Max@Math Revolution wrote:[GMAT math practice question]

a O b is defined as 1/(a+b) - 1/a.
If x = (1+a) O (1-a) and y = (1-a) O (1+a), then what is the value of x*y?

A. 1/4
B. 1/3
C. 1/2
D. 1
E. 3/4

Let a=0.

x = (1+a)â—‹(1-a) = (1+0)â—‹(1-0) = 1â—‹1 = 1/(1+1) - 1/1 = 1/2 - 1 = -1/2

y = (1-a)â—‹(1+a) = (1-0)â—‹(1+0) = 1â—‹1 = 1/(1+1) - 1/1 = 1/2 - 1 = -1/2

Thus:
xy = -1/2 * -1/2 = 1/4

The correct answer is A.

$$x=\left(1+a\right)O\left(1-a\right)$$
$$y=\left(1-a\right)O\left(1+a\right)$$
Question=> Find the product of xy.
$$The\ relationship\ aOb=>\frac{1}{a+b}-\frac{1}{a}$$
Let a=any positive integer e.g 2
$$x=\left(1+a\right)O\left(1-a\right)\ where\ a=2$$
$$x=\left(1+2\right)O\left(1-2\right)$$
$$x=3\ O\ -1\ \ \ comparing\ this\ to\ aOb$$
a=3 and b=-1
Inserting these values into the expression
$$\frac{1}{a+b}-\frac{1}{a}$$
$$\frac{1}{3+\left(-1\right)}-\frac{1}{3}=>\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$$
$$y=\left(1-a\right)O\left(1+a\right)\ where\ a=2$$
$$=\left(1-2\right)O\left(1+2\right)$$
$$=-1\ O\ 3\ compare\ this\ with\ aOb$$ $$a=-1\ and\ b=3\ $$
$$Hence;\ \frac{1}{a+b}-\frac{1}{a}=>\frac{1}{-1+3}-\frac{1}{-1}=\frac{1}{2}+\frac{1}{1}=\frac{3}{2}$$
$$Therefore,\ product\ of\ x=\frac{1}{6}\cdot\frac{3}{2}=\frac{1}{4}$$

Answer = option A

=>

x = (1+a) o (1-a) = 1/(1+a+1-a) - 1/(1+a) = 1/2 - 1/(1+a) = (a-1)/(2(a+1)).
y = (1-a) o (1+a) = 1/(1-a+1+a) - 1/(1-a) = 1/2 - 1/(1-a) = (-a-1)/(2(1-a)).
So, xy = [(a-1)/(2(a+1))]*[(-a-1)/(2(1-a))] = [-(a-1)(a+1)]/[-4(a+1)(a+1))] = ¼.

Therefore, A is the answer.
Answer: A