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TheUltimateWinner
Joined: 23 Feb 2015
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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: 2,465
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) \(2x-y+x^2=y^2\)
firas92
Current Student
Joined: 16 Jan 2019
Last visit: 02 Dec 2024
Posts: 616
Given Kudos: 142
Location: India
Concentration: General Management
Schools: Tepper '23 (M$) Foster '23 (D)
GMAT 1: 740 Q50 V40
WE:Sales (Other)
Products:
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!
General Discussion
29 May 2024, 23:15
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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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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.
This post was generated automatically.
