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Given distinct positive integers 1, 11, 3, x, 2, and 9, which of the following could be the median?

3 5 7 8 9

Can some please explain the concept behind solving such a question.

The median of a set with even number of terms is the average of two middle terms when arranged in ascending (or descending) order.

Arrange numbers in ascending order: 1, 2, 3, 9, 11, and x.

Now, x can not possibly be less than 3 as given that all integers are positive and distinct (and we already have 1, 2, and 3).

Next, if x is 3<x<9 then the median will be the average of 3 and x. As all answers for the median are integers, then try odd values for x: If x=5, then median=(3+5)/2=4 --> not among answer choices; If x=7, then median=(3+7)/2=5 --> OK;

Answer: B.

P.S. If x is more than 9 so 10 or more then the median will be the average of 3 and 9 so (3+9)/2=6 (the maximum median possible).
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Re: Given distinct positive integers 1, 11, 3, x, 2, and 9, whic [#permalink]

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11 Nov 2014, 04:25

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Re: Given distinct positive integers 1, 11, 3, x, 2, and 9, whic [#permalink]

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28 Nov 2015, 18:21

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Re: Given distinct positive integers 1, 11, 3, x, 2, and 9, whic [#permalink]

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25 Dec 2015, 11:32

Arranging the numbers, except x, the series is 1,2,3,9,11

Its a 6 term series. The median would be calculated by taking the mean of the middle two terms i.e. the 3rd and the 4th term.

so (3 + x)/2 is the median. Also, at the same time the number should be less than 9. The numbers could be 4,5,6,7,8. (distinct integers) Had 6 been an option, then x would be any other term but in the first 4.

Only one of the options suffice the conditions given.

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