Quantitative Comparisons: Algebra Flashcards

Compare the quantities in Column A and Column B. **Column A:** The value of x when 3x - 4 = 11 **Column B:** The value of y when 2y + 1 = 13

Answer: Quantity B is greater.

To solve for x in Column A: 3x - 4 = 11. Add 4 to both sides to get 3x = 15. Divide by 3 to get x = 5. To solve for y in Column B: 2y + 1 = 13. Subtract 1 from both sides to get 2y = 12. Divide by 2 to get y = 6. Since 6 is greater than 5, Quantity B is greater.

Compare the quantities in Column A and Column B. Given: n is a positive integer. **Column A:** (-1)^(2n) **Column B:** (-1)^(2n + 1)

Answer: Quantity A is greater.

For any positive integer n, the expression 2n will always result in an even number. Any negative number raised to an even power results in a positive number, so (-1) raised to any even power is 1. Thus, Quantity A is 1. The expression 2n + 1 will always result in an odd number. Any negative number raised to an odd power results in a negative number, so (-1) raised to any odd power is -1. Thus, Quantity B is -1. Since 1 > -1, Quantity A is greater.

Compare the quantities in Column A and Column B. Given: x = 3 and y = -2 **Column A:** (x + y)² **Column B:** x² + y²

Answer: Quantity B is greater.

Substitute the given values into the expression for Quantity A: (3 + (-2))² = (1)² = 1. Substitute the values into the expression for Quantity B: (3)² + (-2)² = 9 + 4 = 13. Since 13 is greater than 1, Quantity B is greater.

Compare the quantities in Column A and Column B. Given: The average (arithmetic mean) of p, q, and 10 is 15. **Column A:** The average of p and q **Column B:** 17.5

Answer: The two quantities are equal.

The average of p, q, and 10 is (p + q + 10)/3. We are given this average is 15. So, (p + q + 10)/3 = 15. Multiplying both sides by 3 gives p + q + 10 = 45. Subtracting 10 from both sides gives p + q = 35. Quantity A asks for the average of p and q, which is (p + q)/2. Substituting the value we found for p + q, we get 35/2 = 17.5. Therefore, Quantity A is 17.5, which is equal to Quantity B.

Compare the quantities in Column A and Column B. Given: k > 0 **Column A:** The area of a square with a perimeter of 12k **Column B:** The area of a rectangle with a width of 2k and a length of 4k

Answer: Quantity A is greater.

For the square in Column A, the perimeter is 4 times the side length. So, 4s = 12k. Dividing by 4, the side length is s = 3k. The area of the square is side squared, so Area_A = (3k)² = 9k². For the rectangle in Column B, the area is length times width. So, Area_B = (4k) * (2k) = 8k². Since k > 0, k² is also positive. Comparing 9k² and 8k², it's clear that 9k² > 8k². Therefore, Quantity A is greater.

Compare the quantities in Column A and Column B. Given: a > 0 and b < 0 **Column A:** a * b **Column B:** a / b

Answer: The relationship cannot be determined from the information given.

Let's test numbers. Case 1: Let a = 2 and b = -1. Quantity A = 2 * (-1) = -2. Quantity B = 2 / (-1) = -2. In this case, they are equal. Case 2: Let a = 2 and b = -2. Quantity A = 2 * (-2) = -4. Quantity B = 2 / (-2) = -1. In this case, Quantity B is greater. Since we get different results depending on the numbers chosen, the relationship cannot be determined from the information given.