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Eight friends, AA, DD, HH, II, KK, NN, SS and UU are seated in a circle facing the centre. SS is between HH and UU. KK is third to the left of II. AA is between NN and II. UU is third to the right of DD. Which of the following must be true?

(A) NN is seated next to KK.

(B) UU is seated next to DD.

(C) II is seated next to DD.

(D) AA is seated third to the right of SS.

Solution

✅ Correct Option: 4

We have 8 friends seated in a circle facing the center. Let's use the clues to find their positions.


Let's place II at position 1 (we can start anywhere in a circle).

Since KK is third to the left of II, count counterclockwise: I→8→7→6I \to 8 \to 7 \to 6

So KK is at position 6.

Positions so far: I(1)I(1), K(6)K(6)


Since AA is between NN and II, and II is at position 1, AA must be adjacent to II.

Place AA at position 8 (to the left of II), then NN must be at position 7 (so AA is between NN and II).

Positions: I(1)I(1), K(6)K(6), N(7)N(7), A(8)A(8)


Remaining people are DD, HH, SS, UU for positions 2, 3, 4, 5.

Since UU is third to the right of DD (clockwise), let's try DD at position 2.

Count clockwise: 2→3→4→52 \to 3 \to 4 \to 5

So UU is at position 5.

Positions: I(1)I(1), D(2)D(2), U(5)U(5), K(6)K(6), N(7)N(7), A(8)A(8)


Since SS is between HH and UU, and UU is at position 5, SS must be adjacent to UU.

Positions 3 and 4 are available. Place SS at position 4.

For SS to be between HH and UU, place HH at position 3.

Final positions: I(1)I(1), D(2)D(2), H(3)H(3), S(4)S(4), U(5)U(5), K(6)K(6), N(7)N(7), A(8)A(8)


Circular arrangement (clockwise): I→D→H→S→U→K→N→A→II \to D \to H \to S \to U \to K \to N \to A \to I


Now check each option:

(A) NN is seated next to KK: NN is at position 7, KK is at position 6. They are adjacent. True.

(B) UU is seated next to DD: UU is at position 5, DD is at position 2. Not adjacent. False.

(C) II is seated next to DD: II is at position 1, DD is at position 2. They are adjacent. True.

(D) AA is seated third to the right of SS: SS is at position 4. Third to the right: 4→5→6→74 \to 5 \to 6 \to 7 gives position 7, which is NN, not AA. False.


Both (A) and (C) must be true.

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