Problem Solving

Q. Which of the following could be the median of a set consisting of 6 different primes?
(A) 2 (B) 3 (C) 9.5 (D) 12.5 (E) 39

MI3 wrote:Q. Which of the following could be the median of a set consisting of 6 different primes?
(A) 2 (B) 3 (C) 9.5 (D) 12.5 (E) 39

Sorry - Ignore the above question, I figured it out, answer is E.

I'm throwing in my analysis even though you already got the correct answer.

Since the median of a set consisting of an even number of integers is the average of the two middle numbers (in this case the 3rd and 4th numbers of the set), the question can be rephrased to "Which of the following could be the average of two prime numbers greater than 3?"

A) Cannot be the median because it is not greater than 3.
B) Cannot be the median because it is not greater than 3.
C) 9.5 is the average of two numbers whose sum is 19, since 9.5 = 19/2. 19 is an odd number, and any pair of available prime numbers will add to an even number since they are all odd (5, 7, 11, 13, 17).
D) 12.5 is the average of two numbers whose sum is 25, since 12.5 = 25/2. 25 is an odd number, and any pair of available prime numbers will add to an even number since they are all odd (5, 7, 11, 13, 17, 19, 23).
E) 39 = 78/2. 78 is even, so there will likely be many odd prime + odd prime combinations that yield this number. Examples are 73 + 5 and 71 + 7.

E

using POE here,
A and B not possible as prime numbers > 0 and 6 numbers are all different.

C and D not possible as medians will be 9 and 12 for prime numbers 7-11 and 11-13.

E is the one left and hence the solution.