Two DS problems..

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Two DS problems..

by moneyman » Mon May 21, 2007 5:22 am
Guys pls help me out on these two questions..

If x&#8800;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!!
Maxx

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by f2001290 » Mon May 21, 2007 7:36 am
1. A

2. C

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by Cybermusings » Mon May 21, 2007 8:37 am
If x&#8800;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
Last edited by Cybermusings on Tue May 22, 2007 11:04 pm, edited 2 times in total.

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Thanks

by moneyman » Tue May 22, 2007 5:56 am
Thank you very much Cybermusings..But doesnt your explanation for the 2nd prob say that the answer should be C??
Maxx

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by Cybermusings » Tue May 22, 2007 11:02 pm
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?