A list of numbers has six positive integers. Three of those integers are known: 4, 5 and 24 and three of those are unknown: x, y and z. The three unknowns are known to be distinct. It is also known that the mean of the list is 10 and the median lies between 7 and 8 (exclusive). Which of the following CANNOT be the value of any one of the unknowns?  Select the correct option
A) 13	B) 12	C) 11	D) 10	E) 5
VERITAS PREPOFFICIAL SOLUTION:   The question gives us concrete information about mean – it is 10 – but not about median – it is between 7 and 8 (exclusive). What can we say about median from this? That it cannot be 7 or 8 but anything in between. But we know that the list has all integers. When we have even number of integers, we know that the median is the average of the middle two numbers – when all are placed in increasing order. So can the average of the two middle numbers be, say, 7.1? Which two positive integers can average to give 7.1? None! Note that if the average of two integers is a decimal, the decimal must be (some number).5 such as 7.5 or 9.5 or 22.5 etc. This happens in case one number is odd and the other is even. In all other cases, the average would be an integer.   Since the median is given to be between 7 and 8, the median of the list of the six positive integers must be 7.5 only.   Now we know that the mean = 10 and median = 7.5   Method 1: Algebra/Logic   Let’s try to solve the question algebraically/logically first.   There are 6 elements in the list. The average of the list is 10 which means the sum of all 6 elements = 6*10= 
<gadget id="calculator">6*10</gadget>
<output>60</output>
60   4 + 5 + 24 + x + y + z= 60   x + y + z = 27   Median of the list = 7.5   So sum of third and fourth elements must be   7.5 * 2= 
<gadget id="calculator">  7.5 * 2</gadget>
<output>15</output>
15   There are two cases possible:   Case 1: Two of the three integers x, y and z could be the third and the fourth numbers. In that case, since already 4 and 5 are less than 7.5, one of the unknown number would be less than 7.5 (the third number) and the other two would be greater than 7.5.   The sum of the third and fourth elements of the list is 15 so   15 + z= 27   z = 12   So, two numbers whose sum is 15 such that one is less than 7.5 and the other greater than 7.5 could be   5 and 10   6 and 9   7 and 8   x, y and z could take values 5, 6, 7, 8, 9, 10 and 12.   Case 2: The known 5 could be the third number in which case one of the unknown numbers is less than 5 and two of the unknown numbers would be more than 7.5.   If the third number is 5, the fourth number has to be 10 to get a median of 7.5. Hence, 10 must be one of the unknown numbers.   The sum of the other two unknown numbers would be 27 – 10 = 17.   One of them must be less than 5 and the other greater than 10. So possible options are   4 and 13   3 and 14   2 and 15   1 and 16   x, y and z could take various values but none of them could be 11   Answer (11)
<result>C</result>
A list of numbers has six positive integers. Three of those integers are known: 4, 5 and 24 and three of those are unknown: x, y and z. The three unknowns are known to be distinct. It is also known that the mean of the list is 10 and the median lies between 7 and 8 (exclusive). Which of the following CANNOT be the value of any one of the unknowns?