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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).

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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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. _________________

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

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).

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).

(2) \(2x+5y-20=0\) --> \(2x=20-5y\) --> \(-20+5y>3y\) --> \(y>10\). 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) \(2x+5y-20=0\) --> \(2x=20-5y\) --> 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.

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

Answer D _________________

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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

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

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 _________________

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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