GRE Quantitative Quiz: Quantitative Comparison Reasoning
20 questions · exam conditions
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Quantitative Comparison ReasoningQuestion 1 of 20

For real numbers xx and yy, x>yx>y.

Compare Quantity A and Quantity B, using additional information centered above the two quantities if such information is given, and select one of the following four answer choices.

Column A: x2x^2

Column B: y2y^2

The quantity in Column A is greater.
The quantity in Column B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
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GRE Quantitative Quiz

GRE Quantitative Quiz: Quantitative Comparison Reasoning

Practice Quantitative Comparison Reasoning in GRE Quantitative with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Quantitative Comparison Reasoning, giving you a quick way to practice the rules, question types, and explanations that matter most for GRE Quantitative.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

For real numbers xx and yy, x>yx>y.

Compare Quantity A and Quantity B, using additional information centered above the two quantities if such information is given, and select one of the following four answer choices.

Column A: x2x^2

Column B: y2y^2

  1. The quantity in Column A is greater.
  2. The quantity in Column B is greater.
  3. The two quantities are equal.
  4. The relationship cannot be determined from the information given. (correct answer)

Explanation: 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.

Question 2

For real number xx, x≠0x\ne 0.

Compare Quantity A and Quantity B, using additional information centered above the two quantities if such information is given, and select one of the following four answer choices.

Column A: x2x\dfrac{x^2}{x}

Column B: xx

  1. The quantity in Column A is greater.
  2. The quantity in Column B is greater.
  3. The two quantities are equal. (correct answer)
  4. The relationship cannot be determined from the information given.

Explanation: 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.

Question 3

For real numbers pp and qq, p+q=0p+q=0.

Compare Quantity A and Quantity B, using additional information centered above the two quantities if such information is given, and select one of the following four answer choices.

Column A: p3+q3p^3+q^3

Column B: 00

  1. The quantity in Column A is greater.
  2. The quantity in Column B is greater.
  3. The two quantities are equal. (correct answer)
  4. The relationship cannot be determined from the information given.

Explanation: 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.

Question 4

For real number xx.

Compare Quantity A and Quantity B, using additional information centered above the two quantities if such information is given, and select one of the following four answer choices.

Given: ∣x−3∣=5|x-3|=5

Column A: xx

Column B: 00

  1. The quantity in Column A is greater.
  2. The quantity in Column B is greater.
  3. The two quantities are equal.
  4. The relationship cannot be determined from the information given. (correct answer)

Explanation: 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.

Question 5

For real number xx, 0<x<10<x<1.

Compare Quantity A and Quantity B, using additional information centered above the two quantities if such information is given, and select one of the following four answer choices.

Column A: 1x\dfrac{1}{x}

Column B: 11−x\dfrac{1}{1-x}

  1. The quantity in Column A is greater.
  2. The quantity in Column B is greater.
  3. The two quantities are equal.
  4. The relationship cannot be determined from the information given. (correct answer)

Explanation: 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.

Question 6

A rectangle has perimeter P=40P=40.

Compare Quantity A and Quantity B, using additional information centered above the two quantities if such information is given, and select one of the following four answer choices.

Column A: The area of the rectangle

Column B: 100100

  1. The quantity in Column A is greater.
  2. The quantity in Column B is greater.
  3. The two quantities are equal.
  4. The relationship cannot be determined from the information given. (correct answer)

Explanation: 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.

Question 7

A sequence is defined by an=3n−2a_n = 3n-2 for integers n≥1n\ge 1.

Compare Quantity A and Quantity B, using additional information centered above the two quantities if such information is given, and select one of the following four answer choices.

Column A: a5a_5

Column B: a4+3a_4 + 3

  1. The quantity in Column A is greater.
  2. The quantity in Column B is greater.
  3. The two quantities are equal. (correct answer)
  4. The relationship cannot be determined from the information given.

Explanation: 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.

Question 8

For real number tt, 0<t<10<t<1.

Compare Quantity A and Quantity B, using additional information centered above the two quantities if such information is given, and select one of the following four answer choices.

Column A: tt

Column B: t2t^2

  1. The quantity in Column A is greater. (correct answer)
  2. The quantity in Column B is greater.
  3. The two quantities are equal.
  4. The relationship cannot be determined from the information given.

Explanation: 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.

Question 9

Let aa and bb be positive integers such that aa is divisible by 66 and bb is divisible by 1515.

Compare Quantity A and Quantity B, using additional information centered above the two quantities if such information is given, and select one of the following four answer choices.

Column A: The least possible value of gcd⁡(a,b)\gcd(a,b)

Column B: 33

  1. The quantity in Column A is greater.
  2. The quantity in Column B is greater.
  3. The two quantities are equal. (correct answer)
  4. The relationship cannot be determined from the information given.

Explanation: 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.

Question 10

A set SS consists of 8 distinct integers. The median of SS is defined as the average of the 4th and 5th greatest elements when the elements are listed in increasing order.

Compare Quantity A and Quantity B, using additional information centered above the two quantities if such information is given, and select one of the following four answer choices.

Given: The mean (arithmetic average) of the elements of SS is 1010

Column A: The median of SS

Column B: 1010

  1. The quantity in Column A is greater.
  2. The quantity in Column B is greater.
  3. The two quantities are equal.
  4. The relationship cannot be determined from the information given. (correct answer)

Explanation: 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.

Question 11

Let kk be an integer.

Compare Quantity A and Quantity B, using additional information centered above the two quantities if such information is given, and select one of the following four answer choices.

Column A: The remainder when 7k+57k+5 is divided by 77

Column B: 55

  1. The quantity in Column A is greater.
  2. The quantity in Column B is greater.
  3. The two quantities are equal. (correct answer)
  4. The relationship cannot be determined from the information given.

Explanation: This question tests Quantitative Comparison reasoning with modular arithmetic. The strategy is to find the remainder when 7k + 5 is divided by 7. We can rewrite 7k + 5 as 7k + 5 = 7(k) + 5. When dividing by 7, the term 7k contributes no remainder (it's divisible by 7), so the remainder comes entirely from the 5. Therefore, the remainder is 5 for any integer k. Column A equals 5, which matches Column B, so the quantities are equal. A potential mistake is thinking the remainder might depend on the value of k, but the 7k term always divides evenly by 7.

Question 12

For positive integer nn, nn is divisible by 66.

Compare Quantity A and Quantity B, using additional information centered above the two quantities if such information is given, and select one of the following four answer choices.

Column A: The remainder when nn is divided by 44

Column B: 22

  1. The quantity in Column A is greater.
  2. The quantity in Column B is greater.
  3. The two quantities are equal.
  4. The relationship cannot be determined from the information given. (correct answer)

Explanation: This question tests Quantitative Comparison reasoning. The key strategy is modular arithmetic with examples of multiples of 6. For Column A, n÷4 remainders alternate: 6 rem 2, 12 rem 0, 18 rem 2, 24 rem 0. For Column B, the value is 2. The relationship cannot be determined because the remainder is sometimes 0 (<2) and sometimes 2 (=2). A tempting incorrect option is to choose C, assuming all multiples of 6 leave remainder 2 when divided by 4. For example, testing only 6 and 18 supports equality, but 12 gives 0, showing variation.

Question 13

Let rr be a real number.

Compare Quantity A and Quantity B, using additional information centered above the two quantities if such information is given, and select one of the following four answer choices.

Given: r≠0r \ne 0

Column A: r2r\frac{r^2}{r}

Column B: rr

  1. The quantity in Column A is greater.
  2. The quantity in Column B is greater.
  3. The two quantities are equal. (correct answer)
  4. The relationship cannot be determined from the information given.

Explanation: This question tests Quantitative Comparison reasoning with algebraic simplification. The key strategy is to simplify the expression in Column A using the given constraint that r ≠ 0. For Column A, we have r²/r = r·r/r = r (since we can cancel one r from numerator and denominator when r ≠ 0). Column B is simply r. Therefore, both columns equal r, making them equal regardless of the specific value of r (as long as r ≠ 0). This is a straightforward algebraic simplification problem where students might be tempted to think the answer depends on whether r is positive or negative, but the algebraic cancellation works for all non-zero values of r. The correct answer is C because the two quantities are always equal.

Question 14

Let nn be an integer.

Compare Quantity A and Quantity B, using additional information centered above the two quantities if such information is given, and select one of the following four answer choices.

Given: nn is divisible by 66

Column A: nn is divisible by 1212

Column B: nn is divisible by 99

  1. The quantity in Column A is greater.
  2. The quantity in Column B is greater.
  3. The two quantities are equal.
  4. The relationship cannot be determined from the information given. (correct answer)

Explanation: This question tests Quantitative Comparison reasoning by examining divisibility relationships. The key strategy is to test specific examples that satisfy the given constraint. We know n is divisible by 6, meaning n = 6k for some integer k. For Column A, n is divisible by 12 if and only if 6k is divisible by 12, which happens when k is even. For Column B, n is divisible by 9 if and only if 6k is divisible by 9, which happens when 6k = 9m for some integer m, or when 2k = 3m. Testing examples: if n = 6, then n is not divisible by 12 or 9; if n = 12, then n is divisible by 12 but not 9; if n = 18, then n is divisible by 9 but not 12. Since we can find cases where Column A is true but B is false, and vice versa, the relationship cannot be determined.

Question 15

In a right triangle, the two legs have lengths aa and bb, where a>0a>0 and b>0b>0.

Compare Quantity A and Quantity B, using additional information centered above the two quantities if such information is given, and select one of the following four answer choices.

Given: a2+b2=25a^2+b^2=25

Column A: a+ba+b

Column B: 77

  1. The quantity in Column A is greater.
  2. The quantity in Column B is greater.
  3. The two quantities are equal.
  4. The relationship cannot be determined from the information given. (correct answer)

Explanation: This question tests Quantitative Comparison reasoning using the Pythagorean theorem and optimization. The key strategy is to find the range of possible values for a + b given the constraint a² + b² = 25. Since a² + b² = 25 represents a circle of radius 5 centered at the origin (restricted to the first quadrant since a, b > 0), we need to maximize a + b. Using Cauchy-Schwarz or Lagrange multipliers, (a + b)² ≤ 2(a² + b²) = 50, so a + b ≤ √50 = 5√2 ≈ 7.07. The maximum occurs when a = b = 5/√2. The minimum value of a + b approaches 5 as one variable approaches 5 and the other approaches 0. Since a + b can be less than, equal to, or greater than 7, the relationship cannot be determined from the given information.

Question 16

A real number pp satisfies −4≤p≤2-4\le p\le 2.

Compare Quantity A and Quantity B, using additional information centered above the two quantities if such information is given, and select one of the following four answer choices.

Column A: ∣p∣|p|

Column B: p+3p+3

  1. The quantity in Column A is greater.
  2. The quantity in Column B is greater.
  3. The two quantities are equal.
  4. The relationship cannot be determined from the information given. (correct answer)

Explanation: This question tests Quantitative Comparison reasoning with absolute values and inequalities. The strategy is to analyze different cases within the given range -4 ≤ p ≤ 2. For Column A, |p| ranges from 0 (when p = 0) to 4 (when p = -4). For Column B, p + 3 ranges from -4 + 3 = -1 (when p = -4) to 2 + 3 = 5 (when p = 2). When p = -4, |p| = 4 > -1 = p + 3; when p = 0, |p| = 0 < 3 = p + 3; when p = 2, |p| = 2 < 5 = p + 3. 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 one quantity is always larger without checking all possible values of p.

Question 17

A rectangle has perimeter PP and area AA. The side lengths are positive real numbers.

Compare Quantity A and Quantity B, using additional information centered above the two quantities if such information is given, and select one of the following four answer choices.

Given: P=20P=20

Column A: The greatest possible value of AA

Column B: 2525

  1. The quantity in Column A is greater.
  2. The quantity in Column B is greater.
  3. The two quantities are equal. (correct answer)
  4. The relationship cannot be determined from the information given.

Explanation: This question tests Quantitative Comparison reasoning using optimization with constraints. The key strategy is to maximize the area of a rectangle given a fixed perimeter of 20. If the rectangle has dimensions length l and width w, then 2(l + w) = 20, so l + w = 10. The area A = lw needs to be maximized subject to l + w = 10. Using calculus or the AM-GM inequality, the maximum area occurs when l = w = 5, giving A = 25. This makes sense geometrically: among all rectangles with the same perimeter, the square has the maximum area. Since the maximum possible value of A is 25, and Column B is also 25, the two quantities are equal. Any other rectangle with perimeter 20 would have area less than 25, confirming that 25 is indeed the maximum.

Question 18

A jar contains only red and blue marbles. The probability of randomly selecting a red marble is 35\frac{3}{5}.

Compare Quantity A and Quantity B, using additional information centered above the two quantities if such information is given, and select one of the following four answer choices.

Column A: The ratio of the number of red marbles to the number of blue marbles

Column B: 32\frac{3}{2}

  1. The quantity in Column A is greater.
  2. The quantity in Column B is greater.
  3. The two quantities are equal. (correct answer)
  4. The relationship cannot be determined from the information given.

Explanation: This question tests Quantitative Comparison reasoning with probability and ratios. The key strategy is to convert the given probability into a ratio. If the probability of selecting a red marble is 3/5, then out of every 5 marbles, 3 are red and 2 are blue (since the jar contains only red and blue marbles). Therefore, the ratio of red marbles to blue marbles is 3:2, which can be written as the fraction 3/2. Column B is also 3/2. Since both columns represent the same value, they are equal. A common error would be to confuse the probability (red marbles/total marbles = 3/5) with the ratio (red marbles/blue marbles = 3/2), but careful analysis shows that if 3/5 of all marbles are red, then 2/5 are blue, giving the ratio 3/2.

Question 19

For real numbers pp and qq.

Compare Quantity A and Quantity B, using additional information centered above the two quantities if such information is given, and select one of the following four answer choices.

Given: p<qp<q

Column A: p2p^2

Column B: q2q^2

  1. The quantity in Column A is greater.
  2. The quantity in Column B is greater.
  3. The two quantities are equal.
  4. The relationship cannot be determined from the information given. (correct answer)

Explanation: This question tests Quantitative Comparison reasoning with inequalities and squares. The key strategy is to consider how squaring affects the relationship between numbers. Given p < q, we need to compare p² and q². If both p and q are positive, then p² < q² since squaring preserves order for positive numbers. If both are negative, then p² > q² since p is more negative than q. If p is negative and q is positive, we cannot determine the relationship without knowing their specific values. For example, if p = -3 and q = 2, then p² = 9 > q² = 4. But if p = -1 and q = 2, then p² = 1 < q² = 4. Since the relationship between p² and q² depends on the signs and magnitudes of p and q, which are not specified, the answer is D.

Question 20

For integer kk, k≥1k\ge 1.

Compare Quantity A and Quantity B, using additional information centered above the two quantities if such information is given, and select one of the following four answer choices.

Column A: k!k!

Column B: 2k2^k

  1. The quantity in Column A is greater.
  2. The quantity in Column B is greater.
  3. The two quantities are equal.
  4. The relationship cannot be determined from the information given. (correct answer)

Explanation: This question tests Quantitative Comparison reasoning. The key strategy is computation for small integer values of k. For Column A, k! grows factorially: 1,2,6,24,... for k=1,2,3,4. For Column B, 2^k grows exponentially: 2,4,8,16,... The relationship cannot be determined because for k=1-3, k!<2^k, but for k≥4, k!>2^k. A tempting incorrect option is to choose B, assuming exponential always exceeds factorial by testing only small k. For example, at k=3, 6<8 supports B greater, but at k=5, 120>32 reverses it.