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23 Nov 2022, 08:55
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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:
65%
(hard)
Question Stats:
25% (01:49) correct
75%
(02:01)
wrong
based on 4
sessions
History
Date
Time
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Not Attempted Yet
If \(s - \frac{1}{s} < \frac{1}{t} - t\), is \(s > t\) ?
(1) \(s > 1\)
(2) \(t > 0\)
(1) \(s > 1\)
(2) \(t > 0\)
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
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23 Nov 2022, 11:27
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Bunuel
The inequality can be transformed as
\(\frac{s^2 - 1}{s}\) < \(\frac{1-t^2}{t}\)
Question
s > t
Inference: Does 's' lie to the right of 't' when plotted on a number line
Statement 1
\(s > 1\)
As s > 1, the RHS of the inequality is +ve as \(\frac{+ve }{ +ve}\) will yield a +ve result.
We know that the LHS > RHS, so the LHS also has to be positive.
Let's find out for what values of 't' is the LHS positive
\(\frac{1-t^2}{t}\) > 0
\(\frac{(1+t)(1-t)}{t}\) > 0
\(\frac{(t+1)(t-1)}{t}\) < 0
Plotting this over a number line
--------- -1 --------- 0 --------- 1 ---------
The highlighted part denotes the possible values of 't'
As both the ranges are less than 1, we have a definite answer to the question is s > t -- Yes !!
Statement 1 is sufficient and we can eliminate B, C and E.
Statement 2
\(t > 0\)
We already know from Statement 1, that when \(0 < t < 1\) and \(s > 1\) we have one possible answer to the question - Is \(s > t\) - Yes !.
So let's see if there is any other possibility.
Say t =1, the LHS = 0. Now for s negative values of s, the base equation still holds true and we now have \(s < t\). Is \(s > t\) - No !.
As we have two possible answers this statement is not sufficient.
Option A
29 Nov 2022, 10:26
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Bunuel
\((\frac{s^2 -1}{s}) -( \frac{1 -t^2}{t}) <0\)
or I -II <0
(1) \(s > 1\)
from this we know I >0 hence for the main condition to hold true II > 0
Therefore \(\frac{1 -t^2}{t} >0 \)
This will only be true when :
\( t <-1\) and \(0< t <1,\) in both the cases \(t < s\)
SUFF.
(2) \(t > 0\)
\(s= \frac{1}{2}\) and \( t =2 \), No to the stem.
\(s=2 \) and \(t = \frac{1}{2}\), Yes to the stem.
INSUFF.
Ans A
Unless we have missed something OA should be A.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
