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Updated on: 10 Aug 2018, 01:08
Originally posted by ankita1211 on 08 Aug 2018, 09:10.
Last edited by Bunuel on 10 Aug 2018, 01:08, edited 3 times in total.
Last edited by Bunuel on 10 Aug 2018, 01:08, edited 3 times in total.
Renamed the topic, edited the question and added the OA.
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
77% (01:30) correct
23%
(01:23)
wrong
based on 168
sessions
History
Date
Time
Result
Not Attempted Yet
Is x^3 > 1?
(1) x > -2
(2) 2x – (b – c) < c – (b – 2)
(1) x > -2
(2) 2x – (b – c) < c – (b – 2)
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
08 Aug 2018, 09:52
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ankita1211
OA:B
Rephrasing the question
Subtracting 1 from both sides, we get
\(x^3-1>0\)
\((x-1)(x^2+x+1)>0\) [Using \(a^3 - b^3 = (a - b)(a^2 + b^2 + ab)\)]
\(x^2+x+1\) is always positive, as its lowest value will occurs \(x=-\frac{1}{2}\) i.e at \(\frac{-b}{{2a}}\) given expression is of form \(ax^2+bx+c\)
At \(x=-\frac{1}{2}\);\(x^2+x+1= {(-\frac{1}{2})}^2 -\frac{1}{2}+1=\frac{3}{4}\)
Lowest value of \(x^2+x+1\) is \(\frac{3}{4}\), for all other values of \(x\) , \(x^2+x+1\) would be greater than \(\frac{3}{4}\)
Question reduces: Is \((x-1)>0?\) or Is \(x>1\)
Statement 1 :\(x > -2\)
\(x\) can be\(-1\)
Is \(x>1\) : No
\(x\) can be \(2\)
Is \(x>1\) : yes
Statement 1 alone is not sufficient
Statement 2: \(2x – (b – c) < c – (b – 2)\)
\(2x -b+c< c-b+2\)
Adding \(b-c\) on both sides, we get \(2x<2\) or \(x<1\)
Is \(x>1\) : No always
Statement 2 alone is sufficient
08 Aug 2018, 12:05
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This question is based on an important Concept 
WhatsApp Image 2018-08-09 at 00.31.35.jpeg [ 111.39 KiB | Viewed 5892 times ]
Attachment:
WhatsApp Image 2018-08-09 at 00.31.35.jpeg [ 111.39 KiB | Viewed 5892 times ]
14 Dec 2025, 00:20
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A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.
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