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# If -2x > 3y, is x negative?

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If -2x > 3y, is x negative? [#permalink]

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23 May 2010, 00:35
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If -2x > 3y, is x negative?

(1) y > 0
(2) 2x + 5y - 20 = 0
[Reveal] Spoiler: OA

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If -2x > 3y, is x negative? [#permalink]

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23 May 2010, 02:49
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If -2x>3y , is x negative

Given: $$-2x>3y$$.
Question: is $$x<0$$? (Note here that if $$y$$ is any positive number then we would have $$-2x>positive$$, and in order that to be true $$x$$ must be some negative number).

(1) $$y>0$$ --> $$-2x>3y>0$$ --> $$x<0$$. Sufficient.

(2) $$2x+5y-20=0$$ --> $$2x=20-5y$$ --> $$-20+5y>3y$$ --> $$y>10$$. The same as above: $$x<0$$. Sufficient.

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Re: If -2x > 3y, is x negative? (1) y > 0 (2) 2x + 5y - 20 = 0 [#permalink]

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29 Jun 2013, 07:45
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fozzzy wrote:

In statement 2 we can write the equation 2x+3y+2y = 20 we know 2x+3y is positive and we get y = 10 hence same as statement 1 is this approach correct?

If -2x > 3y, is x negative?

(1) y > 0
-2x > +ve number, hence x is negative.
Sufficient

(2) 2x + 5y - 20 = 0
The area defined by -2x > 3y is the area under the red line. If we know that $$2x + 5y - 20 = 0$$ (blue line) (given the initial condition) we can say that x is negative because they intersect when x is negative. (refer to the image)
Sufficient

Your approach is correct. We know that 2x+3y is negative (typo I think), so $$2x + 3y +2y= 20$$ can be seen as $$-ve +2y=20$$ so y is positive for sure as $$2y=20+(+ve)$$
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Re: If -2x > 3y, is x negative? [#permalink]

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13 Jan 2016, 22:59
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If -2x > 3y, is x negative?

(1) y > 0
(2) 2x + 5y - 20 = 0

In the original condition, there are 2 variables(x,y) and 1 equation(-2x>3y), which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer. For 1), when y>0, it becomes 3y>2y. That is, -2x>3y>2y, -2x>2y. -x>y --> -x>y>0, -x>0 therefore x<0, which is yes and sufficient.
For 2), substitute y=(-2/5)x+4 to the equation. It becomes -2x>3(-2/5)x+4 and multiply 5 to both equations. Divide -10x>-6x+20, -4x>20 with -4 and x<-5<0 is also yes and sufficient. Therefore, the answer is D.

-> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: If -2x > 3y, is x negative? [#permalink]

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21 Jun 2016, 07:42
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Shrivathsan wrote:
If -2x > 3y, is x negative?
(1) y > 0
(2) 2x + 5y - 20 = 0

Hi,
-2x > 3y...
(a)If y<0, x can be both +ive and -ive..
(b)if y>0, x will have to be +ive as 3y is positive and -2x , to be positive, has to have x as -ive..

now lets see the choices..

(1) y > 0
If y>0, x is -ive as proved in (b) above... suff

(2) 2x + 5y - 20 = 0..
this can be written as 2x+3y + 2y -20=0..
now 2x+3y<0, so 2y>20... or y is +ive and therefore x is -ive.... suff

ans D
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Re: is X negative [#permalink]

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23 May 2010, 09:54
Bunuel wrote:
dimitri92 wrote:
If -2x>3y , is X negative

1) y>0
2) 2x+5y-20=0

Given: $$-2x>3y$$. Q: is $$x<0$$? (Note here that if $$y$$ is any positive number than we would have $$-2x>positive$$, and in order that to be true $$x$$ must be some negative number).

(1) $$y>0$$ --> $$-2x>3y>0$$ --> $$x<0$$. Sufficient.

(2) $$2x+5y-20=0$$ --> $$2x=20-5y$$ --> $$-20+5y>3y$$ --> $$y>10$$. Same as above: $$x<0$$. Sufficient.

Can you please explain stmt. 2 again.
Unable to understand the following stmt---

$$-20+5y>3y$$
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Re: is X negative [#permalink]

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23 May 2010, 10:08
onedayill wrote:
Bunuel wrote:
dimitri92 wrote:
If -2x>3y , is X negative

1) y>0
2) 2x+5y-20=0

Given: $$-2x>3y$$. Q: is $$x<0$$? (Note here that if $$y$$ is any positive number then we would have $$-2x>positive$$, and in order that to be true $$x$$ must be some negative number).

(1) $$y>0$$ --> $$-2x>3y>0$$ --> $$x<0$$. Sufficient.

(2) $$2x+5y-20=0$$ --> $$2x=20-5y$$ --> $$-20+5y>3y$$ --> $$y>10$$. Same as above: $$x<0$$. Sufficient.

Can you please explain stmt. 2 again.
Unable to understand the following stmt---

$$-20+5y>3y$$

(2) $$2x+5y-20=0$$ --> $$2x=20-5y$$ --> given $$-2x>3y$$, substitute $$2x$$ --> $$-(20-5y)>3y$$ --> $$-20+5y>3y$$ --> $$y>10$$ --> $$y=positive$$, as discussed above if $$y$$ is any positive number then $$x$$ must be some negative number: $$x<0$$. Sufficient.

Hope it's clear.
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Re: If -2x > 3y, is x negative [#permalink]

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25 Aug 2013, 00:04
SUNGMAT710 wrote:
If -2x > 3y, is x negative?
(1) y > 0
(2) 2x + 5y - 20 = 0

-2x > 3y
2x + 3y<0 -----(1)

Statement 1
If y>0
& 2x + 3y<0

Then x must be Negative.
Sufficient

Statement 2
2x + 5y - 20 = 0
2x + 5y = 20
(2x + 3y) + 2y=20
We can write
2y + some negative no = 20
2y = 20 + some Positiveno
y = 10 + some Positiveno/2
This mean that y>10

2x + 3y<0
2x< -3y
x < -1.5 (Positive no) because y is positive

Then x must be Negative.
Sufficient

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Re: If -2x > 3y, is x negative [#permalink]

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25 Aug 2013, 00:06
SUNGMAT710 wrote:
If -2x > 3y, is x negative?
(1) y > 0
(2) 2x + 5y - 20 = 0

From F.S 1, we know that -2x>0. Thus, x<0. Sufficient.

From F.S 2, we know that $$y = \frac{20-2x}{5}$$ , replacing this value , $$-2x>3*\frac{20-2x}{5} \to$$$$-10x>60-6x \to -4x>60$$. Again, x<0.

D.
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Re: If -2x > 3y, is x negative? [#permalink]

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Re: If -2x > 3y, is x negative? [#permalink]

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10 Jan 2016, 05:17
We are given
-2x > 3y
need to find if x is negative?

(1) y > 0
x has to be negative to ensure that -2x > 3y
Suff

(2) 2x + 5y - 20 = 0
substituting y = (20-2x)/5
gives
x>15
Suff

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Re: If -2x > 3y, is x negative? [#permalink]

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10 Jan 2016, 05:18
NoHalfMeasures wrote:
If -2x > 3y, is x negative?
(1) y > 0
(2) 2x + 5y - 20 = 0

Hi,
-2x > 3y...
(a)If y<0, x can be both +ive and -ive..
(b)if y>0, x will have to be +ive as 3y is positive and -2x , to be positive, has to have x as -ive..

now lets see the choices..
(1) y > 0
If y>0, x is -ive as proved in (b) above... suff

(2) 2x + 5y - 20 = 0..
this can be written as 2x+3y + 2y -20=0..
now 2x+3y<0, so 2y>20... or y is +ive and therefore x is -ive.... suff

ans D
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Re: If -2x > 3y, is x negative? [#permalink]

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10 Jan 2016, 05:58
chetan2u wrote:
NoHalfMeasures wrote:
If -2x > 3y, is x negative?
(1) y > 0
(2) 2x + 5y - 20 = 0

Hi,
-2x > 3y...
(a)If y<0, x can be both +ive and -ive..
(b)if y>0, x will have to be +ive as 3y is positive and -2x , to be positive, has to have x as -ive..

now lets see the choices..
(1) y > 0
If y>0, x is -ive as proved in (b) above... suff

(2) 2x + 5y - 20 = 0..
this can be written as 2x+3y + 2y -20=0..
now 2x+3y<0, so 2y>20... or y is +ive and therefore x is -ive.... suff

ans D

how can you say
2x+3y<0?
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Re: If -2x > 3y, is x negative? [#permalink]

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10 Jan 2016, 06:20
paidlukkha wrote:
chetan2u wrote:
NoHalfMeasures wrote:
If -2x > 3y, is x negative?
(1) y > 0
(2) 2x + 5y - 20 = 0

Hi,
-2x > 3y...
(a)If y<0, x can be both +ive and -ive..
(b)if y>0, x will have to be +ive as 3y is positive and -2x , to be positive, has to have x as -ive..

now lets see the choices..
(1) y > 0
If y>0, x is -ive as proved in (b) above... suff

(2) 2x + 5y - 20 = 0..
this can be written as 2x+3y + 2y -20=0..
now 2x+3y<0, so 2y>20... or y is +ive and therefore x is -ive.... suff

ans D

how can you say
2x+3y<0?

Hi,
2x + 3y <0 comes from -2x>3y..
-2x>3y..
add 2x to both sides..
2x-2x>3y+2x..
0>2x+3y...
hope it helps
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If -2x > 3y, is x negative? [#permalink]

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10 Jan 2016, 06:29
Hi,
2x + 3y <0 comes from -2x>3y..
-2x>3y..
add 2x to both sides..
2x-2x>3y+2x..
0>2x+3y...
hope it helps[/quote]

Aye, it does!
Thanks

Also, if I understand, <> sign changes in multiplication only!

Btw,
St2 gives me x as +ve
What am I doing wrong?
(2) 2x + 5y - 20 = 0
substituting y = (20-2x)/5
gives
x>15
Suff
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Re: If -2x > 3y, is x negative? [#permalink]

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10 Jan 2016, 06:32
paidlukkha wrote:
Hi,
2x + 3y <0 comes from -2x>3y..
-2x>3y..
add 2x to both sides..
2x-2x>3y+2x..
0>2x+3y...
hope it helps

Aye, it does!
Thanks

Also, if I understand, <> sign changes in multiplication only![/quote]

hi,
yes you are right , whenever you multiply two sides on either side of equality with a -ive sign or -ive quantity, you are required to change the greater/lesser than sign..
-2x>3y..
2x<-3y..
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If -2x > 3y, is x negative? [#permalink]

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13 Jan 2016, 07:02
If -2x > 3y, is x negative?
(1) y > 0
(2) 2x + 5y - 20 = 0
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If -2x > 3y, is x negative? [#permalink]

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13 Jan 2016, 07:42
Sash143 wrote:
If -2x > 3y, is x negative?
(1) y > 0
(2) 2x + 5y - 20 = 0

Given that -2x > 3 y---> x<-1.5 y and the question asks whether x<0.

Per statement 1, y>0 ---> from this statement and the fact that x<-1.5y, clearly we can see that x must be <0. Sufficient.

Per statement 2, 2x + 5y - 20 = 0 ---> y = (20-2x)/5 and from the given fact, y<-2x/3 ---> x < -15. Again sufficient to answer the question asked.

Hence both statements are sufficient to answer the question asked. D is thus the correct answer.

Hope this helps.
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If -2x > 3y, is x negative? [#permalink]

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13 Jan 2016, 07:59
Sash143 wrote:
If -2x > 3y, is x negative?
(1) y > 0
(2) 2x + 5y - 20 = 0

It is clear that if x negative, -2x > 0, so we will wish to see whether 3y >0, if so -2x > 3y > 0.

(1) y >0 => clearly, -2x > 3y > 0 => x negative

(2) 2x + 5y - 20 = 0. => 2x + 5y =20
Because -2x > 3y => 2x < -3y
=> 2x + 5y < -3y + 5y =2y
or 20 < 2y => 10<y => -2x > 3y > 30 => x negative

So, answer is D
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Re: If -2x > 3y, is x negative? [#permalink]

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13 Jan 2016, 08:32
Sash143 wrote:
If -2x > 3y, is x negative?
(1) y > 0
(2) 2x + 5y - 20 = 0

Please search before posting. Thank you.
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Re: If -2x > 3y, is x negative?   [#permalink] 13 Jan 2016, 08:32

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