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Re: In a family of four people, none of the people have the same [#permalink]
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vndnjn wrote:
Two of the people are less than 12 years old, possible age: 2, 3, 5, and 7
the other two people are between 40 and 52 years old, possible age: 41, 43, 47
If the average of their four ages is also a prime number,
min possible avg of ages = (2+2+41+41)/4 = 86/4 = 21.5
max possible avg of ages = (7+7+47+47)/4 = 108/4 = 27
as mentioned, none of the people has the same age and the average of their four ages is also a prime number, it means the avg is in between 21.5 and 27 and a prime no. i.e. 23
the sum of all four ages = 4*23 = 92
only 3, 5, 41 and 43 can give a sum 92



Why you have not taken 11 as max. possible avg age?
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Re: In a family of four people, none of the people have the same [#permalink]
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kanny wrote:
Why you have not taken 11 as max. possible avg age?


If I understood your que correctly; We have to select two no from 2, 3, 5, and 7 and two no from 41, 43, 47.
With any combination, we can not have 11 as vag value of four of these no. The min possible avg is 21.5.
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Re: In a family of four people, none of the people have the same [#permalink]
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vndnjn wrote:
kanny wrote:
Why you have not taken 11 as max. possible avg age?


If I understood your que correctly; We have to select two no from 2, 3, 5, and 7 and two no from 41, 43, 47.
With any combination, we can not have 11 as vag value of four of these no. The min possible avg is 21.5.


I think he meant, 11 is also less than 12 so this is also the possible age
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Re: In a family of four people, none of the people have the same [#permalink]
avg is 21.5?
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Re: In a family of four people, none of the people have the same [#permalink]
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I used the solution that was already provided here, but I added some changes.

We have two sets.
Set A --- 2, 3, 5, 11
Set B --- 41, 43, 47

We need to choose two numbers from each set.
Also the sum of these four numbers should be even. If we take four numbers and sum them, we have three odd numbers, and therefore, the fourth number must be also odd to make the case be even.
odd+odd+odd+[must be odd to make the sum be even].
So we can remove 2 from set A.
Set A --- 3, 5, 11
Set B --- 41, 43, 47

Now we want to find the smallest and the largest possible average:
Smallest: (3+5+41+43)\4 = 23
Largest: (5+11+43+47)\4= 26.5

So we need to find all prime numbers between 23 and 26.5.
23 is the the prime number, and we know the is was created by 3,5,41,43, so these four numbers are the answer.
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Re: In a family of four people, none of the people have the same [#permalink]
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