In aptitude (quadratic comparison and simplification), two quadratic equations are given: I. x^2 + 30x + 81 = 0 and II. y^2 - 9y - 162 = 0. After solving both, compare x and y and choose the correct relationship between their possible values.

Introduction / Context:
This problem is a classic quadratic comparison question. Instead of asking for explicit numerical roots, the examiner wants to know the nature of the relationship between any possible value of x and any possible value of y. Such questions are common in data comparison sections of aptitude exams, where candidates must reason about all roots of both equations and decide whether one variable is always greater, always smaller, always equal, or if no consistent comparison can be made.

Given Data / Assumptions:
- Equation I: x^2 + 30x + 81 = 0.
- Equation II: y^2 - 9y - 162 = 0.
- x and y are real roots of their respective equations.
- We must compare all possible combinations of x and y and judge the relationship.

Concept / Approach:
The concept involves solving both quadratic equations, finding their roots, and then listing possible pairs (x, y). If x is greater than y for every possible combination, we choose x > y, and similarly for other options. If in some cases x > y and in others x < y, the relationship cannot be determined. Factorisation is a convenient method here because both equations have integer coefficients and neat factorisation patterns.

Step-by-Step Solution:
Step 1: Factor Equation I: x^2 + 30x + 81 = 0.Step 2: The discriminant is 30^2 - 4 * 1 * 81 = 900 - 324 = 576, which is 24^2, so roots are real and rational.Step 3: Factor: x^2 + 30x + 81 = (x + 3)(x + 27) = 0. Thus, x = -3 or x = -27.Step 4: Factor Equation II: y^2 - 9y - 162 = 0.Step 5: We need factors of -162 that add up to -9. These are -18 and +9, giving (y - 18)(y + 9) = 0.Step 6: Therefore, y = 18 or y = -9.Step 7: List all possible (x, y) pairs: (-27, 18), (-27, -9), (-3, 18), (-3, -9).Step 8: Compare: for (-27, 18), x < y; for (-27, -9), x < y; for (-3, 18), x < y; for (-3, -9), x > y.

Verification / Alternative check:
From the comparison, we see three cases where x < y and one case where x > y. There is no case where x equals y, and there is no consistent ordering. An alternative check is to note that the set of x-values is {-27, -3} and the set of y-values is {-9, 18}. Since -27 is less than both -9 and 18, while -3 is greater than -9 but less than 18, there is no single inequality that holds for all roots. Therefore, no definitive order can be assigned between x and y in general.

Why Other Options Are Wrong:
- x > y: False, because when x = -27 and y = 18, clearly x < y.
- x < y: Also false, because for x = -3 and y = -9, we have x > y.
- x = y: There is no pair of equal roots between the two equations.
- x >= y: This would require x >= y in all cases, but we have multiple cases where x < y.

Common Pitfalls:
Students sometimes compare only the larger root of one equation with the larger root of the other equation, or compare only one arbitrary pair, and then assume that relationship holds universally. Another mistake is to forget that quadratic equations can have two distinct roots and that any of them is a possible value of the variable. For comparison questions like this, it is essential to consider all combinations of roots and not rely on partial inspection.

Final Answer:
The relationship between x and y varies depending on which roots are chosen, so the relationship cannot be determined.