GSBae wrote:
To solve these types of inequality questions, it's useful to see the ranges where x>y, x<y, and where x=y.
1) To start off, we know that \(x = 4y^2\). If we wanna see where x=y, set both variables equal to each other.
x = 4x^2
1 = 4x (dividing by y means we have to consider the case when x = 0. Notice in this case that x = y.
x = 1/4.
So our critical values here are y = 0 and x = 1/4.
<-------(0)--------(1/4)-------->
If x<0 (for instance, -1, then y = 4), x<y
if 0<x<1/4 (for instance, 1/8, then y =1/16) x>y
If x>1/4 (for instance, x = 1, y= 4) x<y
So, for x to be greater than y, x has be greater than 0 and less than 1/4. This is not enough information on its own. Insufficient.
2) is obviously not enough information on its own. Insufficient.
However, if we know that x>1, we know that it's outside this range and therefore that x<y.
We need both pieces of information. Therefore answer: C
The highlighted part is incorrect. The variables X & Y are interchanged in the post above. Please find below the corrected form.
y = 4y^2
1 = 4y (dividing by y means we have to consider the case when y = 0. Notice in this case that x = y.
y = 1/4.
So our critical values here are y = 0 and y = 1/4.
<-------(0)--------(1/4)-------->
If y<0 (for instance,y= -1, then x= 4), x>y
if 0<y<1/4 (for instance, y = 1/8, then x =1/16) x<y...
PS: So when 0<x<1/4, x<y... (Statement 2 tells us that x>1, so, x>y.. Both statement sufficient)If y>1/4 (for instance, y = 1, x= 4)x>y
So, for x to be greater than y, y has to be greater than 1/4 or less than 0. This is not enough information on its own. Insufficient.