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27 Jul 2010, 07:27
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Is |x| < 1 ?
(1) \(x^4 - 1 > 0\)
(2) \(\frac{1}{1-|x|} > 0\)
(1) \(x^4 - 1 > 0\)
(2) \(\frac{1}{1-|x|} > 0\)
One more basic doubt i ve regarding DS ?
Say If statement 1 . gives values of y as 0 , 1 , 2 , 3
Say If statement 2 . gives values of y as 1 , 2 , 3
Then while checking for C do we have to include 0 OR we just have to take common values i.e. 1,2,3 and not 0.. i hope i am able to make my Q clear.. I am missing somewhere..
Thanks
Say If statement 1 . gives values of y as 0 , 1 , 2 , 3
Say If statement 2 . gives values of y as 1 , 2 , 3
Then while checking for C do we have to include 0 OR we just have to take common values i.e. 1,2,3 and not 0.. i hope i am able to make my Q clear.. I am missing somewhere..
Thanks
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27 Jul 2010, 08:20
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as per cond 1
x^4 -1 > 0
-> X^4 > 1
i.e x can be -2, -3 .... ,2 , 3 , 4.....
mod x is always greater than 1
sufficient
as per cond 2
1/ (1-MOD X ) >0
i.e 1-mod x> 0
i.e 1> mod x
hence sufficient
answer D
x^4 -1 > 0
-> X^4 > 1
i.e x can be -2, -3 .... ,2 , 3 , 4.....
mod x is always greater than 1
sufficient
as per cond 2
1/ (1-MOD X ) >0
i.e 1-mod x> 0
i.e 1> mod x
hence sufficient
answer D
27 Jul 2010, 08:49
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sag
Not a good question.
Is \(|x| < 1\)? --> is \(-1<x<1\)?
(1) \(x^4-1>0\) --> \(x^4>1\) --> \(x<-1\) or \(x>1\). So \(x\) is not in the range (-1,1). Sufficient.
(2) \(\frac{1}{1-|x|}>0\) --> nominator is positive thus denominator must also be positive for fraction to be positive --> \(1-|x|>0\) --> \(|x|<1\). Sufficient.
Answer: D.
But: From (1) we have that \(x\) is NOT in the range (-1,1) and from (2) that \(x\) is in the range (-1,1). Two statements contradict each other.
This will never occur on GMAT as: on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other.
sag
Consider the following question (I just made it up):
If \(y\) is an integer, is \(|y+1|<3\)?
\(|y+1|<3\) means is \(-4<y<2\) (-3, -2, -1, 0, 1)?
(1) \(-3<y^3<10\) --> \(y\) can be: -1, 0, 1, or 2. Not sufficient.
(2) \((y^2+4y)(y-1)=0\) --> \(y\) can be: -4, 0, or 1. Not sufficient.
(1)+(2) Intersection of the values from (1) and (2) are \(y=0\) and \(y=1\), both these values satisfy inequality \(|y+1|<3\). Sufficient.
Answer: C.
So if statement (1) gives one set of values for x and statement (2) gives another set of values for x, then when considering statements together we should take only the values which satisfy both statements.
Hope it helps.
