).

   Schedule: C _ D B G A F

7. Fill the remaining slot: The only person left is E, and the only day left is Day 2 (Tuesday).

   Final Schedule: C E D B G A F

The final schedule is:

Analyzing the Pairs

Now, let's examine the given pairs based on their scheduled days and the time difference between them:

• GA: G is on Friday (Day 5), A is on Saturday (Day 6). The difference is $6 - 5 = 1$ day.
• CE: C is on Monday (Day 1), E is on Tuesday (Day 2). The difference is $2 - 1 = 1$ day.
• FG: G is on Friday (Day 5), F is on Sunday (Day 7). The difference is $7 - 5 = 2$ days.
• EG: E is on Tuesday (Day 2), G is on Friday (Day 5). The difference is $5 - 2 = 3$ days.
• DB: D is on Wednesday (Day 3), B is on Thursday (Day 4). The difference is $4 - 3 = 1$ day.

Identifying the Odd One Out

We observe the following time differences between the persons in each pair:

• GA: 1 day
• CE: 1 day
• DB: 1 day
• FG: 2 days
• EG: 3 days

The pairs GA, CE, and DB all have a time difference of exactly 1 day between them. The pair FG has a difference of 2 days. The pair EG has a difference of 3 days.

Comparing the differences (1, 1, 1, 2, 3), the difference of 3 days for the pair EG is the largest and deviates most from the common difference of 1 day seen in the majority of the pairs. Therefore, EG is the odd one out.