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

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Is xy > 0? 1. x - y > -2 2. x - 2y < -6 [#permalink] New post 29 Jan 2006, 22:31
Is xy > 0?

1. x - y > -2
2. x - 2y < -6
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Re: Is xy > 0 [#permalink] New post 29 Jan 2006, 22:57
Bhai wrote:
Is xy > 0?

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


my answer E.

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

therefore, AD ruled out

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.
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 [#permalink] New post 30 Jan 2006, 01:08
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
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 [#permalink] New post 30 Jan 2006, 01:34
allabout wrote:
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.
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 [#permalink] New post 30 Jan 2006, 06:12
its c.

Y > 4 and x > 2.
By solving the given conditions we get the above.
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 [#permalink] New post 30 Jan 2006, 08:02
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 .........
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 [#permalink] New post 30 Jan 2006, 08:04
I did this for y

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

The answer is C.
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 [#permalink] New post 30 Jan 2006, 08:27
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.
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 [#permalink] New post 30 Jan 2006, 08:51
Quote:
The most important thing is to simplify or rearrange the inequality


Point noted allabout :)

Thanx
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 [#permalink] New post 30 Jan 2006, 15:55
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

A little note about inequality
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.
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 [#permalink] New post 30 Jan 2006, 17:30
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
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 [#permalink] New post 30 Jan 2006, 18:09
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


:beer Goood solution!
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 [#permalink] New post 31 Jan 2006, 10:45
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.
  [#permalink] 31 Jan 2006, 10:45
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