GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Oct 2019, 12:19 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

we will pick new questions that match your level based on your Timer History

Track

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.  If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Manager  Joined: 25 Nov 2009
Posts: 50
Location: India
If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

Show Tags

2
10 00:00

Difficulty:   75% (hard)

Question Stats: 53% (01:57) correct 47% (01:42) wrong based on 495 sessions

HideShow timer Statistics

If $$x ≠ -y$$, then is $$\frac{x-y}{x + y} > 1$$?

(1) $$x > 0$$

(2) $$y < 0$$
Math Expert V
Joined: 02 Sep 2009
Posts: 58427
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

Show Tags

13
6
gmatcracker2010 wrote:
Please explain the attached problem.

My take on it:

is (X+y)/ (x-y) > 1

ie. x + y > x -y
ie. 0 > y

which is B but OA is E.

If $$x\neq{-y}$$ is $$\frac{x-y}{x+y}>1$$?

Is $$\frac{x-y}{x+y}>1$$? --> Is $$0>1-\frac{x-y}{x+y}$$? --> Is $$0>\frac{x+y-x+y}{x+y}$$? --> Is $$0>\frac{2y}{x+y}$$?

(1) $$x>0$$ --> Not sufficient.

(2) $$y<0$$ --> Not sufficient.

(1)+(2) $$x>0$$ and $$y<0$$ --> numerator (y) is negative, but we cannot say whether the denominator {positive (x)+negative (y)} is positive or negative. Not sufficient.

The problem with your solution is that when you are then writing $$x-y>x+y$$, you are actually multiplying both sides of inequality by $$x+y$$: never multiply an inequality by variable (or expression with variable) unless you know the sign of variable (or expression with variable). Because if $$x+y>0$$ you should write $$x-y>x+y$$ BUT if $$x+y<0$$, you should write $$x-y<x+y$$ (flip the sign when multiplying by negative expression).

So again: given inequality can be simplified as follows: $$\frac{x-y}{x+y}>1$$ --> $$0>1-\frac{x-y}{x+y}<0$$ --> $$0>\frac{x+y-x+y}{x+y}$$ --> $$0>\frac{2y}{x+y}$$ --> we can drop 2 and finally we'll get: $$0>\frac{y}{x+y}$$.

Now, numerator is negative ($$y<0$$), but we don't know about the denominator, as $$x>0$$ and $$y<0$$ can not help us to determine the sign of $$x+y$$. So the answer is E.

Hope it helps.
_________________
General Discussion
Intern  Joined: 08 Jan 2010
Posts: 2
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

Show Tags

This is my first post to GMATCLUB & this is how i do it.

is (x-y)/ (x + y) > 1?
==>is x-y>x+y
==>boils down to ..is y<0?
and does not depend on x at all

so ans is (B)
CEO  B
Joined: 17 Nov 2007
Posts: 3041
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40 Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

Show Tags

$$\frac{x-y}{x+y}>1$$ ---> $$\frac{x+y-2y}{x+y}>1$$ ---> $$1 - \frac{2y}{x+y}>1$$ ---->

$$\frac{y}{x+y}<0$$

Even if both statements are right, x+y could be either positive or negative. So, E
_________________
HOT! GMAT Club Forum 2020 | GMAT ToolKit 2 (iOS) - The OFFICIAL GMAT CLUB PREP APPs, must-have apps especially if you aim at 700+ | Limited Online GMAT/GRE Math tutoring
Intern  Joined: 08 Jan 2010
Posts: 2
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

Show Tags

But i don't understand how this assumption is incorrect?

(x-y)/ (x + y) > 1
==> x-y>x+y

I was able to conclude to E only after substituting numbers

There must be a faster way!!
Math Expert V
Joined: 02 Sep 2009
Posts: 58427
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

Show Tags

9
5
anujairaj770 wrote:

But i don't understand how this assumption is incorrect?

(x-y)/ (x + y) > 1
==> x-y>x+y

I was able to conclude to E only after substituting numbers

There must be a faster way!!

Given: $$\frac{x-y}{x+y}>1$$. When you are then writing $$x-y>x+y$$, you are actually multiplying both sides of inequality by $$x+y$$: never multiply an inequality by variable (or expression with variable) unless you know the sign of variable (or expression with variable). Because if $$x+y>0$$ you should write $$x-y>x+y$$ BUT if $$x+y<0$$, you should write $$x-y<x+y$$ (flip the sign when multiplying by negative expression).

Given inequality can be simplified as follows: $$\frac{x-y}{x+y}>1$$ --> $$0>1-\frac{x-y}{x+y}<0$$ --> $$0>\frac{x+y-x+y}{x+y}$$ --> $$0>\frac{2y}{x+y}$$ --> we can drop 2 and finally we'll get: $$0>\frac{y}{x+y}$$.

Now, numerator is negative ($$y<0$$), but we don't know about the denominator, as $$x>0$$ and $$y<0$$ can not help us to determine the sign of $$x+y$$. So the answer is E.

Hope it helps.
_________________
Intern  Joined: 18 Aug 2010
Posts: 4
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

Show Tags

1
Rephrasing the question
$$\frac{(x-y)}{(x+y)} > 1$$
$$(x-y) > (x+y)$$ , Since x <> -y, we can multiply (x+y) both sides
Adding -x both sides
-y > y
Adding y both sides
0 > y , which is y < 0, given by statement 2, so why Answer is E(both together not sufficient)? Am I missing something?
_________________
"Learning never exhausts the mind."
--Leonardo da Vinci
Retired Moderator Joined: 02 Sep 2010
Posts: 724
Location: London
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

Show Tags

2
1
thirst4edu wrote:
Rephrasing the question
$$\frac{(x-y)}{(x+y)} > 1$$
$$(x-y) > (x+y)$$ , Since x <> -y, we can multiply (x+y) both sides
Adding -x both sides
-y > y
Adding y both sides
0 > y , which is y < 0, given by statement 2, so why Answer is E(both together not sufficient)? Am I missing something?

What you did is only true is (x+y)>0.
If you multiply both sides of an inequality with a number, the sign remains same if the number or expression is positive, whereas it flips if it is negative.

A counter example to show answer is indeed (e) is taking x=5 and y=-20 ... you'll get 25/-15 = -5/3 which is less than 1 ... on the other hand x=5 & y=-1 would get you 6/4 which is greater than 1.
_________________
Senior Manager  Joined: 25 Jul 2010
Posts: 279
Location: United States
Concentration: Finance
Schools: Cornell (Johnson) - Class of 2014
WE: Engineering (Telecommunications)
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

Show Tags

Thanks Bunuel, that explains it perfectly
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9701
Location: Pune, India
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

Show Tags

2
4
whichscore wrote:
if x is different from -y, in (x-y)/(x+y) greater than 1 ?
1. x > 0
2. y< 0

Even though fluke has already provided the detailed solution, there are a couple of points I would like to reinforce here.

People often get confused when they read 'if x is different from -y'. Why do they give this information? They do that because you have (x+y) in the denominator and hence it cannot be 0. That is, x + y = 0 or x = -y is not valid. Hence, they are just clarifying that the fraction is indeed defined.

Also, we are used to having 0 on the right of an inequality. What do we do when we have a 1? We take the 1 to the left hand side and get a 0 on the right.
Is $$\frac{(x-y)}{(x+y)} > 1$$? (Don't forget it is a question, not given information)
Is $$\frac{(x-y)}{(x+y)} - 1 > 0$$?
which simplifies to: Is y/(x+y) < 0 ?

We know how to deal with this inequality! Note here that we have no information about x and y as yet (except that x is not equal to -y which is more of a technical issue rather than actual information)

1. x > 0
No information about y so not sufficient.
2. y< 0
No information about x so not sufficient.

Both together, we know that y is negative. We need the sign of (x+y) now. But (x+y) may be positive or negative depending on whether x or y has greater absolute value. Hence not sufficient.
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9701
Location: Pune, India
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

Show Tags

3
2
MackyCee wrote:
From OG11...

Q139) if "x" is not equal to "-y", is x-y / x+y > 1?

(1) X > 0
(2) Y < 0

We know that we cannot multiply an inequality by an unknown if we don't know whether the unknown is a positive value or a negative value. The inequality behaves differently in the two cases. If the unknown is positive, the inequality stays as it is. If the unknown is negative, the inequality flips sign.
Therefore, x-y > x+y is not an option.
Next, '>1' is much more complicated that '>0' where we just need to consider whether the variables are positive or negative. Therefore, bring whatever is on the right hand side to the left hand side.

Question: Is $$\frac{(x - y)}{(x + y)} > 1$$
Is $$\frac{(x - y)}{(x + y)} - 1 > 0$$
Is $$\frac{-2y}{(x + y)} > 0$$

For this expression to be positive, y/(x+y) should be negative. Therefore, either 'y should be negative and (x+y) should be positive' or 'y should be positive and (x+y) should be negative'.

Statement 1: X > 0
No info about y so not sufficient.
Statement 2: Y < 0
No info about x so not sufficient.

Both together, we know that y is negative. We now need to know the sign of (x+y). Just knowing that x is positive doesn't tell us the sign of (x+y) because we don't know which one out of x and y has greater absolute value. If absolute value of x is greater than that of y, x+y is positive. If absolute value of x is less than that of y, x+y is negative. We do not know the sign of x+y so we still cannot say whether $$\frac{-2y}{(x + y)}$$ is positive. Not sufficient.
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9701
Location: Pune, India
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

Show Tags

1
RaviChandra wrote:
If x is not equal to –y, is (x – y) / (x + y) > 1?
(1) x > 0 (2) y < 0

Working with 1 on the right hand side is hard. It is better in case we have 0 on the right hand side.

Is (x – y) / (x + y) - 1 > 0 ?
Is -2y/(x+y) > 0 ?

For -2y/(x+y) to be positive, either both (-2y) and (x+y) should be positive or both should be negative.

Both (-2y) and (x+y) positive
y should be negative and x should be positive with greater absolute value than that of y (so that x+y is positive)
OR
Both (-2y) and (x+y) negative
y should be positive and x should be negative with greater absolute value than that of y (so that x+y is negative)

Both statements together tell us that x is positive and y is negative but they still do not tell us whether absolute value of x is greater than that of y. Hence, both statements together are not sufficient.

_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9701
Location: Pune, India
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

Show Tags

Responding to a pm:

Question: Is (x -y)/(x+y) >1?

What do you do when you are given >1 ? It is hard to find implications of '>1'. It is much easier to handle '>0'

Is $$\frac{(x -y)}{(x+y)} -1 > 0$$ ?
Is $$\frac{-2y}{(x+y)} > 0$$ ?

When will this be positive? In 2 cases:
1. When y is negative and (x+y) is positive i.e. x is positive and x has greater absolute value than y's absolute value.
2. When y is positive and (x+y) is negative i.e. x is negative and x has greater absolute value than y's absolute value.

Both statements together tell us that y is negative and x is positive but we don't know whether 'x's absolute value is greater than y's absolute value'.
Hence (E)
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Intern  Joined: 07 Sep 2014
Posts: 17
Location: United States (MA)
Concentration: Finance, Economics
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

Show Tags

in your explanation i just dont see the progression how the "1-" goes away and we change the numerator to "x+y-x+y" to arrive at 2y
Math Expert V
Joined: 02 Sep 2009
Posts: 58427
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

Show Tags

bsmith37 wrote:
in your explanation i just dont see the progression how the "1-" goes away and we change the numerator to "x+y-x+y" to arrive at 2y

$$1-\frac{x-y}{x+y}$$;

$$\frac{x+y}{x+y}-\frac{x-y}{x+y}$$;

$$\frac{(x+y)-(x-y)}{x+y}$$;

$$\frac{2y}{x+y}$$.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8027
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

Show Tags

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If x#-y is (x-y)/(x+y)>1?

(1) x>0
(2) y<0

If we modify the original condition, as squared numbers are always positive, the inequality sign does not change even if we multiply certain positive integers.
So if we multiply (x+y)^2, (x-y)(x+y)>(x+y)^2, --> (x-y)(x+y)-(x+y)^2>0, (x+y)(x-y-x-y)>0, (x+y)(-2y)>0, and if we divide both sides by -2,
the inequality sign changes, and ultimately we want to know whether (x+y)y<0.
There are 2 variables, so we need 2 equations, which are provided by the 2 conditions. If we look at the conditions together, the answer to what we want to know becomes 'yes' for x=2, y=-1, but 'no' for x=2,y=-3. Hence, the conditions are insufficient and the answer becomes (E).

For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
_________________
Intern  Joined: 05 Jun 2015
Posts: 24
Location: Viet Nam
GMAT 1: 740 Q49 V41 GPA: 3.66
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

Show Tags

Hi Karishma,
I solved this DS question incorrectly and still cannot figure out my mistake. Please help.

Here is how I solved the problem:

$$\frac{(x-y)}{(x+y)}>1$$

$$x-y>x+y>0$$
or
$$x-y<x+y<0$$

For case 1, solve $$x-y>x+y$$ we get $$y<0$$. Then solve $$x+y>0$$ we get $$x>(-y)$$. Together, we have $$x>0>y$$.

For case 2, solve $$x-y<x+y$$ we get $$y>0$$. Then solve $$x+y<0$$ we get $$x<(-y)$$. Together, we have $$x<0<y$$.

The two cases imply that x and y have opposite signs. Since statement (1) and statement (2) prove that x and y have oposite signs, hence sufficient.

After reading the solutions of Bunuel and you multiple times, I still cannot understand where I went wrong. Please point out the mistake.
Thank you very much!
Math Expert V
Joined: 02 Aug 2009
Posts: 8006
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

Show Tags

truongynhi wrote:
Hi Karishma,
I solved this DS question incorrectly and still cannot figure out my mistake. Please help.

Here is how I solved the problem:

$$\frac{(x-y)}{(x+y)}>1$$

$$x-y>x+y>0$$
or
$$x-y<x+y<0$$

For case 1, solve $$x-y>x+y$$ we get $$y<0$$. Then solve $$x+y>0$$ we get $$x>(-y)$$. Together, we have $$x>0>y$$.

For case 2, solve $$x-y<x+y$$ we get $$y>0$$. Then solve $$x+y<0$$ we get $$x<(-y)$$. Together, we have $$x<0<y$$.

The two cases imply that x and y have opposite signs. Since statement (1) and statement (2) prove that x and y have oposite signs, hence sufficient.

After reading the solutions of Bunuel and you multiple times, I still cannot understand where I went wrong. Please point out the mistake.
Thank you very much!

Hi,
I'll work beyond your solution..
after you have find some range, then substitute values to see if it fits in..
you have found out that x and y are of opposite sign..
say x is -ive, then y is positive..

1) let x= -3 and y=5..
$$\frac{(x-y)}{(x+y)}>1$$...
$$\frac{(-3-5}{(-3+5)}>1$$...
$$\frac{-8}{2}>1$$... ans NO

2) let x= -3 and y=2..
$$\frac{(x-y)}{(x+y)}>1$$..
$$\frac{(-3-2}{(-3+2)}>1$$..
$$\frac{-5}{-1}>1$$... ans YES

so different answers possible

Insuff even when combined
_________________
CEO  D
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2978
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

Show Tags

jusjmkol740 wrote:
If x not equal to (-y), then is (x-y)/ (x + y) > 1

(1) x > 0
(2) y < 0

Please find the solution as attached
Attachments

File comment: www.GMATinsight.com 2.jpg [ 122.69 KiB | Viewed 5576 times ]

_________________
Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8027
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

Show Tags

gmatcracker2010 wrote:
If x ≠ -y is $$\frac{x-y}{x+y}>1$$?

(1) x > 0
(2) y < 0

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

(x-y)/(x+y) > 1
⇔ (x-y)(x+y) > ( x + y )^2, by multiplying both sides by (x+y)^2
⇔ x^2 - y^2 > x^2 + 2xy + y^2
⇔ 2y^2 + xy < 0
⇔ y(2y+x)<0

Since we have 2 variables (x and y) and 0 equations,C is most likely to be the answer. So, we should consider 1) & 2) first.

Conditions 1) & 2):
x = 3, y = -1 : Yes
x = -1, y = -1 : No

They are not sufficient.

Therefore, the answer is E.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
_________________ Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0   [#permalink] 09 Jan 2018, 19:24

Go to page    1   2    Next  [ 24 posts ]

Display posts from previous: Sort by

If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  