a £ b means a ba ↑ b means a < ba ● b means a > ba â–² b means a = bThe Statement is (i) W ● X (ii) X £ Y (iii) Y ↑ Zi.e, W > X, X W > X and Z > Y > XConclusion I: W â–² Y => W = Y cannot be concludedHence (I) does not follow.Conclusion II: W ↑ Z => W < Zcannot be concludedHence, (II) does not follow.

a £ b means a ≤ ba $ b means a ≥ ba ↑ b means a < ba ● b means a > ba â–² b means a = b
The Statement is (i) a ● b (ii) b ▲ C(iii) c $ di.e. a > b, b = c and c ≥ d=> a > b = c ≥ dConclusion I: d ↑ b => d < b or b < dcannot be concludedHence, (I) does not follow.Conclusion II: b ▲ d => b =d cannot be concluded.Hence, (II) does not follow.But one of them will always be true.

p ۞ q means p > qp @ q means p < qp # q means p ≥ qp ∆ q means p ≤ qp □ q means p = qThe Statement is (i) p # q (ii) q ۞ r (iii) r ∆ sFrom (i) p # q => p ≥ q or q ≤ pFrom (ii) q ۞ r => q > rFrom (iii) r ∆ s => r ≤ s or s ≥ rp ≥ q > r and s ≥ rConclusion I: s ۞ p => s > p cannot be concluded.Hence, I does not follow.Conclusion II: q @ s => q < s cannot be concluded.Hence, II does not follow.

P @ Q => P < QP # Q => P ≥ QP $ Q => P = QP © Q => P ≤ QP % Q => P > QJ % M => J > MJ © R => J ≤ RS $ R => S = RS © T => S ≤ TBy combining the abovestatementswe get,
M < J ≤ R = S ≤ TConclusion I: J @ S => J < S, does not follow.Conclusion II: T % J => T > J, does not follow.Conclusion III: J $ T => J = T, does not follow.But conclusion II and III are contradictory to each other.Either II or III follows.

a £ b means a ≤ ba $ b means a ≥ ba ↑ b means a < ba ● b means a > ba â–² b means a = bThe statement is (i) m â–² n (ii) n ₤ o (iii) o ● pi.e, m = n, n ≤ o and o > p=> o ≥ n = m and o > pConclusion I : p ● m => p > m cannot be concludedHence, I does not follow.Conclusion II: m ₤ o => m ≤ o is true.Hence, II follows.

a + b means a > b, a - b means a = b, a = b means a ≥ b, a * b means a ≤ b and a / b means a < b.According to the statementX + Y => X > YY * Z => Y ≤ Z => Z ≥ YW = Z => W ≥ Z => Z ≤ WW ≥ Z ≥ Y and X > YConclusion (I): Y * W => Y ≤ W, followsConclusion (II): X + Z => X > Z.
Which cannot be definitely determined.Conclusion (I) follows.

The following are the expressions which the symbols are used.A $ B => A < B.A © B => A > B.A @ B => A ≤ B.A * B => A ≥ B.A # B => A = B.
C * D => C > D.D @ F => D ≤ F.F © G => F ≥ G.G $ H => G < H.By combining the above statements
We get,C > D ≤ F ≥ G < H.Conclusion I: C © F => C ≥ F, does not follow.Conclusion II: H © F => H ≥ F, does not follow.Conclusion III: G $ D => G < D, does not follow.Conclusion IV: D $ H => D < H, does not follow.None follows.

M 4 C @ F 7 1 $ A E N 9 H > 5 ↓ K ʘ 3 ? B J # G 8 D 6 I L 2M⁺³, @⁺⁵, A⁺⁷, ↓⁺⁹, 8

The following are the expressions which the symbols are used.A $ B => A < B.A © B => A > B.A @ B => A ≤ B.A * B => A ≥ B.A # B => A = B.
M $ K => M > K.K @ P => K ≤ P.P # Q => P = Q.Q * R => Q ≥ R.By combining the above statements We get,M < K ≤ P = Q ≥ RConclusion I: M $ P => M < P, follows.Conclusion II: K # Q => K = Q, does not follow.Conclusion III: K $ Q => K < Q, does not follow.Conclusion IV: P © R => P > R, does not follow.But conclusion II and III are contradictory to each other.
Only I and either II or III follows.

The given sequence isR K 5 9 # B 2 % * E ? A 8 L $ I 4 S V 7 ! C 6 N @ H 1 3 & DExcept BLI, in all others, the three letters are consecutive alphabets in the given sequence.