GRE - Graduate Record Examinations Quantitative Comparison Strategies Questions and Answers

Question 1

Given that x is an integer and x > 1.

**Quantity A:** (x + 1) / (x - 1)

**Quantity B:** x / (x + 1)

Accepted Answer

The quantity in Column A is greater.

Answer

A powerful strategy is to pick numbers. Let's test a value for x that satisfies the condition x > 1, such as x = 2. 
Quantity A = (2 + 1) / (2 - 1) = 3/1 = 3. 
Quantity B = 2 / (2 + 1) = 2/3. In this case, Quantity A is greater. 
Let's try another value, x = 3. 
Quantity A = (3 + 1) / (3 - 1) = 4/2 = 2. 
Quantity B = 3 / (3 + 1) = 3/4. Again, Quantity A is greater. 
Alternatively, we can use algebraic manipulation. Since x > 1, both denominators (x-1) and (x+1) are positive, so we can cross-multiply without changing the inequality direction. We are comparing (x+1)(x+1) with x(x-1). 
(x+1)² = x² + 2x + 1
x(x-1) = x² - x
Comparing x² + 2x + 1 to x² - x, we can subtract x² from both sides to get 2x + 1 and -x. Since x > 1, 2x + 1 is always positive and -x is always negative. Therefore, Quantity A is always greater.

Question 2

y < 0

**Quantity A:** y²

**Quantity B:** y³

Accepted Answer

The relationship cannot be determined from the information given.

Answer

The 'plugging in numbers' strategy is effective here, especially when dealing with inequalities. We need to test different types of numbers that fit the condition y < 0. Case 1: Test a negative integer, like y = -2. Quantity A: (-2)² = 4 Quantity B: (-2)³ = -8 In this case, Quantity A is greater. Case 2: Test a negative fraction between -1 and 0, like y = -1/2. Quantity A: (-1/2)² = 1/4 Quantity B: (-1/2)³ = -1/8 In this case, Quantity A (1/4) is also greater than Quantity B (-1/8). Let's reconsider our number choices. The prompt asks to test different *types* of numbers. What if y is between -1 and 0 versus y < -1? Let's re-test y = -1/2: Quantity A = (-1/2)² = 1/4 Quantity B = (-1/2)³ = -1/8 Here A > B. Let's test y = -3: Quantity A = (-3)² = 9 Quantity B = (-3)³ = -27 Here A > B. Let's re-evaluate. The strategy is to prove D by finding two different outcomes. Let's compare y^2 and y^3. We can divide both quantities by y^2. Since y^2 is always positive, this is a valid operation that won't flip the inequality. This simplifies the comparison to 1 vs y. Since we are given that y < 0, it means 1 is always greater than y. Therefore, Quantity A is always greater. Let's re-check the initial premise. Ah, there was a mistake in the logic of trying to prove D. Let's stick with the algebraic simplification. Comparing y² and y³. We can factor both: y² and y*y². We can simplify by dividing by y², which must be positive since y is not 0. This leaves us comparing 1 and y. Since y < 0, 1 is always greater than y. Therefore, Quantity A is always greater. My initial assessment of 'D' was incorrect. The correct answer is A. Let's construct a new question where the answer is D. **New Question:** x is a non-zero number. **Quantity A:** x² **Quantity B:** x **Explanation:** Use the strategy of plugging in different types of numbers that satisfy the condition. Case 1: Test a positive integer, x = 2. Quantity A: 2² = 4 Quantity B: 2 Here, Quantity A is greater. Case 2: Test a positive fraction between 0 and 1, x = 1/2. Quantity A: (1/2)² = 1/4 Quantity B: 1/2 Here, Quantity B is greater. Since we have found one case where A > B and another where B > A, the relationship cannot be determined from the information given.

Question 3

A circle is inscribed in a square with a side length of 10.

**Quantity A:** The area of the circle.

**Quantity B:** 25π

Accepted Answer

The two quantities are equal.

Answer

The key is to understand the relationship between the inscribed circle and the square. If a circle is inscribed in a square, the diameter of the circle is equal to the side length of the square. In this case, the side length is 10, so the diameter of the circle is also 10. The radius of the circle is half the diameter, which is 5. The formula for the area of a circle is A = πr². 
For Quantity A, the area is π * (5)² = 25π. 
Quantity B is given as 25π. 
Therefore, the two quantities are equal.

Question 4

A list of numbers has an average (arithmetic mean) of 15. A new list is created by multiplying each number in the original list by 2 and then adding 5.

**Quantity A:** The average of the new list of numbers.

**Quantity B:** 35

Accepted Answer

The two quantities are equal.

Answer

This question tests your understanding of how changes to a set of numbers affect its average. If every number in a set is multiplied by a constant, the average is also multiplied by that constant. If a constant is added to every number, the average is increased by that constant. In this scenario, the original average is 15. First, each number is multiplied by 2, so the new average becomes 15 * 2 = 30. Then, 5 is added to each number, so the final average becomes 30 + 5 = 35. Therefore, Quantity A is 35, which is equal to Quantity B.

Question 5

p is a prime number greater than 2.

**Quantity A:** The remainder when p is divided by 2.

**Quantity B:** 1

Accepted Answer

The two quantities are equal.

Answer

This question is about number properties. A prime number is a positive integer greater than 1 that has no positive integer divisors other than 1 and itself. The only even prime number is 2. The question states that p is a prime number greater than 2. This means p must be an odd number (e.g., 3, 5, 7, 11, ...). Any odd integer, when divided by 2, will have a remainder of 1. Therefore, Quantity A is always 1, which is equal to Quantity B.

GRE Practice Test

GRE Practice Test

GRE - Graduate Record Examinations Quantitative Comparison Strategies Questions and Answers