Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 27 Mar 2015, 16:37

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5079
Location: Singapore
Followers: 21

Kudos [?]: 180 [0], given: 0

Is |x| = y-z? 1)x+y=z 2) x < 0 |x| = -x or x. If -x = [#permalink]  11 Sep 2004, 01:09
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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.
Director
Joined: 08 Jul 2004
Posts: 604
Followers: 2

Kudos [?]: 78 [0], given: 0

Re: DS from OG [#permalink]  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.
S
Manager
Joined: 02 Apr 2004
Posts: 224
Location: Utrecht
Followers: 1

Kudos [?]: 12 [0], given: 0

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 canÂ´t be the OA.

Regards,

Alex
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5079
Location: Singapore
Followers: 21

Kudos [?]: 180 [0], given: 0

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.
Manager
Joined: 21 Aug 2004
Posts: 136
Followers: 1

Kudos [?]: 14 [0], given: 0

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
Director
Joined: 20 Jul 2004
Posts: 593
Followers: 1

Kudos [?]: 44 [0], given: 0

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
Manager
Joined: 21 Aug 2004
Posts: 136
Followers: 1

Kudos [?]: 14 [0], given: 0

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.
Manager
Joined: 31 Aug 2004
Posts: 169
Followers: 1

Kudos [?]: 4 [0], given: 0

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.
Senior Manager
Joined: 07 Oct 2003
Posts: 358
Location: Manhattan
Followers: 2

Kudos [?]: 6 [0], given: 0

i agree with C
thanks for explanation
Similar topics Replies Last post
Similar
Topics:
4 x^2 - x < 0 4 13 May 2012, 11:54
9 If x#0, is x^2/|x| < 1? 12 08 Sep 2010, 10:51
7 If x 0, is x^2/|x| < 1? 6 03 Jan 2010, 06:14
4 If x 0, is x^2 / |x| < 1? 10 14 Dec 2009, 00:03
What is the value of 5x/yz? (1) xyz = 40 (2) yz/x=10 4 21 Mar 2006, 17:33
Display posts from previous: Sort by