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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)

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

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.

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

median is average of the middle 2 numbers(if the number of the items is even). sum of even, so the median will definitely be integer.

if another choice is 40, then i belive we got to pick up the numbers to see which one is the correct median.

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

odd+odd is even and even divided by 2 is always an integer. Also it can't be 2 or 3 as all primes numbers are different. Hence best possible choices is 39.

Re: Can you please explain this !!!!! [#permalink]

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30 Mar 2010, 22:20

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!!!!!

Re: Can you please explain this !!!!! [#permalink]

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30 Mar 2010, 22:34

divakarbio7 wrote:

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

Re: Can you please explain this !!!!! [#permalink]

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01 Apr 2010, 05:59

Hi ,

Thank you for the explanation.

Have 2 more questions: what is the value of n in the list k, n, 12, 6, 17? a> k< n b> The median of the numbers in the list is 10

What is the median number of employees per project for the projects at company Z? a> 25 percent of the projects at company Z have 4 or more employees assigned to each project b> 35 percent of the projects at company Z have 2 or fewer employees assigned to each project.

i am not able to understand the logic for 2nd problem..answer says C - but note sure why?

Re: Which of the following could be the median of a set consisti [#permalink]

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19 Apr 2014, 23:47

2

This post received KUDOS

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) _________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Re: Which of the following could be the median of a set consisti [#permalink]

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24 Jun 2015, 08:16

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Re: Which of the following could be the median of a set consisti [#permalink]

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22 Dec 2015, 12:37

Countdown wrote:

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

Can somebody plz explain how to proceed.

Thanks in advance.

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.

gmatclubot

Re: Which of the following could be the median of a set consisti
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22 Dec 2015, 12:37

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