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|x| >= |x-y| + |y|, is y > x? 1. x > 0 2. y > 0

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Manager
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|x| >= |x-y| + |y|, is y > x? 1. x > 0 2. y > 0 [#permalink] New post 16 Sep 2004, 07:43
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A
B
C
D
E

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|x| >= |x-y| + |y|, is y > x?

1. x > 0
2. y > 0

This question has already been posted earlier. I somehow dont agree with the OA.
Director
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 [#permalink] New post 16 Sep 2004, 07:50
Do not remember this one. I choose E
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 [#permalink] New post 16 Sep 2004, 08:05
x value, y value, |x| >= |x-y| + |y|, is Y > X?

x=3, y=2, 3 >= 1 + 2, No
x=2, y=3, 2 >= 1 + 3, Doesn't satisfy
x=-2, y=-3, 2 >= 1 + 3, Doesn't satisfy
x=-3, y=-2, 3 >= 1 + 2, Yes
x=-3, y=2, 3 >= 5 + 2, Doesn't satisfy
x=-2, y=3, 2 >= 5 + 3, Doesn't satisfy
x=3, y=-2, 3 >= 5 + 2, Doesn't satisfy
x=2, y=-3, 2 >= 5 + 3, Doesn't satisfy

Ans should be C. Since from above we need to know the sign of both x and y to give an answer.
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 [#permalink] New post 16 Sep 2004, 08:40
I think it is D.

Consider condition 1:
When x>0 , y can be negative or positive
Let y = -5 x =2, Then 2 < 7 + 5 Does not satisfy
Let y = -2, Then 5 < does not satisfy

When y is positive:
Let y = 5 ; x = 2
2 < 8 does not satisfy
Let y=2; x = 5
5 = 5 satisfies In this case y<x

Hence y<x. Condition 1 Sufficient.

Consider Condition 2 only.
y>0; x can be either pos / neg
Let x = -2, y =5
2 < 12 does not satisf
5 < 7 does not satisf
Let x = 2, y =5
2 < 8 does not satisf
Let x=5 y =2
5 = 5 satisfies In this case y<x

Hence y<x. Condition 2 Sufficient.
Either of these are sufficient by themselves, Hence D
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Re: DS - Mod & Ineq [#permalink] New post 16 Sep 2004, 09:49
mba4me wrote:
|x| >= |x-y| + |y|, is y > x?

1. x > 0
2. y > 0

This question has already been posted earlier. I somehow dont agree with the OA.


By 1)

(5,2)

|5| >= |5-2| + |2|
5 >= 3 + 2, satisfy the equation, y=2, which is smaller than x, our answer to is y>x? is NO

(5,6)
|5| >= |5-6| + |6|
5 >= 1 + 6, does not satisfy the equstion

(5,-1)
|5| >= |5-(-1)| + |-1|
5 >= 6 + 1, does not satisfy the equation

(5, -5)
|5| >= |5-(-5)| + |-5|
5 >= 0 + 5, satisfy the equation, y=-5, which is smaller than x, our answer to is y>x? is NO

(5, 0)
|5| >= |5-0| + |0|
5 >= 5 + 0, satisfy the equation, y=0, which is smaller than x, our answer
to is y>x? is NO

(5, -6)
|5| >= |5-(-6)| + |-6|
5 >= 11 + 6, does not satisfy the equation

By 1) alone, all the answers we get is NO, so it is sufficient.

By 2)

(5,2)
|5| >= |5-2| + |2|
5 >= 3 + 2, satisfy the equation, x=5, which is greater than y, our answer to is y>x? is YES

(-5,2)
|-5| >= |-5-2| + |2|
5 >= 7 + 2, does not satisfy the equation

(0,2)
|0| >= |0-2| + |2|
0 >= 2 + 2, does not satisfy the equation

(-1, 2)
|-1| >= |-1-2| + |2|
1 >= 3 + 2, does not satisfy the equation

(2,2)
|2| >= |2-2| + |2|
2 >= 0 + 2, satisfy the equation, x=0, our answer to is y>x? is NO

(1,2)
|1| >= |1-2| + |2|
1 >= 1 + 2, does not satisfy the equation

By 2) alone, we get both YES and NO for the answer, so it is not sufficient.

My answer is A
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 [#permalink] New post 16 Sep 2004, 13:57
In your analysis everything looks good except the first x,y pair for st. 2
(5,2)
|5| >= |5-2| + |2|
5 >= 3 + 2, satisfy the equation, x=5, which is greater than y, our answer to is y>x? is YES

here 2 > 5 is a NO

So, you can consistently answer the question with a NO for St.2.

So, the answer would be D.
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 [#permalink] New post 16 Sep 2004, 14:10
Hardworker,
The constraints stipulate that either X or Y is always positive and if that is the case we can answer the question with a definite NO for y > x for both st1 & st2, independently.

So, I think the answer is D.

To all,
Is there a quicker way to solve these number properties problems without exhaustively picking numbers?. How can one be smart about picking numbers, if that is the only choice, so that we can definitively answer these kinds of questions.

Thoughts?.

Cheers,
scoobee...
Manager
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 [#permalink] New post 16 Sep 2004, 15:25
A very important identity for inequalities is used to solve the problem

it is |A+B| <= |A| + |B| absolute value of a sum is less or equal than the sum of the absolute values of the. ........(1)

In the given he gives |x| >= |x-y| + |y|
the given can be written as |x-y +y| >= |x-y| + |y|

but we know that |x-y +y| <= |x-y| + |y| from (1)

this gives |x|= |x-y| + |y| or |x-y|=|x| - |y|

|x-y| is always positve . Implies |x| - |y| >=0

Implies |x| >= |y|. We are still in the given phase.

if x and y are positive . Implies x>=y and it is solved. C


Awaiting any comments.
Manager
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 [#permalink] New post 16 Sep 2004, 15:29
simpler approach

In the given he gives |x| >= |x-y| + |y|

implies |x|>= |y| ( |x-y| is a positive number)

and the problem is simpler.
Manager
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 [#permalink] New post 16 Sep 2004, 16:34
Scoobee

You are right....My silly mistake! The answer would be D.

Anyone else can provide a simpler way to solve? I guarantee if this
question shows in the real test, I have no way to solve it with 1.5 minutes.

:shock:
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 [#permalink] New post 16 Sep 2004, 17:49
From 1, say x = 4 and y = 7 (x<y)
then |4| = 4
|x-y| = 3
|y| = 7
So |x| is not >= |x-y| + |y|

If x = 20 and y = 7 (x > y)
then |x| = 20
|x-y| = 13
|y| = 7
Now |x| is = |x-y| + |y|

If x = 10 and y = 7
then |x| = 10
|x-y| = 3
|y| = 7
Again |x| = |x-y|+|y|

If x = 6, and y = 7
then |6| is not >=|1| + |7|

So 1 is sufficient. As long |x| >= |x-y| + |y|, y > x


From 2, if y = 10, x = 5
|x| = 5
|x-y| = 5
|y| = 10

|x| < |x-y|+|y|

If y=10, x= 40
|x| = 40
|x-y| = 30
|y| = 10

Now |x| = |x-y|+|y|

If y = 30, x = 100
|x| = 100
|x-y| = 70
|y| = 30

Again |x| = |x-y|+|y|

So 2 is also sufficient.

(D) is the answer.
Manager
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 [#permalink] New post 16 Sep 2004, 22:31
The answer is D. I dont remember now why i felt that the answer should be E yesterday. :shock: I must have been half asleep.


Thanks.
  [#permalink] 16 Sep 2004, 22:31
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