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Re: DS: Inequalities
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Updated on: 19 Feb 2012, 22:57
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.
Originally posted by
omerrauf on 19 Feb 2012, 20:25.
Last edited by
omerrauf on 19 Feb 2012, 22:57, edited 1 time in total.