Data Sufficiency

Is |x| = y - z?

(1) x + y = z
(2) x < 0

Are statmnts (1) & (2) not saying the same thing?
Cos if x + y = z; it implies x<0;

Else, x = y - z ..for x>0

OA C

Your opinions pls as i thought OA should b D
I feel so cos both statements seem to say the same thing.

peter456 wrote:Is |x| = y - z?

(1) x + y = z
(2) x < 0

Are statmnts (1) & (2) not saying the same thing?
Cos if x + y = z; it implies x<0;

Else, x = y - z ..for x>0

OA C

Statement 1 and 2 are not saying the same thing.
Statement 1 is saying: x = z - y; x can have any value (negative, positive or 0)
Statement 2 is saying: x < 0; x can only have negative values

Is |x| = y - z?

(1) x + y = z
(2) x < 0

Rephrased question is (y-z) >= 0?

1) x + y = z
y-z = -x: we dont know if x is positive or negative
If x >0; y-z is negative
If x <0; y-z is positive
Insufficent

2) x<0
We know nothing about y or z so statement is insufficent

Combined: x + y = z -> y - z = -x and x<0
Therefore y - z > 0; which means we can answer the question

Ans = C

peter456 wrote:Is |x| = y - z?

(1) x + y = z
(2) x < 0

Statement 1: x = z-y
Case 1: x = z-y = 0, which case y-z = 0.
In this case, |x| = y-z.

Case 2: x = z-y = 1, in which case y-z = -1.
In this case, |x| ≠ y-z.
INSUFFICIENT.

Statement 2: x < 0
No information about y-z.
INSUFFICIENT.

Statements combined:
Case 3: x = z-y = -1, in which case y-z = 1.
In this case, |x| = y-z.

Case 4: x = z-y = -10, in which case y-z = 10.
In this case, |x| = y-z.

Case 5: x = z-y = -1/2, in which case y-z = 1/2.
In this case, |x|= y-z.

Since |x|= y-z in every case, SUFFICIENT.

The correct answer is C.

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) x < 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 the answer.
Looking at the conditions together,
|x|=x(when x>0), and this proves true for x+z=y, 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.

S1 gives

x = z - y

If x > 0, then |x| = x, and we have |x| = z - y, which is not necessarily = y - z.

But if x < 0, then |x| = -x, and we have -x = z - y, or x = y - z. So the question becomes "Is x < 0?"