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Divisibility by ‘9’: Sum of the digits must be divisible by ‘9’. So, (4 + 7 + 9 + 3 + 8 + 5 + A + 2 + 6 + 7) must be divisible by ‘9’ Or, (51 + A) must be divisible by 9. So, A = 3
Statement: B > C = J; B > S > E; B < N
Conclusion: I. E < C II. J ≤ E
Statements: Q > W > X; J > W; Z < X < P
Conclusions:
I. P > Z
II. J > Q
III. W < P
Statements: P < Q = R ≥ S = T; R < U; R = W
Conclusion: I. W ≥ T II. U < P
Statements:
A ≥ Z > B ≥ Y; C > B ≥ W
Conclusion:
I. A > W
II. C > Y
Statements: A $ B @ D & E @ G % H, F & A, G $ J
Conclusions: I.F & E II. J # B
...Statements: U > H ≥ W; S > T ≥ B; S < H; C ≤ D = U
Conclusions:
I. D > B
II. T < U
III. W ≤ D
Statements: D > E > G ≤ H < I; G > P > F
Conclusions:
I. D > F
II. P < I
III. D > I
Statements: H > S ≥ V ≥ I; T ≤ G = I; U < J ≤ T
Conclusions:
I. S > J
II. U < I
III. H ≥ G
