In this aptitude (simplification: quadratic comparison) question, two quadratic equations for x and y are given. Solve each equation to find all roots of x and y, then compare these values and select the option that correctly describes the relationship between x and y. I. x^2 - 11x + 28 = 0 II. y^2 - 18y + 81 = 0

Introduction / Context:
This problem gives two quadratic equations and asks for the relationship between x and y. Unlike earlier questions where sets of roots overlapped, here the roots are positioned in a way that delivers a clear inequality. Such tasks check understanding of factorisation and the standard relationships between roots and coefficients while also training students to compare multiple values correctly.

Given Data / Assumptions:
Equation I: x^2 - 11x + 28 = 0.
Equation II: y^2 - 18y + 81 = 0.
We treat all real roots of each equation as possible values of x and y.

Concept / Approach:
We first factor each quadratic to get explicit roots. After that we compare every x root with every y root. If every possible x is less than every possible y, the correct relation is x < y. This approach guarantees that we do not miss any combination and supports a logically sound inequality conclusion.

Step-by-Step Solution:
Factor equation I: x^2 - 11x + 28 = 0. We get (x - 7)(x - 4) = 0. So x = 7 or x = 4. Now factor equation II: y^2 - 18y + 81 = 0. This is a perfect square: (y - 9)^2 = 0. Hence y = 9 is the only (repeated) root. Now compare values: x can be 4 or 7, while y is always 9. Both 4 and 7 are strictly less than 9, so for all valid roots we have x < y.

Verification / Alternative check:
Using the sum and product of roots confirms our factorisation. For equation I, sum of roots is 11 and product is 28, matching 7 and 4. For equation II, sum is 18 and product is 81, which matches a double root at 9. When we place the values on the number line, we see 4 and 7 lying to the left of 9. Therefore in every allowable combination, x is less than y and no combination violates this ordering.

Why Other Options Are Wrong:
Option x > y is wrong because no x value exceeds 9. Option x = y fails as neither 4 nor 7 equals 9. Option x >= y would require at least equality or greater than in some case, but that never occurs. Relationship cannot be determined would apply only if there were inconsistent comparisons across pairs, which does not happen here.

Common Pitfalls:
Students sometimes misfactor the second quadratic and overlook that it is a perfect square, leading to incorrect roots. Another pitfall is to forget to test all roots for x, comparing only one of them with y. Writing down the complete sets {4, 7} and {9} and then checking each combination avoids such mistakes. Drawing a quick number line diagram with marks at 4, 7, and 9 can further support intuitive understanding.

Final Answer:
Since both possible values of x are strictly less than the single value of y, the correct relationship is x < y.