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TheUltimateWinner
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Schools: Harvard '24 Judge '23 LBS '24 McCombs '24 Mendoza Tepper '24 Ross '24 Cox Marshall '24 Fisher Kelley '24 Jones McDonough '24 Owen '24 Terry BU Simon '24 Katz GWU Mason Babson Gabelli Foster '24 Goizueta '24 Scheller Stern '23 CBS '23 Booth '23 Haas '25 Wharton '25 Stanford '26 Johnson'23 Fuqua '23 Kellogg '23 Kenan-Flagler'23 INSEAD Aug 22 intake Olin '23 Said '23 Sloan '23 Carey Arizona "22 Yale '23 UC Davis Madison '23 Anderson '23 Carroll Carlson '23 Darden '23
Posts: 2,464
03 Jul 2019, 08:50
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
50% (01:50) correct
50%
(01:44)
wrong
based on 139
sessions
History
Date
Time
Result
Not Attempted Yet
If \(x\) and \(y\) are positive integers, is \(x\) odd?
1) \(y\) is odd
2) \(2x-y+x^2=y^2\)
1) \(y\) is odd
2) \(2x-y+x^2=y^2\)
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
firas92
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Joined: 16 Jan 2019
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03 Jul 2019, 09:14
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(1) \(y\) is odd
No info about \(x\)
Insufficient
(2) \(2x-y+x^2=y^2\)
This can be rearranged as \(2x+x^2=y^2+y\)
\(2x+x^2=y(y+1)\)
\(2x\) is always even, \(y\) and \((y+1)\) are consecutive integers so their product is definitely even
So, we have
\(Even + x^2 = Even\)
Since only \(Even + Even = Even\)
\(x^2\) is even and so, \(x\) is even
Sufficient
Answer is (B)
Hit Kudos if this helped!
No info about \(x\)
Insufficient
(2) \(2x-y+x^2=y^2\)
This can be rearranged as \(2x+x^2=y^2+y\)
\(2x+x^2=y(y+1)\)
\(2x\) is always even, \(y\) and \((y+1)\) are consecutive integers so their product is definitely even
So, we have
\(Even + x^2 = Even\)
Since only \(Even + Even = Even\)
\(x^2\) is even and so, \(x\) is even
Sufficient
Answer is (B)
Hit Kudos if this helped!
29 May 2024, 23:15
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