In the following comparison question, two equations are given for the variables x and y. I. x^2 = 529 II. y = 529 Solve both equations, determine all possible real values of x and y, and then choose the correct relationship between x and y from the options given below.

Introduction / Context:
This question tests basic algebraic manipulation and comparison of two variables defined by separate equations. Such comparison questions are very common in aptitude exams, where examinees must solve each equation correctly, list all possible values of the variables, and then compare them logically. Careful handling of squares and square roots is important so that no valid solution is missed.

Given Data / Assumptions:

• Equation I: x^2 = 529.
• Equation II: y = 529.
• x and y are real numbers.
• We must compare x and y and select a single correct relationship.

Concept / Approach:
To solve the problem, we first find all real solutions for x from the quadratic equation x^2 = 529. Then we identify the unique value of y from the linear equation y = 529. After listing all possible values of x and y, we compare the ranges or sets of values. The key idea is that squaring a number leads to two real roots, one positive and one negative, whenever the right side is positive and we are working over real numbers.

Step-by-Step Solution:
Step 1: Solve equation I, x^2 = 529. Since 529 is 23^2, we have x^2 = 23^2. Step 2: Take square roots on both sides. For real numbers, x can be positive or negative. So x = 23 or x = -23. Step 3: Solve equation II, y = 529. This is already solved and gives a single value. Therefore y has a unique real value y = 529. Step 4: Compare each possible x with y. If x = 23, then compare x and y: 23 versus 529 gives 23 < 529, so x < y. If x = -23, then compare x and y: -23 versus 529 gives -23 < 529, so again x < y. Step 5: In every allowed case, x is less than y. There is no situation where x is greater than or equal to y.

Verification / Alternative check:
An alternative way is to focus on magnitudes and signs. The number 529 is a large positive integer. The real solutions of x^2 = 529 are 23 and -23, whose absolute values are 23 only. Both possible x values are much closer to zero, and one is even negative. A very large positive number like 529 is certainly greater than both 23 and -23, so the only consistent relationship is x < y.

Why Other Options Are Wrong:

• Option a (x > y) is impossible because both 23 and -23 are less than 529.
• Option c (x = y) is wrong since x never equals 529 for the given equation x^2 = 529.
• Option d (relationship cannot be determined) is incorrect because we have a definite comparison once we list all possible x values.
• Option e (x ≥ y) would require x to be at least 529, which never happens here.

Common Pitfalls:
Many learners forget that a positive square has two real roots, one positive and one negative. Another mistake is to take the square root and assume x = 23 only. Even if that mistake is made, x would still be less than 529, but forgetting negative roots can cause errors in more complex comparison questions. Some learners also try to compare x^2 and y directly instead of first solving for x and y separately, which can lead to confusion when signs are involved.

Final Answer:
The correct relationship is x < y, so option b is correct.