In the following inequality based reasoning question, assume all statements are true and then determine which of the given conclusions must follow. Statements: A > C; G > E; G ≤ C; R ≤ I; K ≤ I Conclusions: I. A > G II. C > E

Introduction / Context:
This problem is another example of symbolic inequality reasoning. Several variables are compared with greater than and less than or equal to symbols. Your goal is to combine these relationships logically and decide which conclusions must be true in every possible arrangement that satisfies the given statements. This evaluates your comfort with chaining inequalities and using transitivity to infer new comparisons.

Given Data / Assumptions:

• A > C
• G > E
• G ≤ C
• R ≤ I
• K ≤ I
• Conclusion I: A > G
• Conclusion II: C > E
• All variables represent quantities on a common ordered scale.

Concept / Approach:
The strategy is to focus on the parts of the data that are relevant to the conclusions. Here, R, I, and K are not involved in the conclusions, so they can be ignored. We then connect the remaining inequalities using transitivity. If we find that a conclusion holds under all possible numeric assignments consistent with the statements, then it is definitely true. If there exists any valid assignment where the conclusion fails, it is not definite.

Step-by-Step Solution:
Step 1: Look at Conclusion I, A > G. We know A > C and G ≤ C. From G ≤ C, G is at most C, and from A > C, A is strictly greater than C. Step 2: Combine these. Since A is greater than C and C is greater than or equal to G, A must also be greater than G. More formally: A > C and C ≥ G implies A > G. So Conclusion I is logically forced by the given inequalities. Step 3: Next, check Conclusion II, C > E. We know G > E and G ≤ C. From G > E, E is less than G. From G ≤ C, C is at least as large as G. Step 4: Apply transitivity again. If C is greater than or equal to G, and G is greater than E, then C must be greater than E. Expressed as a chain, C ≥ G > E implies C > E. This directly supports Conclusion II. Step 5: Confirm there is no counterexample. To see that both conclusions always hold, consider arbitrary numbers satisfying the original inequalities. For example, let C = 10, A = 12, G = 9, E = 5. Then:

• A > C becomes 12 > 10, true.
• G > E becomes 9 > 5, true.
• G ≤ C becomes 9 ≤ 10, true.

Now check the conclusions: A > G is 12 > 9 (true) and C > E is 10 > 5 (true). Trying other consistent choices gives the same pattern: both conclusions remain true.

Verification / Alternative check:
Another way to verify is to reason symbolically. For Conclusion I:

• From A > C and C ≥ G, substitute C with any value at least G. A is greater than that value, so A must be greater than G.

For Conclusion II:

• From G > E and G ≤ C, G is sandwiched between E and C. Because G is greater than E and C is no smaller than G, C must be greater than E.

Since these chains rely only on transitivity and the definitions of the inequality symbols, there is no room to assign values in a way that breaks the conclusions while keeping the premises true.

Why Other Options Are Wrong:
Option b: “Only conclusion I is true” is incorrect because it ignores the strong chain C ≥ G > E that guarantees C > E. Option c: “Only conclusion II is true” is wrong since A > C ≥ G forces A > G. Option d: “Either conclusion I or conclusion II is true” suggests that sometimes only one might hold, but we have seen that both must hold together. Option e: “None of the conclusions is true” is clearly false because both conclusions are directly supported by the premises.

Common Pitfalls:
Learners sometimes fail to spot that when a variable is bounded on both sides, as G is between E and C, it can create strong relationships between the outer variables. Another common mistake is to think that a non-strict inequality like ≤ or ≥ is too weak to support strict conclusions. In fact, when combined correctly with strict inequalities, they often do yield strict results, as in C ≥ G and G > E giving C > E. Always trace the direction of inequalities carefully and remember that transitivity is your main tool in such questions.

Final Answer:
Both conclusions I and II are true.