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If xy > 0 and both x and y are even numbers, is x > y?

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If xy > 0 and both x and y are even numbers, is x > y?  [#permalink]

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New post 10 Sep 2018, 20:29
12
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A
B
C
D
E

Difficulty:

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Question Stats:

42% (02:29) correct 58% (02:30) wrong based on 204 sessions

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If xy > 0 and both x and y are even numbers, is x > y?

(1) x > y - 2
(2) |x - y| > 4


New question!!!..

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If xy > 0 and both x and y are even numbers, is x > y?  [#permalink]

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New post 10 Sep 2018, 21:26
chetan2u wrote:
If xy>0 and both x and y are even numbers, is x>y?
(I) x>y-2
(II) |x-y|>4

New question!!!..

OE to follow..


xy>0.....
x and y are even...

1) x>y-2
x+2>y
So y <x in all cases except when x=y..
Example say x=6, y<x+2 or y<6+2 that is y<8 but y cannot be 7 as y is even,
so (I) y will be 6 or x=y OR
(II) y can be 4,2 etc or X>y
Insufficient

2) |x-y|>4
We cannot say if y>x or x>y but surely \(x\neq{y}\)
Insufficient

Combined..
We know from statement I that either x=y or X>y
But statement II tells us that \(x\neq{y}\)
So only possibility x>y
Sufficient

C
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Re: If xy > 0 and both x and y are even numbers, is x > y?  [#permalink]

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New post 10 Sep 2018, 21:51
chetan2u wrote:
If xy>0 and both x and y are even numbers, is x>y?
(I) x>y-2
(II) |x-y|>4

New question!!!..


The answer is C . since xy>0; here is my list of values

from statement 1:
x=-2; y= -8
x=4; y =4

statement 1 is insufficient

from statement 2:
x=-2; y=-8
x=-8; y=-2
x=2; y=8
x=8; y=2

statement 2 is insufficient

combining both makes it sufficient. C
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Re: If xy > 0 and both x and y are even numbers, is x > y?  [#permalink]

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New post 10 Sep 2018, 22:31
Xy>0 means either x ND y r both -ve or both are positive.
1) not sufficient.as x can be bigger/smaller than y or equal
2) sufficient: x-y >4 or y-x <- 4; for both +ve ND -ve x> y

AnswerB


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Re: If xy > 0 and both x and y are even numbers, is x > y?  [#permalink]

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New post 10 Sep 2018, 23:01
deepverma wrote:
Xy>0 means either x ND y r both -ve or both are positive.
1) not sufficient.as x can be bigger/smaller than y or equal
2) sufficient: x-y >4 or y-x <- 4; for both +ve ND -ve x> y

AnswerB


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you will have to recheck highlighted portion

|x-y|>4
two cases
1) \(x-y\geq{0}\)
x-y>0
2) \(x-y<{0}\)
-(x-y)>4.......y-x>4

both x-y >4 and y-x <- 4 are SAME
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Re: If xy > 0 and both x and y are even numbers, is x > y?  [#permalink]

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New post 10 Sep 2018, 23:38
1
chetan2u wrote:
If xy>0 and both x and y are even numbers, is x>y?
(I) x>y-2
(II) |x-y|>4

New question!!!..


Given:
both x and y are even numbers
xy>0: This means both x & y have same sign.

Statement 1: x>y-2

Case 1 both x,y >0
eg 8>6-2
x>y Yes

Case 2: both x,y<0
eg -8>-8-2
x>y No
Hence Insuficient

Statement 1: |x-y|>4
that means the distance between x & y is 4, but we dont know their relative positions.
Hence Insufficient.

Considering both Statement 1 &2
10>6-2

Sufficient.

Answer: C
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Re: If xy > 0 and both x and y are even numbers, is x > y?  [#permalink]

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New post 10 Sep 2018, 23:42
Ya got it. thanks. Should be x-y >4 or x-y < -4 ND not sufficient.
Combining 1)ND 2) gives ultimate x-y > 4 which means x>y . So answer is C

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Re: If xy > 0 and both x and y are even numbers, is x > y?  [#permalink]

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New post 17 Sep 2018, 03:33
How does the "even" information come into play?

Once I consider 2 conditions together, it was hard to see if it satisfies x > y.
x -y > -2 AND |x- y| > 4

I had to guess it is C because I picked a pair of different sign x and y ( x=-5, y =1) and it could not satisfy both conditions -> probably they must be same sign. Kinda not sure about this one.
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Re: If xy > 0 and both x and y are even numbers, is x > y?  [#permalink]

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New post 17 Sep 2018, 04:37
In the posted solutions above where A is determined as insufficient, there was the assumption that X can equal y.
My question is , when you are told x and y are even number(s), doesn't that mean that these numbers are different numbers?
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Re: If xy > 0 and both x and y are even numbers, is x > y?  [#permalink]

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New post 17 Sep 2018, 13:03
GmatDaddy wrote:
chetan2u wrote:
If xy>0 and both x and y are even numbers, is x>y?
(I) x>y-2
(II) |x-y|>4

New question!!!..


Given:
both x and y are even numbers
xy>0: This means both x & y have same sign.

Statement 1: x>y-2

Case 1 both x,y >0
eg 8>6-2
x>y Yes

Case 2: both x,y<0
eg -8>-8-2
x>y No
Hence Insuficient

Statement 1: |x-y|>4
that means the distance between x & y is 4, but we dont know their relative positions.
Hence Insufficient.

Considering both Statement 1 &2
10>6-2

Sufficient.

Answer: C


Understood why A and B is not an answer but still not able to understand why C is an answer, GmatDaddy could you please elaborate how both equation together are sufficient.
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Re: If xy > 0 and both x and y are even numbers, is x > y?  [#permalink]

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New post 17 Sep 2018, 13:37
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parthos wrote:
Understood why A and B is not an answer but still not able to understand why C is an answer, GmatDaddy could you please elaborate how both equation together are sufficient.


We have to judge if X>Y such that both the conditions mentioned in Statement 1 and 2 are met..
If you will try to choose values accordingly, you will see that X>Y

Eg: We know from 2 that the difference between X and Y is 4.
Now select values of X and Y such that X>Y-2
ie Y<X+2

let y be 6
now x can take either 2 or 10
Lets try both the cases
6<10+2, the answer to our ultimate question X>Y is Yes
6<2+2, this equality is absurd.

similarly the other case when both X and Y are -ve
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If xy > 0 and both x and y are even numbers, is x > y?  [#permalink]

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New post 17 Sep 2018, 18:45
Given xy>0 (Same signs). we have to check x>y or x-y>0?

St 1: x>y-2, x-y>-2. Insufficient as x-y could be negative, zero or greater than 0.

St 2: x-y>4 or x-y<-4. Insufficient as Inequality 1 gives us Yes answer while Inequality 2 gives us No Answer.

Combining st 1 & 2, we have three inequalities:
x-y>-2
x-y>4
x-y<-4

From here on, if anyone can explain how both statements together are sufficient it would help me in error correction. Please explain algebraic approach.
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If xy > 0 and both x and y are even numbers, is x > y?  [#permalink]

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New post 18 Sep 2018, 04:29
chetan2u wrote:
If xy > 0 and both x and y are even numbers, is x > y?

(1) x > y - 2
(2) |x - y| > 4


New question!!!..


xy>0.....
x and y are even...

1) x>y-2
x+2>y
So y <x in all cases except when x=y..
Example say x=6, y<8 but y cannot be 7 as y is even, so y will be 6 or <6..
Insufficient as x can be y AND X can be >y

2) |x-y|>4
We cannot say if y>x or x>y but \(x\neq{y}\)
Insufficient

Combined..
We know from statement I that either x=y or X>y
But statement II tells us that \(x\neq{y}\)
So only possibility x>y
Sufficient

C
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If xy > 0 and both x and y are even numbers, is x > y?  [#permalink]

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New post 20 Oct 2018, 04:40
chetan2u wrote:
chetan2u wrote:
If xy > 0 and both x and y are even numbers, is x > y?

(1) x > y - 2
(2) |x - y| > 4


New question!!!..


xy>0.....
x and y are even...

1) x>y-2
x+2>y
So y <x in all cases except when x=y..
Example say x=6, y<8 but y cannot be 7 as y is even, so y will be 6 or <6..
Insufficient as x can be y AND X can be >y

2) |x-y|>4
We cannot say if y>x or x>y but \(x\neq{0}\)
Insufficient

Combined..
We know from statement I that either x=y or X>y
But statement II tells us that \(x\neq{0}\)
So only possibility x>y
Sufficient

C


Hi chetan2u
From St.2, how did you derive x<>0?
If x = 0 & y = -6,
Modulus (0-(-6)) = 6 ==> 6>4

x<>0 can be derived from the question stem itself, because xy>0.
So what value is statement 2 adding here?
Didn't quite get it...
Awaiting your explanation..
Thanks.
Also, is this a 700+ level question?
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Re: If xy > 0 and both x and y are even numbers, is x > y?  [#permalink]

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New post 20 Oct 2018, 06:40
ankitamundhra28 wrote:
chetan2u wrote:
chetan2u wrote:
If xy > 0 and both x and y are even numbers, is x > y?

(1) x > y - 2
(2) |x - y| > 4


New question!!!..


xy>0.....
x and y are even...

1) x>y-2
x+2>y
So y <x in all cases except when x=y..
Example say x=6, y<8 but y cannot be 7 as y is even, so y will be 6 or <6..
Insufficient as x can be y AND X can be >y

2) |x-y|>4
We cannot say if y>x or x>y but \(x\neq{0}\)
Insufficient

Combined..
We know from statement I that either x=y or X>y
But statement II tells us that \(x\neq{0}\)
So only possibility x>y
Sufficient

C


Hi chetan2u
From St.2, how did you derive x<>0?
If x = 0 & y = -6,
Modulus (0-(-6)) = 6 ==> 6>4

x<>0 can be derived from the question stem itself, because xy>0.
So what value is statement 2 adding here?
Didn't quite get it...
Awaiting your explanation..
Thanks.
Also, is this a 700+ level question?


Hi...
Although combined too I have worked on \(x\neq{y}\) but have mentioned 0 instead of y...
\(x\neq{y}\) is correct as x-y cannot be 0
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Re: If xy > 0 and both x and y are even numbers, is x > y?  [#permalink]

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New post 21 Oct 2018, 07:05
amresh09 wrote:
Given xy>0 (Same signs). we have to check x>y or x-y>0?

St 1: x>y-2, x-y>-2. Insufficient as x-y could be negative, zero or greater than 0.

St 2: x-y>4 or x-y<-4. Insufficient as Inequality 1 gives us Yes answer while Inequality 2 gives us No Answer.

Combining st 1 & 2, we have three inequalities:
x-y>-2
x-y>4
x-y<-4

From here on, if anyone can explain how both statements together are sufficient it would help me in error correction. Please explain algebraic approach.



I arrived in the same point with the 3 equations and I combined the two first equations:
x-y>-2
x-y>4

>>> 2x - 2y > 2
>>> x - y > 1
>>> x > y +1

The only way to satisfy x > y +1 is when x>y.

I think I didn't use those information: :?
x and y are even and xy > 0, so we know y and x are different to zero and or both are positive or both are negative.

Or am I missing some step and should I had used ?
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Re: If xy > 0 and both x and y are even numbers, is x > y?   [#permalink] 21 Oct 2018, 07:05
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