Direction : Read the following information carefully and answer the questions…

Concept

This is a two-stage arrangement puzzle. Stage 1 is a fixed linear (row) arrangement governed by a deterministic placement rule; Stage 2 is two circular arrangements derived from the class split. The governing rule is the class division: a name is placed in Class A if and only if the first letter of its English spelling is a vowel (A, E, I, O, U) — so Allen, Alpha, Emma, Ivory, Olive and Unique are Class A; every other name is Class B. Within each row, Class A names are seated first in dictionary (A-to-Z) order in alternate seats starting from the 2nd seat, and Class B names follow in reverse alphabetical (Z-to-A) order. The class split also fixes the circular tables: Class A sits at Table 1, Class B at Table 2. In a circular table of 6, “opposite” means 3 seats apart, and the number of persons “between” two people is counted along the shorter arc between them.

Applying the class rule

List the twelve names and test the first letter of each against the vowel rule:

• Class A (vowel-initial), dictionary order: Allen, Alpha, Emma, Ivory, Olive, Unique.

• Class B (consonant-initial), reverse-alphabetical order: Nick, London, Jack, Fame, Den, Coco.

So there are six students in each class. Table 1 (Class A) seats Allen, Alpha, Emma, Ivory, Olive and Unique; Table 2 (Class B) seats Coco, Den, Fame, Jack, London and Nick.

Building the rows and finding N

Each row has 8 seats and the occupied seats are the alternate seats 2, 4, 6, 8 (4 seats per row × 3 rows = 12 seats for 12 students). Class A fills these seats first, dictionary order, then Class B continues in reverse-alphabetical order:

1. Row 1 (seats 2, 4, 6, 8): Allen, Alpha, Emma, Ivory.

2. Row 2 (seats 2, 4, 6, 8): Olive, Unique, Nick, London.

3. Row 3 (seats 2, 4, 6, 8): Jack, Fame, Den, Coco.

London sits at seat 8 of Row 2, which is the last (extreme-right) seat of the row, so there is no one seated to her right. Hence N = 0: the number of persons sitting to the right of London in the row arrangement is 0. The required count in the circular arrangement must therefore also be 0.

Fixing Table 1 (Class A)

Read “second from the left end in row 2” as the physical second seat of the row, i.e. seat 2 — Olive; this is the only reading that keeps the two Table-1 clues consistent with each other (the alternative reading, “second occupied person,” makes Allen sit both opposite and two seats from the same person, which is impossible).

1. “Allen sits opposite the one who sits second from the left end in Row 2” → Allen sits opposite Olive.

2. “The one who sits right of Olive in Row 2 sits second to the right of Allen” → the occupant immediately to Olive's right in Row 2 is Unique, so Unique sits second to the right of Allen.

3. Placing Allen at position 0 of the 6-seat circle: Unique is at position 2, and Allen's opposite (position 3) is Olive — consistent with step 1.

4. “Alpha sits opposite Emma and is not adjacent to Olive” → the only unused opposite pair is (position 1, position 4); Alpha cannot take position 4 (adjacent to Olive at position 3), so Alpha is at position 1 and Emma at position 4.

5. The one remaining seat, position 5, goes to Ivory.

Table 1 seating (position 0 = Allen, going the same direction used throughout):

Counting the gap from Allen

• Allen–Alpha (0 and 1): adjacent — 0 persons between.

• Allen–Olive (0 and 3): opposite — 2 persons between.

• Allen–Emma (0 and 4): 1 person between, via position 5 (the shorter way round).

• Allen–Ivory (0 and 5): adjacent — 0 persons between.

Only Alpha and Ivory give a gap of 0, matching N = 0 from the row arrangement. Olive (2) and Emma (1) do not match, so options naming Olive alone or Emma alone are ruled out, and ‘Alpha alone’ or ‘Ivory alone’ are each only half of the correct pair.

Cross-check with Table 2 (Class B)

The Table-2 clues confirm the puzzle is fully solvable (so the row/table readings used above are the intended ones, not an arbitrary guess). “Adjacent to Unique” and “at the extreme end” only make sense as row-arrangement references, since Unique's own circular neighbours (Alpha, Olive) both sit at Table 1, not Table 2:

1. Unique's row neighbours are Olive (Table 1) and Nick (Table 2); only Nick is a Table-2 person, so “the one who sits adjacent to Unique” is Nick, and London sits second to the right of Nick.

2. “The one who sits at the extreme end” is Coco, the occupant of the last physical seat (Row 3, seat 8); Coco sits third to the left of London.

3. Placing Nick at position 0: London is at position 2, and Coco (third to London's left) is at position 5.

4. The remaining opposite pair (positions 1 and 4) must hold Jack and Den (“Jack opposite Den”); Jack cannot be at position 1 (adjacent to Nick), so Jack is at 4 and Den at 1 — and Jack (4) is indeed not adjacent to Nick (0).

Table 2 resolves to a single, fully consistent seating (Nick, Den, London, Fame, Jack, Coco in order), with no leftover contradiction — confirming the row layout and the “adjacent/extreme end” readings used for both tables are correct.

Table 2 seating (position 0 = Nick):

Answer

The number of persons to the right of London in the row arrangement is 0, and exactly two Table-1 members — Alpha and Ivory — have 0 persons between themselves and Allen. So the correct choice is: Both Alpha and Ivory.