Each family in a locality has at most two adults, and no family has fewer than 3 children. Considering all the families together, there are more adults than boys, more boys than girls, and more girls than families. Then the minimum possible number of families in the locality is

Let n_f be number of families, n_g be number of girls, n_b be number of boys and n_a be number of adults.

 

number of adults more than number of boys, number of boys more than number of girls and number of girls more than number of families.

 

hence we have,  n_a > n_b > n_g > n_f  ..........................(1)

 

we check the above relation (1) with different values for number of families, starting with n_f = 1

 

Let n_f = 1 , then n_b + n_g = 3  , if n_g =2 then n_b = 1 . But number of boys is greater than number of girls.

due to this contradiction, n_f ≠ 1

Let n_f = 2 , then n_b + n_g = 6  , if n_g =3 then n_b = 3 . But number of boys is greater than number of girls.

due to this contradiction, n_f ≠ 2

 

Let n_f = 3 , then n_b + n_g = 9  , if n_g =4 then n_b = 5 . number of adults n_a = 6.

 

hence the relation n_a > n_b > n_g > n_f  is satisfied for the above numbers,  6>5>4>3

hence minimum number of families is 3