GRE Prep 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.
x > y > 0
[#permalink]
16 Apr 2018, 15:32
16 Apr 2018, 15:32
Expert Reply
00:00
Question Stats:
73% (00:52) correct
26% (01:01) wrong based on 53 sessions
Hide Show timer Statistics
\(x > y > 0\)
Quantity A |
Quantity B |
\(x - y\) |
\(\frac{x}{3} + \frac{y}{3}\) |
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Re: x > y > 0
[#permalink]
18 Apr 2018, 02:05
18 Apr 2018, 02:05
2
simplify the given quantities
\(x -y\) \(?\) \(x+y/3\)
\(3x - 3y ? x + y\)
subtracting x from both sides
\(2x - 3y ? y\)
subtracting y from both sides
\(2x -4y ? 0\)
adding 4y to both sides
\(2x ? 4y\)
Dividing both sides by \(2\)
\(x ? 2y\)
we know that \(x > y\) but we do not know by how much so
it could be the case that \(x =4\) and \(y =2\) so both quantities are equal or
\(x = 100\) and \(y = 10\) and as such \(2y =20\)
A>B
Therefore option D
\(x -y\) \(?\) \(x+y/3\)
\(3x - 3y ? x + y\)
subtracting x from both sides
\(2x - 3y ? y\)
subtracting y from both sides
\(2x -4y ? 0\)
adding 4y to both sides
\(2x ? 4y\)
Dividing both sides by \(2\)
\(x ? 2y\)
we know that \(x > y\) but we do not know by how much so
it could be the case that \(x =4\) and \(y =2\) so both quantities are equal or
\(x = 100\) and \(y = 10\) and as such \(2y =20\)
A>B
Therefore option D
Senior Manager
Joined: 23 Jan 2021
Posts: 294
Given Kudos: 81
Concentration: , International Business
Re: x > y > 0
[#permalink]
23 May 2021, 22:39
23 May 2021, 22:39
amorphous wrote:
simplify the given quantities
\(x -y\) \(?\) \(x+y/3\)
\(3x - 3y ? x + y\)
subtracting x from both sides
\(2x - 3y ? y\)
subtracting y from both sides
\(2x -4y ? 0\)
adding 4y to both sides
\(2x ? 4y\)
Dividing both sides by \(2\)
\(x ? 2y\)
we know that \(x > y\) but we do not know by how much so
it could be the case that \(x =4\) and \(y =2\) so both quantities are equal or
\(x = 100\) and \(y = 10\) and as such \(2y =20\)
A>B
Therefore option D
\(x -y\) \(?\) \(x+y/3\)
\(3x - 3y ? x + y\)
subtracting x from both sides
\(2x - 3y ? y\)
subtracting y from both sides
\(2x -4y ? 0\)
adding 4y to both sides
\(2x ? 4y\)
Dividing both sides by \(2\)
\(x ? 2y\)
we know that \(x > y\) but we do not know by how much so
it could be the case that \(x =4\) and \(y =2\) so both quantities are equal or
\(x = 100\) and \(y = 10\) and as such \(2y =20\)
A>B
Therefore option D
This is bit easy and less time consuming.
