Question

The length of rectangle x is 20% greater than the length of rectangle y. The width of rectangle x is 20% less than the width of rectangle y.
Quantity A: The area of rectangle x.
Quantity B: The area of rectangle y.

• A. if Quantity A is greater;
• B. if Quantity B is greater;
• C. if the two quantities are equal;
• D. if the relationship cannot be determined from the information given.
• E.

Hint:

For any problem where a quantity is increased by x% and then decreased by x%, the net result is always a decrease. The final value will be \((1 - (\frac{x}{100})^2)\) times the original value. In this case, the area is multiplied by \((1+0.2)(1-0.2) = 1 - 0.2^2 = 1 - 0.04 = 0.96\), which is a 4% decrease.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem involves calculating the area of a rectangle after its dimensions have been altered by given percentages. This is a classic successive percentage change problem.
Step 2: Key Formula or Approach:
Let the length and width of rectangle y be \(L_y\) and \(W_y\), respectively. Its area is \(A_y = L_y \times W_y\). We will express the dimensions of rectangle x in terms of \(L_y\) and \(W_y\) and then calculate its area, \(A_x\).
An increase of 20% is equivalent to multiplying by \(1 + 0.20 = 1.2\).
A decrease of 20% is equivalent to multiplying by \(1 - 0.20 = 0.8\).
Step 3: Detailed Explanation:
Let the dimensions for rectangle y be: \[ \text{Length} = L_y, \quad \text{Width} = W_y \] The area of rectangle y (Quantity B) is: \[ A_y = L_y \times W_y \] Now, let's find the dimensions for rectangle x. The length of rectangle x, \(L_x\), is 20% greater than \(L_y\): \[ L_x = L_y + 0.20 L_y = 1.2 L_y \] The width of rectangle x, \(W_x\), is 20% less than \(W_y\): \[ W_x = W_y - 0.20 W_y = 0.8 W_y \] The area of rectangle x (Quantity A) is: \[ A_x = L_x \times W_x = (1.2 L_y) \times (0.8 W_y) \] \[ A_x = (1.2 \times 0.8) \times (L_y \times W_y) \] \[ A_x = 0.96 \times A_y \] Step 4: Final Answer:
We are comparing Quantity A (\(A_x\)) and Quantity B (\(A_y\)).
We found that \(A_x = 0.96 A_y\). Since areas must be positive, this means that the area of rectangle x is 96% of the area of rectangle y.
Therefore, \(A_x \textless A_y\), which means Quantity B is greater than Quantity A.