In the following questions assuming the given statement to be true, find which…

Concept:

In a coded-inequality problem, a relationship between two letters is definitely true only when an unbroken chain of inequalities of the same direction connects them. Two rules govern this: (1) you may combine A > B and B > C into A > C only when the inequality signs do not reverse; (2) a letter that sits at a "peak" — where the signs point away from it on both sides, as in X < R > Y — cannot be used to relate the two letters on its sides, because R is larger than both X and Y and tells you nothing about how X and Y compare.

Given relations:

M > A ≥ P > U; E < R > U; E ≥ L > W.

Conclusion I — is M > L definitely true?

1. Trace M: M > A ≥ P > U. So M is linked downward only as far as U.

2. Trace L: E ≥ L > W, and E < R > U. So L is linked to E, R, W.

3. The only letter the two groups share is U, but U is a low point reached from both sides (U < P ≤ A < M on one side, and U < R on the other). There is no same-direction chain running from M down to L.

4. Therefore the order of M and L is not fixed — M > L is not definitely true.

Conclusion II — is W ≤ A definitely true?

1. Trace W: W < L ≤ E, and E < R. So W < E < R. W sits below R.

2. Trace A: M > A ≥ P > U, so A > U. A sits above U.

3. The bridge would have to pass through R and U, but the statement E < R > U makes R a peak: R > E and R > U both point away from R, so R cannot link the E-side (where W lives) to the U-side (where A lives).

4. Therefore the order of W and A is not fixed — W ≤ A is not definitely true.

Cross-check ("either / or" test):

An "Either I or II" answer would require the same two letters to form a complementary pair such as M > L and M ≤ L. Here the two conclusions involve different letter pairs (M,L and W,A), so the either/or case cannot apply. Each pair independently has no fixed order.

Result:

Neither Conclusion I nor Conclusion II is definitely true.