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16 Dec 2009, 02:37
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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
OPEN DISCUSSION OF THIS QUESTION IS HERE: given-distinct-positive-integers-1-11-3-x-2-and-9-whic-109801.html
A. 3
B. 5
C. 7
D. 8
E. 9
OPEN DISCUSSION OF THIS QUESTION IS HERE: given-distinct-positive-integers-1-11-3-x-2-and-9-whic-109801.html
23 Nov 2016, 02:27
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delta09
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
A. 3
B. 5
C. 7
D. 8
E. 9
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).
OPEN DISCUSSION OF THIS QUESTION IS HERE: given-distinct-positive-integers-1-11-3-x-2-and-9-whic-109801.html
General Discussion
16 Dec 2009, 03:01
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delta09
Given distinct positive integers 1, 11, 3, x, 2, and 9, which of the following could be the median?
a)3
b)5
c)9
d)8
e)7
a)3
b)5
c)9
d)8
e)7
arranging the series we have 1,2,3,9,11 and x after 3 and anywhere b/w 3,9,11 or after 11.
a-3 cannot be the median as we have 6 elements in set
b -5 possible if x= 7
c- 9 not possible
d- 8 not possible
e- 7 not possible
