Set 8

For convenience, let the army officer be represented by A, the lawyer by B, the trainer by C and the IT engineer by D. Also, let the preferred traits be represented as T for tall, F for fair and H for handsome. NT for not tall, NF for not fair and NH for not handsome, then, the following cases arise:

Case I: Only A is tall and dark between A and D
->A = T+NF…. from ii
Then, from iii, two possibilities arise I.1 and I.2
I.1 -> B = NT+H —– (from iii)
Further, two possibilities arise; I.1.(i) & I.1.(ii)
I.1.(i) -> A = T and C = T —– (from iv)
Further, two possibilities arise, I.1.(i).α and I.1.(ii).β
I.1.(i).α -> B = F and D = F —– (from v)
If this is true, then the following is true:
A = T+NF
B = NT + H + F
C = T
D = F
This is not possible because from condition i, one of them has no trait, one has one trait, one has two traits and the remaining one has all three traits. But in I.1.(i).α, none of them has no traits.
I.1.(ii).β -> B = NF and D = NF —– (from v)
If this is true, then the following is true:
A = T + NF
B = NT + H + NF
C = T
D = NF
This possibility can’t be overruled.
I.1.(ii) -> A = NT and C = NT —– (from iv)
However, this is not possible because this case already establishes A to be tall.
I.2 -> C = NT+H —– (from iii)
Further, two possibilities arise; I.2.(i) & I.2.(ii)
I.2.(i) -> A = T and C = T —– (from iv)
This is not possible because this case already established C to be NT.
I.2.(ii) -> A = NT and C = NT —– (from iv)
This is not possible because this case already establishes A to be T.

Case II: Only D is tall and dark between A and D
-> D = T = NF —– (from ii)
Then, two possibilities arise: II.1 and II.2
II.1 -> B = NT + H —– (from iii)
Further, two possibilities arise: II.1.(i) and II.1.(ii)
II.1.(i) -> A = T and C = T —– (from iv)
This is not possible because from condition i, at least one of them has to have no traits.
II.1.(ii) -> A = NT and C = NT —– (from iv)
This is also not possible because in this case all of them lack at least one trait, however from i, one person has all these traits.
II.2 -> C = NT+H —– (from iii)
Further, two possibilities arise: II.2.(i) and II.2.(ii)
II.2.(i) -> A = T and C = T —– (from iv)
This is not possible because II.2 assumes that C = NT
II.2.(ii) -> A = NT and C = NT —– (from iv)
Further, two possibilities arise: II.2.(ii).α and II.2.(ii).β
II.2.(ii).α -> B = F and D = F
This is not possible because II assumes that D is NF.
II.2.(ii).β -> B = NF and D = NF
If this is true, then the following has to be true:
A = NT
B = NF
C = NT
D = NF
This is not possible because one of them has all three traits.
By exploring all possibilities, we conclude that only I.1.(i).β is possible, where:
A = T+NF
B = NT+H+NF
C = T
D = NF
->D is the only person who does not possess even a single trait -> D = NF+NT+NH
Also, B is the person with one trait i.e. H
So, between A and C, one person has 2 traits and the other has all three. Since, A is NF, he can’t have all 3 traits; so, A = T+NF+H and C = T+F+H.
Therefore, the trainer is the person who is tall, fair and handsome

 

C’s sister-in-law has to be a female, A is a married person and B has a sibling.

Two possibilities arise:

Case I: C’s sister-in-law is A – Further two possibilities:

1. A and B are married -> B is the married man and B’s brother is C, since B’s sibling (i.e., C) and A’s spouse (i.e., B) are of the same gender. This is possible.
2. B and C are married -> This is not possible because A has to be one of the married persons.

Case II: C’s sister-in-law is B – Further two possibilities:

1. A and B are married -> A is the married man. But this is not possible because C cannot be B’s sibling and A is also ruled out.
2. A and C are married -> This is also not possible because in this A’s spouse (i.e., C) and B’s sibling (i.e. A) are of the same gender and can’t be married.

So, the only possibility is Case I – 1, where B is the married man.

• D loses the fourth game, so he doubles the amount with A, B and C -> A, B and C all had Rs. 48 each at the end of third game and D had (48*3)+96 = 240.
• C loses the third game, so he doubles the amount with A, B and D -> A and B had Rs. 24 each at the end of second game; D has Rs. 120 and C had 48+(120+48) = 216
• B loses the second game, so he doubles the amount with A, C and D -> A had Rs. 12 at the end of 1^st game; C had Rs. 108(=216/2); D had Rs. 60 (=120/2) and B had 24+(12+108+60) = 204
• A loses the first game, so he doubles the amount with B, C and D -> B had Rs. 102(=204/2) at the beginning; C had Rs. 54 (=108/2); D had Rs. 30 (=60/2) and A had 12+(102+54+30) = 198 at the beginning. The information is shown in the following table.

	Amount
	A	B	C	D	Total
End of 4th game	96	96	96	96	384
End of 3rd game	48	48	48	240	384
End of 2nd game	24	24	216	120	384
End of 1st game	12	204	108	60	384
Beginning	198	102	54	30	384