This question tests Quantitative Comparison reasoning with inequalities and squares. The key strategy is to consider different cases for the signs of x and y, since squaring can change relative magnitudes. For Column A, we have x², and for Column B, we have y². While we know x > y, this doesn't determine which square is larger: if x = 2 and y = 1, then x² = 4 > y² = 1; but if x = -1 and y = -2, then x² = 1 < y² = 4. Since we can find cases where Column A is greater and cases where Column B is greater, the relationship cannot be determined. A common error is assuming that x > y implies x² > y², which only holds when both numbers are positive.

This question tests Quantitative Comparison reasoning. The key strategy is algebraic simplification of the expression. For Column A, x²/x simplifies to x for x≠0. For Column B, the value is x. The quantities are equal because x²/x = x algebraically. A tempting incorrect option is to choose D, assuming it depends on the sign of x. For example, testing x=-1 gives 1/-1=-1, which equals x, and x=1 gives 1=1, showing it always holds.

This question tests Quantitative Comparison reasoning. The key strategy is algebraic substitution using p+q=0. For Column A, p³ + q³ = p³ + (-p)³ = p³ - p³ = 0. For Column B, the value is 0. The quantities are equal because the expression simplifies to 0 regardless of p. A tempting incorrect option is to choose D, assuming it depends on specific p and q values. For example, testing p=1, q=-1 gives 1-1=0, and p=2, q=-2 gives 8-8=0, but it always holds.

This question tests Quantitative Comparison reasoning with absolute value equations. The key strategy is to solve |x - 3| = 5 and determine all possible values of x. The equation |x - 3| = 5 means that x - 3 = 5 or x - 3 = -5. Solving the first equation gives x = 8, and solving the second gives x = -2. So x can be either 8 or -2. Comparing to Column B (which is 0): when x = 8, we have 8 > 0; when x = -2, we have -2 < 0. Since x can be greater than or less than 0 depending on which solution we consider, the relationship cannot be determined from the given information. A common error would be to assume x must be positive because of the absolute value, but absolute value equations often have both positive and negative solutions.

This question tests Quantitative Comparison reasoning. The key strategy is algebraic comparison and testing values in the interval. For Column A, 1/x >1 since 0<x<1. For Column B, 1/(1-x) >1 as 1-x is between 0 and 1. The relationship cannot be determined because at x=0.5 they equal 2, but for x=0.2, 5>1.25, and for x=0.8, 1.25<5. A tempting incorrect option is to choose A, assuming 1/x is always larger by testing only x<0.5. For example, x=0.4 gives 2.5>1.666, but x=0.6 gives 1.666<2.5, showing it varies.

This question tests Quantitative Comparison reasoning with geometric constraints. The strategy is to find the range of possible areas for a rectangle with perimeter 40. For a rectangle with perimeter P = 40, we have 2(length + width) = 40, so length + width = 20. The area equals length × width, and by the AM-GM inequality, this is maximized when length = width = 10, giving area = 100. The area can be as small as we want by making the rectangle very thin (e.g., length = 19.9, width = 0.1 gives area = 1.99). Since the area can equal 100 or be less than 100, we cannot determine whether Column A is greater than, less than, or equal to Column B. A tempting error is to assume the rectangle must be a square, which would make the quantities equal.

This question tests Quantitative Comparison reasoning. The key strategy is direct computation using the sequence formula. For Column A, a5 = 3×5 - 2 = 13. For Column B, a4 = 3×4 - 2 = 10, plus 3 is 13. The quantities are equal because both compute to 13. A tempting incorrect option is to choose A or B by misapplying the formula, like forgetting the -2. For example, computing a5 as 15 and a4+3 as 15 ignores -2, but correctly it matches at 13.

This question tests Quantitative Comparison reasoning with inequalities and exponents. The strategy is to compare t with t² when 0 < t < 1. For Column A, we have t, and for Column B, we have t². When 0 < t < 1, multiplying t by itself (which is less than 1) makes it smaller, so t² < t. For example, if t = 0.5, then t² = 0.25 < 0.5 = t. This relationship holds for all values in the given range, so Column A is always greater than Column B. A common error is confusing this with the case when t > 1, where t² > t instead.

This question tests Quantitative Comparison reasoning with greatest common divisors. The strategy is to analyze the prime factorizations and find the minimum possible gcd. Since a is divisible by 6 = 2 × 3, we can write a = 6k for some positive integer k. Since b is divisible by 15 = 3 × 5, we can write b = 15m for some positive integer m. The gcd(a,b) = gcd(6k, 15m) = gcd(6, 15) × gcd(k, m) = 3 × gcd(k, m). The least possible value occurs when gcd(k, m) = 1, giving gcd(a,b) = 3. Therefore, Column A equals 3, making the two quantities equal. A common mistake is thinking the gcd could be 1, forgetting that both numbers share the factor 3.

This question tests Quantitative Comparison reasoning about mean and median relationships. The key strategy is to understand that for a set of 8 distinct integers with mean 10, the sum of all elements is 80. The median is the average of the 4th and 5th elements when listed in order. While the mean equals 10, the median can vary depending on the distribution of the values. For example, if S = {1, 2, 3, 4, 16, 17, 18, 19}, the mean is 10 but the median is (4 + 16)/2 = 10. However, if S = {1, 2, 3, 9, 11, 17, 18, 19}, the mean is still 10 but the median is (9 + 11)/2 = 10. We can also construct sets where the median differs from 10, such as S = {1, 2, 3, 4, 5, 6, 7, 52} where the median is 4.5. Since the median can equal or differ from the mean, the relationship cannot be determined.