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

 It is currently 05 Dec 2019, 22:50

### 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

# Is x^2 - y^2 > 0?

Author Message
TAGS:

### Hide Tags

Senior SC Moderator
Joined: 14 Nov 2016
Posts: 1347
Location: Malaysia
Is x^2 - y^2 > 0?  [#permalink]

### Show Tags

21 Mar 2017, 18:52
1
5
00:00

Difficulty:

95% (hard)

Question Stats:

38% (02:31) correct 62% (02:26) wrong based on 105 sessions

### HideShow timer Statistics

Is $$x^2-y^2>0$$ ?

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

_________________
"Be challenged at EVERY MOMENT."

“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”

"Each stage of the journey is crucial to attaining new heights of knowledge."

Chat Moderator
Joined: 07 Mar 2016
Posts: 49
Re: Is x^2 - y^2 > 0?  [#permalink]

### Show Tags

09 Jan 2019, 11:30
3
Here's how I solved this - please critique!

Is $$x^{2}$$-$$y^{2}$$>0?

Rewriting the stem: Is (x-y)(x+y)>0?

Statement 1) $$x^{2}$$-$$y^{2}$$>(x-y)

Rewriting: (x-y)(x+y)>(x-y)

If (x-y)>0
Dividing both sides by (x-y) we are left with (x+y)>1.

If (x-y)<0
Dividing both sides by (x-y) we are left with (x+y)<1.

Because we don't know if (x-y) is positive, negative or zero, Statement 1 is insufficient.

Statement 2) $$x^{2}$$-$$y^{2}$$<(x+y)

Rewriting: (x-y)(x+y)<(x+y)

If (x+y)>0
Dividing both sides by (x+y) we are left with (x-y)<1.

If (x+y)<0
Dividing both sides by (x+y) we are left with (x-y)>1.

Because we don't know if (x+y) is positive, negative or zero, Statement 2 is insufficient.

Statement 1 & 2) We are left with a bunch of possibilities and no way to come to 1 final answer; as such, "C" is insufficient.

_________________
...once in a while you get shown the light, in the strangest of places if you look at it right...
SVP
Joined: 26 Mar 2013
Posts: 2343
Is x^2 - y^2 > 0?  [#permalink]

### Show Tags

10 Jan 2019, 00:53
2
ocelot22 wrote:
hazelnut wrote:
Is $$x^2-y^2>0$$ ?

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

Notice that x^2-y^2 factors to (x+y)(x-y) for this reason, and all of the confusion that can occur from writing so much, missing a sign etc, I like to define a =(x+y), b=(x-y)

then the question becomes ab>0 Yes/NO?

(1) ab>b if b>0, a>1. if b <0, a<1. Suppose b=1,a=2 then ab>0 Yes. Suppose b=-1/2 a=1/2, then ab<0 NO.

Since we get a Yes and a No. (1) is NS

(2) ab<a if a>0,b <1. if a<0, b>1. Suppose a=1, b=1/2, then ab>0 Yes. Suppose a=-1 b=2, then ab<0 No Since we get a Yes and a No NS

(1) and (2) if b>0,a>1, if b<0, a<1, if b>0, a>1, if b<0, a <1, if a>0, b<1, if a<0, b>1. Suppose a=1/2, b=-1 then ab<0 No. Suppose a=1/2, b=1/2, then ab>0 YES. Since we get a Yes and a NO (1) and (2) are NS E.

Hi ocelot22

Nice solution. However, I would take your concept in simplicity by the following:

ab>b
ab<a
-----------Subtract the inequalities as different signs

ab -ab > b -a

0 > b -a

a > b....Form here it is easy to judge quickly

If a = 10 & b= 4...Then answer is Yes

If a = -10 & b = 4....Then answer No

hazelnut
Can you please post Veritas solution
Manager
Joined: 16 Oct 2011
Posts: 109
GMAT 1: 570 Q39 V41
GMAT 2: 640 Q38 V31
GMAT 3: 650 Q42 V38
GMAT 4: 650 Q44 V36
GMAT 5: 570 Q31 V38
GPA: 3.75
Re: Is x^2 - y^2 > 0?  [#permalink]

### Show Tags

09 Jan 2019, 13:05
1
hazelnut wrote:
Is $$x^2-y^2>0$$ ?

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

Notice that x^2-y^2 factors to (x+y)(x-y) for this reason, and all of the confusion that can occur from writing so much, missing a sign etc, I like to define a =(x+y), b=(x-y)

then the question becomes ab>0 Yes/NO?

(1) ab>b if b>0, a>1. if b <0, a<1. Suppose b=1,a=2 then ab>0 Yes. Suppose b=-1/2 a=1/2, then ab<0 NO.

Since we get a Yes and a No. (1) is NS

(2) ab<a if a>0,b <1. if a<0, b>1. Suppose a=1, b=1/2, then ab>0 Yes. Suppose a=-1 b=2, then ab<0 No Since we get a Yes and a No NS

(1) and (2) if b>0,a>1, if b<0, a<1, if b>0, a>1, if b<0, a <1, if a>0, b<1, if a<0, b>1. Suppose a=1/2, b=-1 then ab<0 No. Suppose a=1/2, b=1/2, then ab>0 YES. Since we get a Yes and a NO (1) and (2) are NS E.
Senior Manager
Joined: 13 Oct 2016
Posts: 354
GPA: 3.98
Re: Is x^2 - y^2 > 0?  [#permalink]

### Show Tags

22 Mar 2017, 02:01
ziyuen wrote:
Is $$x^2-y^2>0$$ ?

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

Hi

In other words "Is $$x^2 > y^2$$ ?

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

$$x(x - 1) > y(y - 1)$$

We can have $$x>y>0$$ -----> $$3>2>0$$

$$3*2>2*1$$ --------> $$6^2 > 2^2$$ Yes.

Or $$0<x<y$$ ---------> $$0<\frac{1}{3}<\frac{1}{2}$$

$$\frac{1}{3}*(-\frac{2}{3}) "?" \frac{1}{2} * (-\frac{1}{2})$$ -------> $$-\frac{2}{9} > -\frac{1}{4}$$ ----------> $$(\frac{1}{3})^2 < (\frac{1}{2})^2$$ No. Insufficient.

(2) $$x^2-y^2<(x + y)$$

$$x(x - 1) < y(y + 1)$$

$$x=2, y=4$$ ----> $$2*1 < 4*5$$ -----> $$2^2 < 4^2$$ No.

$$x=\frac{1}{2}, y=\frac{1}{3}$$ ---> $$\frac{1}{2}*(-\frac{1}{2}) "?" \frac{1}{3}*\frac{4}{3}$$ -----> $$-\frac{1}{4} < \frac{4}{9}$$ -------> $$(\frac{1}{2})^2 > (\frac{1}{4})^2$$ Yes. Insufficient.

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

or $$y^2 - x^2 < y - x$$ & $$x^2 - y^2 < x + y$$

Adding we'll get: 0 < 2y ----> y>0. But we already have different answers when y is positive (x can be >0, <0, integer or a fraction). Insufficient.

What's the source of the question?
Director
Joined: 28 Jul 2016
Posts: 670
Location: India
Concentration: Finance, Human Resources
GPA: 3.97
WE: Project Management (Investment Banking)
Re: Is x^2 - y^2 > 0?  [#permalink]

### Show Tags

09 Jan 2019, 20:13
emockus wrote:
Here's how I solved this - please critique!

Is $$x^{2}$$-$$y^{2}$$>0?

Rewriting the stem: Is (x-y)(x+y)>0?

Statement 1) $$x^{2}$$-$$y^{2}$$>(x-y)

Rewriting: (x-y)(x+y)>(x-y)

If (x-y)>0
Dividing both sides by (x-y) we are left with (x+y)>1.

If (x-y)<0
Dividing both sides by (x-y) we are left with (x+y)<1.

Because we don't know if (x-y) is positive, negative or zero, Statement 1 is insufficient.

Statement 2) $$x^{2}$$-$$y^{2}$$<(x+y)

Rewriting: (x-y)(x+y)<(x+y)

If (x+y)>0
Dividing both sides by (x+y) we are left with (x-y)<1.

If (x+y)<0
Dividing both sides by (x+y) we are left with (x-y)>1.

Because we don't know if (x+y) is positive, negative or zero, Statement 2 is insufficient.

Statement 1 & 2) We are left with a bunch of possibilities and no way to come to 1 final answer; as such, "C" is insufficient.

Thanks for the explanation
I dont think we can divide by x-y. since its no where mentioned that x is not equal to y
similarly for the next given statement x can be equal to -y.
Intern
Joined: 25 Jun 2017
Posts: 36
GMAT 1: 660 Q45 V36
GPA: 3.5
WE: Human Resources (Consumer Products)
Re: Is x^2 - y^2 > 0?  [#permalink]

### Show Tags

24 May 2019, 05:05
Bunuel, could you provide your insight on this question? I'm stuck on why E is the answer.

Per Statement 1, when x-y >0, x + y >1 (x=3, y=2) and |x|>|y|. When x-y<0, x+y<1 (x=(-2.5), y=3), and |x|<|y|. NS.

Per Statement 2, when x+y>0, x-y<1 (x=3, y=2.5) and |x|>|y|. When x+y<0, x-1>1 (x=(-2), y=(-4), and |x|<|y|. NS.

x=3 and y = 2, works for both statements. I can't find values that work for |x|<|y| fr both statements.
Intern
Joined: 19 Apr 2019
Posts: 35
Location: Indonesia
GMAT 1: 690 Q49 V34
GPA: 3.8
WE: Project Management (Commercial Banking)
Re: Is x^2 - y^2 > 0?  [#permalink]

### Show Tags

25 May 2019, 08:52
hazelnut wrote:
Is $$x^2-y^2>0$$ ?

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

Hi, I answered E using the following approach.
(1)
$$x^2-y^2>(x - y)$$
$$(x + y)(x - y)>(x - y)$$
$$(x + y)>1$$
Insufficient.

(2)
$$x^2-y^2<(x + y)$$
$$(x + y)(x - y)<(x + y)$$
$$(x - y)<1$$
Insufficient.

Together both statements are still insufficient as the possibility for (x - y) can range from positive fraction < 1 to unlimited negatives.
Let me know if that makes sense.
Re: Is x^2 - y^2 > 0?   [#permalink] 25 May 2019, 08:52
Display posts from previous: Sort by