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
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Does GMAT RC seem like an uphill battle? e-GMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days. Sat., Oct 19th at 7 am PDT
If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y
[#permalink]
Show Tags
12 Nov 2009, 10:14
25
16
Economist wrote:
If xy ≠ 0, is x > y?
(1) 4x = 3y
(2) |y - x| = x - y
This is a good one. +1 Economist for it.
If xy ≠ 0, is x > y?
Question asks whether \(x>y\)?
(1) 4x = 3y --> \(x=\frac{3}{4}y\), well this one is relatively easy. This statement only tells that x and y have the same sign. When they are both positive then \(x<y\), BUT when they are both negative \(y<x\). Not sufficient.
(2) \(|y - x| = x - y\) --> \(|y - x| = -(y-x)\). This means that \(y - x\leq{0}\) --> \(x\geq{y}\). Thus, this statement says that x can be more than or equal to y. Not sufficient.
(1)+(2) From (1) (\(x=\frac{3}{4}y\)) it follows that \(x\) is not equal to \(y\) (bearing in mind that xy ≠ 0), hence from (2): \(x>y\). Sufficient.
\(|y - x| = x - y\). Look at the LHS it's absolute value, it's never negative, hence RHS or \(x-y\) is never negative, \(x-y>=0\) or \(y-x<=0\). If so then only one possibility for \(|y - x|\), it must be \(-y+x\).
So we'll have: Condition: \(y-x<=0\), which is the same as \(x>=y\)
The above means that \(|y - x| = x - y\) is always true for \(x>=y\). But as we concluded earlier it's not enough. We need to be sure that \(x>y\). And \(x>=y\) leaves the possibility that they are equal, which is removed with the statement (1) when considering together. Hence C.
_________________
st 1) y=1, x=3/4 - false y=-1, x=-3/4 - true Not sufficient st 2) |y-x| =x-y--> x>y Sufficient coz |-x| = -(-x) = x -- basically a negative numbers should be negated again to get the result for absolute value all the cases as cay that x>y but fails when x=y
combining we can answer the question
C
<didnt consider x=y option, so answered as B >
Originally posted by chix475ntu on 10 Feb 2010, 22:16.
Last edited by chix475ntu on 11 Feb 2010, 10:06, edited 1 time in total.
st 1) y=1, x=3/4 - false y=-1, x=-3/4 - true Not sufficient st 2) |y-x| =x-y--> x>y Sufficient coz |-x| = -(-x) = x -- basically a negative numbers should be negated again to get the result for absolute value
B
B alone is not sufficient as we dont know whether x=y?
So A gives information that x is not = y.
So Final ans is C This was nice question...even I would have done the same mistake in the exam..Its really important to think of all the possibilities on the G day.
_________________
(1) \(4x = 3y\) --> if \(x\) and \(y\) are both positive, then \(x<y\) BUT if they are both negative then \(x>y\). Not sufficient. But from this statement we can grasp an important property we'll use while evaluating statements together: as \(xy\neq0\) then from statement (1) \(x\neq{y}\).
(2) \(|y-x|=x-y\). Now as LHS is absolute value, which is never negative, RHS must also be \(\geq0\) --> so \(x-y\geq0\) --> then \(|y-x|=-(y-x)\), and we'll get \(-(y-x)=x-y\) --> \(0=0\), which is true. This means that equation \(|y-x|=x-y\) holds true when \(x-y\geq0\) or, which is the same, when \(x\geq{y}\). But this not enough as \(x=y\) is still possible. Not sufficient.
(1)+(2) From (1) we got that \(x\neq{y}\) and from (2) \(x\geq{y}\), hence \(x>y\). Sufficient.
(1) \(4x = 3y\) --> if \(x\) and \(y\) are both positive, then \(x<y\) BUT if they are both negative then \(x>y\). Not sufficient. But from this statement we can grasp an important property we'll use while evaluating statements together: as \(xy\neq0\) then from statement (1) \(x\neq{y}\).
(2) \(|y-x|=x-y\). Now as LHS is absolute value, which is never negative, RHS must also be \(\geq0\) --> so \(x-y\geq0\) --> then \(|y-x|=-(y-x)\), and we'll get \(-(y-x)=x-y\) --> \(0=0\), which is true. This means that equation \(|y-x|=x-y\) holds true when \(x-y\geq0\) or, which is the same, when \(x\geq{y}\). But this not enough as \(x=y\) is still possible. Not sufficient.
(1)+(2) From (1) we got that \(x\neq{y}\) and from (2) \(x\geq{y}\), hence \(x>y\). Sufficient.
Answer: C.
Hope it helps.
Dear Bunuel:
I am a bit confused with this problem. I don't have a problem to understand the first statement, but I do have a problem trying to understand the second one..
I saw a user on this thread was solving the second statement as this:
2. |y - x| = x - y
y - x = x - y 2x = 2y x=y
-y + x = x-y 0=0
Isn't his way of solving the correct way to solve this kind of problems? I have seen some other exercises and they always negate the LHS that has the absolute value on it regarding of the inequality symbol or equality. I would like to learn why we are not doing this on this problem.
If I am wrong, would you be so kind to revamp the way of solving statement 2, I already read the way you solved it, but it was not 100% clear for me.
(1) \(4x = 3y\) --> if \(x\) and \(y\) are both positive, then \(x<y\) BUT if they are both negative then \(x>y\). Not sufficient. But from this statement we can grasp an important property we'll use while evaluating statements together: as \(xy\neq0\) then from statement (1) \(x\neq{y}\).
(2) \(|y-x|=x-y\). Now as LHS is absolute value, which is never negative, RHS must also be \(\geq0\) --> so \(x-y\geq0\) --> then \(|y-x|=-(y-x)\), and we'll get \(-(y-x)=x-y\) --> \(0=0\), which is true. This means that equation \(|y-x|=x-y\) holds true when \(x-y\geq0\) or, which is the same, when \(x\geq{y}\). But this not enough as \(x=y\) is still possible. Not sufficient.
(1)+(2) From (1) we got that \(x\neq{y}\) and from (2) \(x\geq{y}\), hence \(x>y\). Sufficient.
Answer: C.
Hope it helps.
Dear Bunuel:
I am a bit confused with this problem. I don't have a problem to understand the first statement, but I do have a problem trying to understand the second one..
I saw a user on this thread was solving the second statement as this:
2. |y - x| = x - y
y - x = x - y 2x = 2y x=y
-y + x = x-y 0=0
Isn't his way of solving the correct way to solve this kind of problems? I have seen some other exercises and they always negate the LHS that has the absolute value on it regarding of the inequality symbol or equality. I would like to learn why we are not doing this on this problem.
If I am wrong, would you be so kind to revamp the way of solving statement 2, I already read the way you solved it, but it was not 100% clear for me.
Thanks in advance.
There can be many correct ways to solve a questions.
Try this one: (2) says that \(|y-x|=-(y-x)\). We know that \(|x|=-x\), when \(x\leq{0}\), thus \(y-x\leq{0}\).
_________________
Re: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y
[#permalink]
Show Tags
20 Oct 2014, 22:26
1
1
If xy ≠ 0, is x > y?
(1) 4x = 3y
(2) |y - x| = x - y
Consider (1) 4x=3y ; x=3/-3 and y =4/-4 satisfy this condition. so x can be > y or x can be <y. insufficient
Consider (2)
|y - x| = x - y x= 4 y =3 satisfy this , and x>y x= 3 y=3 satisfy this but x =y x= 3 and y = 4 doesnt satisfy x=-3 and y =-4 satisfy this and x >y hence insufficient
combining 1 and 2 the only option that satisfy 1 and 2 is x=-3 and y=-4 and hence x>y
Re: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y
[#permalink]
Show Tags
09 Aug 2016, 23:33
1
Bunuel wrote:
Given: \(xy\neq0\). Question: is \(x>y\) true?
(1) \(4x = 3y\) --> if \(x\) and \(y\) are both positive, then \(x<y\) BUT if they are both negative then \(x>y\). Not sufficient. But from this statement we can grasp an important property we'll use while evaluating statements together: as \(xy\neq0\) then from statement (1) \(x\neq{y}\).
(2) \(|y-x|=x-y\). Now as LHS is absolute value, which is never negative, RHS must also be \(\geq0\) --> so \(x-y\geq0\) --> then \(|y-x|=-(y-x)\), and we'll get \(-(y-x)=x-y\) --> \(0=0\), which is true. This means that equation \(|y-x|=x-y\) holds true when \(x-y\geq0\) or, which is the same, when \(x\geq{y}\). But this not enough as \(x=y\) is still possible. Not sufficient.
(1)+(2) From (1) we got that \(x\neq{y}\) and from (2) \(x\geq{y}\), hence \(x>y\). Sufficient.
Answer: C.
Hope it helps.
Hello Bunuel,
Thank you for your answer. Although , as per mod property, isn't it the case that |x| = x when x>=0 and -x when x<0 ? I would like to know why you are considering |x| = -x when x<=0, because this is where the answer is really hinged on.
Re: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y
[#permalink]
Show Tags
10 Aug 2016, 00:58
ramblersm wrote:
Bunuel wrote:
Given: \(xy\neq0\). Question: is \(x>y\) true?
(1) \(4x = 3y\) --> if \(x\) and \(y\) are both positive, then \(x<y\) BUT if they are both negative then \(x>y\). Not sufficient. But from this statement we can grasp an important property we'll use while evaluating statements together: as \(xy\neq0\) then from statement (1) \(x\neq{y}\).
(2) \(|y-x|=x-y\). Now as LHS is absolute value, which is never negative, RHS must also be \(\geq0\) --> so \(x-y\geq0\) --> then \(|y-x|=-(y-x)\), and we'll get \(-(y-x)=x-y\) --> \(0=0\), which is true. This means that equation \(|y-x|=x-y\) holds true when \(x-y\geq0\) or, which is the same, when \(x\geq{y}\). But this not enough as \(x=y\) is still possible. Not sufficient.
(1)+(2) From (1) we got that \(x\neq{y}\) and from (2) \(x\geq{y}\), hence \(x>y\). Sufficient.
Answer: C.
Hope it helps.
Hello Bunuel,
Thank you for your answer. Although , as per mod property, isn't it the case that |x| = x when x>=0 and -x when x<0 ? I would like to know why you are considering |x| = -x when x<=0, because this is where the answer is really hinged on.
Thanks!
|0| = 0 = -0, so you can include = sign in either of the cases: |x| = -x, when x<=0 and |x| = x, when x >= 0.
_________________
Re: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y
[#permalink]
Show Tags
31 Jan 2018, 08:28
Bunuel wrote:
Economist wrote:
If xy ≠ 0, is x > y?
(1) 4x = 3y
(2) |y - x| = x - y
This is a good one. +1 Economist for it.
If xy ≠ 0, is x > y?
Question asks whether \(x>y\)?
(1) 4x = 3y --> \(x=\frac{3}{4}y\), well this one is relatively easy. This statement only tells that x and y have the same sign. When they are both positive then \(x>y\), BUT when they are both negative \(y>x\). Not sufficient.
(2) \(|y - x| = x - y\) --> \(|y - x| = -(y-x)\). This means that \(y - x\leq{0}\) --> \(x\geq{y}\). Thus, this statement says that x can be more than or equal to y. Not sufficient.
(1)+(2) From (1) (\(x=\frac{3}{4}y\)) it follows that \(x\) is not equal to \(y\) (bearing in mind that xy ≠ 0), hence from (2): \(x>y\). Sufficient.
Answer: C.
For (1), isn't it the opposite? when both +ve x<y, when both -ve x>y.
Re: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y
[#permalink]
Show Tags
31 Jan 2018, 08:36
urvashis09 wrote:
Bunuel wrote:
Economist wrote:
If xy ≠ 0, is x > y?
(1) 4x = 3y
(2) |y - x| = x - y
This is a good one. +1 Economist for it.
If xy ≠ 0, is x > y?
Question asks whether \(x>y\)?
(1) 4x = 3y --> \(x=\frac{3}{4}y\), well this one is relatively easy. This statement only tells that x and y have the same sign. When they are both positive then \(x>y\), BUT when they are both negative \(y>x\). Not sufficient.
(2) \(|y - x| = x - y\) --> \(|y - x| = -(y-x)\). This means that \(y - x\leq{0}\) --> \(x\geq{y}\). Thus, this statement says that x can be more than or equal to y. Not sufficient.
(1)+(2) From (1) (\(x=\frac{3}{4}y\)) it follows that \(x\) is not equal to \(y\) (bearing in mind that xy ≠ 0), hence from (2): \(x>y\). Sufficient.
Answer: C.
For (1), isn't it the opposite? when both +ve x<y, when both -ve x>y.
Yes but that doesn't change the answer anyway.
_________________
If my Post helps you in Gaining Knowledge, Help me with KUDOS.. !!
Re: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y
[#permalink]
Show Tags
14 Apr 2019, 07:29
Hello from the GMAT Club BumpBot!
Thanks to another GMAT 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.
_________________
close
20% (01:47) correct 
>
