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Re: If -2x > 3y, is x negative? (1) y > 0 (2) 2x + 5y - 20 = 0 [#permalink]
29 Jun 2013, 06:45
This post received KUDOS
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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It is beyond a doubt that all our knowledge that begins with experience.
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.
Statement 1: y > 0 In other words, y is POSITIVE This means that 3y is POSITIVE It is given that -2x > 3y Since 3y is POSITIVE, we can write: -2x > SOME POSITIVE # If -2x is greater than SOME POSITIVE #, we know that -2x is POSITIVE If -2x is POSITIVE, then x must be negative Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: 2x + 5y - 20 = 0 IMPORTANT: It is given that -2x > 3y So, let's take 2x + 5y - 20 = 0 and rewrite it as 5y - 20 = -2x [I have isolated -2x, just like we have in the GIVEN information] Now, we'll take -2x > 3y, and replace -2x with 5y - 20 to get: 5y - 20 > 3y Subtract 3y from both sides: 2y - 20 > 0 Add 20 to both sides: 2y > 20 Solve: y > 10 This means that y is POSITIVE We already saw in statement 1, that when y is positive, x must be negative Since we can answer the target question with certainty, statement 2 is 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---
(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.
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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..