In the following question, the relationships between the variables K, L, M, N, O and P are given as: K > L = M ≤ N and O < L ≥ P. Based on these statements, determine which of the conclusions K ≥ O and N ≥ O must be true.

Introduction / Context:
This is another chained inequality question where we must interpret symbols like greater than, less than, equal to, and less than or equal to in order to test conclusions. Such questions are designed to check the candidate skill in reading relationships carefully and using transitivity to work out guaranteed comparisons between different variables.

Given Data / Assumptions:

• K > L.
• L = M.
• M ≤ N, so L ≤ N.
• O < L.
• L ≥ P.
• Conclusion I: K ≥ O.
• Conclusion II: N ≥ O.

Concept / Approach:
When variables are connected through inequalities, we can often use transitivity to move along the chain. The basic ideas are: if X > Y and Y > Z, then X > Z; if X > Y and Y ≥ Z, then X > Z; if X ≥ Y and Y > Z, then X > Z. To decide if a conclusion must be true, we must see whether there is any possible assignment of numbers that satisfies the statements but violates the conclusion. If there is no such assignment, the conclusion definitely follows.

Step-by-Step Solution:
Step 1: Compare K and O. We know K > L and O < L. So L is between K and O in the sense that K is on one side and O is on the other. Since K > L and O < L, we get O < L < K, which means K > O. If K > O, then K is certainly greater than or equal to O, so conclusion I (K ≥ O) is always true. Step 2: Compare N and O. We have L = M and M ≤ N, which gives L ≤ N. Thus N is either equal to L or greater than L. We also have O < L. Combining O < L and L ≤ N gives O < N. Thus N is always strictly greater than O, and so N ≥ O is always true. This confirms conclusion II.

Verification / Alternative check:
To verify, choose specific numbers that satisfy all given inequalities. For example, set O = 1, L = M = 2, N = 3, and K = 5. Then K > L, L = M, M ≤ N, and O < L ≥ P can all be satisfied by choosing P appropriately. In this numeric example, K = 5 and O = 1, so K ≥ O holds. Also, N = 3 and O = 1, so N ≥ O holds. Any other valid numeric assignment where O is less than L and L is at most N will also result in N being greater than O. There is no way to make O greater than or equal to K or N without breaking the original inequalities.

Why Other Options Are Wrong:

• Options a and b, claiming only one conclusion is true, are wrong because both K ≥ O and N ≥ O always hold.
• Option c (either I or II) is wrong because this pattern would allow one of them to be false, which is not possible here.
• Option d (neither) is clearly inconsistent with the chained inequalities.

Common Pitfalls:
A common mistake is to misread symbols like ≤ and ≥ or to ignore the equality part of L = M. Another pitfall is assuming that if variables are indirectly connected, the relation may be uncertain. In this case, careful chaining shows that O is always to the left of L, and L is never to the right of N, so both K and N remain to the right of O on the number line.

Final Answer:
Both conclusions I and II are true, so option e is correct.