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Is |x| = y-z? 1)x+y=z 2) x < 0 |x| = -x or x. If -x =

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Is |x| = y-z? 1)x+y=z 2) x < 0 |x| = -x or x. If -x = [#permalink] New post 11 Sep 2004, 01:09
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A
B
C
D
E

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Is |x| = y-z?

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

|x| = -x or x.

If -x = y-z. then x+y = z
If x = y-z, then y-x = z

So from 1) x+y = z, so -x = y-z and |x| must be equals to y - z. So 1) is sufficient.

From 2) all we know is x is less than 0. Nothing about y or z is given. So 2) is insufficient.

OA is not my answer (which is A). What do you think of my working ? I'm having the thought that satisfying x+y=z OR y-x=z is sufficient to say if |x| = y-z.
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Re: DS from OG [#permalink] New post 11 Sep 2004, 02:20
ywilfred wrote:
Is |x| = y-z?

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

|x| = -x or x.

If -x = y-z. then x+y = z
If x = y-z, then y-x = z

So from 1) x+y = z, so -x = y-z and |x| must be equals to y - z. So 1) is sufficient.

From 2) all we know is x is less than 0. Nothing about y or z is given. So 2) is insufficient.

OA is not my answer (which is A). What do you think of my working ? I'm having the thought that satisfying x+y=z OR y-x=z is sufficient to say if |x| = y-z.


From 1 , one get x=z-y. If x is +ve then, it is z-y; with x negative, it is y-z
From 2 , you get x is negative, so C.
Let me know what do u feel.
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 [#permalink] New post 11 Sep 2004, 02:36
Yes I do get C too.

st.1
x + y = z

x = z - y
if x = 7, z could be - 10 and y could be 17

|7| is not 17 -- 10

if x = 3, z could be 5 and y could be 2

|3| is 2 - 5

I guess this proves that A can´t be the OA.

Regards,

Alex
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 [#permalink] New post 11 Sep 2004, 04:41
I am a little confused when it got to |x| = -x = y-z. How can an absolute value of x be equals to its negative value ? Absolute values, I thought, have no negatives.
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 [#permalink] New post 11 Sep 2004, 22:52
Is |x| = y-z?

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

From 1 x = z-y
the value of x will depend on the magnitude of both z and y. if z> Y
then we get x to be positive and x = Z-Y. Iand abs(X) = y -z, right?

Now if Z < Y then X = - (z-y). If we take the absolute value, we get
abs(x) = abs(-(z-y)) = z -y, right. so A is sufficient.

Thus A is the answer and not C
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 [#permalink] New post 12 Sep 2004, 13:50
ywilfred wrote:
I am a little confused when it got to |x| = -x = y-z. How can an absolute value of x be equals to its negative value ? Absolute values, I thought, have no negatives.


You are right and wrong.
Absolute values have no sign. So |x| is always postive and represents the magnitude of x.
But in this problem, we are talking about x. x, the actual number, does have sign apart from magnitude.

From statement I alone, you cannot say "Sufficient". I is sufficient if the number x is proved to be negative, which is given in statement II. Hence C. The problem with your approach is that you missed the OR part.

|x| = -x or x.

If -x = y-z. then x+y = z
OR
If x = y-z, then y-x = z

So from 1) x+y = z, so -x = y-z and x must be equals to y - z. So (1) is sufficient if x is negative. Nothing is known about the if x is positive. Insufficient.

From 2) all we know is x is less than 0. Insufficient.

Together (2) says that x is negative and 1 proves the negative part of stem. Sufficient
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 [#permalink] New post 12 Sep 2004, 16:13
Is |x| = y-z?

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


Hello, I think you are missing something here. wht you shd know is that the value of x is dependent on z and Y for instance let say z = 2 and y = 5
then
From statement one, X = 5-2 = 3, abs(x) = 3
Now flip it around and assume that y = 2 and Z = 5
then x = 2 - 5 = -3, abs(x) = 3.

So why do you say A is not sufficient

Now consider it this way, abs(x) could mean that x is positive or negative.
if x is positive
then we have
from one we have
X = y -z and abs(x) = abs(y-z) and abs(x) = y -z)
If X <0,
then from one it would mean
-x = z-y
X = -(z-y), take the absolute value of x and you get z-y.
meaning A is sufficient.

I don't think you need B to conclude that abs(x) will always be equal to
y -z or z - y.
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 [#permalink] New post 21 Oct 2004, 11:13
Is |x|=y-z?

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

|x|=y-z gives 2 possibilities:

y-z=x or y-z=-x

From 1), x+y=z, first we can rearrange this equation to

-x=y-z, which matches one of the 2 possibilities; second, we

can rearrange the equation to x=z-y, which does not matches

another possibility. So we cannot say by 1) alone is sufficient

for the equation |x|=y-z to hold.

From 2), x<0, we have no clue to determine the validity of

|X|=y-z

If we take 1) and 2) together, and ask if x+y=z and x<0, is

|x|=y-z valid? We can get rid of one of the possibilities from

the equation |x|=y-z, and only -x=y-z left. And x+y=z is

sufficient to make -x=y-z hold. So 1) and 2) together are

sufficient, and my answer is C.
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 [#permalink] New post 21 Oct 2004, 12:37
i agree with C
thanks for explanation
  [#permalink] 21 Oct 2004, 12:37
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