This question combines elementrary concepts of number properties including prime numbers and properties of odd and even numbers with statistics. Elementary statistics concepts including range, mean, and median of a set of numbers are tested in this question.

Question 11: 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.

- I only
- I and II only
- I and III only
- III only
- I, II, and III

@ INR

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 < x < 40 and 0 < y < 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 = 35. It is greater than 33.

**Statement I is true.** So, eliminate choices that do not 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 \\frac{\text{57 + x}}{6}) = 9.5 + \\frac{x}{6})

If x = 3, median = 5 and average = 10. Average > median.

If x = 5, median = 5.5 and average = 10.33. Average > median

If x = 7, median = 6.5 and average = 10.66. Average > medain

If x = 11 or greater, the median = 7. Average will be definitely greater than 10. So, Average > 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.

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