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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]  08 Dec 2009, 08:07
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Is xy>0?

(1) x-y>-2
(2) x-2y<-6
[Reveal] Spoiler: OA

Last edited by Bunuel on 26 May 2012, 02:08, edited 1 time in total.
Edited the question and added the OA
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Re: Gprep DS: Is xy>o? [#permalink]  08 Dec 2009, 08:43
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4test1 wrote:
Can't seem to find this in the forum. From the Gprep test:

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

Can't seem to get my head around this. Appreciate any help.

answer c
Statement 1:
x = 5 and y = 4 answer is yes
x = -1 and y = 0 answer is no

Statement 2:
x = -7 and y = 0 answer is no
x = -10 and y = -1 answer is yes

Combined:
x - y > -2
x - 2y < -6

or
-x + y < 2
x - 2y < - 6
-y < -4
y > 4 so you know y is positive

What about x..plug a value of y > 4 in the equation and you'll get x > 0
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Re: Gprep DS: Is xy>o? [#permalink]  18 Dec 2009, 13:14
burnttwinky wrote:
How did you get X > 0? I got X< 2. Can you please show me the math when you plug y > 4 back in to get x?

I just meant that x is positive if you plug in y>4 in equation 1 and 2
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Re: Gprep DS: Is xy>o? [#permalink]  25 Dec 2009, 04:03
lagomez wrote:
4test1 wrote:
Can't seem to find this in the forum. From the Gprep test:

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

Can't seem to get my head around this. Appreciate any help.

answer c
Statement 1:
x = 5 and y = 4 answer is yes
x = -1 and y = 0 answer is no

Statement 2:
x = -7 and y = 0 answer is no
x = -10 and y = -1 answer is yes

Combined:
x - y > -2
x - 2y < -6

or
-x + y < 2
x - 2y < - 6
-y < -4
y > 4 so you know y is positive

What about x..plug a value of y > 4 in the equation and you'll get x > 0

hi lagomez,

nice method, but i m afraid there is some mistake in ur explanation.

if y>4. say y=5
now plugging this value in eq2, i.e. option 2

x-2y < -6 ===> x-10<-6 ===> x<4 ===> so x can also be -ve... in this case xy<0

plugging the value of y=5 in eq1, i.e. option 1

x-y > -2 ===> x>y+2 ===> x>7 ===> x=+ve.... in this case xy>0

two answers. hence data insuff.
Answer:E

here is my method..
as u know, 1 & 2 are individually insuff..
taking 1&2 together

x-y>-2 ===> x>y+2
x-2y<-6 ===> x<2y-6

so combining both the above equations (y+2)< x < (2y-6)

by plugging values, if y= 2 ==> 4<x<-2 ==> x<-2 & x>4
two cases arise. so, data insuff.
Answer: E
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Re: Gprep DS: Is xy>o? [#permalink]  25 Dec 2009, 05:22
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logan wrote:
hi lagomez,

nice method, but i m afraid there is some mistake in ur explanation.

if y>4. say y=5
now plugging this value in eq2, i.e. option 2

x-2y < -6 ===> x-10<-6 ===> x<4 ===> so x can also be -ve... in this case xy<0

plugging the value of y=5 in eq1, i.e. option 1

x-y > -2 ===> x>y+2 ===> x>7 ===> x=+ve.... in this case xy>0

two answers. hence data insuff.
Answer:E

here is my method..
as u know, 1 & 2 are individually insuff..
taking 1&2 together

x-y>-2 ===> x>y+2
x-2y<-6 ===> x<2y-6

so combining both the above equations (y+2)< x < (2y-6)

by plugging values, if y= 2 ==> 4<x<-2 ==> x<-2 & x>4
two cases arise. so, data insuff.
Answer: E

Not so. Lagomez's answer is correct.

Is xy>0?

(1) x-y > -2
(2) x-2y <-6

Note that question basically asks whether $$x$$ and $$y$$ have the same sign.

Statements alone are not sufficient as was shown above.

Now, remember we can subtract inequalities with the signs in opposite direction.

$$x-y-(x-2y)>-2-(-6)$$ --> $$y>4$$. As $$y>4$$ and (from 1) $$x>y-2$$, hence $$x>2$$ (we can add inequalities when their signs are in the same direction, so $$y+x>4+y-2$$ --> $$x>2$$) --> both $$x$$ and $$y$$ are positive. Sufficient.

Answer: C.

When you got that for y=5 --> x<4, it's correct, but it doesn't mean that x can be negative, as this is not ALL information you have. If you plug the same value y=5 in (1) you'll get --> x>3, which is also correct.

Basically when y=5 --> 3<x<4.

Hope it's clear.
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Re: Gprep DS: Is xy>o? [#permalink]  25 Dec 2009, 05:51
Expert's post
chetan2u wrote:
i think ans should be E...
agreed y>4..
substitute in eq 1...x>2...
substitute in eq 2... x<2y-6...x<8-6...x<2..... combining the two , can we get a value of x, which is both < 2 and>2,, i dont think

When you substitute in equation 2, you take the value of y=4 (which can not be as y>4).

C is correct.
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Re: Gprep DS: Is xy>o? [#permalink]  25 Dec 2009, 05:51
This question is best seen graphically. It is a very quick method.

See the graphs below. They only meeting in 1st Quadrant. So both equations together tell us that x>0 y>0.

Attachment:

x-ygr-2.GIF [ 3.63 KiB | Viewed 3886 times ]

Attachment:

eqn2.GIF [ 2.88 KiB | Viewed 3882 times ]

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Re: Gprep DS: Is xy>o? [#permalink]  25 Dec 2009, 10:34
Bunuel wrote:
Not so. Lagomez's answer is correct.

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

Statements alone are not sufficient as was shown above.

Now, remember we can subtract inequalities with the signs in opposite direction.

x-y-(x-2y)>-2-(-6) --> y>4. As y>4 and (from 1) x>y-2, hence x>2. Sufficient.

Answer: C.

When you got that for y=5 --> x<4, it's correct, but it doesn't mean that x can be negative, as this is not ALL information you have. If you plug the same value y=5 in (1) you'll get --> x>3, which is also correct.

Basically when y=5 --> 3<x<4.

Hope it's clear.

oh yeah... too bad that i forgot to plug the value of y=5 in both the equations...
thanx a ton Bunuel...
kudos..
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Re: Gprep DS: Is xy>o? [#permalink]  25 Dec 2009, 15:25
I am still having some confusion about this problem. Let me show you my steps. I solved for X first. First I aligned the inequalities:

y < x + 2
1/2x + 3 < y

so by combining the equations, we get:

1/2x + 3+ y < x+y+2

the y's cancel on both sides, and we have 1/2x > 1

so x>2

My question is, how do you plug in greater than 2 into an inequality? y < (Greater than 2) + 2

we get y < Greater than 4, so doesn't that mean y can still be negative while X is greater than 2 (always positive)?
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Re: Gprep DS: Is xy>o? [#permalink]  31 Dec 2009, 21:59
Draw the X,Y graph like what msunny did.

Hence, the answer is D
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Re: Gprep DS: Is xy>o? [#permalink]  01 Jan 2010, 11:28
Bunuel wrote:
chetan2u wrote:
i think ans should be E...
agreed y>4..
substitute in eq 1...x>2...
substitute in eq 2... x<2y-6...x<8-6...x<2..... combining the two , can we get a value of x, which is both < 2 and>2,, i dont think

When you substitute in equation 2, you take the value of y=4 (which can not be as y>4).

C is correct.

If I substituted x=2 into each of the equations to solve for Y, i got two different answers. Y>4 and Y<4, depending on which equation you substitute it into.
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Re: Gprep DS: Is xy>o? [#permalink]  01 Jan 2010, 16:04
lagomez wrote:
4test1 wrote:
Can't seem to find this in the forum. From the Gprep test:

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

Can't seem to get my head around this. Appreciate any help.

answer c
Statement 1:
x = 5 and y = 4 answer is yes
x = -1 and y = 0 answer is no

Statement 2:
x = -7 and y = 0 answer is no
x = -10 and y = -1 answer is yes

Combined:
x - y > -2
x - 2y < -6

or
-x + y < 2
x - 2y < - 6
-y < -4
y > 4 so you know y is positive

What about x..plug a value of y > 4 in the equation and you'll get x > 0

Correct answer is C...

As pointed by Bunuel, when u plug in y > 4 in both the above equations, you get x range as 3<x<4.... Therefore xy > 0!

You shouldn't consider only x - 2y < -6 for determining the value of x. Both the equations should be considered!

Cheers!
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Re: Gprep DS: Is xy>o? [#permalink]  01 Jan 2010, 16:08
Expert's post
burnttwinky wrote:
Bunuel wrote:
chetan2u wrote:
i think ans should be E...
agreed y>4..
substitute in eq 1...x>2...
substitute in eq 2... x<2y-6...x<8-6...x<2..... combining the two , can we get a value of x, which is both < 2 and>2,, i dont think

When you substitute in equation 2, you take the value of y=4 (which can not be as y>4).

C is correct.

If I substituted x=2 into each of the equations to solve for Y, i got two different answers. Y>4 and Y<4, depending on which equation you substitute it into.

I don't quite understand what's your point. Answer to the original question is C.

xy>0 means that the point (x,y) is either in I quadrant or in III.

(1) x-y > -2 the points satisfying this inequality can be in ALL four quadrants, refer to the first graph msunny provided (blue area). Hence insufficient.

(2) x-2y <-6 the points satisfying this inequality can be in I, II or in III quadrants, refer to the second graph msunny provided (blue area). Hence insufficient.

(1)+(2) Points satisfying BOTH inequalities are only in I quadrant: intersection of blue areas. Hence sufficient.

Hope it's clear.
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Re: Gprep DS: Is xy>o? [#permalink]  01 Jan 2010, 16:49
I apologize if my point is unclear, but my question is that, after I calculated X>2 (I solved for X first), I plugged in X=2 into the first equation, and I got Y<4. What I am confused about is that if Y<4, that means Y can be negative; however, x>2, so if Y is negative and X is always positive, then you would be in quadrant IV.

If plugged in X=2 into the second equation, then you get y>4, which is sufficient since X>2 and Y>4, so you will always be in quadrant I.
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Re: Gprep DS: Is xy>o? [#permalink]  01 Jan 2010, 17:01
burnttwinky wrote:
I apologize if my point is unclear, but my question is that, after I calculated X>2 (I solved for X first), I plugged in X=2 into the first equation, and I got Y<4. What I am confused about is that if Y<4, that means Y can be negative; however, x>2, so if Y is negative and X is always positive, then you would be in quadrant IV.

If plugged in X=2 into the second equation, then you get y>4, which is sufficient since X>2 and Y>4, so you will always be in quadrant I.

Your X value is > 2, do not substitute 2 as X in the equation. Substituting X = 3 would be a right approach. On doing that for first eq, you get:
3-y>-2 i.e. y < 5 --- [1]
Substituting X = 3 in equation 2, you get:

3-2y<-6 i.e. y > 4.5....

Therefore if X = 3, then y range is from 4.5 to 5 which means both X and Y are positive. Hence XY > 0.

Hope this is clear..

Thanks,
JT
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Re: Gprep DS: Is xy>o? [#permalink]  01 Jan 2010, 17:16
Expert's post
burnttwinky wrote:
I apologize if my point is unclear, but my question is that, after I calculated X>2 (I solved for X first), I plugged in X=2 into the first equation, and I got Y<4. What I am confused about is that if Y<4, that means Y can be negative; however, x>2, so if Y is negative and X is always positive, then you would be in quadrant IV.

If plugged in X=2 into the second equation, then you get y>4, which is sufficient since X>2 and Y>4, so you will always be in quadrant I.

First of all when you solved system of inequalities and got x>2 and you want to continue by number plugging you should plug correct numbers. You got x>2, hence you should plug x more than 2. Any value more than 2 for x will give you the RANGE for y which is positive.

x=3:
x-y > -2 --> 3-y>-2 --> y<5
and
x-2y <-6 --> 3-2y<-6 --> y>4.5

So, for x=3, y is in the range {4.5, 5}.

You'll get only positive range for y, when you test ANY value of x>2, as the solution of this system of inequalities is x>2 and y>4 (refer to the solution in my first post in this topic). And vise-versa: if you try values of y>4 you'll get only positive range for x.

Hope it's clear.
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Re: Gprep DS: Is xy>o? [#permalink]  01 Jan 2010, 17:25
That makes much more sense. Thanks for the explanation.
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Re: Gprep DS: Is xy>o? [#permalink]  20 Dec 2010, 01:54
you are wrong.

we CAN NOT ADD INEQUALITY, though we can add equality.

a>b and c>d

we can not have a+c>b+d . because there is some redundant

we use this method

y-2<x<2y-6
y-2<2y-6
4<y
that means x>2, C is correct.
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Re: Gprep DS: Is xy>o? [#permalink]  20 Dec 2010, 02:01
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thangvietnam wrote:
you are wrong.

we CAN NOT ADD INEQUALITY, though we can add equality.

a>b and c>d

we can not have a+c>b+d . because there is some redundant

we use this method

y-2<x<2y-6
y-2<2y-6
4<y
that means x>2, C is correct.

The red part is not correct.

You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

Hope it's clear.
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xy>0 ? [#permalink]  25 Jan 2011, 13:28
Is xy>0?

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

Help!!?? )
xy>0 ?   [#permalink] 25 Jan 2011, 13:28

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