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Joined: 10 Feb 2017
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Is x < y ? (1) x^2 - y^2 < 0 (2) -x - y < 0  [#permalink]

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28 00:00

Difficulty:   55% (hard)

Question Stats: 61% (01:45) correct 39% (01:58) wrong based on 278 sessions

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Is x < y ?

(1) x^2 - y^2 < 0
(2) -x - y < 0
Target Test Prep Representative V
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Re: Is x < y ? (1) x^2 - y^2 < 0 (2) -x - y < 0  [#permalink]

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4
1
LamaSalah wrote:
Is x < y ?

(1) x^2 - y^2 < 0
(2) -x - y < 0

Statement One Alone:

x^2 - y^2 < 0

We see that x^2 > y^2; however we still do not have sufficient information to determine whether x > y.

For instance, if x = -2 and y = 1, then x is less than y. However, if x = 2 and y = 1, then x is greater than y. Statement one alone is not sufficient to answer the question.

Statement Two Alone:

-x - y < 0

We can multiply the inequality in statement two by -1, remembering to reverse the inequality sign, and obtain:

x + y > 0, or equivalently, 0 < x + y. However, we still cannot determine whether x > y.

For instance if x = 2 and y = 1, then x is greater than y; however if x = 1 and y = 2, then x is less than y. Statement two alone is not sufficient to answer the question.

Statements One and Two Together:

Notice that x^2 - y^2 in statement one is a difference of two squares: x^2 - y^2 = (x + y)(x - y). So x^2 - y^2 < 0 means (x + y)(x - y) < 0. From statement two, we know that x + y > 0. That means x - y < 0, since x + y is positive. Therefore, x - y must be negative in order for their product (x + y)(x - y) to be negative.

Since x - y < 0, we have that x < y.

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##### General Discussion
Marshall & McDonough Moderator D
Joined: 13 Apr 2015
Posts: 1675
Location: India
Re: Is x < y ? (1) x^2 - y^2 < 0 (2) -x - y < 0  [#permalink]

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1
2
St1: x^2 - y^2 < 0 --> |x| < |y|. Not sufficient as we do not know the signs of x and y.

St2: x + y > 0
x and y are positive --> x > y or y > x
or
either x or y is positive --> The positive value must be greater than the other value.
Not Sufficient.

Combining St1 and St2, we know that |y| > |x| and x + y > 0.
If |y| > |x| and x and y are positive, then y > x
If |y| > |x| and one value is positive, then the value, which has a greater number, must be positive --> y > x.
In both the cases we get a definite yes answer for the question is x < y.
Sufficient.

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

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1
1
LamaSalah wrote:
Is x < y ?

(1) x^2 - y^2 < 0
(2) -x - y < 0

(1) \(x^2 < y^2 \iff |x| < |y|\)

We have \(3^2 < 5^2 \implies 3 < 5\)
However \(3^2 < (-5)^2 \implies 3 > - 5\).

Hence, insufficient.

(2) \(-x-y < 0 \implies x+y > 0\)

If x=5, y=2 then \(x+y>0\) and \(x>y\)
If x=-2, y=5 then \(x+y>0\) and \(x<y\).

Hence, insufficient.

Combine (1) and (2):
\((2) \implies x+y > 0\)
\((1) \implies (x-y)(x+y) < 0 \implies x-y < 0 \implies x<y\). Sufficient

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

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x < y?

stmt-1:
x^2 - y^2 < 0
|x| < |y|
here x or y or both can be postitive or negative. so we cannot decide if x < y or not.

stmt-2:
-x-y < 0
-x < y

say x=2, y=1 then -2 < 1, so x>y. so answer to the given question is NO.
say x=-1, y=2 then 1 < 2, but here x < y. so answer to the given question is Yes.

stmt-1 + stmt-2:
|x| < |y| AND -x < y then is x < y?
x=1, y=2 satisfies both |x| < |y| AND -x < y. in this case x < y (1<2). so answer is YES.
x=-1,y=2 satisfies both |x| < |y| AND -x < y. in this case x < y (-1<2). so answer is YES.

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

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1
broall wrote:
LamaSalah wrote:
Is x < y ?

(1) x^2 - y^2 < 0
(2) -x - y < 0

(1) \(x^2 < y^2 \iff |x| < |y|\)

We have \(3^2 < 5^2 \implies 3 < 5\)
However \(3^2 < (-5)^2 \implies 3 > - 5\).

Hence, insufficient.

(2) \(-x-y < 0 \implies x+y > 0\)

If x=5, y=2 then \(x+y>0\) and \(x>y\)
If x=-2, y=5 then \(x+y>0\) and \(x<y\).

Hence, insufficient.

Combine (1) and (2):
\((2) \implies x+y > 0\)
\((1) \implies (x-y)(x+y) < 0 \implies x-y < 0 \implies x<y\). Sufficient

-----------------------------------------

Can we write the first statement as -y<x<y
since (x+y)(x-y)<0, x falls in the range of -y and y.

This would make statement 1 alone sufficient to answer the question.
Math Expert V
Joined: 02 Sep 2009
Posts: 64150
Re: Is x < y ? (1) x^2 - y^2 < 0 (2) -x - y < 0  [#permalink]

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1
broall wrote:
LamaSalah wrote:
Is x < y ?

(1) x^2 - y^2 < 0
(2) -x - y < 0

(1) \(x^2 < y^2 \iff |x| < |y|\)

We have \(3^2 < 5^2 \implies 3 < 5\)
However \(3^2 < (-5)^2 \implies 3 > - 5\).

Hence, insufficient.

(2) \(-x-y < 0 \implies x+y > 0\)

If x=5, y=2 then \(x+y>0\) and \(x>y\)
If x=-2, y=5 then \(x+y>0\) and \(x<y\).

Hence, insufficient.

Combine (1) and (2):
\((2) \implies x+y > 0\)
\((1) \implies (x-y)(x+y) < 0 \implies x-y < 0 \implies x<y\). Sufficient

-----------------------------------------

Can we write the first statement as -y<x<y
since (x+y)(x-y)<0, x falls in the range of -y and y.

This would make statement 1 alone sufficient to answer the question.

x^2 - y^2 < 0;

x^2 < y^2;

|x| < |y|.

This means that y is further from 0 than x is. Which is not enough to get whether x < y. Basically we can have the following cases:

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

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LamaSalah wrote:
Is x < y ?

(1) x^2 - y^2 < 0
(2) -x - y < 0

i will also go with c as by using both the statement we can conclude whether x and y are positive and negative and then we can say x<y.
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Re: Is x < y ? (1) x^2 - y^2 < 0 (2) -x - y < 0  [#permalink]

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Hi ScottTargetTestPrep and Bunuel, thanks for your explanations.

What values could we plug into statement 2 to make it agree with the constraints?

I didn't think to multiply by -1, making it easier to eliminate statement 2, so I'm trying to see what I missed in my plug-in approach
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Joined: 02 Sep 2009
Posts: 64150
Re: Is x < y ? (1) x^2 - y^2 < 0 (2) -x - y < 0  [#permalink]

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dcummins wrote:
Hi ScottTargetTestPrep and Bunuel, thanks for your explanations.

What values could we plug into statement 2 to make it agree with the constraints?

I didn't think to multiply by -1, making it easier to eliminate statement 2, so I'm trying to see what I missed in my plug-in approach

For (2) x = 1 and y = 2 gives an YES answer, while x = 2 and y = 1 gives a NO answer.
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WE: Management Consulting (Consulting)
Re: Is x < y ? (1) x^2 - y^2 < 0 (2) -x - y < 0  [#permalink]

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Bunuel wrote:
dcummins wrote:
Hi ScottTargetTestPrep and Bunuel, thanks for your explanations.

What values could we plug into statement 2 to make it agree with the constraints?

I didn't think to multiply by -1, making it easier to eliminate statement 2, so I'm trying to see what I missed in my plug-in approach

For (2) x = 1 and y = 2 gives an YES answer, while x = 2 and y = 1 gives a NO answer.

Damn. I realise how silly this mistake was. I didn't write "-" in front of the x i.e. -x-y so was wondering how the hell you could get x-y<0
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Here's how I went from 430 to 710, and how you can do it yourself:
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Re: Is x < y ? (1) x^2 - y^2 < 0 (2) -x - y < 0  [#permalink]

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dcummins wrote:
Hi ScottTargetTestPrep and Bunuel, thanks for your explanations.

What values could we plug into statement 2 to make it agree with the constraints?

I didn't think to multiply by -1, making it easier to eliminate statement 2, so I'm trying to see what I missed in my plug-in approach

Statement Two Alone:

-x - y < 0

We can multiply the inequality in statement two by -1, remembering to reverse the inequality sign, and obtain:

x + y > 0, or equivalently, 0 < x + y. However, we still cannot determine whether x > y.

For instance if x = 2 and y = 1, then x is greater than y; however if x = 1 and y = 2, then x is less than y. Statement two alone is not sufficient to answer the question.
_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Senior Manager  P
Joined: 09 Jan 2017
Posts: 314
Location: India
Re: Is x < y ? (1) x^2 - y^2 < 0 (2) -x - y < 0  [#permalink]

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ScottTargetTestPrep wrote:
dcummins wrote:
Hi ScottTargetTestPrep and Bunuel, thanks for your explanations.

What values could we plug into statement 2 to make it agree with the constraints?

I didn't think to multiply by -1, making it easier to eliminate statement 2, so I'm trying to see what I missed in my plug-in approach

Statement Two Alone:

-x - y < 0

We can multiply the inequality in statement two by -1, remembering to reverse the inequality sign, and obtain:

x + y > 0, or equivalently, 0 < x + y. However, we still cannot determine whether x > y.

For instance if x = 2 and y = 1, then x is greater than y; however if x = 1 and y = 2, then x is less than y. Statement two alone is not sufficient to answer the question.

@ScottTargetTestPrep@Bunuel

Please evaluate the attached problem, the work done on this same question. I tried not to do it by putting values.
Attachments IMG_20190821_222100.jpg [ 1.66 MiB | Viewed 3633 times ]

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

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ScottTargetTestPrep wrote:
dcummins wrote:
Hi ScottTargetTestPrep and Bunuel, thanks for your explanations.

What values could we plug into statement 2 to make it agree with the constraints?

I didn't think to multiply by -1, making it easier to eliminate statement 2, so I'm trying to see what I missed in my plug-in approach

Statement Two Alone:

-x - y < 0

We can multiply the inequality in statement two by -1, remembering to reverse the inequality sign, and obtain:

x + y > 0, or equivalently, 0 < x + y. However, we still cannot determine whether x > y.

For instance if x = 2 and y = 1, then x is greater than y; however if x = 1 and y = 2, then x is less than y. Statement two alone is not sufficient to answer the question.

Thanks Scott, I must have overlooked this.
_________________
Here's how I went from 430 to 710, and how you can do it yourself:
Target Test Prep Representative V
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
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Re: Is x < y ? (1) x^2 - y^2 < 0 (2) -x - y < 0  [#permalink]

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dcummins wrote:
ScottTargetTestPrep wrote:
dcummins wrote:
Hi ScottTargetTestPrep and Bunuel, thanks for your explanations.

What values could we plug into statement 2 to make it agree with the constraints?

I didn't think to multiply by -1, making it easier to eliminate statement 2, so I'm trying to see what I missed in my plug-in approach

Statement Two Alone:

-x - y < 0

We can multiply the inequality in statement two by -1, remembering to reverse the inequality sign, and obtain:

x + y > 0, or equivalently, 0 < x + y. However, we still cannot determine whether x > y.

For instance if x = 2 and y = 1, then x is greater than y; however if x = 1 and y = 2, then x is less than y. Statement two alone is not sufficient to answer the question.

Thanks Scott, I must have overlooked this.

My pleasure.
_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

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

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Detail Solution -

1) x2- y2 < 0 ----->(x-y)(x+y) <0, -----------> either x-y>0 and x+y<0 or x-y<0 and x+y>0 -----------> hen x>y or x<y both possible----Not Sufficient - A,D rejected
2)-x-y<0 -------------> x+y>0 Not ufficient----B rejected.

combining 1+2 ---------> x+y>0 hence x-y<0 ---->x<y ...sufficient ----option C is correct.

you can also solve by putting random -ve +ve values.
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Re: Is x < y ? (1) x^2 - y^2 < 0 (2) -x - y < 0  [#permalink]

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Official explanation

Explanation: (1) The values 2 and 3 will give us a YES answer to the question stem, but the values 2 and -3 will give us a response of NO. Another way to approach this is to realize that you can factor the left hand side of the inequality to get (x - y)(x + y) < 0. This means that either the first term, (x - y), is negative or the second term, (x + y), is negative, but not both. If the first term is negative, then YES is answer to the question. However, if the second term is negative, the answer could be YES or NO. Accordingly, this statement is insufficient.

(2) You can simplify this by multiplying both sides by -1, which means you need to flip the direction of the inequality. You'll now have x + y > 0. This, however, doesn't tell us which value is larger, and this statement is insufficient.

Together, we know that (x - y)(x + y) < 0 and x + y > 0. This means that x - y < 0 . Remember, for the product of two values to be negative, one of them and only one of them must be negative. Because we know that (x + y) must be positive, then (x - y) < 0. Simply adding to each side of that inequality, we find that x < y, and know that the statements together are sufficient. Accordingly, the answer is C. Re: Is x < y ? (1) x^2 - y^2 < 0 (2) -x - y < 0   [#permalink] 18 Nov 2019, 04:00

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