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# M14 #5

Author Message
Forum Moderator
Status: mission completed!
Joined: 02 Jul 2009
Posts: 1391

Kudos [?]: 953 [1], given: 621

GPA: 3.77

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04 Nov 2010, 11:57
1
KUDOS
If set consists of even number of integers, is the median of set negative?

Exactly half of all elements of set are positive.
The largest negative element of set is -1.

[Reveal] Spoiler:
С

I found the most efficient way to solve this question.

Is is asked whether the median is negative? for sufficiency we have to demonstrate one Yes and one No.

lets say that in S are 4 numbers, $$A1, A2, A3, A4$$. and median is equal to $$(A2+A3)/2$$ - for this to be negative, absolute value of A2 must be grater than A3. |A2|>|A3|. you need to check whether it |A2|>|A3| is always true or false.

1)Exactly half of all elements of set are positive.
Remember here that 0 is neither positive nor negative.
For insuficiency we need to find this |A2|>|A3| to be true and false.
True |A2|>|A3|, S {-2;-2;1;2}, |-2|>|1|
False |A2|>|A3|, S {-2;0;1;2}, |0|<|1|

2)The largest negative element of set is -1.
True |A2|>|A3|, {-2;-1;0;0} |-1|>|0|
False |A2|>|A3, false {-2;-1;3;4} , |-1|<|3|

1)&2) two positive elements and the largest negative integer is -1
so {-3;-1;1;2}, because the second number must be -1 and the smallest postitive out of two is 1, thus (-1+1)/2=0 - which is not negative , for other cases when third element is grater than 1, answer is NO, not negative (-1+4)/2=positive.
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Kudos [?]: 953 [1], given: 621

Math Expert
Joined: 02 Sep 2009
Posts: 41912

Kudos [?]: 129370 [0], given: 12197

### Show Tags

04 Nov 2010, 19:33
Pkit wrote:
If set consists of even number of integers, is the median of set negative?

Exactly half of all elements of set are positive.
The largest negative element of set is -1.

[Reveal] Spoiler:
С

I found the most efficient way to solve this question.

Is is asked whether the median is negative? for sufficiency we have to demonstrate one Yes and one No.

lets say that in S are 4 numbers, $$A1, A2, A3, A4$$. and median is equal to $$(A2+A3)/2$$ - for this to be negative, absolute value of A2 must be grater than A3. |A2|>|A3|. you need to check whether it |A2|>|A3| is always true or false.

1)Exactly half of all elements of set are positive.
Remember here that 0 is neither positive nor negative.
For insuficiency we need to find this |A2|>|A3| to be true and false.
True |A2|>|A3|, S {-2;-2;1;2}, |-2|>|1|
False |A2|>|A3|, S {-2;0;1;2}, |0|<|1|

2)The largest negative element of set is -1.
True |A2|>|A3|, {-2;-1;0;0} |-1|>|0|
False |A2|>|A3, false {-2;-1;3;4} , |-1|<|3|

1)&2) two positive elements and the largest negative integer is -1
so {-3;-1;1;2}, because the second number must be -1 and the smallest postitive out of two is 1, thus (-1+1)/2=0 - which is not negative , for other cases when third element is grater than 1, answer is NO, not negative (-1+4)/2=positive.

Also discussed here: ds-set-14343.html
_________________

Kudos [?]: 129370 [0], given: 12197

Re: M14 #5   [#permalink] 04 Nov 2010, 19:33
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# M14 #5

Moderator: Bunuel

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