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DS: SET

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15 Apr 2005, 08:12
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PLZ EXPL
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Senior Manager
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Re: DS: SET [#permalink]

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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.

Hence the answer is D.
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VP
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Re: DS: SET [#permalink]

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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.

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Director
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Re: DS: SET [#permalink]

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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.

Hence the answer is D.

Thanks for a decent explanation Kaps

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Senior Manager
Joined: 02 Feb 2004
Posts: 344

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Re: DS: SET [#permalink]

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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.

Hence the answer is D.

Thanks for a decent explanation Kaps

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

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VP
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Re: DS: SET [#permalink]

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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?

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Senior Manager
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Re: DS: SET [#permalink]

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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.

Hence the answer is D.

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 !!
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Re: DS: SET   [#permalink] 16 Apr 2005, 23:56
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