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

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.

Answer: E.


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.
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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.
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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)

\(\frac{x-y}{x+y}>1\) ---> \(\frac{x+y-2y}{x+y}>1\) ---> \(1 - \frac{2y}{x+y}>1\) ---->

\(\frac{y}{x+y}<0\)

Even if both statements are right, x+y could be either positive or negative. So, E

Correct..my bad!!

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!!

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.
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Thanks Bunuel, that explains it perfectly

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.
Answer (E)
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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.
Answer E.
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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.

Answer (E)
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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)
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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

\(1-\frac{x-y}{x+y}\);

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

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

\(\frac{2y}{x+y}\).
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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.

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
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Answer: option E

Please find the solution as attached

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.