median of a set containing 6 different primes
median will be average of 3rd and 4th prime.
All prime greater than 2 are odd. So 3th and 4th primes are odd and their sum will be even and the median will be an integer. 3rd prime will be greater than 2 or 3(first 2 primes)

E

Thanks for the reply.

Had an another doubt, say if the answer choice had "40" as one of the option. Then the answer 39 makes sense?
( Why I am confused is as per the rule stated, sum of the odd numbers is even, however 39 is not even)..
Correct me if I am wrong


Thanks Again
Divakar KN

The median is calculated as the average of the two middle numbers. If the sum of the two numbers is even, average of those numbers can be even or odd. In this case the median is an odd number. Hence 39.

Is this the correct Reasoning?

A> 2 - first prime, hence cant be a median
B> 3 - this option is out, because 3 is not a prime number
Option C & D - 9.5 & 12.5 - cant be the median because in 6 different prime the average for median shouldnt be a decimal (odd+ odd = even) so these options are out

E> 39 - only option left out is this - so it should be the answer!!!!!

Approach is correct. Some modification to your explanation:
A: 2 - first prime, hence cant be a median
B: 3 - Second prime, hence cant be a median
C & D: 9.5 & 12.5 - cannot be the median because in case of 6 different prime the average of two odd numbers cannot be decimal.
E: 39 - only option left out is this - so it should be the answer

Please post the official explaination E seems to be ambiguis as 39 is not average of any two primes

Actually it is: 37 and 41 --> average = 39.
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Median of 6 primes :\(\frac{odd+odd}{2}\) = Integer. Options C and D are discarded immediately.

Options A and B can never be the median of 6 primes.

Only option E remains.
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This can be done by hit and trial.

(A) 2 (B) 3 (C) 9.5 (D) 12.5 (E) 39

Since all primes should be different, we can take one set as below:

2..3..5..7..11..13

Median (5+7)/2 = 6

Hence, median cannot be less than 6, so reject options A and B

Since primes are odd and the number of terms is 6. So, we will always have median in the below form

(odd+odd)/2 = Even/2 = Integer.

Reject (C) and (D)

Hence (E)

let the consecutive primes be a,b,c,d,e,f
So the median will be (c+d)/2

Looking options we have to find 2*(any options) as sum of two consecutive primes.

becoz median here is avg. of two primes.

1) 2 (ie 2*2=4) so 4 cannot be the sum of middle two primes in a set of six consecutive primes.
2) 3 (3*2=6) same as option 1.
3)9.5(2*9.5=19) so sum can be 17+2 but 17 and 2 were not consecutive.
4)12.5 (2*12.5=25) same as option 3 ie we cannot find any two consecutive primes adding to 25.
5)39 (2*39=78) so here we find 78=37+41 are the two consecutive primes so the set is{29,31,37,41,43,47}

so E is our ans.

The median of 6 numbers will be the average of the 3rd and 4th Number.
Prime numbers start with 2 hence option a & b are ruled out
for option c to be valid the sum of 3rd and 4th prime number should be 21 (2 and 19 are the only primes to sum to21)
for option d to be valid the sum of 3rd and 4th prime number should be 25 (2 and 23 are the only primes to sum to23)
Option E is left and that is the correct answer

Rule: 2 is the only Even Prime No. - every other Prime No. is a (+)Positive ODD Integer.

A set consisting of 6 Different Prime Numbers placed in Ascending Order will have the Middle 2 Values both = ODD

Rule: in a Set consisting of an EVEN COUNT of Terms, the Median can be found by taking the 2 Middle Terms (here the 3rd Term and 4th Term) and finding the MEAN

(Odd Prime + Odd Prime) / 2 = Even /2 = Integer Median

Eliminate C and D

Since A (2) is the 1st Prime No. and B (3) is the 2nd Prime No. they can NEVER be the Median in a Set of 6 Distinct Prime Nos.

E - 39

2 Middle Prime Nos. in the Set = 37 and 41

3 is a prime number ! however the way you have calculated the answer is correct

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