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).
