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

Is x>0?

Per statement 1, z<0 and with the given equation, (x+y)/z = -2 ---> z = -(x+y)/2 <0 ---> x+y >0 . Not sufficient (x=-5,y =6 NO but if x=5 , y =6 YES).

Per statement 2, y>0 and with the given equation, y=-2z-x >0 ---> -2z>x. Dont know the sign of z. Not sufficient.

Combining both statements, you get,

z<0 and y >0 ---> y=6, z=-1 will give you x = -4, and with y=6 and z=-4 will give you x=2. Thus both statements combined are not sufficient as well.

E is thus the correct answer.
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Target question: Is x positive?

Given: (x + y)/z = -2
Since the question asks about the variable x, let's solve this equation for x.
(x + y)/z = -2
So, x + y = -2z
And, x = -2z - y

This allows us to REPHRASE the target question....
REPHRASED target question: Is -2z - y positive?

Aside: We have a free video with tips on rephrasing the target question: https://www.gmatprepnow.com/module/gmat-data-sufficiency?id=1100

Statement 1: z is negative
This statement doesn't FEEL sufficient, so I'll TEST some values.
There are several values of z that satisfy statement 1. Here are two:
Case a: z = -1 and y = 1. In this case -2z - y = -2(-1) - 1 = 1 = a POSITIVE NUMBER
Case b: z = -1 and y = 3. In this case -2z - y = -2(-1) - 3 = -1 = a NEGATIVE NUMBER
Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Aside: For more on this idea of plugging in values when a statement doesn't feel sufficient, you can read my article: https://www.gmatprepnow.com/articles/dat ... lug-values

Statement 2: y is positive
This statement doesn't FEEL sufficient either, so I'll TEST some values.
There are several values of y that satisfy statement 2. Here are two:
Case a: z = -1 and y = 1. In this case -2z - y = -2(-1) - 1 = 1 = a POSITIVE NUMBER
Case b: z = -1 and y = 3. In this case -2z - y = -2(-1) - 3 = -1 = a NEGATIVE NUMBER
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
There are several values of y and z that satisfy BOTH statements. Here are two:
Case a: z = -1 and y = 1. In this case -2z - y = -2(-1) - 1 = 1 = a POSITIVE NUMBER
Case b: z = -1 and y = 3. In this case -2z - y = -2(-1) - 3 = -1 = a NEGATIVE NUMBER
Since we cannot answer the REPHRASED target question with certainty, the combined statements are NOT SUFFICIENT

Answer = E
Cheers,
Brent

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.
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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

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