Question

From 1992 to 1993, the price of a home increased by \(x%\). From 1993 to 1994, the price of the home then decreased by \(x%\).
Quantity A: The price of the home at the beginning of 1992.
Quantity B: The price of the home at the beginning of 1994.

• 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:

A common mistake is to think that an \(x%\) increase followed by an \(x%\) decrease returns you to the starting value. This is incorrect. The percentage decrease is calculated on a larger base amount, making the monetary decrease larger than the initial monetary increase. This always results in a net loss.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem deals with successive percentage changes. A key concept is that a percentage change is always calculated based on the current value, not the original value.
Step 2: Key Formula or Approach:
Let the initial price be \(P\). An increase of \(x%\) is equivalent to multiplying by \((1 + \frac{x}{100})\). A decrease of \(x%\) is equivalent to multiplying by \((1 - \frac{x}{100})\). We will apply these multipliers sequentially to find the final price.
Step 3: Detailed Explanation:
Let \(P_{1992}\) be the price of the home at the beginning of 1992. This is Quantity A.
From 1992 to 1993, the price increased by \(x%\). The price at the beginning of 1993 (\(P_{1993}\)) is: \[ P_{1993} = P_{1992} \times \left(1 + \frac{x}{100}\right) \] From 1993 to 1994, the price decreased by \(x%\). This decrease is applied to the new, higher price \(P_{1993}\). The price at the beginning of 1994 (\(P_{1994}\)) is: \[ P_{1994} = P_{1993} \times \left(1 - \frac{x}{100}\right) \] Now, substitute the expression for \(P_{1993}\) into the equation for \(P_{1994}\): \[ P_{1994} = \left(P_{1992} \times \left(1 + \frac{x}{100}\right)\right) \times \left(1 - \frac{x}{100}\right) \] \[ P_{1994} = P_{1992} \times \left(1 + \frac{x}{100}\right)\left(1 - \frac{x}{100}\right) \] Using the difference of squares formula, \((a+b)(a-b) = a^2 - b^2\): \[ P_{1994} = P_{1992} \times \left(1^2 - \left(\frac{x}{100}\right)^2\right) \] \[ P_{1994} = P_{1992} \times \left(1 - \frac{x^2}{10000}\right) \] This is Quantity B.
Step 4: Final Answer:
We are comparing Quantity A (\(P_{1992}\)) with Quantity B (\(P_{1992} \times \left(1 - \frac{x^2}{10000}\right)\)). Assuming there was a change, \(x \textgreater 0\). This means \(x^2 \textgreater 0\), and \(\frac{x^2}{10000}\) is a positive value. The multiplier \(\left(1 - \frac{x^2}{10000}\right)\) is therefore less than 1. Multiplying the original price \(P_{1992}\) by a factor less than 1 results in a smaller final price. Thus, \(P_{1994} \textless P_{1992}\). Quantity B is less than Quantity A. Therefore, Quantity A is greater.