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Another Absolute value question

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Another Absolute value question

by gmat009 » Thu Oct 09, 2008 11:01 am
Is |x| = y - z ?
(1) x + y = z
(2) x < 0

here is how how approached :-

To prove:- x=y-z
OR -x=y-z or x=z-y

a) x+y=z
or x=z-y {sufficient}

b) x<0 {insufficient}

Plz. let me know if I am wrong somewhere.....
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by rtfact » Thu Oct 09, 2008 11:15 am
the answer is C since:

(1) x+y=z --> -x=y-z thus |x|=y-z is true only if x<=0.
for example, if x=1 then according to (1): -1=y-z which contradicts: 1=y-z
insufficient.

(2) x<0 insufficient as you said.

(1+2) sufficient. x<0, therefore in statement (1): -x=y-z, x is positive, which leads us to conclude that |x|=y-z is true.
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by Gmatss » Thu Oct 09, 2008 11:18 am
is it C?
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by gmat009 » Thu Oct 09, 2008 11:34 am
rtfact wrote:the answer is C since:

(1) x+y=z --> -x=y-z thus |x|=y-z is true only if x<=0.
for example, if x=1 then according to (1): -1=y-z which contradicts: 1=y-z
insufficient.

(2) x<0 insufficient as you said.

(1+2) sufficient. x<0, therefore in statement (1): -x=y-z, x is positive, which leads us to conclude that |x|=y-z is true.
Thanks......
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by vivek.kapoor83 » Thu Oct 09, 2008 12:07 pm
i have same ques.
If in any ques it is asked mod x = mod y
and nothing about -ve and +ve.
Do we need to prove , x=y AND -x =-y........
I am putting essence on AND...it will be AND (both condition)or OR...means any1 of condition
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by cubicle_bound_misfit » Thu Oct 09, 2008 2:04 pm
vivek.kapoor83,

there can be different cases for |x| =|y|

x + -

y + -

check how many combos you can have.
Cubicle Bound Misfit
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by raju232007 » Thu Oct 09, 2008 5:46 pm
Is |x|=y-z


statement 1:

x+y=z
x=z-y( if x>0)
x=y-z( if x<0)

Insufficient

statement 2

x<0

Insufficient

Combining both the statements we have x=y-z for x<0


So the ans should be C..Let me know if you still have any doubts
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