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

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

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New post 27 Feb 2017, 07:17
10
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A
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C
D
E

Difficulty:

  55% (hard)

Question Stats:

61% (01:40) correct 39% (01:50) wrong based on 113 sessions

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Is x < y ?

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

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New post 27 Feb 2017, 08:02
1
1
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.

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

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New post 27 Feb 2017, 08:32
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

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

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New post 28 Feb 2017, 07:20
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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New post 02 Mar 2017, 17:59
3
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.

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

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New post 21 Feb 2018, 13:40
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

The answer is C


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

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

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New post 21 Feb 2018, 21:17
1
manimadhuri wrote:
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

The answer is C


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

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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New post 22 Feb 2018, 23:50
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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Re: Is x < y ? (1) x^2 - y^2 < 0 (2) -x - y < 0   [#permalink] 23 Mar 2019, 05:14
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