Guys pls help me out on these two questions..
If x≠0, is |x| <1?
(1) xsq<1
(2) |x| < 1/x
Does x + y = 0?
(1) xy<0
(2) xsq = ysq
Note:xsq is x squared and ysq is y squared.
Thanks!!
Two DS problems..
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If x≠0, is |x| <1?
(1) xsq<1
(2) |x| < 1/x
Does x + y = 0?
(1) xy<0
(2) xsq = ysq
Statement I : x^2 < 1
It means that x is a fraction...otherwise x^2 cannot be less than 1
Hence absolute value of x is less than 1
So sufficient
Statement II : This is also possible only when x is a fraction...Hence sufficient
So it should be D for the first question
Q2) Statement I : xy< 0 ; it means either x is negative or y is negative....however we cannot deduce that x+y = 0 (this is true only when x = -y)Hence insufficient
Statement II : x^2 = y^2 ; -2^2 = 2^2; here x+y = 0; 2^2 = 2^2 ; here x+y = 4; hence insufficient
Since xy<0 we know that x and y have opposite signs...and now x^2 = y^2 ; which means both numbers have the same absolute values but with opposite signs....hence x+y = 0
So sufficient
Hence C
(1) xsq<1
(2) |x| < 1/x
Does x + y = 0?
(1) xy<0
(2) xsq = ysq
Statement I : x^2 < 1
It means that x is a fraction...otherwise x^2 cannot be less than 1
Hence absolute value of x is less than 1
So sufficient
Statement II : This is also possible only when x is a fraction...Hence sufficient
So it should be D for the first question
Q2) Statement I : xy< 0 ; it means either x is negative or y is negative....however we cannot deduce that x+y = 0 (this is true only when x = -y)Hence insufficient
Statement II : x^2 = y^2 ; -2^2 = 2^2; here x+y = 0; 2^2 = 2^2 ; here x+y = 4; hence insufficient
Since xy<0 we know that x and y have opposite signs...and now x^2 = y^2 ; which means both numbers have the same absolute values but with opposite signs....hence x+y = 0
So sufficient
Hence C
Last edited by Cybermusings on Tue May 22, 2007 11:04 pm, edited 2 times in total.
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Oops....sorry....I meant C....that's how i have solved....B alone is insufficient....since the two numbers can have either opposite signs or the same sign!!
What's the OA?
What's the OA?