Question

In a certain store, computer X costs 30 percent more than computer Y, and computer Y costs 30 percent more than computer Z.
 

Column A	Column B
The cost of computer X minus the cost of computer Y.	The cost of computer Y minus the cost of computer Z.

 

Hint:

In chained percentage increase problems, the absolute increase becomes larger at each step because the base for the percentage calculation is increasing. A 30% increase on a larger number (Cost of Y) is a bigger amount than a 30% increase on a smaller number (Cost of Z).

Answer 1

Step 1: Understanding the Concept:

This is a percentage problem involving chained increases. We need to calculate the difference in costs between the computers and compare them. It's a common mistake to assume the differences are the same.

Step 2: Key Formula or Approach:

Let's use variables to represent the costs. A 30 percent increase means the new price is 130% of the original, or 1.3 times the original.
Let \(C_Z\) be the cost of computer Z.
Cost of Y (\(C_Y\)) = \(C_Z + 0.30 C_Z = 1.3 C_Z\).
Cost of X (\(C_X\)) = \(C_Y + 0.30 C_Y = 1.3 C_Y\).

Step 3: Detailed Explanation:

Let's pick a simple value for the cost of computer Z to make the calculations easier. Let \(C_Z = \$100\).
Cost of Y: \(C_Y = 1.3 \times C_Z = 1.3 \times 100 = \$130\).
Cost of X: \(C_X = 1.3 \times C_Y = 1.3 \times 130 = \$169\).

Column A:
Cost of X minus cost of Y = \(C_X - C_Y = \$169 - \$130 = \$39\).

Column B:
Cost of Y minus cost of Z = \(C_Y - C_Z = \$130 - \$100 = \$30\).

Comparison:
Column A is $39 and Column B is $30. Since 39 > 30, the quantity in Column A is greater.

Algebraic Approach:
Column B: \(C_Y - C_Z = 0.30 C_Z\).
Column A: \(C_X - C_Y = 0.30 C_Y = 0.30 \times (1.3 C_Z) = 0.39 C_Z\).
Since \(0.39 C_Z > 0.30 C_Z\), Column A is greater.

Step 4: Final Answer:
The difference in cost between X and Y is greater than the difference in cost between Y and Z. The quantity in Column A is greater.