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.
Nov 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
03 Apr 2020, 08:20
Kudos
Bookmarks
C
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:
61% (02:49) correct
39%
(03:00)
wrong
based on 236
sessions
History
Date
Time
Result
Not Attempted Yet
Given that \(x^{2018} y^{2017} = \frac{1}{2}\) and \(x^{2016} y^{2019} = 8\), what is the value of \(x^2 + y^3\) ?
A. 37/4
B. 35/4
C. 33/4
D. 31/4
E. 29/4
Are You Up For the Challenge: 700 Level Questions
A. 37/4
B. 35/4
C. 33/4
D. 31/4
E. 29/4
Are You Up For the Challenge: 700 Level Questions
03 Apr 2020, 12:56
Kudos
Bookmarks
Given that \(x^{2018}y^{2017}=\frac{1}{2}\) and \(x^{2016}y^{2019}=8\), what is the value of \(x^2+y^3\) ?
--> Since nothing tells us about what x and y are, the number picking approach is best way to get the value of it:
\(x= \frac{1}{2}\) and \(y =2\)
--> \(\frac{1}{2^{2018}}* 2^{2017} = \frac{1}{2}\)
--> \(\frac{1}{2^{2016}}* 2^{2019} = 2^3 =8\)
well, \(x^{2}+ y^{3}= (\frac{1}{2})^{2} + 2^{3}= \frac{1}{4}+ 8 = \frac{33}{4} \)
Answer (C).
--> Since nothing tells us about what x and y are, the number picking approach is best way to get the value of it:
\(x= \frac{1}{2}\) and \(y =2\)
--> \(\frac{1}{2^{2018}}* 2^{2017} = \frac{1}{2}\)
--> \(\frac{1}{2^{2016}}* 2^{2019} = 2^3 =8\)
well, \(x^{2}+ y^{3}= (\frac{1}{2})^{2} + 2^{3}= \frac{1}{4}+ 8 = \frac{33}{4} \)
Answer (C).
yashikaaggarwal
Senior Moderator - Masters Forum
Joined: 19 Jan 2020
Last visit: 17 Jul 2025
Posts: 3,086
Given Kudos: 1,510
Location: India
Schools: Cranfield (A) Warwick MiM "23 (M)
GPA: 4
WE:Analyst (Internet and New Media)
03 Apr 2020, 08:46
Kudos
Bookmarks
X^2018*Y^2017=1/2------(1)
X^2016*Y^2019=8--------(2)
Divide equation 1 by 2
(X^2018*Y^2017)/(X^2016*Y^2019)=(1/2)/8
X^2/Y^2=1/16
X^2*16=Y^2
By taking square root
4X = Y ------(3)
Putting value in equation 1
X^2018*Y^2017=1/2
X^2018*(4X)^2017=1/2
X^2018*x^2017*4^2017=1/2
X^4035*2^4034=1/2
X^4035=1/2*2^4034
X^4035=(1/2)^4035
X=1/2
And Y = 4*1/2
Y=2
Therefore, Value of X^2+Y^3 is
(1/2)^2+2^3=33/4
So OA is C
Posted from my mobile device
X^2016*Y^2019=8--------(2)
Divide equation 1 by 2
(X^2018*Y^2017)/(X^2016*Y^2019)=(1/2)/8
X^2/Y^2=1/16
X^2*16=Y^2
By taking square root
4X = Y ------(3)
Putting value in equation 1
X^2018*Y^2017=1/2
X^2018*(4X)^2017=1/2
X^2018*x^2017*4^2017=1/2
X^4035*2^4034=1/2
X^4035=1/2*2^4034
X^4035=(1/2)^4035
X=1/2
And Y = 4*1/2
Y=2
Therefore, Value of X^2+Y^3 is
(1/2)^2+2^3=33/4
So OA is C
Posted from my mobile device
