BTGmoderatorDC wrote:Is $$\frac{x+1}{y+1}$$ > $$\frac{x}{y}$$ ?
A. 0 < x < y
B. xy > 0
Target question: Is (x+1)/(y+1) > x/y ?
Statement 1: 0 < x < y
This tells us that y is POSITIVE, which means y+1 is also POSITIVE.
This means we can safely take the inequality
(x+1)/(y+1) > x/y and safely multiply both sides by y
When we do so, we get:
(y)(x+1)/(y+1) > x
We can also multiply both sides by y+1 to get:
(y)(x+1) > (x)(y+1)
Expand to get:
xy + y > xy + x
Subtract xy from both sides to get:
y > x
So, with the help of statement 1, our original target question
Is (x+1)/(y+1) > x/y ? becomes
Is y > x ?
Since statement 1 tells us that y > x, the answer to the target question is a definitive
YES
Since we can answer the
target question with certainty, statement 1 is SUFFICIENT
Statement 2: xy > 0
Let's TEST some values
There are several values of x and y that satisfy statement 2 (xy > 0). Here are two:
Case a: x = 1 and y = 1. In this case, (x+1)/(y+1) = (1+1)/(1+1) = 2/2 = 1, and x/y = 1/1 = 1. So, the answer to the target question is
NO, (x+1)/(y+1) is NOT greater than x/y ?
Case b: x = -3 and y = -2. In this case, (x+1)/(y+1) = (-3 +1)/(-2 +1) = -2/-1 = 2, and x/y = -3/-2 = 3/2. So, the answer to the target question is
YES, (x+1)/(y+1) IS greater than x/y ?
Since we cannot answer the
target question with certainty, statement 2 is NOT SUFFICIENT
Answer: A
Cheers,
Brent
