For me C, but i have sure as haven't started DS yet.

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

1. x + y = z

2. x < 0

IMO C

1) x=z-y
|x| = y-z, when z<y
|x| = z-y when z>=y, NOT SUFF

2) x<0, no info about y and z, not suff
combined
x=z-y<0
z<y
SUFF

When you have modulus |x| = a, then x can be
X = +a
Or
X = -a

I’m not sure but the answer should be A. The question can be rephrased as:

X = y – z
OR
X = -y + z

From 1 we do get x = -y + z

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

1. x + y = z

2. x < 0

mehravikas wrote:When you have modulus |x| = a, then x can be
X = +a
Or
X = -a

I’m not sure but the answer should be A. The question can be rephrased as:

X = y – z
OR
X = -y + z

From 1 we do get x = -y + z

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

1. x + y = z

2. x < 0

Lets say question is

Is |x| =a

1. x=-a

2. x <0

Don't you think we both require 1 and 2. Since with 1 only we are proving that x=-a for both x<0 and x>0, which is wrong.

madhur_ahuja wrote:
Lets say question is

Is |x| =a

1. x=-a

2. x <0

Don't you think we both require 1 and 2. Since with 1 only we are proving that x=-a for both x<0 and x>0, which is wrong.

only statement 1 should be enough

stment 1: x=-a
=>|x|=|-a|=a
sufficient

stment 2: x<0 not sufficient

scoobydooby wrote:

madhur_ahuja wrote:
Lets say question is

Is |x| =a

1. x=-a

2. x <0

Don't you think we both require 1 and 2. Since with 1 only we are proving that x=-a for both x<0 and x>0, which is wrong.

only statement 1 should be enough

stment 1: x=-a
=>|x|=|-a|=a
sufficient

stment 2: x<0 not sufficient

scooby,
I don't think so
stment 1: x=-a
=>|x|=|-a|=a
thats when a is positive
when a is negative, it will not hold true.
what you are doing wrong is this:

x=-a
|x|=|-a|=a (no, this is wrong)

|x|=|-1|*|a| =|a|
and this will be a or -a depending upon whether a is positive or negative

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

1. x + y = z

2. x < 0

would go with C.

Is |x| = y-z ?
|x| always >0
=>y-z>0
=>y>z?

stment 1. x + y = z
if x=1, y=2, z=3 => No
if x=0, y=1, z=1 => yes
not sufficient

stmnt 2. x>0 not sufficient

together,
x=-1, y=5, z=4 =>yes
z=-1, y=-5, z=-6=> yes
(y will always be greater than z)
sufficient, hence C

If you know that x = -a, why do you need to know that x < 0. Lets say the question is |x| = 5

i.e. x = 5 or x = -5

if statement 1 says that x = -5, then why do we need to know that x < 0.

please explain.

madhur_ahuja wrote:

mehravikas wrote:When you have modulus |x| = a, then x can be
X = +a
Or
X = -a

I’m not sure but the answer should be A. The question can be rephrased as:

X = y – z
OR
X = -y + z

From 1 we do get x = -y + z

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

1. x + y = z

2. x < 0

Lets say question is

Is |x| =a

1. x=-a

2. x <0

Don't you think we both require 1 and 2. Since with 1 only we are proving that x=-a for both x<0 and x>0, which is wrong.

The answer should be C.

In 1st case, x + y = z.
That means, x = -(y-z).
If y is 3, z is 2, then x = -1 and |x| = 1 = (y-z)
If y is 2, z is 3, then x = 1, and |x| = 1 is not equal to (y-z).
so 1st is not sufficient.

In 2nd case, no information about y and z,
so 2nd is not sufficient.

Taking both together, if x is always less than 0, then y should always be greater than z, so in the above example in which y is 3 and z is 2, it will give x = -1, and |x| = 1 = (y-z). So, taking both the conditions together, it is sufficient to answer the question.

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

1. x + y = z

2. x < 0

X = Y-Z WHEN X>0 OR X = Z-Y WHEN X<0

FROM 1

X+Y=Z IE: X = Z-Y , HOWEVER THIS IS ONLY VALID IF X<0...WE DONT KNOW THAT

FROM 2

INSUFF

BOTH ................C