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mirhaque
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PLZ EXPL



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mirhaque
PLZ EXPL
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(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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MA
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DS: SET Updated on: 15 Apr 2005, 18:38
Originally posted by MA on 15 Apr 2005, 18:27.
Last edited by MA on 15 Apr 2005, 18:38, edited 1 time in total.
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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.
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mirhaque
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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Thanks for a decent explanation Kaps
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mirhaque
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mirhaque
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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Good try. OA is A. for S2, x could be 0 or 2
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MA
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Good try. OA is A. for S2, x could be 0 or 2
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Mirhaque, did not you notice my posting?
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mirhaque
gmat2me2
kapslock
[quote="mirhaque"]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| 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[/quote]

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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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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