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 W

The length and width of rectangle A are 10 percent greater and 10 percent less, respectively, than the length and width of rectangle C
So, the length of rectangle A = L + (10% of L) = L + (0.1L) = 1.1L
And, the width of rectangle A = W - (10% of W) = W - (0.1L) = 0.9W

The length and width of rectangle B are 20 percent greater and 20 percent less, respectively, than the length and width of rectangle C
So, the length of rectangle B = L + (20% of L) = L + (0.2L) = 1.2L
And, the width of rectangle B = W - (20% of W) = W - (0.2L) = 0.8W

Area 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

Let, For C , Length 100, width 50, Area=5000
so For A, Length 110,width 45, Area=4950
and For B, Length 120, width 40, Area=4800

So A is greater.

Maybe can try to draw the table to understand


Item:----A-------------B-----------C
-----------------------------------------
length:-1.1X----------1.2X----------X
-----------------------------------------
width:--0.9Y----------0.8Y----------Y
-----------------------------------------
Area: --0.99XY-------0.96XY--------XY

Area A > Area B

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