In the inequality chain L ≥ K ≥ P = N = J @ U ≤ A % R, which pair of symbols should replace (@) and (%) respectively so that L ≥ U and R > J are definitely true?

Introduction / Context:
This question is about symbol substitution in a chain of inequalities. Some comparison signs in the expression are hidden behind special symbols, and you must choose the correct pair so that two target relations, L ≥ U and R > J, will always be true whenever the whole chain holds.

Given Data / Assumptions:

• Base expression: L ≥ K ≥ P = N = J @ U ≤ A % R.
• The symbols @ and % must be replaced by one of the standard inequality symbols given in the options.
• The final chain must make both L ≥ U and R > J definitely true.
• There is no extra numerical information about the letters.

Concept / Approach:
The method is to plug in each candidate pair of symbols into the expression and then see whether both required relations necessarily follow. If we can find even one counterexample, then that pair is rejected. The correct pair must force L to be at least as large as U and R to be strictly larger than J in every valid arrangement of the letters.

Step-by-Step Solution:
Step 1: Try option D with @ = = and % = <. The chain becomes L ≥ K ≥ P = N = J = U ≤ A < R. Step 2: Since J = U and L ≥ K ≥ P = N = J, we obtain L ≥ J and thus L ≥ U. So L ≥ U always holds for this choice. Step 3: From U ≤ A and A < R, and J = U, we get J ≤ A < R. Hence R is strictly greater than J in every possible case, so R > J is forced. Step 4: Check other options quickly. Choices that make J ≤ U or J > U without equality do not guarantee both L ≥ U and R > J together; counterexamples can be built. Only the pair =, < gives both required conclusions in all cases.

Verification / Alternative check:
Pick sample values that satisfy option D. Let J = U = N = P = 2, K = 3, L = 4, A = 3 and R = 5. Then L ≥ K ≥ P = N = J = U ≤ A < R becomes 4 ≥ 3 ≥ 2 = 2 = 2 = 2 ≤ 3 < 5, which is true. Here L ≥ U becomes 4 ≥ 2 and R > J becomes 5 > 2, both satisfied. Trying to alter any value while respecting the chain cannot break these two relations.

Why Other Options Are Wrong:
When @ is ≤, J ≤ U is allowed, and L can be less than U even though the chain is true, so L ≥ U is not guaranteed. When @ is >, R > J is not forced because R may only be greater than or equal to A, which can still be less than or equal to J. For these pairs, one of the required conditions fails in some valid arrangements, so they cannot be chosen.

Common Pitfalls:
Students often verify only one target relation and forget to check the other. Another pitfall is to look at the chain locally without writing it cleanly after substitution. Writing the full expanded version of the chain for each option makes it easier to see whether both requirements are always met.

Final Answer:
The correct replacement is @ = = and % = <, that is, the pair '=, <'.