Given distinct positive integers 1, 11, 3, x, 2, and 9, whic : GMAT Problem Solving (PS)
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# Given distinct positive integers 1, 11, 3, x, 2, and 9, whic

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

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21 Feb 2011, 13:20
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Given distinct positive integers 1, 11, 3, x, 2, and 9, which of the following could be the median?

A. 3
B. 5
C. 7
D. 8
E. 9
[Reveal] Spoiler: OA

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Ajit

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21 Feb 2011, 13:39
1, 11, 3, x, 2, and 9

Arrange them in ascending order;

1,2,3,9,11 Keep X aside;

Since number of elements are even

Median = Average of (n/2)th element and ((n/2)+1)th element

If x is anywhere but between 3 and 9; 3 and 9 will become n/2th and ((n/2)+1)th number
Median would be;

(3+9)/2=6

However; if x is between 3 and 9.
x can be 4,5,6,7,8; because all numbers are distinct; they can't be 3 or 9.

if x=4; Median = (3+4)/2=3.5
if x=5; Median = (3+5)/2 = 4
if x=6; Median = 4.5
if x=7; Median=5
x=8; Median = 5.5.

So; possible values of Median=3.5,4,4.5,5,5.5,6

Only option given 5.

x can't be between 2 and 3 as it is an integer and has to be different from 3 and 2. Not possible.

Ans: "B"
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21 Feb 2011, 13:43
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ajit257 wrote:
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;

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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21 Feb 2011, 14:06
Bunuel.... is it possible if you could point me to more questions like these on the forum.
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Ajit

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21 Feb 2011, 14:08
ajit257 wrote:
Bunuel.... is it possible if you could point me to more questions like these on the forum.

DS: search.php?search_id=tag&tag_id=34
PS: search.php?search_id=tag&tag_id=55
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21 Feb 2011, 14:13
thanks a ton.... Bunuel
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Ajit

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

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15 Dec 2016, 16:10
This is one great Question on the median.
Here is my solution

Number of terms = 6
Hence median = 3rd+4th/2

Lets use Brute Force here.

Option 1 =>
Median =3
Hence 3rd term + 4th term = 6

This can only occur when x=3
But if x= 3 => This will violate the original condition that all elements are distinct.
Hence rejected.

Option 2-->

Median = 5
Hence 3rd term +4th term = 10

This can happen when x=7

Hence Acceptable case

Hence B

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Re: Given distinct positive integers 1, 11, 3, x, 2, and 9, whic   [#permalink] 15 Dec 2016, 16:10
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