Discussion
Statements: T * Y, S # M, Y $ S, M @ K
Conclusions:
I. K # S
II. Y @ M
III.T # M
IV.Y @ K
Answer: A
Correct answer: None is true
This set’s code reads \(\$\) as \(\ge\), \(@\) as \(>\), \(\#\) as \(<\), \(\&\) as \(=\) and \(*\) as \(\le\). The statements become \( T \le Y \), \( S < M \), \( Y \ge S \) and \( M > K \). Drawn out, this is two loose fragments joined only at S: T and Y stand above S, while M stands above S as well and K below M. Nothing runs from one fragment through to the other in a single direction.
I. \( K < S \): K and S both sit below M, so their order is free. Not determined.
II. \( Y > M \): Y and M both sit above S. Not determined.
III. \( T < M \): T is only bounded above, by Y, and Y itself is unordered against M. Not determined.
IV. \( Y > K \): Y stands above S and K below M, and the two are three links apart with a reversal between them. Not determined.
None of the four follows. Every failure here is the same one - two elements attached to a common third from the same side.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development