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# Is xy > 0? 1. x - y > -2 2. x - 2y < -6

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SVP
Joined: 16 Oct 2003
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Is xy > 0? 1. x - y > -2 2. x - 2y < -6 [#permalink]  29 Jan 2006, 22:31
Is xy > 0?

1. x - y > -2
2. x - 2y < -6
Director
Joined: 26 Sep 2005
Posts: 578
Location: Munich,Germany
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Kudos [?]: 10 [0], given: 0

Re: Is xy > 0 [#permalink]  29 Jan 2006, 22:57
Bhai wrote:
Is xy > 0?

1. x - y > -2
2. x - 2y < -6

stmt 1,

x= -6
y= -5
-6-(-5)= -1 which is greater than -2 xy>0

x=1
y=0
1-0=1 which is greater than -2 again but xy=0

stmt2,
x=1
y=4
1-8=-7 is lesser than -6 xy>0

x=-2
y=5
-2-10=-12 is lesse than -6 xy<0

therefore B ruled out

combning
x=10
y=11
satisfies both conditions and xy<0

x=19
y=18
satisfies both conditions but xy>0

Hence, E.
Director
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Hmmm...

xy>0 ?

1) x-y>-2 or x>y-2

2) x-2y<-6 or x<2y-6

Either statement alone is insuff.

1+2)

y-2 < x < 2y-6

This is only true if y>4

and then x is also positive

Maybe I've made a logical mistake here, not sure
VP
Joined: 21 Sep 2003
Posts: 1066
Location: USA
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Hmmm...

xy>0 ?

1) x-y>-2 or x>y-2

2) x-2y<-6 or x<2y-6

Either statement alone is insuff.

1+2)

y-2 < x < 2y-6

This is only true if y>4

and then x is also positive

Maybe I've made a logical mistake here, not sure

Agree. I also got to the same conclusion but was not sure. I believe it is C also.
_________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

- Bernard Edmonds

Intern
Joined: 27 Jun 2005
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Kudos [?]: 0 [0], given: 0

its c.

Y > 4 and x > 2.
By solving the given conditions we get the above.
Manager
Joined: 30 Jan 2005
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guys wat is the best way to solve such diffcult problems, other than picking numbers??

the confusion starts when u have to decide whether it is C or E .........
_________________

"Nothing is Impossible"

SVP
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I did this for y

(x + 6)/2 < y < x + 2 and plugged in few negative and positive values.

Director
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Kudos [?]: 11 [0], given: 0

jinesh wrote:
guys wat is the best way to solve such diffcult problems, other than picking numbers??

the confusion starts when u have to decide whether it is C or E .........

Picking numbers is often necessary in DS inequality problems.

The most important thing is to simplify or rearrange the inequality, in order to ease picking numbers. Look at this case:

If you have

y-2 < x < 2y-6

you instantly see that y>4, since otherwise the inequality won't be fullfilled. If you don't arrange it like this, its hard and time consuming to find the actual set; And the problem beteen E and C may become more apparent.
Manager
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Quote:
The most important thing is to simplify or rearrange the inequality

Thanx
_________________

"Nothing is Impossible"

Senior Manager
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This question will be easier if you know how to sketch linear graph.

Basic rule for plotting linear graph

1) know basic form of linear graph y = ax + b
2) give x = 0, y = b
3) and give y = 0, x = -b/a
4) When you have (0,b) and (-b/a,0), you can sketch graph quickly

1) y > ax + b means all area above ax + b graph
2) y < ax + b means all area below ax + b graph

Ok! now go to our question.

xy > ?

1) x - y > -2 or y < x + 2
sketch graph and see the area below this graph, the area will include quadrant 4 which x is always positive and y is always negative (eg. (5, -4) (2, -3))

xy < 0 in quadrant 4; therefore, 1) is insuff.

2) x - 2y < -6 or y > (x/2) + 3
Sketch graph and notice the area above this graph, the area will include quadrant 2 which x is always negative and y is always positive (eg. (-2, 3) (-1, 8)

xy < 0 in quadrant 2; therefore, 2) is insuff.

3) consider 1) and 2) you will get y < x + 2 and y > (x/2) + 3
these two graph will cross each other at point (2, 4) in quadrant 1
the area of graph 1) and 2) will intersect. The intersect area will appear only in quadrant 1 which x and y are always positive

xy > 0 always; therefore, answer is C)

P.S. Sorry if you have a hard time understand my explanation, English is my second language.
SVP
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Should be "C"

this can be solved using simultaneous equations, since we have two equations,

1. x - y > -2
2. x - 2y < -6

we get X > 2 & Y > 4
VP
Joined: 21 Sep 2003
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vivek123 wrote:
Should be "C"

this can be solved using simultaneous equations, since we have two equations,

1. x - y > -2
2. x - 2y < -6

we get X > 2 & Y > 4

Goood solution!
_________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

- Bernard Edmonds

VP
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Posts: 1023
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Kudos [?]: 27 [0], given: 0

C....here are the steps.
I
x-y > -2 => y-x < 2.....INSUFF
II
x-2y < -6. ... INSUFF

I + II gives,

Changing the inequalities to point in the same direction with positive numbers on the right, we get
2 > y-x ....(a)
and
2y -x > 6....(b)

(a) + (b) gives, y>4.

Since y>4 => y-2>2 and since y-2 < x, we get 2 < y-2 < x => x >2.

Since y>4 and x >2 , xy > 0.....SUFF.
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