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M is a list of 12 consecutive integers and N is a list of 31 consecutive integers. The median of N is equal to the greatest integer of M. If the two lists are combined into one list of 43 integers, how many integers are repeated?
A. 0
B. 6
C. 12
D. 15
E. 31
A. 0
B. 6
C. 12
D. 15
E. 31
08 Feb 2022, 11:18
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The median of the list N must be the 16th consecutive integer in the list. If this number is also the largest number of the list M, then all numbers of list M overlap with the 4th-16th numbers of list N. So all numbers of list M are repeated in list N. C)
08 Feb 2022, 11:47
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Bunuel
M is a list of 12 consecutive integers and N is a list of 31 consecutive integers. The median of N is equal to the greatest integer of M. If the two lists are combined into one list of 43 integers, how many integers are repeated?
A. 0
B. 6
C. 12
D. 15
E. 31
A. 0
B. 6
C. 12
D. 15
E. 31
Let the numbers in N are: n, n+1, n+2, ... n+30
Thus, the median is the (31+1)/2 th term = 16th term = n+15
Thus, the largest number in M is n+15
Since M has 12 consecutive numbers, the numbers are: n+4, ... n+15
Thus, the number of terms common to M and N are: n+4, n+5, ... n+15, i.e. all 12 numbers
Answer C
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