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22 May 2018, 03:47
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
33% (01:29) correct
67%
(01:46)
wrong
based on 286
sessions
History
Date
Time
Result
Not Attempted Yet
If \(r,s\) and \(t\) are integers, is \(r+s =\sqrt{t}\) ?
[1] \(r<-s\)
[2] \((r+s)^2=t\)
[1] \(r<-s\)
[2] \((r+s)^2=t\)
Princ
(1) \(r<-s\)
\(r+s<0\). Since, \(\sqrt{t}\) is non-negative
(2) \(|r+s|=\sqrt{t}\). Not sufficient.
Answer: A
General Discussion
22 May 2018, 05:34
Kudos
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OA: A
Statement 1 : \(r<-s\)
\(r+s<0\)
\(\sqrt{t}\) can not be -ve.
So answer for is \(r+s=\sqrt{t}\) :no.
Statement 1 is alone sufficient.
Statement 2 : \((r+s)^2=t\)
\(|r+s|=\sqrt{t}\)
Case a :When \(r+s <0\)
Then statement 2 becomes \(-(r+s)=\sqrt{t}\)
Answer for is \(r+s=\sqrt{t}\) :no
Case b : When \({r+s}\geq{0}\),
then statement 2 becomes \(r+s=\sqrt{t}\).
answer for is \(r+s=\sqrt{t}\) :yes.
So statement 2 alone is insufficient.
Statement 1 : \(r<-s\)
\(r+s<0\)
\(\sqrt{t}\) can not be -ve.
So answer for is \(r+s=\sqrt{t}\) :no.
Statement 1 is alone sufficient.
Statement 2 : \((r+s)^2=t\)
\(|r+s|=\sqrt{t}\)
Case a :When \(r+s <0\)
Then statement 2 becomes \(-(r+s)=\sqrt{t}\)
Answer for is \(r+s=\sqrt{t}\) :no
Case b : When \({r+s}\geq{0}\),
then statement 2 becomes \(r+s=\sqrt{t}\).
answer for is \(r+s=\sqrt{t}\) :yes.
So statement 2 alone is insufficient.
