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?
For Solution SCROLL DOWN...
Solution:
There are 7 females and 6 males.
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)
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.
16 comments:
Females : 7
J and K have to be girls, else there will be more men...
E is a male, because only then can J and K be girls (same sex as I)...
And hence, D is a male...
yes there are 7 females and A,D,E,C,G are males in this question.
No. of females - 7
Females - B, F, I, J, K, H, M
Males - D, C, E, G, A, L
To Solve this problem, there could be two possible branches out of this,
Case 1: "B" is Female and "D" is Male
Follow this case and you will easily deduce that there are 7 females and 6 males.
Case 2: "B" is Male & "D" is Female
Workout this case and you will come up with 6 Females and 7 Males.
Since, its been already given that there are more females than males, therefore, case 1 holds.
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Dina : 8 am
Bunnie : 9 am
Candy : 10 am
Annie : 11 am
This is right solution might be..
see 1 st: Candy did not visit Edy between Bunnie and Dina.
2nd : At least one female visited Edy between Annie and Bunnie and that is candy
3rd :Annie did not visit Edy before both Candy and Dina so annie is lastone to meet edy.
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