imo A..not sure tho...

IMO

C both are needed

IMO B.

Ques is Is (x-2)^2 > x^2.

after simplifying this we get is (X < 1) ?

So from statement A we get x^2 >x

that means x could -ve or could be +ve.
Not Sufficient.

From statement B we get 1/x > 0.
that x is less than 1 ,such as 0<x<1.
Sufficient.

IMO C

(x-2)^2 > x^2

x^2 - 4x + 4 > x^2

4 - 4x > 0

4(1-x) > 0

Choice 1 states that x^2 >x

x can be positive or negative (5,-5) as example: Insufficient

Choice 2 states that 1/x > 0

x can be any positive number(including fractions) and still be greater than zero (1/2, 2) as an example.

together it limits x to be a positive INteger.

If you plug in any positive integer, the result is not > 0

So I think it is (C)

What is the OA????

OA is C

I have certain isuues with range for stmt 1.

x^2> x

implies

x can be positive/negative integer also x CAN BE negative fraction

so x can be 5, -5 or -1/5


stmt 2 says 1/x>0 now x can be +ve fraction or integer in both cases 1/x >0

therfore together,
x can be only positive INteger hence the answer to is x<1 is 'NO'

choose C.

PAB2706 wrote:Is (x – 2)^2 > x^2?

(1) x^2 > x

(2) 1/x > 0

number properties DS are my major drawback. pls suggest some good material i can download from the internet.

thanks cheers.

Step 1 of the Kaplan method for DS: focus on the question stem.

Here, we definitely want to simplify the question. ssmiles08 has done a great job getting us to:

Is 4(1-x) > 0?

but ideally we want to go a couple of steps further.

First, divide both sides by 4:

Is 1 - x > 0?

Finally, add x to both sides to get our ultimate question:

Is 1 > x?

Step 2 of the Kaplan method for DS: consider each statement by itself.

(1) x^2 > x

When determining the sufficiency of a statement, the two major options are algebra and picking numbers.

Algebraically:

x^2 > x
x^2 - x > 0
x(x-1) > 0

To get a positive product, both terms must have the same sign. We can see that large positive values will give us two positives; big negative values will give us two negatives. Therefore, x could be either greater than or less than 1: insufficient.

Picking numbers:

If x = 10, we get 100 > 10, which is certainly true. Is 10>1? YES
if x = -10, we get 100 > -10, which is certainly true. Is -10 > 1? NO

We can get both a yes and a no answer: insufficient.

(2) 1/x > 0

Any positive value of x will make 1/x positive. Are some positive numbers greater than 1? yes; are some positive numbers less than 1? also yes. Therefore, x may be less than or greater than 1: insufficient.

Picking numbers:

If x = 10, we get 1/10 > 0, which is certainly true. Is 10>1? YES

If x = 1/10, we get 10 > 0, which is certainly true. Is 1/10 > 1? NO

We can get both a yes and a no answer: insufficient.

Step 3 of the Kaplan method for DS: if necessary (i.e. neither statement was good enough by itself), combine.

Now we have two rules we must live by:

x(x-1) > 0
and
1/x > 0

From statement (2), we know that x must be positive.

If x is positive, and x(x-1) is positive, then (x-1) must also be positive, or:

x - 1 > 0

and therefore

x > 1.

So, if both statements are true, is x > 1? Definitely YES: choose (C).