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vcbabu
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Whether xy<1? 1). x + y = 1 2). x^2 + y^2 = 1 27 Jun 2009, 22:17
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Whether xy<1?
1). x + y = 1 2). x^2 + y^2 = 1



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Whether xy<1? 1). x + y = 1 2). x^2 + y^2 = 1 28 Jun 2009, 18:47
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very good explanation. +1 to you.
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skpMatcha
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Whether xy<1? 1). x + y = 1 2). x^2 + y^2 = 1 19 Jul 2009, 05:08
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IMO 'C'

Quote:
very good explanation. +1 to you.
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I dont see any explanation !

I cant explain why B is not sufficient in itself.

COming to C

ya\((x+y)^2 = x^2+y^2+2xy\)

substitute stmt1 & 2, we get xy = 0 , hence < 1
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Whether xy<1? 1). x + y = 1 2). x^2 + y^2 = 1 19 Jul 2009, 10:12
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vcbabu
Whether xy<1?
1). x + y = 1 2). x^2 + y^2 = 1
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xy< 1 only if xy = fraction ,-ve or 0

from 1

x+y =1 we can tell that xy is not -ve but it could = 0 and we can't tell if xy is a fraction or not.......insuff

from 2

xy could be -ve or 0 or fraction but we can be sure.........insuff

both toegther

the only possible way for x+y =1 , x^2+y^2 = 1 is that one of the variables is 0 and the other is = 1

thus xy = 0 then C is the answer
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Whether xy<1? 1). x + y = 1 2). x^2 + y^2 = 1 19 Jul 2009, 10:13
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vcbabu
Whether xy<1?
1). x + y = 1 2). x^2 + y^2 = 1
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xy< 1 only if xy = fraction ,-ve or 0

from 1

x+y =1 we can tell that xy is not -ve but it could = 0 and we can't tell if xy is a fraction or not.......insuff

from 2

xy could be -ve or 0 or fraction but we can be sure.........insuff

both toegther

the only possible way for x+y =1 , x^2+y^2 = 1 is that one of the variables is 0 and the other is = 1

thus xy = 0 then C is the answer
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vcbabu
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Whether xy<1? 1). x + y = 1 2). x^2 + y^2 = 1 22 Jul 2009, 11:21
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OA IS D.

By picking numbers , each is suff.
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rashminet84
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Whether xy<1? 1). x + y = 1 2). x^2 + y^2 = 1 22 Jul 2009, 11:33
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vcbabu
Whether xy<1?
1). x + y = 1 2). x^2 + y^2 = 1
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D is correct

xy< 1 only when
1) exactly one of x and y is negative
2) they are both fractions <1 (can be negative or positive)


stmt1: x+y cannot be equal to 1 if both are positive integers, or both are negative, or one is negative but other is a positive integer so that xy >1, So this confirms that x+y must be satisfying atleast one of the 2 conditions given above

stmt 2: Again x^2 + y+2 will never be =1 if both are positive/negative integers.
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Whether xy<1? 1). x + y = 1 2). x^2 + y^2 = 1 22 Jul 2009, 12:28
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thanks for the explanation ! As I get more aquainted with DS I get more terrified :)



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