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? Choices:
A) 13  B) 12  C) 11  D) 10  E) 5
Mean = 10   Therefore sum of the 6 digits = 6*10= 
<gadget id="calculator">6*10</gadget>
<output>60</output>
60 -------- (1)   The 3 digits are 4,5 &amp; 24 -------------- (2)   Now, 7 &lt; median &lt; 8   -&gt; 7 &lt; (sum of 2 digits)/2 &lt; 8   -&gt; 14 &lt; (sum of 2 digits) &lt; 16   Therefore the sum of 2 digits= 15.   The 2 digits whose sum is 15 &amp; whose median is b/w 7 &amp; 8 can be :   1. 7, 8,   2. 6, 9,   3. 5, 10, or   But as soon as we take 4, 11 -&gt; the series becomes 4,4,5,11,12,24 &amp;   11 pairs up with 5 to give median as 8.   Hence the number can't 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?