j_shreyans wrote:Is (x−2)^2 > x^2?

(1) x^2>x

(2) 1/x>0

OAC

(x-2)² > x²
x² - 4x + 4 > x²
-4x > -4
x < 1.

Question stem rephrased:
Is x < 1?

Statement 1: x² > x
Here, x can be any value other than 0, 1, or a positive fraction between 0 and 1.
In other words, x<0 or x>1.
Since x can be greater than 1 or less than 1, INSUFFICIENT.

Statement 2: 1/x > 0.
Here, x can be any positive value.
In other words, x>0.
Since x can be greater than 1 or less than 1, INSUFFICIENT.

Statements combined:
The only range that satisfies both statements is x>1.
SUFFICIENT.

The correct answer is C.

Hi Guru ,

Pls correct me if i am wrong and suggest some technique to solve this kind of questions.

Is (x-2)^2 > x^2?

(1) x^2>x

(2) 1/x>0

(x-2)^2>x^2

if we solve the above will get x<1 this is our main target right?

statement 1 - X^2>x by solving this will get 1<x so x must be positive greater than 1 right?

So can this statement be sufficient by saying that x can not less than 1?

Statement 2 - 1/x>0

if we multiply both side by x so will get 1>0 right?

Now what next should i do?

Pls help me because i really need a help on this kind of questions.

Hi ,

Can you guys pls review my problem and pls correct me..

j_shreyans wrote:Hi Guru ,

Pls correct me if i am wrong and suggest some technique to solve this kind of questions.

Is (x-2)^2 > x^2?

(1) x^2>x

(2) 1/x>0

(x-2)^2>x^2

if we solve the above will get x<1 this is our main target right?

statement 1 - X^2>x by solving this will get 1<x so x must be positive greater than 1 right?

So can this statement be sufficient by saying that x can not less than 1?

Your solution assumes that x is positive.
If x is negative, then dividing each side of x² > x by x requires that the direction of the inequality flip from > to <.
A safer approach is to subtract x from each side, as follows:
x² - x > 0
x(x-1) > 0.

The CRITICAL POINTS are the values of x such that x(x-1) = 0:
x=0 and x=1.
To determine the ranges where x(x-1) > 0, test one value to the left and right of each critical point.

x<0:
If x=-1, then x(x-1) = (-1)(-2) = 2, which is greater than 0.
Thus, x<0 is a valid range.

0<x<1:
If x=1/2, then x(x-1) = (1/2)(-1/2) = -1/4, which is NOT greater than 0.
Thus, 0<x<1 is NOT a valid range.

x>1:
If x=2, then x(x-1) = (2)(1) = 2, which is greater than 0.
Thus, x>1 is a valid range.

Result:
The valid ranges are x<0 and x>1.
Since it's possible that x<1 or that x>1, INSUFFICIENT.

Statement 2 - 1/x>0

if we multiply both side by x so will get 1>0 right?

Now what next should i do?

Pls help me because i really need a help on this kind of questions.

Here, algebra makes the situation unnecessarily confusing.
Rather than use algebra, I suggest the following line of reasoning:
1/x > 0
1/x = positive
(positive)/x = positive.

Since (positive)/(positive) = positive, the equation above holds true only if x>0.
Since it's possible that x<1 or that x>1, INSUFFICIENT.

Statements combined:
The only range that satisfies both statements is x>1.
SUFFICIENT.

The correct answer is C.