sandy wrote:
A, B, and C are three rectangles. The length and width of rectangle A are 10 percent greater and 10 percent less, respectively, than the length and width of rectangle C. The length and width of rectangle B are 20 percent greater and 20 percent less, respectively, than the length and width of rectangle C.
Quantity A |
Quantity B |
The area of rectangle A |
The area of rectangle B |
Sandy has already provided a fast solution by assigning some nice dimensions to rectangle C.
So, let's examine an algebraic approach.
Let rectangle C have length L and width WThe length and width of rectangle A are 10 percent greater and 10 percent less, respectively, than the length and width of rectangle CSo, the length of rectangle A = L + (10% of L) = L + (0.1L) =
1.1LAnd, the width of rectangle A = W - (10% of W) = W - (0.1L) =
0.9WThe length and width of rectangle B are 20 percent greater and 20 percent less, respectively, than the length and width of rectangle CSo, the length of rectangle B = L + (20% of L) = L + (0.2L) =
1.2LAnd, the width of rectangle B = W - (20% of W) = W - (0.2L) =
0.8WArea or rectangle = (length)(width)We get:
Quantity A: (
1.1L)(
0.9W)
Quantity B: (
1.2L)(
0.8W)
Simplify to get:
Quantity A: 0.99LW
Quantity B: 0.96LW
Since L and W are POSITIVE, we can be certain that 0.99LW > 0.96LW
Answer: A
Cheers,
Brent