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# Is x > 0? (1) (x + y)^2 < (x - y)^2 (2) x + y < x - y

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Joined: 05 Jul 2016
Posts: 16
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Concentration: Finance, Entrepreneurship
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Is x > 0? (1) (x + y)^2 < (x - y)^2 (2) x + y < x - y  [#permalink]

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29 Aug 2017, 16:37
1
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Difficulty:

35% (medium)

Question Stats:

69% (01:26) correct 31% (02:03) wrong based on 85 sessions

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Is x > 0?

(1) (x + y)^2 < (x - y)^2
(2) x + y < x - y
Math Expert
Joined: 02 Aug 2009
Posts: 8256
Re: Is x > 0? (1) (x + y)^2 < (x - y)^2 (2) x + y < x - y  [#permalink]

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29 Aug 2017, 18:51
4
1
guireif wrote:
Is x > 0?
(1) (x + y)^2 < (x - y)^2
(2) x + y < x - y

Hi...
70% going wrong as per timer, so this may help few..
Open the equation, it may give you something to work on.

1) $$(x+y)^2<(x-y)^2$$..
$$x^2+y^2+2xy<x^2+y^2-2xy......4xy<0....xy<0$$..
So we can say, none of the two is 0 and x and y are of OPPOSITE sign
Insufficient

2)$$x+y<x-y$$
2y<0......y<0
Insufficient

Combined..
Both x and y are of OPPOSITE sign and y<0..
So x >0
Sufficient

C
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Joined: 05 Jul 2016
Posts: 16
Location: Brazil
Concentration: Finance, Entrepreneurship
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Re: Is x > 0? (1) (x + y)^2 < (x - y)^2 (2) x + y < x - y  [#permalink]

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29 Aug 2017, 19:04
chetan2u wrote:
guireif wrote:
Is x > 0?
(1) (x + y)^2 < (x - y)^2
(2) x + y < x - y

Hi...
70% going wrong as per timer, so this may help few..
Open the equation, it may give you something to work on.

1) $$(x+y)^2<(x-y)^2$$..
$$x^2+y^2+2xy<x^2+y^2-2xy......4xy<0....xy<0$$..
So we can say, none of the two is 0 and x and y are of OPPOSITE sign
Insufficient

2)$$x+y<x-y$$
2y<0......y<0
Insufficient

Combined..
Both x and y are of OPPOSITE sign and y<0..
So x >0
Sufficient

C

Thanks. I just don't get why in this question you can simplify the inequalities to something like xy<0 or 2y<0. Can I always simplify inequalities that way?
Math Expert
Joined: 02 Aug 2009
Posts: 8256
Re: Is x > 0? (1) (x + y)^2 < (x - y)^2 (2) x + y < x - y  [#permalink]

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29 Aug 2017, 19:11
1
1
guireif wrote:
chetan2u wrote:
guireif wrote:
Is x > 0?
(1) (x + y)^2 < (x - y)^2
(2) x + y < x - y

Hi...
70% going wrong as per timer, so this may help few..
Open the equation, it may give you something to work on.

1) $$(x+y)^2<(x-y)^2$$..
$$x^2+y^2+2xy<x^2+y^2-2xy......4xy<0....xy<0$$..
So we can say, none of the two is 0 and x and y are of OPPOSITE sign
Insufficient

2)$$x+y<x-y$$
2y<0......y<0
Insufficient

Combined..
Both x and y are of OPPOSITE sign and y<0..
So x >0
Sufficient

C

Thanks. I just don't get why in this question you can simplify the inequalities to something like xy<0 or 2y<0. Can I always simplify inequalities that way?

Yes, you can cancel out 4 in 4xy<0 without changing the inequalities because 4 is POSITIVE..
1) If say I knew that x is POSITIVE, I can cancel out 4x in 4xy<0 and get y<0..
2) if say x is NEGATIVE, then you can cancel out the 4x but change the INEQUALITY sign.
That is 4xy<0...x<0 so y>0..

But if you don't know the signs, don't touch the INEQUALITY sign

Hope it helps
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Re: Is x > 0 ? (1) (x + y)^2 > (x - y)^2 (2) x + y > x - y  [#permalink]

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10 Apr 2018, 03:37
1
According to statement 1,
reduce the given equation to get 4xy > 0
Now, this could be possible only when x and y both are positive or x and y both are negative. So, we cannot say whether x is greater than zero.

Insufficient.

Statement 2 also gives 2y > 0 or y > 0. Insufficient again.

Combining both, we know that
y >0, there for 4xy > 0 to hold true x >0 has to be true. Hence, the answer is a clear Yes.

Thus, C...both are together sufficient.

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Re: Is x > 0 ? (1) (x + y)^2 > (x - y)^2 (2) x + y > x - y  [#permalink]

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12 Apr 2018, 00:55
=>

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.

Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
Condition 1) tells us that
$$(x+y)^2 > (x-y)^2$$
$$=> x^2 + 2xy + y^2 > x^2 - 2xy + y^2$$
$$=> 2xy > -2xy$$
$$=> 4xy > 0$$
$$=> xy > 0$$

Condition 2) tells us that
$$x + y > x –$$y
$$=> y > -y$$
$$=> 2y > 0$$
$$=> y > 0.$$

Since $$xy > 0$$ and $$y > 0$$, we have $$x > 0$$.

Thus, both conditions 1) & 2) together are sufficient.

In general, there are many questions involving integers and statistics to which we need to apply CMT(Common Mistake Type) 4.

Condition 1):
If $$x = -2$$ and $$y = -1$$, then the answer is “yes”.
If $$x = 1$$ and $$y = 2$$, then the answer is “no”.

Thus, condition 1) is not sufficient.

Condition 2):
If $$x = -3$$ and $$y = -1$$, then the answer is “yes”.
If $$x = -1$$ and $$y = -3$$, then the answer is “no”.
Thus condition 2) is not sufficient.

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.
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Re: Is x > 0 ? (1) (x + y)^2 > (x - y)^2 (2) x + y > x - y  [#permalink]

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12 Apr 2018, 08:43
1
Top Contributor
MathRevolution wrote:
[GMAT math practice question]

Is $$x>0$$?

$$1) (x+y)^2 > (x-y)^2$$
$$2) x+y > x-y$$

Target question: Is x POSITIVE?

Statement 1: (x + y )² > (x - y)²
Take: (x + y )² > (x - y)²
Expand and simplify both sides to get: x² + 2xy + y ² > x² - 2xy + y ²
Subtract x² from both sides: 2xy + y ² > -2xy + y ²
Subtract y² from both sides: 2xy > -2xy
Add 2xy both sides: 4xy > 0
Divide both sides by 4 to get: xy > 0
If the product xy> 0, then there are two possibilities:
Case a: x is positive and y is positive. In this case, the answer to the target question is YES, x is positive
Case b: x is negative and y is negative. In this case, the answer to the target question is NO, x is not positive
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x + y > x - y
Take: x + y > x - y
Add y to both sides: x + 2y > x
Subtract x from both sides: 2y > 0
Divide both sides by 2 to get: y > 0
So, we know that y is positive, but we have no information about x
So, we cannot answer the target question with certainty.
Statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that there are two possible cases (case a: x and y are both positive OR case b: x and y are both negative)
Statement 2 tells us that y is positive
So, statement 2 ELIMINATES case b, which means x and y are both positive
The answer to the target question is YES, x is positive
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

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Re: Is x > 0? (1) (x + y)^2 < (x - y)^2 (2) x + y < x - y  [#permalink]

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02 Feb 2020, 10:13
I don't understand why (1) is not sufficient.
After FOIL we have 2xy < -2xy
Divided by 2y this yields x < -x which is only true for negative numbers, so we can answer the question with "No".

Where am I wrong?
Re: Is x > 0? (1) (x + y)^2 < (x - y)^2 (2) x + y < x - y   [#permalink] 02 Feb 2020, 10:13
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