Consider a set S = {2, 4, 6, 8, x, y} with distinct elements. If x and y are both prime numbers and 0 < x < 40 and 0 < y < 40, which of the following MUST be true?
I. The maximum possible range of the set is greater than 33.
II. The median can never be an even number.
III. If y = 37, the average of the set will be greater than the median.
Choose the most appropriate option.
A) I only  B) I and II only  C) I and III only  D) III only  E) I, II, and III
Solution   Key Data from the Question Stem   Set S has 6 elements.   The elements of set S are distinct.   x and y are prime numbers. Because 2 is already an element in S, both x and y have to be odd.   0 &lt; x &lt; 40 and 0 &lt; y &lt; 40   I. The maximum possible range of the set is greater than 33   The key word in this entire statement is maximum. We have to determine whether the maximum value possible for the range exceeds 33.   We know x and y are prime numbers. The largest prime number less than 40 is 37.   If either x or y is 37, the largest number in the set will be 37 and the smallest number is 2.   Therefore, the maximum range of the set will be   37 - 2= 
<gadget id="calculator">  37 - 2</gadget>
<output>35</output>
35. It is greater than 33.   Statement I is true. So, eliminate choices that donot contain I. Eliminate choice D   II. The median can never be an even number   There are 6 numbers in the set. Therefore, the median is the arithmetic mean of the 3rd and the 4th term when the numbers are written in ascending or descending order.   The elements are {2, 4, 6, 8, x, y}, where x and y are prime numbers.   If x and y take 3 and 5 as values, the median is 4.5   If x= 3, y = 7 or greater, the median is 5   If x = 5, y = 7 or greater, the median is 5.5   If x = 7, y = 11 or greater, the median is 6.5   If x = 11 or greater and y = 13 or greater, the median is 7.   It is quite clear that the median is either an odd number or is not an interger. So, the median can never be an even integer.   Statement II is true. Eliminate choices that do not contain II. Eliminate choices A and C as well.   III. If y = 37, the average of the set will be greater than the median   If y = 37, the set will be {2, 4, 6, 8, x, 37}, where x is a prime number greater than 2 and less than 37.   The average will be 57+x657+x6 = 9.5 + x6x6   If x = 3, median = 5 and average = 10. Average &gt; median.   If x = 5, median = 5.5 and average = 10.33. Average &gt; median   If x = 7, median = 6.5 and average = 10.66. Average &gt; medain   If x = 11 or greater, the median = 7. Average will be definitely greater than 10. So, Average &gt; Median.   It is true that the average is greater than the median if y = 37.   Statement III is also true.   Statements I, II, and III are true. Choice I, II, and III
<result>E</result>
Consider a set S = {2, 4, 6, 8, x, y} with distinct elements. If x and y are both prime numbers and 0 < x < 40 and 0 < y < 40, which of the following MUST be true?
I. The maximum possible range of the set is greater than 33.
II. The median can never be an even number.
III. If y = 37, the average of the set will be greater than the median.