³P1 ways and remaining 4 digits 4!/2! (as 0 occurs twice) ways ³P1 × 4!/2! = (3×4×3×2!)/2! = 36 ways
Statements: F ≥ O = Z, Z ≥ P > L
Conclusions:
I. O > P
II. F > L
III. O ≥ L
IV. P > F
...Statements:
A ≥ Z > B ≥ Y; C > B ≥ W
Conclusion:
I. A > W
II. C > Y
Which of the following symbols should be placed in the blank spaces respectively (in the same order from left to right) to complete the given expression...
Statements:
P > Q < R; T ≤ U ≤ Q > S; V ≤ W < T
Conclusions:
I). P > V
II). T = S
III). R > V...
Statements: O< V ≤ N = P < S, R = Z ≥ Y = X > O
Conclusions:
I. R > N
II. X > P
III. R > O
Statements: P < Q ≤ R ≤ S; P > T = V ≥ X; S ≤ W = Y < U
Conclusions:
I. U > R
II. W > X
III. Q < Y
Statements: A > B ≤ D; G < C ≤ B; F < C ≤ E
Conclusions:
I. F < D
II. E ≤ B
III. A > G
Statements:
A > B ≥ D > C ≤ V < L; C > Z > Q
Conclusions:
I) B > Q
II) L < Z
...If we arrange all the letters of the word ‘SUPPLIERS’ in alphabetical order from the left end, then, which letter will remain at its initial positio...
Statement: P = Q ≥ S ≥ T < O < Z ≥ X; T > H
Conclusion:
I. H ≥ X
II. P > H
...
