Bunuel wrote:
thapliya wrote:
2) r < |s|
doesn't it mean that r lies between s and -s -s < r < s ... so r definitely is smaller than s ?
That's not correct.
(2) \(r<|s|\) --> either \(r<s\) or \(r<-s\) (for example \(r=1\) and \(s=2\) OR \(r=1\) and \(s=-2\)). Not sufficient.
Bunuel, please correct my below understanding, if i am wrong:
Example: If we have an equation, |x| >1 it means simply x>1 or x<-1;
Coming back to the question, it states that s>r or s<-r
Case 1. s>r => r<s
Case 2. s<-r => -s>r => r<-s
combining the 2 cases we have r<-s;
Now in order for r<-s to be true we have further 2 cases:
s=+ve; then, r must be negative => s=1 then r<-1 clearly r is not greater than s
s=-ve; then, r can be wither + or -; s=-2 then r=1 or -3 in such case we don't know r>s or not. Hence it is insufficient to answer.
Is my reasoning correct?
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