positive integers

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gmatnmein2010: Master | Next Rank: 500 Posts; Posts: 185; Joined: Thu Feb 04, 2010 3:06 am; Thanked: 6 times

by gmatnmein2010 » Sat Feb 20, 2010 3:47 am

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

1)3

2)5

3)7

4)8

5)9

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Source: Beat The GMAT — Problem Solving | Sat Feb 20, 2010 3:47 am

ajith: Legendary Member; Posts: 1275; Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members; Location: Arabian Sea; Thanked: 125 times

by ajith » Sat Feb 20, 2010 4:00 am

gmatnmein2010 wrote:Given distinct positive integers 1, 11, 3, x, 2, and 9, which of the following could be the median?

1)3

2)5

3)7

4)8

5)9

1,2,3,x,9,11 is a possibility when 4<x<9

and then the median = (x+3)/2

When x = 7; the median could be (7+3)/2 = 5
Hence C

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shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pmFollowed by: 2 members; Location: Gandhinagar; Thanked: 41 times

by shashank.ism » Sat Feb 20, 2010 9:11 am

ajith wrote:
gmatnmein2010 wrote:Given distinct positive integers 1, 11, 3, x, 2, and 9, which of the following could be the median?
1)3
2)5
3)7
4)8
5)9
1,2,3,x,9,11 is a possibility when 4<x<9

and then the median = (x+3)/2

When x = 7; the median could be (7+3)/2 = 5
Hence C

we try to arrange the numbers n sequence 1,2,3,9,11
if 4<= x<=8 median = 3.5,4,4.5,5,5.5
if x=10 median = [9+3]/2 =6
if x>11 median = [9+3]/2 =6

[spoiler]so Ans(2) ie 5[/spoiler]

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GMATinsight: Legendary Member; Posts: 1100; Joined: Sat May 10, 2014 11:34 pmFollowed by: 24 members; Location: New Delhi, India; Thanked: 205 times

by GMATinsight » Tue Nov 22, 2016 11:33 pm

gmatnmein2010 wrote:Given distinct positive integers 1, 11, 3, x, 2, and 9, which of the following could be the median?

1)3

2)5

3)7

4)8

5)9

Arranging in increasing order

1, 2, 3, 9, 11 and x has to come somewhere in the sets of numbers

If x<2, then Median = Average of 2 and 3 = 2.5
If x>9, then Median = Average of 3 and 9 = 6

So median must lie between 2.5 and 6

But since numbers are distinct so median can never be 3 (for median to be 3 x should be 3 as well which is not acceptable due to terms being distinct)

Hence Answer: Option B

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by Matt@VeritasPrep » Fri Nov 25, 2016 3:15 am

Arrange the five that you know first:

1, 2, 3, 9, 11

Since x is a positive integer, it must be > 0. It must be distinct from the other integers in the set, so it must be > 3.

From here, it's easiest just to try x's. If x = 4, the median = (3+4)/2 = 3.5. If x = 5, the median = (3+5)/2 = 4. If x = 6, we get another decimal. If x = 7, we get 5, so that's our median: 5.

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8138; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Sun Nov 27, 2016 5:48 pm

gmatnmein2010 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.

Answer:B

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