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x > y > 0
[#permalink]
16 Apr 2018, 15:32
16 Apr 2018, 15:32
Expert Reply
00:00
Question Stats:
72% (00:49) correct
27% (01:02) wrong based on 47 sessions
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\(x > y > 0\)
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
_________________
Need Practice? 20 Free GRE Quant Tests available for free with 20 Kudos
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.
_________________
Senior Manager at gre.myprepclub.com
Re: x > y > 0
[#permalink]
19 Jan 2023, 04:17
19 Jan 2023, 04:17
Hello from the GRE Prep Club BumpBot!
Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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