dynmoz
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:ALet 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