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A group of six friends noticed that the sum of their ages is the squar

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A group of six friends noticed that the sum of their ages is the squar  [#permalink]

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New post 15 Aug 2018, 00:08
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A group of six friends noticed that the sum of their ages is the square of a prime number. What is the average age of the group?

(1) All members are between 50 and 85 years of age.
(2) The standard deviation of their ages is 4.6
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Re: A group of six friends noticed that the sum of their ages is the squar  [#permalink]

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New post 15 Aug 2018, 01:11
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dynmoz wrote:
A group of six friends noticed that the sum of their ages is the square of a prime number. What is the average age of the group?

(1) All members are between 50 and 85 years of age.
(2) The standard deviation of their ages is 4.6


Given: Sum of the ages of 6 members = 9 or 25 or 49 or 121 or 169 or 289 or 361 or 529 or 841 ... etc.

Question : Sum of the ages of 6 members / 6 = ?

Statement 1: All members are between 50 and 85 years of age.

i.e. Sum of the ages will vary from (50*6) to (85*6) i.e. from 300 to 510

There is only one square of prime number in this range which is 529 (23^2)

i.e. sum of the ages of members = 529

i.e. Average = 529/6 hence
SUFFICIENT

Statement 2: The standard deviation of the ages is 4.6 [/b]

There are infinite many sets with the same value of standard deviation hence multiple possible values of ages are possible hence

NOT SUFFICIENT

Answer: Option A
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A group of six friends noticed that the sum of their ages is the squar  [#permalink]

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New post 15 Aug 2018, 09:19
dynmoz wrote:
A group of six friends noticed that the sum of their ages is the square of a prime number. What is the average age of the group?

(1) All members are between 50 and 85 years of age.
(2) The standard deviation of their ages is 4.6


OA:A
Let the age of six friends be \(a,b,c,d, e\) and \(f\)

\(a+b+c+d+e+f = {P}^2\) where \(P\)=prime number

What is Average age \(= \frac{a+b+c+d+e+f}{6}\)?

(1) All members are between \(50\) and \(85\)years of age.
Lower limit of sum of ages of \(6\) friends \(= 6*50 =300\)
Upper limit of sum of ages of \(6\) friends \(= 6*85 =510\)
As per question stem, Sum of their ages is the square of prime number.
\(17^2 =289, 19^2=361, 23^2=529\)

Only \(19^2\) falls in between \(300\) and \(510\).

so Sum of their ages is \(361\).
Average\(= \frac{361}{6}\)
So statement 1 alone is sufficient

(2)The standard deviation of their ages is 4.6
\(A_v\) be average age of six friends,
\(S.D = \sqrt{\frac{(a - A_v )^2+(b-A_v )^2+(c-A_v )^2+(d-A_v )^2+(e-A_v )^2}{6}}\)
\(4.6 = \sqrt{\frac{(a - P^2 )^2+(b-P^2 )^2+(c-P^2)^2+(d-P^2 )^2+(e-P^2 )^2}{6}}\)
We will not be able to calculate either the sum of ages of 6 friends,or their individual ages. \(P\) is also unknown.
So statement 2 alone is not sufficient
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A group of six friends noticed that the sum of their ages is the squar &nbs [#permalink] 15 Aug 2018, 09:19
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