In the following question, relationships between several elements are given by the statements: M < S ≤ T = R ≥ D > E ≥ F and G ≤ S < H. From these statements, determine which of the conclusions G = R and G < R must be true.

Introduction / Context:
This question combines several inequalities and asks about the precise relationship between G and R. The key idea is to see whether G must be equal to R, must be less than R, or could be either depending on the exact values chosen. If more than one relationship is possible without contradicting the given statements, then we can only say that one of the two is true, not which one in particular.

Given Data / Assumptions:

• M < S.
• S ≤ T.
• T = R.
• R ≥ D.
• D > E.
• E ≥ F.
• G ≤ S.
• S < H.
• Conclusion I: G = R.
• Conclusion II: G < R.

Concept / Approach:
We first try to place all variables roughly on a number line using the inequalities. After that, we focus on how close G can be to S and how S relates to T and R. Since G ≤ S and S ≤ T = R, we get G ≤ R, but we must check whether equality is forced or if G can be strictly less than R. If both equality and strict inequality are possible, then exactly one of conclusions I or II must hold in any specific case, but we cannot decide which one, so the correct option is the one that says either I or II is true.

Step-by-Step Solution:
Step 1: From S ≤ T and T = R, we get S ≤ R. Step 2: From G ≤ S and S ≤ R, we derive G ≤ R. Step 3: Thus G can never be greater than R. The only possibilities are G = R or G < R. Step 4: Check if G = R is possible. If we choose G = S = T = R, all inequalities G ≤ S, S ≤ T, and T = R remain valid, and we can assign remaining variables to satisfy other inequalities. So G = R is possible. Step 5: Check if G < R is possible. Take G < S and S ≤ R. For example, let G = 1, S = 2, T = R = 3. Then G ≤ S and S ≤ T are satisfied, and G < R. Hence G < R is also possible. Step 6: Since both G = R and G < R are possible depending on the exact numeric assignments, neither conclusion I nor conclusion II alone is forced. But in every valid arrangement, at least one of these will hold, because we have already shown that G ≤ R.

Verification / Alternative check:
Focus only on the relations directly involving G, S, T, and R. We have G ≤ S and S ≤ T and T = R. This gives us G ≤ R with no further restriction. The inequalities involving M, D, E, F, and H do not restrict G more tightly than this. There is no statement that forces G to be strictly less than S or strictly equal to S, so G may coincide with S or lie strictly below it. Hence equality and strict inequality between G and R are both compatible with the original information.

Why Other Options Are Wrong:

• Option a (only conclusion I) is wrong because G can also be strictly less than R.
• Option b (only conclusion II) is wrong for the opposite reason, since G can equal R.
• Option d (neither) is wrong because G can never be greater than R, so at least one of equality or less than must hold.
• Option e (both) is logically impossible for a single value of G and R, since a number cannot be simultaneously equal to and strictly less than another number.

Common Pitfalls:
A typical mistake is to think that if G ≤ S and S ≤ R then G < R automatically, forgetting the equality possibility. Another pitfall is trying to read too much into unrelated parts of the inequality chain and concluding that G must be placed strictly at one position. Always separate what is forced from what is optional.

Final Answer:
Either conclusion I or conclusion II is true, so option c is correct.