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Is x + y > xy? (1) x > 0 > y (2) |y| = x I know

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Is x + y > xy? (1) x > 0 > y (2) |y| = x I know [#permalink] New post 20 Sep 2008, 05:39
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Is x + y > xy?

(1) x > 0 > y

(2) |y| = x




I know that this can be simple when we experiment with numbers. However, I want to avoid experimenting because I can tell that there is a subtle message in the inequality Is x + y > xy. Would anyone please show how we could abstractly interpret it without doing much math? I know that it's possible, but I just couldn't figure it out in this question.

Thanks
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Re: DS: Is x + y > xy? [#permalink] New post 20 Sep 2008, 07:21
tarek99 wrote:
Is x + y > xy?

(1) x > 0 > y

(2) |y| = x




I know that this can be simple when we experiment with numbers. However, I want to avoid experimenting because I can tell that there is a subtle message in the inequality Is x + y > xy. Would anyone please show how we could abstractly interpret it without doing much math? I know that it's possible, but I just couldn't figure it out in this question.

Thanks


1) x > 0 > y means x is +ve and y is -ve so xy will be -ve
also from x > 0 > y we get x+(-y)>0
so x-y > xy ( +ve > -ve)

1 is sufficient

2) |y| = x

now x can be y or -y

if x=y then x+y is 2x or 2y and xy = x^2 or y^2
2x < x^2 or x+y < xy
if x=-y then -y+y = 0 and xy is -ve
x+y > xy

2 not sufficient

Answer is A ? What am I missing ?
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Re: DS: Is x + y > xy? [#permalink] New post 20 Sep 2008, 23:13
tarek99 wrote:
Is x + y > xy?

(1) x > 0 > y

(2) |y| = x




I know that this can be simple when we experiment with numbers. However, I want to avoid experimenting because I can tell that there is a subtle message in the inequality Is x + y > xy. Would anyone please show how we could abstractly interpret it without doing much math? I know that it's possible, but I just couldn't figure it out in this question.

Thanks


1)x>0>y => x+y>xy say x=1,y=-2 => x+y=-1 > -2
x+y<xy say x=0.5 y=-2 => x+y=-1.5 < -1
INSUFFI
2) |y|=x => x is +ve and and x=value of y where y is +ve or -ve
say y=2 x+y =xy y=x=3 => x+y<xy => INSUFFI
1) and 2) now y<0 and x=|y| => y=-1 x=1 x+y>-1
y=-0.5 x=0.5 x+y>xy always since x+y=0 xy <0
IMO C
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Re: DS: Is x + y > xy? [#permalink] New post 22 Sep 2008, 01:04
The two statements are essentially saying that x and y are equidistant from 0 on the number line and y is to the left and x is to the right.

Hence, x + y will always be zero and xy will be -ve.

However, if x and y are not equidistant then depending upon the absolute value of x or y (whether less than 1 or more than 1), the values of x+y and xy will be different (less or more).

Hence, C is the answer.

tarek99 wrote:
Is x + y > xy?

(1) x > 0 > y

(2) |y| = x




I know that this can be simple when we experiment with numbers. However, I want to avoid experimenting because I can tell that there is a subtle message in the inequality Is x + y > xy. Would anyone please show how we could abstractly interpret it without doing much math? I know that it's possible, but I just couldn't figure it out in this question.
Thanks
SVP
SVP
avatar
Joined: 21 Jul 2006
Posts: 1551
Followers: 8

Kudos [?]: 236 [0], given: 1

Re: DS: Is x + y > xy? [#permalink] New post 22 Sep 2008, 03:03
scthakur wrote:
The two statements are essentially saying that x and y are equidistant from 0 on the number line and y is to the left and x is to the right.

Hence, x + y will always be zero and xy will be -ve.

However, if x and y are not equidistant then depending upon the absolute value of x or y (whether less than 1 or more than 1), the values of x+y and xy will be different (less or more).

Hence, C is the answer.

tarek99 wrote:
Is x + y > xy?

(1) x > 0 > y

(2) |y| = x





I know that this can be simple when we experiment with numbers. However, I want to avoid experimenting because I can tell that there is a subtle message in the inequality Is x + y > xy. Would anyone please show how we could abstractly interpret it without doing much math? I know that it's possible, but I just couldn't figure it out in this question.
Thanks



exactly! statement 2 tells us that x can ONLY be a zero and positive, while y can be either positive, a zero, or even negative. When you look at statement 1, when combining the 2 statements, it says that x and y aren't zero, while statement 2 says that both x and y have the same absolute value, which helps us to approach this inequality.
Re: DS: Is x + y > xy?   [#permalink] 22 Sep 2008, 03:03
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