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19 Feb 2012, 20:05
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
66% (01:36) correct
34%
(01:31)
wrong
based on 170
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19 Feb 2012, 20:23
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dvinoth86
1) NS. Take x=-10 & y=-12. & take x= 10 & y=-32
2) NS. Check different values.
Together. Sufficient. Just put the value y=-22-x into 2nd statement. U will get x>0.
So the answer is no. Sufficient.
Answer should be C. Here is how:
Question: Is \(x\) negative?
Statement A: \(y + x = -22\)
\(x=-22-y\) so let \(y\) be \(-30\), then \(x=8\) so \(x>0\) , Let \(y=30\), then \(x=-52\) so \(x<0\) , Two different answers Hence Insufficient.
Statement B: \(y-x<{-22}\)
Rearranging, \(x>y+22\) , Let \(y=8\) , then \(x>30\) & \(x=+ve\) , let \(y=-30\), then \(x=-8\) & \(x=-ve\) , Two different answers Hence Insufficient.
Combined (Statements A & B):
We know that \(y=-22-x\) , lets substitute this in Statement B:
so \((-22-x)-x<-22\) so \(-2x<0\) which boils down to \(x>0\) (\(x\) has to be \(+ve\) for it to multiply with \(-2\) and result in a \(-ve\) number).
Hence Sufficient. Answer C.
Question: Is \(x\) negative?
Statement A: \(y + x = -22\)
\(x=-22-y\) so let \(y\) be \(-30\), then \(x=8\) so \(x>0\) , Let \(y=30\), then \(x=-52\) so \(x<0\) , Two different answers Hence Insufficient.
Statement B: \(y-x<{-22}\)
Rearranging, \(x>y+22\) , Let \(y=8\) , then \(x>30\) & \(x=+ve\) , let \(y=-30\), then \(x=-8\) & \(x=-ve\) , Two different answers Hence Insufficient.
Combined (Statements A & B):
We know that \(y=-22-x\) , lets substitute this in Statement B:
so \((-22-x)-x<-22\) so \(-2x<0\) which boils down to \(x>0\) (\(x\) has to be \(+ve\) for it to multiply with \(-2\) and result in a \(-ve\) number).
Hence Sufficient. Answer C.
