Patricia purchased x meters of fencing. She originally intended to use all of the...

[AAPL]

Official Guide

Patricia purchased x meters of fencing. She originally intended to use all of the fencing to enclose a square region, but later decided to use all of the fencing to enclose a rectangular region with length y meters greater than its width. In square meters, what is the positive difference between the area of the square region and the area of the rectangular region?

1) xy = 256
2) y = 4

OA B

[deloitte247]

Length of the fencing = x
Originally intended to enclose a square region.
Let the side of square = s
Perimeter = S * (no. of sides) = s * 4 = 4s
Therefore, x = 4s because all of the fencings were to be used in enclosing the square region.
s = x/4, and $$area=\left(\frac{x}{4}\right)^2=\frac{x^2}{16}$$
Patricia later decided to use all of the fencing to enclose a rectangular region ith length 'y' meters greater than its width.
Let the width = w; length = w+y
Perimeter = 2 (length + width) = 2 (w+w+y) = 2 (2w + y) = 4w + 2y
Therefore, x = 4w + 2y (make 'w' subject of formula to get the width of rectangle)
4w = x - 2y
$$w=\frac{\left(x-2y\right)}{4}$$
Length = width + y
$$=\frac{\left(x-2y\right)}{4}+y$$
$$=\frac{x+2y}{4}$$
$$Area\ of\ rec\tan gle=\frac{x-2y}{4}\cdot\frac{x+2y}{4}=\frac{x^2+2yx-2yx-4y^2}{16}$$
$$=\frac{x^2-4y^2}{16}$$
The positive difference between the area of the square region, and rectangular region
$$=\frac{x^2}{16}-\frac{x^2-4y^2}{16}=\frac{x^2-x^2+4y^2}{16}=\frac{4y^2}{16}$$
To find the positive difference between these two regions, we only nee to know the value of y and evaluate
$$\frac{4y^2}{16}$$
Statement 1: xy = 256
y = 256/x
Since the value of 'x' os unknown, we cannot use it to find 'y', and we cannot estimate
$$\frac{4y^2}{16};\ hence,\ statement\ 1\ is\ NOT\ SUFFICIENT.$$

Statement 2: y=4
$$Therefore,\ \frac{4y^2}{16}=>\ \frac{4\left(4\right)^2}{16}=>\frac{4\cdot16}{16}=4$$
Their positive difference = 4
Statement 2 alone is SUFFICIENT, answer = option B