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06 Jun 2022, 06:54
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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:
95%
(hard)
Question Stats:
46% (02:36) correct
54%
(02:39)
wrong
based on 98
sessions
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If \(s - \frac{1}{s} < \frac{1}{t} - t\), then is \(s > t\) ?
(1) \(s > 1\)
(2) \(t > 0\)
(1) \(s > 1\)
(2) \(t > 0\)
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Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
We are told \(s - \frac{s}{1} < \frac{t}{1} - t \).
We can rearrange that to be: \(\frac{s^2-1}{s}<\frac{1-t^2}{t}\)
(1)
Because \(s>1\), we know then that \(t\) must be a positive fraction where \(0<t<1\) in order for \(\frac{s^2-1}{s}<\frac{1-t^2}{t}\) to hold. Therefore \(s>t\)
If \(t\) is negative, \(\frac{1-t^2}{t}\) will be negative because of the denominator.
If \(t\) is greater than \(1\), \(\frac{1-t^2}{t}\) will be negative because of the numerator.
This leaves \(0<t<1\).
SUFFICIENT
(2)
The problem here is that we know \(t>1\), however, we do not know enough about \(s\). As we do not know whether \(s\) is positive or negative we cannot cross multiply. However, we can plug in values to see if we can deduce that \(s>t\) or \(t<s\).
Values plugged into \(\frac{s^2-1}{s}<\frac{1-t^2}{t}\):
Let s = -2 and t = 2
\(\frac{-3}{2}<\frac{1}{2}\). The statement we are plugging into holds and here and s<t.
Let s = 1 and t = 0.5
\(0 < \frac{3}{2}\)
The statement we are plugging into holds, however, in this case s>t.
INSUFFICIENT
Answer A
We can rearrange that to be: \(\frac{s^2-1}{s}<\frac{1-t^2}{t}\)
(1)
Because \(s>1\), we know then that \(t\) must be a positive fraction where \(0<t<1\) in order for \(\frac{s^2-1}{s}<\frac{1-t^2}{t}\) to hold. Therefore \(s>t\)
If \(t\) is negative, \(\frac{1-t^2}{t}\) will be negative because of the denominator.
If \(t\) is greater than \(1\), \(\frac{1-t^2}{t}\) will be negative because of the numerator.
This leaves \(0<t<1\).
SUFFICIENT
(2)
The problem here is that we know \(t>1\), however, we do not know enough about \(s\). As we do not know whether \(s\) is positive or negative we cannot cross multiply. However, we can plug in values to see if we can deduce that \(s>t\) or \(t<s\).
Values plugged into \(\frac{s^2-1}{s}<\frac{1-t^2}{t}\):
Let s = -2 and t = 2
\(\frac{-3}{2}<\frac{1}{2}\). The statement we are plugging into holds and here and s<t.
Let s = 1 and t = 0.5
\(0 < \frac{3}{2}\)
The statement we are plugging into holds, however, in this case s>t.
INSUFFICIENT
Answer A
22 Jun 2022, 04:21
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Nidzo
Hmmm I see a mistake here on cross multiplication that brings you nevertheless to the official correct answer.
Can you please correct it and explain again how is Statement 1 Sufficient?
