Events & Promotions
|
|
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.
May 3102:00 PM PDT
-03:00 PM PDT
Video explanations + diagnosis of 10 weakest areas + 150+ short videos + study plan!
May 2110:00 AM EDT
-11:59 PM EDT
Make the most of your break with the most realistic GMAT™ prep. Take up to $700 off select products.
May 2910:00 AM IST
-11:00 PM IST
Start your journey with a fully customized action plan and work with a dedicated mentor to achieve a 735+ score.
May 3101:00 AM EDT
-02:00 AM EDT
Looking for your GMAT motivation to break through the score plateau? Pragati improved her score by massive 160 points with strategic guidance and hard-work! Find out how personalized mentorship and a strong mindset can turn GMAT struggles into success.
Jun 0210:00 AM PDT
-11:00 AM PDT
Probability is one of the most important GMAT Quant topics because it often combines logic, counting, set theory, and permutations & combinations. Many students try to solve probability questions by listing every possible case, but GMAT probability...- Jun 10
06:00 AM PDT
-06:15 PM PDT
Register for the GMAT Club Virtual MBA Spotlight Fair – the world’s premier event for serious MBA candidates. This is your chance to hear directly from Admissions Directors at nearly every Top 30 MBA program..
28 Jan 2021, 01:15
Kudos
Bookmarks
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:
36% (02:08) correct
64%
(02:14)
wrong
based on 88
sessions
History
Date
Time
Result
Not Attempted Yet
What is the value of \(x^3y^2 +\frac{2098}{xy^2}+\frac{1}{x}-x^y\), given x, and y are positive integer?
(1) \(x^3 + y^ 3 + \frac{512}{x^3y^3}= 24\)
(2) \(x=y\)
Are You Up For the Challenge: 700 Level Questions
(1) \(x^3 + y^ 3 + \frac{512}{x^3y^3}= 24\)
(2) \(x=y\)
Are You Up For the Challenge: 700 Level Questions
29 Jan 2021, 12:42
Kudos
Bookmarks
What is the value of x^3*y^2+2098/xy^2+1/x− x^y, given x, and y are positive integer?
Stat1: x^3+ y^3+ 512/(x^3*y^3)= 24
We know that, (a^+ b^3+ c^3) - 3abc = (a+b+c)*(a^2+b^2+c^2-ab-bc-ca)
where, a= x, b = y, c = 8/xy
So, given, (a^+ b^3+ c^3) = 3abc, which is possible, if (a+b+c)*(a^2+b^2+c^2-ab-bc-ca) = 0, or, (x+y+8/xy)* {x^2 + y^2+ 64/(x^2*y^2) - xy - 8/x - 8/y} =0,
as x and y are +ve integers, so, {x^2 +y^2+ 64/(x^2*y^2) -xy -8/x -8/y} =0 or,
x^2 + y^2+ 64/(x^2*y^2) = xy + 8/x + 8/y, which is only possible, x=y=2. Sufficient to get value of the req. expression.
Stat2: x=y; It is not sufficient.
So, I think A.
Stat1: x^3+ y^3+ 512/(x^3*y^3)= 24
We know that, (a^+ b^3+ c^3) - 3abc = (a+b+c)*(a^2+b^2+c^2-ab-bc-ca)
where, a= x, b = y, c = 8/xy
So, given, (a^+ b^3+ c^3) = 3abc, which is possible, if (a+b+c)*(a^2+b^2+c^2-ab-bc-ca) = 0, or, (x+y+8/xy)* {x^2 + y^2+ 64/(x^2*y^2) - xy - 8/x - 8/y} =0,
as x and y are +ve integers, so, {x^2 +y^2+ 64/(x^2*y^2) -xy -8/x -8/y} =0 or,
x^2 + y^2+ 64/(x^2*y^2) = xy + 8/x + 8/y, which is only possible, x=y=2. Sufficient to get value of the req. expression.
Stat2: x=y; It is not sufficient.
So, I think A.
12 Apr 2021, 23:18
Kudos
Bookmarks
Statement 1 is enough.
The below screenshot shows how I approached the problem.
The below screenshot shows how I approached the problem.
Attachments
1B9EC2A3-F82D-4EA4-AC24-950533F1FA7D.jpeg [ 3.38 MiB | Viewed 1939 times ]
