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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
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If 2x > 3y, is x negative? [#permalink]
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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+5y20=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) \(2x+5y20=0\) > \(2x=205y\) > \(20+5y>3y\) > \(y>10\). Same as above: \(x<0\). Sufficient. Answer: D. 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+5y20=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) \(2x+5y20=0\) > \(2x=205y\) > \(20+5y>3y\) > \(y>10\). Same as above: \(x<0\). Sufficient. Answer: D. Can you please explain stmt. 2 again. Unable to understand the following stmt \(20+5y>3y\) (2) \(2x+5y20=0\) > \(2x=205y\) > given \(2x>3y\), substitute \(2x\) > \((205y)>3y\) > \(20+5y>3y\) > \(y>10\) > \(y=positive\), 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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27 Nov 2011, 19:48
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Patcheko80 wrote: I got this question in the GMATPrep. I just not sure how Statement B is also valid. Please help. Here is it.
if 2X > 3Y, is X negative? (1) Y > 0 (2) 2X + 5Y  20 = 0 The key here is knowing whether Y is positive or negative. If Y is positive, then X MUST be negative.If Y=1, then in order for 2x = 3(1) = 3, then X must be a negative number. If Y is negative, well  X could go either way. For example, if Y = 2, then x could = 2, in which case you would get 2X > 3Y 2X > 3(2) 2X > 6 x < 3 But the major point here is that if Y is positive, then X MUST be negative.We already know (1) is good. But with (2), what info do we know?Well, if you combine 2X > 3Y with 2X + 5Y > 20 then the 2X cancels the 2X, bring the 3Y to the left and negate it and combine it with 5Y. 5Y  3Y gets you to 2Y So you get 2Y > 20 Y>10 OK, so what does that tell you? Well, it tells you that Y is positive! It's essentially a subset of statement (1) where Y>0. So both (1) and (2) basically say that Y is positive. That alone is enough info to answer the original question. Therefore, when both (1) and (2) are good, we pick answer choice (D).See more GMAT Pill material for Data Sufficiency.



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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 > 02x > +ve number, hence x is negative. Sufficient (2) 2x + 5y  20 = 0The 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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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 1If y>0 & 2x + 3y<0 Then x must be Negative. Sufficient Statement 22x + 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 Answer D
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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 \(y = \frac{202x}{5}\) , replacing this value , \(2x>3*\frac{202x}{5} \to\)\(10x>606x \to 4x>60\). Again, x<0. D.
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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 = (202x)/5 gives x>15 Suff
Answer D



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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.. 2x2x>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.. 2x2x>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 = (202x)/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.. 2x2x>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, 07:42
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 = (202x)/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, 09:32
robu wrote: Bunuel wrote: 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) \(2x+5y20=0\) > \(2x=205y\) > \(20+5y>3y\) > \(y>10\). The same as above: \(x<0\). Sufficient.
Answer: D. we have 2 equation and two variables so we can calculate . so answer is D. That is a very dangerous way to look at inequalities question. Same number of equations and variables MAY OR MAY NOT give you unique values. Example, 2x+3y=10 and 4x+6y=20, you have 2 equation and 2 variables but still you can not determine UNIQUELY the values of x and y. Refer to solutions above to understand the correct logic behind D being the correct answer.



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Re: If 2x > 3y, is x negative? [#permalink]
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13 Jan 2016, 22:59
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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14 May 2016, 00:09
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Given: 2x > 3y
St1: y > 0 If y is positive > RHS is positive > For the condition, 2x > 3y, to hold true LHS must be positive > x must be negative Sufficient
St2: 2x + 5y  20 = 0 > x = (20  5y)/2 Substitute x in the given equation 2((20  5y)/2) > 3y
5y  20 > 3y
2y > 20
y > 10
Since y > 10, 2x > 3y will hold true only if x is negative. Sufficient
Answer: D



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Re: If 2x > 3y, is x negative? [#permalink]
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01 Jun 2016, 22:27
Given: 2x > 3y
St1: y > 0 > 3y is positive 2x > 3y > So 2x must be positive > This is possible only if x is negative Sufficient
St2: 2x + 5y = 20 > This means 2x + 5y > 0 ........(1) From the given statement, we have 2x  3y > 0 ........(2) Add (1) and (2) > 2y > 0 > Hence y > 0 If y > 0 then x is negative Sufficient
Answer: D




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