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26 May 2017, 04:43
Kudos
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Is the median of the 11 different numbers greater than 11?
(1) The average (arithmetic mean) of the smallest 5 numbers of them is 11
(2) The average (arithmetic mean) of the 11 numbers is greater than 11
(1) The average (arithmetic mean) of the smallest 5 numbers of them is 11
(2) The average (arithmetic mean) of the 11 numbers is greater than 11
26 May 2017, 04:45
Kudos
Bookmarks
Official Solution:
If you modify the original condition and the question, the median of 11 different numbers is the 6th number.
In the original condition, there are 11 variables. In order to match the number of variables to the number of equations, there should be 11 equations. Therefore, E is most likely to be the answer. By solving con 1) & con 2), if you put 11 numbers in an ascending order, it is impossible to list the numbers such that: 11, 11, 11, 11, 11, median(=11), 11, 11, 11, 11, 11. This is because they are all different numbers, and the average (arithmetic mean) of 5smallestnumbers is greater than 11, and the average (arithmetic mean) of the total 11 numbers is also greater than 11. The 5th smallest number should be always greater than 11 in order to get median>11, hence yes, it is sufficient. The answer is C. However, this is a statistics question, one of the key questions, so if you simply apply "CMT 4(A: if you get C too easily, consider A or B)",
you can just solve con 1) and you would still find that it is impossible to list the 11 numbers in an ascending order and get 11, 11, 11, 11, 11, median. This is because the average (arithmetic mean) of 5 smallest numbers is greater than 11 and they are all different numbers. Thus, just by solving con 1), the 5th smallest number should always be greater than 11, and since the median of the 11 different numbers is always greater than 11, it is yes and sufficient.
In the case of con 2), if it is 1,2,3,4,5,6,7,8,9,10,100, you get no, and if it is 1,2,3,4,5,12,13,14,15,16,17 you get yes, hence it is not sufficient. Therefore the answer is A.
Answer: A
If you modify the original condition and the question, the median of 11 different numbers is the 6th number.
In the original condition, there are 11 variables. In order to match the number of variables to the number of equations, there should be 11 equations. Therefore, E is most likely to be the answer. By solving con 1) & con 2), if you put 11 numbers in an ascending order, it is impossible to list the numbers such that: 11, 11, 11, 11, 11, median(=11), 11, 11, 11, 11, 11. This is because they are all different numbers, and the average (arithmetic mean) of 5smallestnumbers is greater than 11, and the average (arithmetic mean) of the total 11 numbers is also greater than 11. The 5th smallest number should be always greater than 11 in order to get median>11, hence yes, it is sufficient. The answer is C. However, this is a statistics question, one of the key questions, so if you simply apply "CMT 4(A: if you get C too easily, consider A or B)",
you can just solve con 1) and you would still find that it is impossible to list the 11 numbers in an ascending order and get 11, 11, 11, 11, 11, median. This is because the average (arithmetic mean) of 5 smallest numbers is greater than 11 and they are all different numbers. Thus, just by solving con 1), the 5th smallest number should always be greater than 11, and since the median of the 11 different numbers is always greater than 11, it is yes and sufficient.
In the case of con 2), if it is 1,2,3,4,5,6,7,8,9,10,100, you get no, and if it is 1,2,3,4,5,12,13,14,15,16,17 you get yes, hence it is not sufficient. Therefore the answer is A.
Answer: A
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