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.
number properties mayhem
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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.
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.
- ssmiles08
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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)
(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)
- cubicle_bound_misfit
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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.
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.
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- Stuart@KaplanGMAT
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Step 1 of the Kaplan method for DS: focus on the question stem.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.
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).
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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