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Founder  Joined: 18 Apr 2015
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|x| > |y| and x + y > 0 [#permalink]
Expert's post 00:00

Question Stats: 67% (00:49) correct 32% (00:55) wrong based on 109 sessions
$$|x| > |y|$$ and $$x + y > 0$$

 Quantity A Quantity B y x

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Kudos for R.A.E
[Reveal] Spoiler: OA

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Kudos [?]: 80  , given: 0

Re: |x| > |y| and x + y > 0 [#permalink]
1
KUDOS
Carcass wrote:
$$|x| > |y|$$ and $$x + y > 0$$

 Quantity A Quantity B y x

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Kudos for R.A.E

Given

$$|x| > |y|$$

$$x^2 > y^2$$................we can do it safely as both x and y are in absolute value sign.

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

(x + y) (x -Y) > 0

We know x + y> 0 , thus x - y> 0

So, x - y >0 or x>y. Intern Joined: 10 Oct 2017
Posts: 9
Followers: 0

Kudos [?]: 4  , given: 0

Re: |x| > |y| and x + y > 0 [#permalink]
1
KUDOS
kaziselim wrote:
Carcass wrote:
$$|x| > |y|$$ and $$x + y > 0$$

 Quantity A Quantity B y x

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Kudos for R.A.E

Given

$$|x| > |y|$$

$$x^2 > y^2$$................we can do it safely as both x and y are in absolute value sign.

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

(x + y) (x -Y) > 0

We know x + y> 0 , thus x - y> 0

So, x - y >0 or x>y.

thats a nice approach
Director Joined: 09 Nov 2018
Posts: 505
Followers: 0

Kudos [?]: 56 , given: 1

Re: |x| > |y| and x + y > 0 [#permalink]
kaziselim wrote:
Carcass wrote:
$$|x| > |y|$$ and $$x + y > 0$$

 Quantity A Quantity B y x

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Kudos for R.A.E

Given

$$|x| > |y|$$

$$x^2 > y^2$$................we can do it safely as both x and y are in absolute value sign.

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

(x + y) (x -Y) > 0

We know x + y> 0 , thus x - y> 0

So, x - y >0 or x>y.

But (x + y) (x -Y) > 0 means either x + y> 0 , or x - y> 0
Given x + y> 0 does not mean that x - y> 0 In this case we are lucky.

Is there any other approach please?
Supreme Moderator
Joined: 01 Nov 2017
Posts: 371
Followers: 12

Kudos [?]: 181 , given: 4

Re: |x| > |y| and x + y > 0 [#permalink]
Expert's post
AE wrote:
Carcass wrote:
$$|x| > |y|$$ and $$x + y > 0$$

 Quantity A Quantity B y x

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Kudos for R.A.E

Given

$$|x| > |y|$$

$$x^2 > y^2$$................we can do it safely as both x and y are in absolute value sign.

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

(x + y) (x -Y) > 0

We know x + y> 0 , thus x - y> 0

So, x - y >0 or x>y.

But (x + y) (x -Y) > 0 means either x + y> 0 , or x - y> 0
Given x + y> 0 does not mean that x - y> 0 In this case we are lucky.

Is there any other approach please?

hi..
(x + y) (x -Y) > 0 means either 'both (x + y) and (x -Y) are greater than 0' or 'both (x + y) and (x -Y) are less than 0' because + * + =+ and -*-=+
so if x=y>0, x-y will also be greater than 0.

Another approach will be logical ..
|x|>|y| can be written as |x-0|>|y-0|, so distance of x from 0, is more than y from 0.
so x+y will depend on the sign of x, if x is negative (x+y) will be negative and if x is positive, (x+y) will be positive..
so we are given x+y>o, which means x>0..
if x>0, |x|=x, so x>|y|..
y will be equal or less than |y|, so x>y
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Some useful Theory.
1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048
2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825
3. Remainders : https://greprepclub.com/forum/remainders-what-you-should-know-11524.html
4. Number properties : https://greprepclub.com/forum/number-property-all-you-require-11518.html
5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html GRE Instructor Joined: 10 Apr 2015
Posts: 3874
Followers: 159

Kudos [?]: 4680  , given: 70

Re: |x| > |y| and x + y > 0 [#permalink]
1
KUDOS
Expert's post
Carcass wrote:
$$|x| > |y|$$ and $$x + y > 0$$

 Quantity A Quantity B y x

The approach used by @kaziselim is great.
However, if you missed that, here's a different approach:

GIVEN: |x| > |y|
This inequality tells us that the MAGNITUDE of x is greater than the MAGNITUDE of y
In other words, on the number line, the distance from x to 0 is greater than the distance from y to 0

This gives us 4 possible cases:
case i: ----x----y----0------ (x and y are both negative)
case ii: ----x--------0--y---- (x is negative and y is positive)
case iii: ---y--0-------x-- (y is negative and x is positive)
case iv: --0---y----x-- (x and y are both positive)
Notice that, for all 4 cases, I've made the distance from x to 0 greater than the distance from y to 0

Now examine the second piece of given information...
GIVEN: x + y > 0
When we examine the 4 cases above, only cases iii and iv satisfy the condition that x + y > 0
In case iii, x > y
In case iv, x > y
So, it MUST be the case that x > y

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com
If you enjoy my solutions, you'll like my GRE prep course.  Re: |x| > |y| and x + y > 0   [#permalink] 22 Feb 2019, 11:21
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