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Manager  Joined: 28 Aug 2010
Posts: 151
Given distinct positive integers 1, 11, 3, x, 2, and 9, whic  [#permalink]

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Question Stats: 64% (01:50) correct 36% (01:33) wrong based on 439 sessions

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 58411
Re: Mean median average  [#permalink]

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5
6
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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Re: Mean median average  [#permalink]

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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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Manager  Joined: 28 Aug 2010
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Re: Mean median average  [#permalink]

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Bunuel.... is it possible if you could point me to more questions like these on the forum.
Math Expert V
Joined: 02 Sep 2009
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Re: Mean median average  [#permalink]

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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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Manager  Joined: 28 Aug 2010
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Re: Mean median average  [#permalink]

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thanks a ton.... Bunuel
Manager  B
Joined: 17 Jun 2015
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GMAT 1: 540 Q39 V26 GMAT 2: 680 Q46 V37 Re: Given distinct positive integers 1, 11, 3, x, 2, and 9, whic  [#permalink]

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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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GRE 1: Q169 V154 Re: Given distinct positive integers 1, 11, 3, x, 2, and 9, whic  [#permalink]

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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]

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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?

A. 3
B. 5
C. 7
D. 8
E. 9

Since the integers are positive and distinct, x must be greater than 3. However, since we don’t know the exact value of x, we can start by ordering the given integers from least to greatest in the following possible scenarios:

Scenario 1: If the ordering is 1, 2, 3, x, 9, 11, then x is between 3 and 9

Scenario 2: If the ordering is 1, 2, 3, 9, x, 11, then x must be 10.

Scenario 3: If the ordering is 1, 2, 3, 9, 11, x, then x is greater than 11.

In the first scenario, the median is (3 + x)/2. In the latter two scenarios, the median is (3 + 9)/2 = 6. Since 6 is not an answer choice, the median must be (x + 3)/2 in which x is an integer between 3 and 9. Because x is an integer between 3 and 9, the median, (x + 3)/2, must be some value between 3 and 6. Thus:

3 < x < 9

6 < x + 3 < 12

3 < (x + 3)/2 < 6

The only number in the choices that satisfies this condition is 5, which is answer choice B.

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_________________ Re: Given distinct positive integers 1, 11, 3, x, 2, and 9, whic   [#permalink] 05 Feb 2019, 05:19
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