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

 It is currently 26 Jun 2016, 09:21

### 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

# DS: SET

Author Message
Senior Manager
Joined: 02 Feb 2004
Posts: 345
Followers: 1

Kudos [?]: 53 [0], given: 0

### Show Tags

15 Apr 2005, 08:12
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

PLZ EXPL
Attachments

0035.jpg [ 6.16 KiB | Viewed 979 times ]

Senior Manager
Joined: 15 Mar 2005
Posts: 419
Location: Phoenix
Followers: 2

Kudos [?]: 21 [0], given: 0

### Show Tags

15 Apr 2005, 18:11
mirhaque wrote:
PLZ EXPL

(1). There're 5 elements, so median is just one element. That element is 0. Since median is the middle element when arranged in order, one of the elements should be 0. But three elements are non zeros anyway. This implies that x and -x both should be 0.

Sufficient.

(2) Median is x/2. Again there're 5 elements, so median is one of the elements. A couple of cases arise:
Case If |x| < 1.
Then the elements are arranged like this
-1 -x x 1 3. If median = x = x/2 => x = 0. Sufficient.

Case If |x| = 1
Then the elements are arranged as
-1 (-x=-1) (x=1) 1 3. If median = x/2, => x = 0. Sufficient.

Case if 3 > |x| > 1
Then the elements are arranged as
-x -1 1 x 3. If median = x/2, => x = 2. Sufficient.

Logically the case of |x| > 3 is no different from the last case considered.
Therefore sufficient.

_________________

Who says elephants can't dance?

VP
Joined: 25 Nov 2004
Posts: 1493
Followers: 6

Kudos [?]: 76 [0], given: 0

### Show Tags

15 Apr 2005, 18:27
from i) x = 0. sufficient
from ii) x could be anything.
suppose median, x/2 = 0, then the set is -1, 0, 0, 1, 3. x = 0.
suppose median, x/2 = 1, then the set is -2, -1, 1, 2, 3. x = 2.
the median can only be 1 and 0. -1, 3 and other values of x cannot be median. therefore, x could be 2 or 0. insufficient.

Last edited by MA on 15 Apr 2005, 18:38, edited 1 time in total.
Director
Joined: 18 Feb 2005
Posts: 673
Followers: 1

Kudos [?]: 3 [0], given: 0

### Show Tags

15 Apr 2005, 18:27
kapslock wrote:
mirhaque wrote:
PLZ EXPL

(1). There're 5 elements, so median is just one element. That element is 0. Since median is the middle element when arranged in order, one of the elements should be 0. But three elements are non zeros anyway. This implies that x and -x both should be 0.

Sufficient.

(2) Median is x/2. Again there're 5 elements, so median is one of the elements. A couple of cases arise:
Case If |x| < 1.
Then the elements are arranged like this
-1 -x x 1 3. If median = x = x/2 => x = 0. Sufficient.

Case If |x| = 1
Then the elements are arranged as
-1 (-x=-1) (x=1) 1 3. If median = x/2, => x = 0. Sufficient.

Case if 3 > |x| > 1
Then the elements are arranged as
-x -1 1 x 3. If median = x/2, => x = 2. Sufficient.

Logically the case of |x| > 3 is no different from the last case considered.
Therefore sufficient.

Thanks for a decent explanation Kaps
Senior Manager
Joined: 02 Feb 2004
Posts: 345
Followers: 1

Kudos [?]: 53 [0], given: 0

### Show Tags

16 Apr 2005, 16:12
gmat2me2 wrote:
kapslock wrote:
mirhaque wrote:
PLZ EXPL

(1). There're 5 elements, so median is just one element. That element is 0. Since median is the middle element when arranged in order, one of the elements should be 0. But three elements are non zeros anyway. This implies that x and -x both should be 0.

Sufficient.

(2) Median is x/2. Again there're 5 elements, so median is one of the elements. A couple of cases arise:
Case If |x| < 1.
Then the elements are arranged like this
-1 -x x 1 3. If median = x = x/2 => x = 0. Sufficient.

Case If |x| = 1
Then the elements are arranged as
-1 (-x=-1) (x=1) 1 3. If median = x/2, => x = 0. Sufficient.

Case if 3 > |x| > 1
Then the elements are arranged as
-x -1 1 x 3. If median = x/2, => x = 2. Sufficient.

Logically the case of |x| > 3 is no different from the last case considered.
Therefore sufficient.

Thanks for a decent explanation Kaps

Good try. OA is A. for S2, x could be 0 or 2
VP
Joined: 25 Nov 2004
Posts: 1493
Followers: 6

Kudos [?]: 76 [0], given: 0

### Show Tags

16 Apr 2005, 16:40
mirhaque wrote:
Good try. OA is A. for S2, x could be 0 or 2

Mirhaque, did not you notice my posting?
Senior Manager
Joined: 15 Mar 2005
Posts: 419
Location: Phoenix
Followers: 2

Kudos [?]: 21 [0], given: 0

### Show Tags

16 Apr 2005, 23:56
mirhaque wrote:
gmat2me2 wrote:
kapslock wrote:
mirhaque wrote:
PLZ EXPL

(1). There're 5 elements, so median is just one element. That element is 0. Since median is the middle element when arranged in order, one of the elements should be 0. But three elements are non zeros anyway. This implies that x and -x both should be 0.

Sufficient.

(2) Median is x/2. Again there're 5 elements, so median is one of the elements. A couple of cases arise:
Case If |x| < 1.
Then the elements are arranged like this
-1 -x x 1 3. If median = x = x/2 => x = 0. Sufficient.

Case If |x| = 1
Then the elements are arranged as
-1 (-x=-1) (x=1) 1 3. If median = x/2, => x = 0. Sufficient.

Case if 3 > |x| > 1
Then the elements are arranged as
-x -1 1 x 3. If median = x/2, => x = 2. Sufficient.

Logically the case of |x| > 3 is no different from the last case considered.
Therefore sufficient.

Thanks for a decent explanation Kaps

Good try. OA is A. for S2, x could be 0 or 2

Alright, mea culpa. Since there're 2 solutions possible with (B), it does NOT lead to a unique solution, and hence the answer is (A).

MA was right.

Thanks MA and MirHaque !!
_________________

Who says elephants can't dance?

Re: DS: SET   [#permalink] 16 Apr 2005, 23:56
Display posts from previous: Sort by