|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
TheUltimateWinner
Joined: 23 Feb 2015
Last visit: 30 Apr 2026
Posts: 2,465
Given Kudos: 1,990
Concentration: Finance, Technology
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
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
22 Feb 2020, 09:41
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
63% (01:40) correct
37%
(01:55)
wrong
based on 223
sessions
History
Date
Time
Result
Not Attempted Yet
If \(y ≠ 0\), what is the value of \(\frac{6^{(x-y)^2}}{6^{(x+3)(x-3)}}\)?
(1) \(y = 2x\).
(2) \(y = 2\).
(1) \(y = 2x\).
(2) \(y = 2\).
22 Feb 2020, 13:08
Kudos
Bookmarks
If \(y ≠ 0\), what is the value of \(\frac{6^{(x-y)^2}}{6^{(x+3)(x-3)}}\)?
\(\frac{6^{(x-y)^2}}{6^{(x+3)(x-3)}}=6^{(x-y)^2-(x+3)(x-3)}=6^{x^2-2xy+y^2-x^2+9}=6^{y^2-2xy+9}\)
(1) \(y = 2x\). The expression above becomes: \(6^{y^2-2xy+9}=6^{4x^2-4x^2+9}=6^9\). Sufficient.
(2) \(y = 2\). The expression above becomes: \(6^{y^2-2xy+9}=6^{4-4x+9}\). No info about x. Not sufficient.
Answer: A.
\(\frac{6^{(x-y)^2}}{6^{(x+3)(x-3)}}=6^{(x-y)^2-(x+3)(x-3)}=6^{x^2-2xy+y^2-x^2+9}=6^{y^2-2xy+9}\)
(1) \(y = 2x\). The expression above becomes: \(6^{y^2-2xy+9}=6^{4x^2-4x^2+9}=6^9\). Sufficient.
(2) \(y = 2\). The expression above becomes: \(6^{y^2-2xy+9}=6^{4-4x+9}\). No info about x. Not sufficient.
Answer: A.
24 Feb 2020, 03:04
Kudos
Bookmarks
Asad
Solution
Step 1: Analyse Question Stem
- • \(y≠0\)
• We need to find the value of \(\frac{6^{(x-y)^2}}{6^{(x+3)(x-3)}}\)
- o Or, \(\frac{6^{(x-y)^2}}{6^{(x+3)(x-3)}} = \frac{6^{(x^2+y^2–2xy)}}{6^{(x^2 – 9)}}= \frac{6^{x^2} *6^{y^2}*6^9}{6^{x^2}*6^{2xy}}= \frac{6^{y^2} *6^9}{6^{2xy}}\)
Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE
Statement 1: \(y = 2x\)
• \(\frac{6^{y^2}*6^9}{6^{2xy}} = \frac{6^{4*x^2}*6^9}{6^{4x^2}} = 6^9\).
Hence, statement 1 is sufficient and we can eliminate answer options B, C and E
Statement 2: \(y = 2\).
• Here, \(\frac{6^{y^2}*6^9}{6^{2xy} }= \frac{6^{2^2}*6^9}{6^{4x}}\)
• However, we don’t know the value of x.
Hence, statement 2 is NOT sufficient.
Thus, the correct answer is Option A.
