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Given distinct positive integers 1, 11, 3, x, 2, and 9, whic [#permalink]
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21 Feb 2011, 14:20
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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
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Re: Mean median average [#permalink]
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21 Feb 2011, 14:39
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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Re: Mean median average [#permalink]
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21 Feb 2011, 14:43
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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; 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: Mean median average [#permalink]
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21 Feb 2011, 15:06
Bunuel.... is it possible if you could point me to more questions like these on the forum.
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Re: Mean median average [#permalink]
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Re: Mean median average [#permalink]
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21 Feb 2011, 15:13
thanks a ton.... Bunuel
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Re: Given distinct positive integers 1, 11, 3, x, 2, and 9, whic [#permalink]
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Re: Given distinct positive integers 1, 11, 3, x, 2, and 9, whic [#permalink]
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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, 12:32
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. Answer: B
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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, 17:10
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
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