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ATQ, Let HCF of two numbers be ‘x’ So, LCM of two numbers = 8 × x = 8x Or, x × 8x = 10240 Or, 8x² = 10240 Or, x² = 1280 Or, x = √1280 = 35.78 (Approx 36) LCM of two numbers = 36 × 8 = 288 Let two numbers be ‘36a’ and ‘36b’, where ‘a’ and ‘b’ are co-prime numbers. So, 36ab = 288 Or, a × b = 8 So, when a = 1 and b = 8, then difference will be maximum Desired difference = 36 × 8 – 36 × 1 = 252
Statements:
B > C = Z ≥ Q ≥ O; X < C ≤ D < O
Conclusions:
I. O > X
II. B > O
Statements: M ≥ O ≥ P ≤ W, N ≥ K ≥ Y = M
Conclusion:
I. N > W
II. Y ≥ P
Statements: W ≤ B = F; H > T; H < U < F; W ≤ X < S
Conclusions:
I. W < U
II. T < B
III. X > H
Statements: X < Y < Z = L, R = S > T, T ≥ U < V = W > X
Conclusions:
I. R > Y
II. W < L
III. S > X
Statements: P # Q @ R & S @ T # W % I, K $ S @ L
Conclusions: I. Q # W II. R & L
...Statement: Y < Z > I < Q > S = M ≤ N
Conclusions:
I. S= N
II. Q > M
Statements: R ≤ K ≤ H = O ≥ D > Q; K > P
Conclusions:I. O ≥ Q II. Q > P
Statements: Q > U = V ≤ X; R ≥ S ≥ X
Conclusions:
I. U = S
II. V < S
Statements: I % C, C & D, D $ K, K # Z
Conclusions: I. I & D II. D # Z
...Statements: B ≥ C > D; B < E > J; G > A ≥ H > J
Conclusion:
I. D ≤ A
II. G > C
