If (x + y)/z = -2, is x positive? (1) z is negative. (2) y is positive

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 (x+y)/z = -2, is x positive?

(1) z is negative.
(2) y is positive.

In the original condition, there are 3 variables(x,y,z) and 1 equation(x+y)/z = -2, which should match with the number of equations. So you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. When 1) & 2), it becomes z<0, y<0. In (x+y)/z = -2<0, x+y>0 when z<0. Even if y>0, it cannot be x>0, which is not sufficient. Therefore, the answer is E.


 For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

1. IF Z is negative, the sum of x+y would have to positive. However, we cannot definitively answer if x is positive because a variety of pairings could serve to ensure that x+y is positive, including negative numbers.
(insufficient)


2.If y is positive, either x,z or both could be negative. Insufficient.


Combined:We do not the extent to which y is positive, so x could be either positive or negative.

Please critique if you see fit!

\(x + y + 2z = 0\,\,\,,\,\,\,z \ne 0\,\,\,\)

\(x\,\,\mathop > \limits^? \,\,0\)


\(\left( {1 + 2} \right)\,\,y > 0\,\,,\,\,z < 0\,\,\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {x,y,z} \right) = \left( {1,1, - 1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {x,y,z} \right) = \left( { - 1,3, - 1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {\rm{E}} \right)\)


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.

We can solve this easily using logic.
1. As z is negative, (X+y) must be positive. But we cannot say whether x is positive or not. If y is positive and larger than x then x can be negative. But x can be positive also. So, insufficient.
2. If y is positive, x can be positive or negative. Insufficient.
Taking both: as already said in statement 1, X can be negative if y is positive and larger than absolute value of x. So, the same result will come if x positive or negative. So Insufficient.

But if y is negative, then x must be positive. In that case the answer will be c.

So the answer is E

Posted from my mobile device

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.