 It is currently 16 Jun 2019, 21:43 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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. |x| > |y| and x + y > 0  Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
Founder  Joined: 18 Apr 2015
Posts: 6874
Followers: 114

Kudos [?]: 1332 , given: 6293

|x| > |y| and x + y > 0 [#permalink]
Expert's post 00:00

Question Stats: 68% (00:50) correct 31% (00:51) wrong based on 61 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

_________________ Manager  Joined: 06 Jun 2018
Posts: 94
Followers: 0

Kudos [?]: 59  , 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: 10
Followers: 0

Kudos [?]: 3 , given: 0

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.

thats a nice approach
Director Joined: 09 Nov 2018
Posts: 508
Followers: 0

Kudos [?]: 27 , 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: 370
Followers: 5

Kudos [?]: 111 , 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
_________________

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: 1962
Followers: 60

Kudos [?]: 1792  , given: 9

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
_________________

Brent Hanneson – Creator of greenlighttestprep.com  Re: |x| > |y| and x + y > 0   [#permalink] 22 Feb 2019, 11:21
Display posts from previous: Sort by

|x| > |y| and x + y > 0  Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®.