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If x + y > 0, is x > |y|? (1) x > y (2) y < 0

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bibha
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If x + y > 0, is x > |y|? (1) x > y (2) y < 0 [#permalink] New post   07 May 2010, 22:05
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If x + y > 0, is x > |y|?

(1) x > y
(2) y < 0
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D
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gmat1220
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Re: If x + y > 0, is x > |y|? (1) x > y (2) y < 0 [#permalink] New post   08 Apr 2011, 10:19
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First strategy
------------------
You can throw in the values of x and y to be sure of the behavior of the inequality x > |y|
1. x > y and x + y > 0
x = 4 y = 3 the answer is yes
x = 4 y = -3 the answer is yes
sufficient

2. y < 0 and x + y > 0
y = -4 x = 4.5 the answer is yes
sufficient

Answer D.

Second strategy : Doing algebra will lead to same inference.
-----------------
Is x > |y| ?
The question can be rephrased as - Is x > y and x > -y ?
Adding the two we have
x + x > y - y
or 2x > 0
or x > 0
So the question becomes Is x > 0?

1. x > y
x + y >0
Adding the above two. 2x +y > y or x > 0 sufficient

2. 0>y
x + y > 0
Adding the above two. x + y > y or x > 0 sufficient

Hence D.
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jamifahad
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Re: If x + y > 0, is x > |y|? (1) x > y (2) y < 0 [#permalink] New post   08 Apr 2011, 05:58
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If x + y >0, is x > |y|?
(1) x > y
(2) y < 0

Please help me with approach. I am having tough time with inequality. What is the better way to solve such questions? Picking number or doing algebraically?

Here how i attempted this:
Given: x+y > 0 , x > -y
To prove: x > |y|

Stmt1: x > y. Now |y| = +y when y > 0 and |y| = -y when y < 0. As can be seen, x > +y and x > -y. Hence x > |y|. Sufficient.

Stmt2: y < 0. |y| = -y . So is x > -y? From what is given, x > -y. Sufficient.

Ans: D.
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If x + y > 0, is x > |y|? (1) x > y (2) y < 0 [#permalink] New post   08 Apr 2011, 06:43
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If x + y > 0, is x > |y|?

    If \(y \geq 0\), then the question becomes is x > y?
    If \(y < 0\), then the question becomes is x > - y? This is given to be true in the stem!

So, we need to determine whether x > y is true.

(1) x > y. Directly answers our rephrased question. Sufficient.

(2) y < 0. This mean that |y| = -y. So, the question becomes:

    Is x > -y?
    Is x + y > 0?

This is given to be true in the stem. Sufficient.

Answer: D.
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Re: If x + y > 0, is x > |y|? (1) x > y (2) y < 0 [#permalink] New post   14 May 2013, 13:39
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If x+y > 0, is x> |y|
i. x>y
ii.y<0

given

x>-y or -x<y..... question is -x<y<x ?? stem provides first part of this inequality (-x<y)

from 1

x>y...suff therefore sure -x<y<x .

from 2

y is -ve and question becomes is x>-y i.e. is -x<y ... provided in the stem
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If x + y > 0, is x > |y|? (1) x > y (2) y < 0 [#permalink] New post   02 Jun 2020, 07:30
Bunuel wrote:
bibha wrote:
If x + y > 0, is x > |y|?

    If \(y \geq 0\), then the question becomes is x > y?
    If \(y < 0\), then the question becomes is x > - y? This is given to be true in the stem!

So, we need to determine whether x > y is true.

(1) x > y. Directly answers our rephrased question. Sufficient.

(2) y < 0. This mean that |y| = -y. So, the question becomes:

    Is x > -y?
    Is x + y > 0?

This is given to be true in the stem. Sufficient.

Answer: D.

VeritasKarishma

Don't we check modulus by saying y>=0 and y<0? why is it y<=0
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KarishmaB
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If x + y > 0, is x > |y|? (1) x > y (2) y < 0 [#permalink] New post   05 Jun 2020, 03:48
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GDT wrote:
Bunuel wrote:
bibha wrote:
If x + y > 0, is x > |y|?

    If \(y \geq 0\), then the question becomes is x > y?
    If \(y < 0\), then the question becomes is x > - y? This is given to be true in the stem!

So, we need to determine whether x > y is true.

(1) x > y. Directly answers our rephrased question. Sufficient.

(2) y < 0. This mean that |y| = -y. So, the question becomes:

    Is x > -y?
    Is x + y > 0?

This is given to be true in the stem. Sufficient.

Answer: D.

VeritasKarishma

Don't we check modulus by saying y>=0 and y<0? why is it y<=0



We need to account for the entire range of values: negative numbers, 0 and positive numbers.

When y > 0,
|y| = y

When y < 0,
|y| = -y

When y = 0,
|y| = y or -y (both are 0 only so it doesn't matter)

The point is that you need to consider the value 0 somewhere. You can do it in either range.

|y| = y when y >= 0
|y| = -y when y < 0

OR

|y| = y when y > 0
|y| = -y when y <= 0
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Re: If x + y > 0, is x > |y|? (1) x > y (2) y < 0 [#permalink] New post   04 Nov 2023, 19:36
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Re: If x + y > 0, is x > |y|? (1) x > y (2) y < 0 [#permalink]
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