Determine the definitely true relations from K ≥ G > H ≤ F. Which of the following is/are definitely true?

Introduction / Context:
The stem provides a mixed chain, K ≥ G > H ≤ F, and asks which relations are guaranteed. Some comparisons are certain (K vs H), others are not (F vs K or F vs G). We minimally repair the options to allow a multi-true outcome consistent with the premise.

Given Data / Assumptions:

• K ≥ G
• G > H
• H ≤ F
• Candidate relations: F ≥ K, H < K, F < G, K ≥ H.

Concept / Approach:
From K ≥ G and G > H, we get K > H (hence K ≥ H as well). However, because we only know H ≤ F and G > H, F could be less than, equal to, or greater than K and G; those comparisons cannot be fixed.

Step-by-Step Solution:
Since K ≥ G and G > H ⇒ K > H ⇒ H < K (definitely true).From K > H we immediately have K ≥ H (also definitely true).Regarding F ≥ K: only H ≤ F is known, so F could be smaller or larger than K (not definite).Regarding F < G: again, with only H ≤ F and G > H, F could be below or above G (not definite).

Verification / Alternative check:
Model 1: H=1, G=3, K=3, F=2 ⇒ H < K and K ≥ H true; F ≥ K false; F < G true (but not guaranteed by all models). Model 2: H=1, G=3, K=4, F=10 ⇒ H < K and K ≥ H true; F ≥ K true; F < G false. Only the K vs H statements persist across models.

Why Other Options Are Wrong:
Options a/c/d assert single relations; e correctly captures both guaranteed relations. Any option claiming F ≥ K or F < G as definite is overreaching.

Common Pitfalls:
Assuming H ≤ F and G > H fix F relative to K or G—they do not.

Final Answer:
Both H < K and K ≥ H are true