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If \(y-z\geq{0}\) then the question becomes is \(x=y-z\)? If \(y-z\leq{0}\) then the question becomes is \(x=-(y-z)\)?

Basically the question asks whether the distance between \(y\) and \(z\) on the number line equals to \(x\). Note that \(x\) as it's equal to the distance (or to the absolute value) cannot be negative.

(1) \(x=y-z\), not sufficient, because we don't know whether \(y-z>\geq{0}\). (2) \(y-z>0\) --> \(|y-z|=y-z\), but still not sufficient as no info about \(x\).

(1)+(2) From (1) \(x=y-z\) and from (2) \(|y-z|=y-z\) --> \(x=|y-z|=y-z\). Sufficient.

Since absolute value can't be negative, for x to be equal to |y - z|, x must be greater than or equal to 0.

(1) If y > z, then x is positive and x = |y - z| = y - z. However, if y < z, then x is negative and cannot be equal to the absolute value. Insufficient.

(2) This tells you nothing about x. Insufficient.

Together: Since y > z, we know that y - z > 0. Therefore |y - z| = y - z, and we know that x = y - z. Sufficient. (C).

So what you are saying is that the stem asks whether positive x (because its set to an absolute value) is = to y-z or z-y but #1 could allow for x to be a negative value? Thus making it insufficient?

So what you are saying is that the stem asks whether positive x (because its set to an absolute value) is = to y-z or z-y but #1 could allow for x to be a negative value? Thus making it insufficient?

Yes. For (1) x could be negative (when y - z is negative). Now if x is negative it cannot equal to absolute value.

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.
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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.

Is x = |y - z|?

(1) x = y - z (2) y > z

|a|=a when a>=0,

There are 3 variables (x,y,z) but only 2 equations are given by the 2 conditions, so there is high chance (E) will be our answer. If we combine the 2 equations, from y-z>0, x=|y-z|? --> x=y-z? this is always 'yes' so this is sufficient and the answer becomes (C).

For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% 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 E 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, C or D.
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Hello Experts, I interpreted the question as "Is X = positive " since the abs value is always positive. Considering the above understanding I concluded that Option 1 is insufficient since we don't know the signs of z or y but I chose Option 2 (B) because it's given y>z. I substituted all combinations of positive and negative values for y and z and found that we can answer the question if X is positive. (eg: y=-1,z=-2 or y=2,z=-5 or y=3,z=2 --- for all of these I got X as positive) Kindly help me understand where is my reasoning incorrect.

Hello Experts, I interpreted the question as "Is X = positive " since the abs value is always positive. Considering the above understanding I concluded that Option 1 is insufficient since we don't know the signs of z or y but I chose Option 2 (B) because it's given y>z. I substituted all combinations of positive and negative values for y and z and found that we can answer the question if X is positive. (eg: y=-1,z=-2 or y=2,z=-5 or y=3,z=2 --- for all of these I got X as positive) Kindly help me understand where is my reasoning incorrect.

Thanks & Regards, Nab

The question cannot be translated as "is x positive". The question asks whether the distance between y and z on the number line equals to x. Check complete solutions above.
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There are 3 variables in the original condition (x, y, and z). In order to match the number of variables to the number of equations, we need 3 equations. Since the condition 1) and the condition 2) each has 1 equation, we lack 1 equation. There is high chance that E is the correct answer. Using both the condition 1) and the condition 2), from y-z>0, we get |y-z|=y-z. Then, the question becomes x=|y-z|?, x=y-z?. The answer is always yes and the conditions are sufficient. Therefore, the correct answer is C.