In this inequality reasoning question, assume all the statements are true and then decide which of the given conclusions must follow. Statements: Z ≤ K < D = S; D < A < O; G ≥ D < R Conclusions: I. G > A II. Z < R

Introduction / Context:
This problem continues the theme of symbolic inequality reasoning. You are given several chains of inequalities relating different variables, and you are asked to determine which conclusions must always be true. The challenge is to merge the information from each chain, track the relative ordering of variables, and then test whether the proposed conclusions are logically forced or only sometimes true.

Given Data / Assumptions:

• Z ≤ K < D = S
• D < A < O
• G ≥ D and D < R
• Conclusion I: G > A
• Conclusion II: Z < R
• All statements describe the relative order of the variables on a common scale.

Concept / Approach:
We rely on transitivity to connect variables that appear in overlapping inequalities. When you see a sequence like Z ≤ K < D and then D < R, you can form a longer chain Z ≤ K < D < R and deduce relationships between Z and R. For each conclusion, ask whether there is enough information to fix the order of the two variables involved. If you can construct a valid assignment of numbers that violates a conclusion while keeping all the statements true, that conclusion is not definite.

Step-by-Step Solution:
Step 1: Start with the first chain: Z ≤ K < D = S. This tells us that K is less than D and Z is less than or equal to K. Hence Z is also less than D. Step 2: From the third chain we know D < R. Combining this with the relationships from Step 1 gives Z ≤ K < D < R. This implies Z is less than R, since everything up to D is below R and Z is no greater than K, which is below D. Step 3: Therefore, Conclusion II (Z < R) is supported by a continuous chain: Z ≤ K < D < R, so Z is strictly less than R. Even if Z = K, Z is still less than D and thus less than R. Step 4: Now examine Conclusion I, G > A. From the data we have G ≥ D and D < A. So A is above D, while G is at least as large as D. There are several possibilities for G:

• Case 1: G = D. Then since D < A, we have G = D < A, which means G is not greater than A.
• Case 2: G > D. In this case G could be either greater than A or still less than A, depending on the actual numbers, because there is no direct comparison between G and A beyond the fact that both are above D.

Because we can choose values where G ≤ A while all statements remain true, Conclusion I is not guaranteed. Step 5: Construct a counterexample for Conclusion I. Let D = 5, S = 5, K = 4, Z = 3, R = 8, A = 7, O = 9, and G = 5. Then:

• Z ≤ K < D = S gives 3 ≤ 4 < 5 = 5 (true).
• D < A < O gives 5 < 7 < 9 (true).
• G ≥ D < R gives 5 ≥ 5 < 8 (true).

Here G = 5 and A = 7, so G > A is false, while Z = 3 and R = 8, so Z < R is true.

Verification / Alternative check:
By testing multiple sets of values, you will find that Z is always less than R because the chain Z ≤ K < D < R cannot be violated without breaking the given statements. In contrast, the relationship between G and A is underdetermined. Sometimes G can be greater than A, but other times it can be equal to or less than A, depending on how much larger than D we choose them to be. This confirms that only Conclusion II is logically forced by the premises.

Why Other Options Are Wrong:
Option a: “None of the conclusions is true” is incorrect because Conclusion II is always true. Option b: “Only conclusion I is true” is wrong because we have produced an example where G is not greater than A. Option d: “Either conclusion I or conclusion II is true” is incorrect because there is no such either or situation; Conclusion II is always true and Conclusion I is not. Option e: “Both conclusions I and II are true” fails because Conclusion I can be false in a valid arrangement.

Common Pitfalls:
A common error is assuming that G ≥ D and D < A automatically makes G > A. This is not valid because G could be exactly equal to D, which might be significantly less than A. Another pitfall is forgetting to propagate the inequalities properly, for example, not extending Z ≤ K < D with the further inequality D < R. Always make sure to connect chains fully before judging the truth of a conclusion.

Final Answer:
Only conclusion II is true.