What fraction of the total money did T have at the beginning of the game?

Let the initial amount with them be
J1 , B1 and T1
Now as per given
J1+B1 =4T1 (1)
T1+B1 = 3J1 (2)
Total amount with them will be 4J1 or 5T1
Now Let final amounts with them be J2 , B2 and T2 respectively .
J2 +B2 = 3T2 (3)
T2 +B2 = 2J2 (4)
Total amount with them will be 4T2 or 3J2
Now 5T1 =4T2 =4J1=3J2
From (1)
B1 = 4T1-J1
= 4T1-5/4 T1
= 11/4 T1
B2 = 3T2 -J2
=3T2 -4/3T2
=5/3 T2
5/3 (5/4) T1
= 25/12 T1
Now B1-B2 = 11/4 T1 -25/12T1 =200
so we get 8/12 T1 = 200
we get T1 = 300
so B1 = 11/4 (300) = 825
B2 = 25/12 (300) = 625
T2 = 5/4(300) = 375
J1 = 375
J2 = 4/3 (375) = 500
Total money = 825+375+300 =1500
Money T had with respect to total = 300/1500 = 1/5

Alternative solution:

It is given that the J and B had 4 times the money as T(in the beginning). Thus, if T had x amount with him, the total amount with them will be 5x.

So, let's assume that the total amount with them is 60x(LCM of 5, 4, and 3).

In the beginning: Now, the amount with T will be 12x, and the sum of the amount with J and B will be 48x.

Since the amount with T and B is 3 times that with J, the amount with J will be 15x. Thus, the amount with B will be 33x.

At the end: Since the amount with J and B is three times that of T, T will have 15x at the end. Also, the amount with T and B is twice that of J; the amount with J will be 20x. Thus, B will have 25x at the end.

Now, the ratio of money with T in the beginning with respect to the total amount = $$\frac{12x}{60x}=\frac{1}{5}$$

Thus, the correct option is D.