Puzzle:

* A is the father of two children - B and D who are of different sexes.
* C is B's spouse.
* E is the same sex as D.
* B and C have the two children - F who is the same sex as B and G who is the same sex as C.
* E's mother, H who is married to L, is the sister of D's mother, M.
* E and E's spouse, I have two children - J and K who are the same sex as I.

Note that no persons have married more than once. Also, there are more number of females than males. Can you tell how many females are there?

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Solution:

There are 7 females and 6 males.

Assume that there are four sexes - male, female, X and Y.

L(m) H(f) M(f) A(m) E(x) I(y) D(x) B(y) C(x) J(y) K(y) F(y) G(x)

It is clear that there are altogether 13 persons - 2 males, 2 females, 4 Xs and 5 Ys. It is given that there are more number of females than male. Hence, all Y must represent female. Thus, there are 7 females and 6 males.